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CHAPTER 1 KEYS TO THE STUDY OF CHEMISTRY FOLLOW–UP PROBLEMS 1.1A

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms and molecules composed of one red atom and one blue atom. The figure on the right shows a change to blue atoms and molecules containing two red atoms. The change is chemical since the substances themselves have changed in composition.

1.1B

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms that are close together, in the solid state. The figure on the right shows red atoms that are far apart from each other, in the gaseous state. The change is physical since the substances themselves have not changed in composition.

1.2A

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Both the solid and the vapor are iodine, so this must be a physical change. b) The burning of the gasoline fumes produces energy and products that are different gases. This is a chemical change. c) The scab forms due to a chemical change.

1.2B

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Clouds form when gaseous water (water vapor) changes to droplets of liquid water. This is a physical change. b) When old milk sours, the compounds in milk undergo a reaction to become different compounds (as indicated by a change in the smell, the taste, the texture, and the consistency of the milk). This is a chemical change. c) Both the solid and the liquid are butter, so this must be a physical change.

1.3A

Plan: We need to find the amount of time it takes for the professor to walk 10,500 m. We know how many miles she can walk in 15 min (her speed), so we can convert the distance the professor walks to miles and use her speed to calculate the amount of time it will take to walk 10,500 m. Solution:       Time (min) = 10,500 m  = 97.8869 = 98 min       Road map: Distance (m) 1000 m = 1 km Distance (km)

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1.609 km = 1 mi Distance (mi) 1 mi = 15 min Time (min) 1.3B

Plan: We need to find the number of virus particles that can line up side by side in a 1 inch distance. We know the diameter of a virus in nm units. If we convert the 1 inch distance to nm, we can use the diameter of the virus to calculate the number of virus particles we can line up over a 1 inch distance. Solution:     7  5 5 No. of virus particles = 1.0 in     = 8.4667 × 10 = 8.5 × 10 virus particles      Road map: Length (in) 1 in = 2.54 cm Length (cm) 7

1 cm = 1 × 10 nm Length (nm) 30 nm = 1 particle No. of particles 1.4A

Plan: The diameter in nm is used to obtain the radius in nm, which is converted to the radius in dm. The volume of 3 the ribosome in dm is then determined using the equation for the volume of a sphere given in the problem. This volume may then be converted to vo Solution: diameter  21.4 nm   1 m  1 dm  = 1.07 × 10–7 dm Radius (dm) = =   110 9 nm   0.1 m    2 2 3 4 3 4 3  3.141591.07107 dm = 5.13145 × 10–21 = 5.13 × 10–21 dm3 Volume (dm ) = 3 3  1 L  1  –15 –15 Volume ( L) = 5.131451021 dm 3  = 5.13145 × 10 = 5.13 × 10 L   (1 dm)3 106 L 

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Road map: Diameter (dm) diameter = 2r Radius (dm) 3

Volume (dm ) 3

1 dm = 1 L 6 1 L = 10

1.4B

Plan: We need to convert gallon units to liter units. If we first convert gallons to dm , we can then convert to L. Solution: 3    Volume (L) = 8400 gal  = 31,794 = 32,000 L 3    Road map: Volume (gal) 1 gal = 3.785 dm

Volume (dm ) 3

1 dm = 1 L Volume (L) 1.5A

Plan: The time is given in hours and the rate of delivery is in drops per second. Conversions relating hours to seconds are needed. This will give the total number of drops, which may be combined with their mass to get the total mass. The mg of drops will then be changed to kilograms. Solution: 3  60 min   60 s 1.5 drops  65 mg 10 g   1 kg   Mass (kg) = 8.0 h       3  = 2.808 = 2.8 kg  1 h  1 min  1 s  1 drop  1 mg  10 g  Road map: Time (hr) 1 hr = 60 min Time (min) 1 min = 60 s

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Time (s) 1 s = 1.5 drops No. of drops 1 drop = 65 mg Mass (mg) of solution 3

1 mg = 10 g Mass (g) of solution 3

10 g = 1 kg Mass (kg) of solution

1.5B

Plan: We have the mass of apples in kg and need to find the mass of potassium in those apples in g. The number of apples per pound and the mass of potassium per apple are given. Convert the mass of apples in kg to pounds. Then use the number of apples per pound to calculate the number of apples. Use the mass of potassium in one apple to calculate the mass (mg) of potassium in the group of apples. Finally, convert the mass in mg to g. Solution:         = 3.4177 = 3.42 g Mass (g) = 3.25 kg    3       Road map: Mass (kg) of apples 0.4536 kg = 1 lb Mass (lb) of apples 1 lb = 3 apples No. of apples 1 apple = 159 mg potassium Mass (mg) potassium 3

10 mg = 1 g Mass (g) potassium

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1.6A

Plan: We know the area of a field in m . We need to know how many bottles of herbicide will be needed to treat 2 that field. The volume of each bottle (in fl oz) and the volume of herbicide needed to treat 300 ft of field are 2 2 given. Convert the area of the field from m to ft (don’t forget to square the conversion factor when converting from squared units to squared units!). Then use the given conversion factors to calculate the number of bottles of 2 herbicide needed. Convert first from ft of field to fl oz of herbicide (because this conversion is from a squared unit to a non-squared unit, we do not need to square the conversion factor). Then use the number of fl oz per bottle to calculate the number of bottles needed. Solution: 2     2  No. of bottles = 2050 m    = 6.8956 = 7 bottles 2 2 2      Road map: 2

Area (m ) 2

(0.3048) m = 1 ft

Area (ft ) 2

300 ft = 1.5 fl oz Volume (fl oz) 16 oz = 1 bottle No. of bottles

1.6B

Plan: Calculate the mass of mercury in g. Convert the surface area of the lake from mi to ft . Find the volume of 3 2 the lake in ft by multiplying the surface area (in ft ) by the depth (in ft). Then convert the volume of the lake to 3 3 3 3 3 mL by converting first from ft to m , then from m to cm , and from cm to mL. Finally, divide the mass in g by the volume in mL to find the mass of mercury in each mL of the lake. Solution:   7  = 7.5 × 10 g Mass (g) = 75,000 kg    6 3  (5280)2 ft 2   0.02832 m 3  110 cm  1 mL      35 ft     = 1.24349 × 1014 mL Volume (mL) = 4.5 mi  2   1 m 3 1 cm 3  1 ft 3  1 mi   2

Mass (g) of mercury per mL =





 7  14

-7

–7

= 6.0314 × 10 = 6.0 × 10 g/mL

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Road map: 2

Area (mi ) 2

1 mi = (5280) ft

Area (ft ) 2

V = area (ft ) × depth (ft) 3

Volume (ft ) 3

1 ft = 0.02832 m

Volume (m ) 3

1 m = 10 cm

Volume (cm ) Mass (kg) 3

1 cm = 1 mL 3

1 kg = 10 g Volume (mL)

Mass (g)

divide mass by volume Mass (g) of mercury in 1 mL of water 1.7A

Plan: Find the mass of Venus in g. Calculate the radius of Venus by dividing its diameter by 2. Convert the radius from km to cm. Use the radius to calculate the volume of Venus. Finally, find the density of Venus by dividing 3 the mass of Venus (in g) by the volume of Venus (in cm ). Solution:  3  24 27  = 4.9 × 10 g Mass (g) = 4.9 × 10 kg      

 Radius (cm) =   3

Volume (cm ) =

4 3

Density (g/cm ) =



4 3

4.91027 g 

   

 8  = 6.05 10 cm



= 9.27587 × 10 cm

= 5.28252 = 5.3 g/cm

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1-6

Road map: Diameter (km) d = 2r Radius (km) 3

1 km = 10 m Radius (m) 2

1 m = 10 cm Mass (kg)

Radius (cm) 3

1 kg = 10 g Mass (g)

4 3

Volume (cm ) divide mass by volume 3 by volume Densitydivide (g/cmmass )

1.7B

Plan: The mass (pounds) of aluminum must be converted to grams, then dividing by the density of aluminum will give the volume of aluminum available to make cans. Finally, dividing by the volume per can will give the number of cans possible. Solution:  453.6 g  1 cm 3   1 can   Number of cans = 16.2 lb  = 533 cans  3  1 lb  2.70 g   5.1 cm  Road map:





Mass (lb) 1 lb = 453.6 g Mass (g) 2.70 g = 1 cm

Volume (cm ) 3

5.1 cm = 1 can Number of cans

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1.8A

Plan: Using the relationship between the Kelvin and Celsius scales, change the Kelvin temperature to the Celsius temperature. Then convert the Celsius temperature to the Fahrenheit value using the relationship between these two scales. Solution: T (in °C) = T (in K) – 273.15 = 234 K – 273.15 = –39.15 = –39°C 9 9 T (in °F) = T (in °C) + 32 = (–39.15°C) + 32 = –38.47 = –38°F 5 5 Check: Since the Kelvin temperature is below 273, the Celsius temperature must be negative. The low Celsius value gives a negative Fahrenheit value.

1.8B

Plan: Convert the Fahrenheit temperature to the Celsius value using the relationship between these two scales. Then use the relationship between the Kelvin and Celsius scales to change the Celsius temperature to the Kelvin temperature. Solution: 5 5 T (in °C) = (T (in °F)  32) = (2325 °F – 32) = 1273.8889 = 1274 °C 9 9 T (in K) = T (in °C) + 273.15 = 1274 °C + 273.15 = 1547.15 = 1547 K Check: Since the Fahrenheit temperature is large and positive, both the Celsius and Kelvin temperatures should also be positive. Because the Celsius temperature is greater than 273, the Kelvin temperature should be greater than 273, which it is.

1.9A

Plan: Determine the significant figures by counting the digits present and accounting for the zeros. Zeros between non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to the left of a decimal point are only significant if the decimal point is present. Solution: a) 31.070 mg; five significant figures b) 0.06060 g; four significant figures c) 850°C; three significant figures — note the decimal point that makes the zero significant. Check: All significant zeros must come after a significant digit.

1.9B

Plan: Determine the significant figures by counting the digits present and accounting for the zeros. Zeros between non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to the left of a decimal point are only significant if the decimal point is present. Solution: 2 a) 2.000 × 10 mL; four significant figures –6 b) 3.90 × 10 m; three significant figures –4 c) 4.01 × 10 L; three significant figures Check: All significant zeros must come after a significant digit.

1.10A

Plan: Use the rules presented in the text. Add the two values in the numerator before dividing. The time conversion is an exact conversion and, therefore, does not affect the significant figures in the answer. Solution: The addition of 25.65 mL and 37.4 mL gives an answer where the last significant figure is the one after the decimal point (giving three significant figures total): 25.65 mL + 37.4 mL = 63.05 (would round to 63.0 if not an intermediate step) When a four significant figure number divides a three significant figure number, the answer must round to three significant figures. An exact number (1 min / 60 s) will have no bearing on the number of significant figures. 63.05 mL = 51.4344 = 51.4 mL/min 1 min    73.55 s    60 s 

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1.10B

Plan: Use the rules presented in the text. Subtract the two values in the numerator and multiply the numbers in the denominator before dividing. Solution: The subtraction of 35.26 from 154.64 gives an answer in which the last significant figure is two places after the decimal point (giving five significant figures total): 154.64 g – 35.26 g = 119.38 g The multiplication of 4.20 cm (three significant figures) by 5.12 cm (three significant figures) by 6.752 cm (four significant figures) gives a number with three significant figures. 3 4.20 cm × 5.12 cm × 6.752 cm = 145.1950 (would round to 145 cm if not an intermediate step) When a three significant figure number divides a five significant figure number, the answer must round to three significant figures. 3

= 0.82220 = 0.822 g/cm

END–OF–CHAPTER PROBLEMS 1.1

Plan: If only the form of the particles has changed and not the composition of the particles, a physical change has taken place; if particles of a different composition result, a chemical change has taken place. Solution: a) The result in C represents a chemical change as the substances in A (red spheres) and B (blue spheres) have reacted to become a different substance (particles consisting of one red and one blue sphere) represented in C. There are molecules in C composed of the atoms from A and B. b) The result in D represents a chemical change as again the atoms in A and B have reacted to form molecules of a new substance. c) The change from C to D is a physical change. The substance is the same in both C and D (molecules consisting of one red sphere and one blue sphere) but is in the gas phase in C and in the liquid phase in D. d) The sample has the same chemical properties in both C and D since it is the same substance but has different physical properties.

1.2

Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. Solution: a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume of a balloon. Helium is a gas. b) At room temperature, the mercury does not completely fill the thermometer. The surface of the liquid mercury indicates the temperature. c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is a liquid, though it is possible that solid particles of food will be present.

1.3

Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. Solution: a) The air fills the volume of the room. Air is a gas. b) The vitamin tablets do not necessarily fill the entire bottle. The volume of the tablets is determined by the number of tablets in the bottle, not by the volume of the bottle. The tablets are solid. c) The sugar has a definite volume determined by the amount of sugar, not by the volume of the container. The sugar is a solid.

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1-9

1.4

Plan: Define the terms and apply these definitions to the examples. Solution: Physical property – A characteristic shown by a substance itself, without interacting with or changing into other substances. Chemical property – A characteristic of a substance that appears as it interacts with, or transforms into, other substances. a) The change in color (yellow–green and silvery to white), and the change in physical state (gas and metal to crystals) are examples of physical properties. The change in the physical properties indicates that a chemical change occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is an example of a chemical property. b) The sand and the iron are still present. Neither sand nor iron became something else. Colors along with magnetism are physical properties. No chemical changes took place, so there are no chemical properties to observe.

1.5

Plan: Define the terms and apply these definitions to the examples. Solution: Physical change – A change in which the physical form (or state) of a substance, but not its composition, is altered. Chemical change – A change in which a substance is converted into a different substance with different composition and properties. a) The changes in the physical form are physical changes. The physical changes indicate that there is also a chemical change. Magnesium chloride has been converted to magnesium and chlorine. b) The changes in color and form are physical changes. The physical changes indicate that there is also a chemical change. Iron has been converted to a different substance, rust.

1.6

Plan: Apply the definitions of chemical and physical changes to the examples. Solution: a) Not a chemical change, but a physical change — simply cooling returns the soup to its original form. b) There is a chemical change — cooling the toast will not “un–toast” the bread. c) Even though the wood is now in smaller pieces, it is still wood. There has been no change in composition, thus this is a physical change, and not a chemical change. d) This is a chemical change converting the wood (and air) into different substances with different compositions. The wood cannot be “unburned.”

1.7

Plan: If there is a physical change, in which the composition of the substance has not been altered, the process can be reversed by a change in temperature. If there is a chemical change, in which the composition of the substance has been altered, the process cannot be reversed by changing the temperature. Solution: a) and c) can be reversed with temperature; the dew can evaporate and the ice cream can be refrozen. b) and d) involve chemical changes and cannot be reversed by changing the temperature since a chemical change has taken place.

1.8

Plan: A system has a higher potential energy before the energy is released (used). Solution: a) The exhaust is lower in energy than the fuel by an amount of energy equal to that released as the fuel burns. The fuel has a higher potential energy. b) Wood, like the fuel, is higher in energy by the amount released as the wood burns.

1.9

Plan: Kinetic energy is energy due to the motion of an object. Solution: a) The sled sliding down the hill has higher kinetic energy than the unmoving sled. b) The water falling over the dam (moving) has more kinetic energy than the water held by the dam.

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1.10

Observations are the first step in the scientific approach. The first observation is that the toast has not popped out of the toaster. The next step is a hypothesis (tentative explanation) to explain the observation. The hypothesis is that the spring mechanism is stuck. Next, there will be a test of the hypothesis. In this case, the test is an additional observation — the bread is unchanged. This observation leads to a new hypothesis — the toaster is unplugged. This hypothesis leads to additional tests — seeing if the toaster is plugged in, and if it works when plugged into a different outlet. The final test on the toaster leads to a new hypothesis — there is a problem with the power in the kitchen. This hypothesis leads to the final test concerning the light in the kitchen.

1.11

A quantitative observation is easier to characterize and reproduce. A qualitative observation may be subjective and open to interpretation. a) This is qualitative. When has the sun completely risen? b) The astronaut’s mass may be measured; thus, this is quantitative. c) This is qualitative. Measuring the fraction of the ice above or below the surface would make this a quantitative measurement. d) The depth is known (measured) so this is quantitative.

1.12

A well-designed experiment must have the following essential features: 1) There must be two variables that are expected to be related. 2) There must be a way to control all the variables, so that only one at a time may be changed. 3) The results must be reproducible.

1.13

A model begins as a simplified version of the observed phenomena, designed to account for the observed effects, explain how they take place, and to make predictions of experiments yet to be done. The model is improved by further experiments. It should be flexible enough to allow for modifications as additional experimental results are gathered.

1.14

Plan: Review the definitions of mass and weight. Solution: Mass is the quantity of material present, while weight is the interaction of gravity on mass. An object has a definite mass regardless of its location; its weight will vary with location. The lower gravitational attraction on the Moon will make an object appear to have approximately one-sixth its Earth weight. The object has the same mass on the Moon and on Earth.

1.15

The unit you begin with (feet) must be in the denominator to cancel. The unit desired (inches) must be in the numerator. The feet will cancel leaving inches. If the conversion is inverted the answer would be in units of feet squared per inch.

1.16

Plan: Density =

mass . An increase in mass or a decrease in volume will increase the density. A decrease volume in density will result if the mass is decreased or the volume increased.

Solution: a) Density increases. The mass of the chlorine gas is not changed, but its volume is smaller. b) Density remains the same. Neither the mass nor the volume of the solid has changed. c) Density decreases. Water is one of the few substances that expands on freezing. The mass is constant, but the volume increases. d) Density increases. Iron, like most materials, contracts on cooling; thus the volume decreases while the mass does not change. e) Density remains the same. The water does not alter either the mass or the volume of the diamond.

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1-11

1.17

Plan: Review the definitions of heat and temperature. The two temperature values must be compared using one temperature scale, either Celsius or Fahrenheit. Solution: Heat is the energy that flows between objects at different temperatures while temperature is the measure of how hot or cold a substance is relative to another substance. Heat is an extensive property while temperature is an intensive property. It takes more heat to boil a gallon of water than to boil a teaspoon of water. However, both water samples boil at the same temperature. 9 9 Convert 65°C to °F: T (in °F) = T (in °C) + 32 = (65°C) + 32 = 149°F 5 5 A temperature of 65°C is 149°F. Heat will flow from the hot water (65°C or 149°F) to the cooler water (65°F). The 65°C water contains more heat than the cooler water.

1.18

There are two differences in the Celsius and Fahrenheit scales (size of a degree and the zero point), so a simple one-step conversion will not work. The size of a degree is the same for the Celsius and Kelvin scales; only the zero point is different so a one-step conversion is sufficient.

1.19

Plan: Review the definitions of extensive and intensive properties. Solution: An extensive property depends on the amount of material present. An intensive property is the same regardless of how much material is present. a) Mass is an extensive property. Changing the amount of material will change the mass. b) Density is an intensive property. Changing the amount of material changes both the mass and the volume, but the ratio (density) remains fixed. c) Volume is an extensive property. Changing the amount of material will change the size (volume). d) The melting point is an intensive property. The melting point depends on the substance, not on the amount of substance.

1.20

1000 m  100 cm  2 2 ; to convert from m to cm , use 2 2 1 km  1 m c) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used. To convert distance, mi to m, use: 2

b) To convert from km to m , use

1.609 km  1000 m  = 1.609 × 103 m/mi   1 mi  1 km 

To convert time, h to s, use:  1 h  1 min  = 1 h/3600 s   60 s   60 min 

1.609103 m  1 h    0.4469 m  h .  Therefore, the complete conversion factor is    1 mi mi  s  3600 s  Do the units cancel when you start with a measurement of mi/h? Copyright

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1-12

1000 g . 2.205 lb 3  1 ft3  3 3  1 in  = 3.531 × 10–5 ft3/cm3. To convert volume from ft to cm use,   3 12 in3  2.54 cm  d) To convert from pounds (lb) to grams (g), use

1.21

Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion factor so that the unit initially given will cancel, leaving the desired unit. Solution: a) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used.  1 in  To convert distance, cm to in, use:    2.54 cm   1 min  To convert time, min to s, use:    60 s 

100 cm  1 in 3 3 ; to convert from cm to in , use 3 3 1 m  2.54 cm  c) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used.  1 km  To convert distance, m to km, use:    1000 m  2 2 To convert time, s to h , use: 2 2  60 s2  60 min   = 3600 s  1 min 2   1 h 2  h2 3

b) To convert from m to cm , use

d) This problem requires two conversion factors: one for volume and one for time. It does not matter which conversion is done first. Alternate methods may be used.  4 qt   1 L   ; to convert qt to L, use:   To convert volume, gal to qt, use:   1 gal   1.057 qt   1 h  To convert time, h to min, use:    60 min  –12

1.22

Plan: Use conversion factors from the inside back cover: 1 pm = 10 Solution: 1012 m  1 nm     = 1.43 nm Radius (nm) = 1430 pm   1 pm  109 m 

1.23

Plan: Use conversion factors from the inside back cover: 10 Solution:  1 pm  0.01   = 2.22 Radius ( ) = 2.22 1010 m 12   10 m  1 pm 

1.24

Plan: Use conversion factors: 0.01 m = 1 cm; 2.54 cm = 1 in. Solution:  1 cm   1 in   3 3 Length (in) = 100. m    = 3.9370 × 10 = 3.94 × 10 in  2.54 cm   0.01 m 

–12

–9

m; 10 m = 1 nm.

m = 1 pm; 1 pm = 0.01 .

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1-13

1.25

Plan: Use the conversion factor 12 in = 1 ft to convert 6 ft 10 in to height in inches. Then use the conversion factors 1 in = 2.54 cm; 1 cm = 10 mm. Solution: 12 in     10 in = 82 in Height (in) = 6 ft   1 ft   2.54 cm 10 mm     = 2.0828 × 103 = 2.1 × 103 mm Height (mm) = 82 in   1 cm   1 in 

  

1.26



Plan: Use conversion factors (1 cm) = (0.01 m) ; (1000 m) = (1 km) to express the area in km . To calculate 2 2 the cost of the patch, use the conversion factor: (2.54 cm) = (1 in) . Solution: 0.01 m2  1 km 2  2 –9 2  a) Area (km ) = 20.7 cm 2   2  = 2.07 × 10 km  1 cm2   1000 m 

 1 in 2  $3.25   = 10.4276 = $10.43  b) Cost = 20.7 cm 2   1 in 2  2.54 cm 2  1.27

–3

Plan: Use conversion factors (1 mm) = (10 m) ; (0.01 m) = (1 cm) ; (2.54 cm) = (1 in) ; (12 in) = (1 ft) to 2 express the area in ft . Solution: 2  103 m 2  2 2    2 2   1 cm   1 in   1 ft   a) Area (ft ) = 7903 mm   2   1 mm 2  0.01 m 2   12 in 2  2.54 cm     –2

–2

= 8.5067 × 10 = 8.507 × 10 ft

 45 s    = 2.634333 × 103 = 2.6 × 103 s b) Time (s) = 7903 mm 2  135 mm 2 





1.28

Plan: Use conversion factors 1 lb = 16 oz, 0.4536 kg = 1 lb, and 1000 g = 1 kg. Solution:  1 lb  0.4536 kg  1000 g      Mass (g) = 6.14 oz   1 lb  1 kg  = 174 kg 16 oz   

1.29

Plan: Use conversion factor 1 short ton = 2000 lb; 2.205 lb = 1 kg; 1000 kg = 1 metric ton. Solution:  2000 lb   1 kg   1 T   15 15 Mass (T) = 2.60 1015 ton   10 3 kg  = 2.35828 × 10 = 2.36 × 10 T  1 ton   2.205 lb  

1.30

Plan: Mass in g is converted to kg in part (a) with the conversion factor 1000 g = 1 kg; mass in g is converted to lb 3 3 in part (b) with the conversion factors 1000 g = 1 kg; 1 kg = 2.205 lb. Volume in cm is converted to m with the 3 3 3 3 3 3 3 conversion factor (1 cm) = (0.01 m) and to ft with the conversion factors (2.54 cm) = (1 in) ; (12 in) = (1 ft) . The conversions may be performed in any order. Solution:  5.52 g  1 cm3  3  1 kg  = 5.52 × 103 kg/m3 a) Density (kg/m ) =    3  3   cm 0.01 m 1000 g 

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1-14

3 3  5.52 g 2.54 cm   3 12 in   1 kg   2.205 lb  = 344.661 = 345 lb/ft3 b) Density (lb/ft ) =     3 3   1 kg  3     cm  1 in   1 ft  1000 g  

1.31

Plan: Length in m is converted to km in part (a) with the conversion factor 1000 m = 1 km; length in m is converted to mi in part (b) with the conversion factors 1000 m = 1 km; 1 km = 0.62 mi. Time is converted using the conversion factors 60 s = 1 min; 60 min = 1 h. The conversions may be performed in any order. Solution:  2.998108 m   60 s   60 min   1 km   9 9 a) Velocity (km/h) =     3  = 1.07928 × 10 = 1.079 × 10 km/h 1s 1 min   1 h  10 m    2.998108 m   60 s   1 km   0.62 mi   7 7 b) Velocity (mi/min) =    3   = 1.11526 × 10 = 1.1 × 10 mi/min 1s 1 min  10 m   1 km  

1.32

Plan:

–6

–3

= (1 × 10 m) ; (1 × 10 m) = (1 mm) to convert to mm . 3 –6 3 –2 3 3 3 = (1 × 10 m) ; (1 × 10 m) = (1 cm) ; 1 cm = 1 mL;

–3

1 mL = 1 × 10 L. Solution: 3  1106 m    1 mm 3     = 2.56 × 10–9 mm3/cell  3 3 3   1     110 m    3 3  1106 m   2.56   1 cm3    3     b) Volume (L) = 10 5 cells  3 3 3   1102 m  1 cm  1 mL  cell  1   

 2.56 a) Volume (mm ) =   cell

= 2.56 × 10

–10

= 10

L –3

1.33

  

–3

Plan: For part (a), convert from qt to mL (1 qt = 946.4 mL) to L (1 mL = 1 × 10 L) to m (1 L = 10 m ). –3 For part (b), convert from gal to qt (1 gal = 4 qt) to mL (1 qt = 946.4 mL) to L (1 mL = 10 L). Solution: 3 3 3  946.4 mL  3 10 L  10 m  = 9.464 × 10–4 m3 a) Volume (m ) = 1 qt     1 L   1 qt  1 mL   4 qt   946.4 mL 103 L   3 3 b) Volume (L) = 835 gal   = 3.160976 × 10 = 3.16 × 10 L     1 gal   1 qt   1 mL 



1.34



Plan: The mass of the mercury in the vial is the mass of the vial filled with mercury minus the mass of the empty vial. Use the density of mercury and the mass of the mercury in the vial to find the volume of mercury and thus the volume of the vial. Once the volume of the vial is known, that volume is used in part (b). The density of water is used to find the mass of the given volume of water. Add the mass of water to the mass of the empty vial. Solution: a) Mass (g) of mercury = mass of vial and mercury – mass of vial = 185.56 g – 55.32 g = 130.24 g  1 cm 3  3  = 9.626016 = 9.626 cm3 Volume (cm ) of mercury = volume of vial = 130.24 g 13.53 g  3

b) Volume (cm ) of water = volume of vial = 9.626016 cm  0.997 g  Mass (g) of water = 9.626016 cm3  = 9.59714 g water  1 cm 3  Mass (g) of vial filled with water = mass of vial + mass of water = 55.32 g + 9.59714 g = 64.91714 = 64.92 g Copyright

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1-15

1.35

Plan: The mass of the water in the flask is the mass of the flask and water minus the mass of the empty flask. Use the density of water and the mass of the water in the flask to find the volume of water and thus the volume of the flask. Once the volume of the flask is known, that volume is used in part (b). The density of chloroform is used to find the mass of the given volume of chloroform. Add the mass of the chloroform to the mass of the empty flask. Solution: a) Mass (g) of water = mass of flask and water – mass of flask = 489.1 g – 241.3 g = 247.8 g  1 cm 3   3  = 247.8 = 248 cm3 Volume (cm ) of water = volume of flask = 247.8 g  1.00 g 





b) Volume (cm ) of chloroform = volume of flask = 247.8 cm 1.48 g   = 366.744 g chloroform Mass (g) of chloroform = 247.8 cm3   cm3 





Mass (g) of flask and chloroform = mass of flask + mass of chloroform = 241.3 g + 366.744 g = 608.044 g = 608 g 1.36

Plan: Calculate the volume of the cube using the relationship Volume = (length of side) . The length of side in 3 mm must be converted to cm so that volume will have units of cm . Divide the mass of the cube by the volume to find density. Solution: 103 m  1 cm    = 1.56 cm (convert to cm to match density unit)  Side length (cm) = 15.6 mm   2  1 mm  10 m  3 3 3 3 Al cube volume (cm ) = (length of side) = (1.56 cm) = 3.7964 cm mass 10.25 g 3 Density (g/cm3 )   = 2.69993 = 2.70 g/cm volume 3.7964 cm3



1.37



Plan: Use the relationship c = 2 to find the radius of the sphere and the relationship V = 4/3 to find the 3 3 volume of the sphere. The volume in mm must be converted to cm . Divide the mass of the sphere by the volume to find density. Solution: c=2 c 32.5 mm = Radius (mm) = = 5.17254 mm 2 2 4 3  4  3 3 Volume (mm ) = =   (5.17254 mm)3 = 579.6958 mm  3 3 103 m   1 cm 3 3  = 0.5796958 cm3 Volume (cm ) = 579.6958 mm 3     1 mm  102 m  3

Density (g/cm3 )  1.38

mass 4.20 g 3  = 7.24518 = 7.25 g/cm 3 volume 0.5796958 cm

Plan: Use the equations given in the text for converting between the three temperature scales. Solution: 5 5 a) T (in °C) = [T (in °F) – 32] = [68°F – 32] = 20.°C 9 9 T (in K) = T (in °C) + 273.15 = 20.°C + 273.15 = 293.15 = 293 K

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1-16

b) T (in K) = T (in °C) + 273.15 = –164°C + 273.15 = 109.15 = 109 K 9 9 T (in °F) = T (in °C) + 32 = (–164°C) + 32 = –263.2 = –263°F 5 5 c) T (in °C) = T (in K) – 273.15 = 0 K – 273.15 = –273.15 = –273°C T (in °F) =

9 9 T (in °C) + 32 = (–273.15°C) + 32 = –459.67 = –460.°F 5 5

1.39

Plan: Use the equations given in the text for converting between the three temperature scales. Solution: 5 5 a) T (in °C) = [T (in °F) – 32] = [106°F – 32] = 41.111 = 41°C 9 9 (106 – 32) = 74 This limits the significant figures. T (in K) = T (in °C) + 273.15 = 41.111°C + 273.15 = 314.261 = 314 K 9 9 b) T (in °F) = T (in °C) + 32 = (3410°C) + 32 = 6170°F 5 5 T (in K) = T (in °C) + 273.15 = 3410°C + 273 = 3683 K 3 3 3 c) T (in °C) = T (in K) –273.15 = 6.1 × 10 K – 273 = 5.827 × 10 = 5.8 × 10 °C 9 9 4 4 T (in °F) = T (in °C) + 32 = (5827°C) + 32 = 1.0521 × 10 = 1.1 × 10 °F 5 5

1.40

Plan: Find the volume occupied by each metal by taking the difference between the volume of water and metal and the initial volume of the water (25.0 mL). Divide the mass of the metal by the volume of the metal to calculate density. Use the density value of each metal to identify the metal. Solution: Cylinder A: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 28.2 mL – 25.0 mL = 3.2 mL mass 25.0 g = Density = = 7.81254 = 7.8 g/mL volume 3.2 mL Cylinder A contains iron. Cylinder B: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 27.8 mL – 25.0 mL = 2.8 mL mass 25.0 g = Density = = 8.92857 = 8.9 g/mL volume 2.8 mL Cylinder B contains nickel. Cylinder C: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 28.5 mL – 25.0 mL = 3.5 mL mass 25.0 g = Density = = 7.14286 = 7.1 g/mL volume 3.5 mL Cylinder C contains zinc.

1.41

Plan: Use 1 nm = 10 m to convert wavelength in nm to m. To convert wavelength in pm to , use 1 pm = 0.01 . Solution: 109 m    = 2.47 × 10–7 m a) Wavelength (m) = 247 nm   1 nm 

–9





 0.01    = 67.6 b) Wavelength ( ) = 6760 pm   1 pm   





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1-17

1.42

Plan: The liquid with the larger density will occupy the bottom of the beaker, while the liquid with the smaller density volume will be on top of the more dense liquid. Solution: a) Liquid A is more dense than water; liquids B and C are less dense than water. b) Density of liquid B could be 0.94 g/mL. Liquid B is more dense than C so its density must be greater than 0.88 g/mL. Liquid B is less dense than water so its density must be less than 1.0 g/mL.

1.43

Plan: Calculate the volume of the cylinder in cm by using the equation for the volume of a cylinder. The 3 3 diameter of the cylinder must be halved to find the radius. Convert the volume in cm to dm by using the 3 –2 3 –1 3 3 conversion factors (1 cm) = (10 m) and (10 m) = (1 dm) . Solution: Radius = diameter/2 = 0.85 cm/2 = 0.425 cm 3 2 2 3 Volume (cm ) = h = (0.425 cm) (9.5 cm) = 5.3907766 cm 3 3 102 m   1 dm   3    = 5.39078 × 10–3 = 5.4 × 10–3 dm3 Volume (dm ) = 5.3907766 cm 3   1 cm  101 m 

1.44

Plan: Use the percent of copper in the ore to find the mass of copper in 5.01 lb of ore. Convert the mass in lb to mass in g. The density of copper is used to find the volume of that mass of copper. Use the volume equation for a cylinder to calculate the height of the cylinder (the length of wire); the diameter of the wire is used to find the radius which must be expressed in units of cm. Length of wire in cm must be converted to m. Solution:  66%  Mass (lb) of copper = 5.01 lb Covellite   = 3.3066 lb copper 100% 





 1 kg  1000 g   3   Mass (g) of copper = 3.3066 lb  1 kg  = 1.49959 × 10 g  2.205 lb    cm 3 Cu   3  = 167.552 cm3 Cu Volume (cm ) of copper = 1.49959 10 3 g Cu  8.95 g Cu  2 V= rh  6.304 103 in  2.54 cm   = 8.00608 × 10–3 cm  Radius (cm) =   2  1 in 

Height (length) in cm =

= 2

167.552 cm3



  8.0060810

3

–5

cm

= 8.3207 × 10 cm

10 m   = 8.3207 × 103 = 8.32 × 103 m Length (m) = 8.320710 5 cm  1 cm  An exact number is defined to have a certain value (exactly). There is no uncertainty in an exact number. An exact number is considered to have an infinite number of significant figures and, therefore, does not limit the digits in the calculation. 2

1.45

1.46

Random error of a measurement is decreased by (1) taking the average of more measurements. More measurements allow a more precise estimate of the true value of the measurement. Calibrating the instrument will allow greater accuracy but not necessarily greater precision.

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1-18

1.47

a) If the number is an exact count then there are an infinite number of significant figures. If it is not an exact count, there are only 5 significant figures. b) Other things, such as number of tickets sold, could have been counted instead. 4 c) A value of 15,000 to two significant figures is 1.5 × 10 . 4 Values would range from 14,501 to 15,499. Both of these values round to 1.5 × 10 .

1.48

Plan: Review the rules for significant zeros. Solution: a) No significant zeros (leading zeros are not significant) b) No significant zeros (leading zeros are not significant) c) 0.0410 (terminal zeros to the right of the decimal point are significant) 4 d) 4.0100 × 10 (zeros between nonzero digits and terminal zeros to the right of the decimal point are significant)

1.49

Plan: Review the rules for significant zeros. Solution: a) 5.08 (zeros between nonzero digits are significant) b) 508 (zeros between nonzero digits are significant) 3 c) 5.080 × 10 (zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant) d) 0.05080 (leading zeros are not significant; zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant)

1.50

Plan: Review the rules for rounding. Solution: (significant figures are underlined) a) 0.0003554: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 0.00036 b) 35.8348: the extra digits are 48. Since the digit to be removed (4) is less than 5, the last digit kept is unchanged: 35.83 c) 22.4555: the extra digits are 555. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 22.5

1.51

Plan: Review the rules for rounding. Solution: (significant figures are underlined) a) 231.554: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 231.6 b) 0.00845: the extra digit is 5 at the end of the number. When the digit to be removed is 5 and that 5 is not followed by nonzero numbers, the last digit kept remains unchanged if it is even and increased by 1 if it is odd: 0.0084 c) 144,000: the extra digits are 4000 at the end of the number. When the digit to be removed (4) is less than 5, the 5 last digit kept remains unchanged: 140,000 (or 1.4 × 10 )

1.52

Plan: Review the rules for rounding. Solution: 19 rounds to 20: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1. 155 rounds to 160: the digit to be removed is 5 and the digit to be kept is an odd number, so that digit kept is increased by 1. 8.3 rounds to 8: the digit to be removed (3) is less than 5 so the digit kept remains unchanged. 3.2 rounds to 3: the digit to be removed (2) is less than 5 so the digit kept remains unchanged. 2.9 rounds to 3: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1. 4.7 rounds to 5: the digit to be removed (7) is greater than 5 so the digit kept is increased by 1.

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1-19

 20 160 8  2  = 568.89 = 6 × 10  33 5  Since there are numbers in the calculation with only one significant figure, the answer can be reported only to one significant figure. (Note that the answer is 560 with the original number of significant digits.) 1.53

Plan: Review the rules for rounding. Solution: 10.8 rounds to 11: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 6.18 rounds to 6.2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 2.381 rounds to 2.38: the digit to be removed (1) is less than 5 so the digit kept remains unchanged. 24.3 rounds to 24: the digit to be removed (3) is less than 5 so the digit kept remains unchanged. 1.8 rounds to 2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 19.5 rounds to 20: the digit to be removed is 5 and the digit to be kept is odd, so that digit kept is increased by 1. 11 6.2  2.38    24  2  20  = 0.1691 = 0.2 Since there is a number in the calculation with only one significant figure, the answer can be reported only to one significant figure. (Note that the answer is 0.19 with original number of significant figures.)

1.54

Plan: Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the final answer. Solution: a)

2.795 m3.10 m = 1.3371 = 1.34 m (maximum of 3 significant figures allowed since two of the original 6.48 m

numbers in the calculation have only 3 significant figures) 3 4 3 b) V =   = 21,620.74 = 21,621 mm (maximum of 5 significant figures allowed)  3 c) 1.110 cm + 17.3 cm + 108.2 cm + 316 cm = 442.61 = 443 cm (no digits allowed to the right of the decimal since 316 has no digits to the right of the decimal point)



1.55



Plan: Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the final answer. Solution: 2.420 g  15.6 g a) = 3.7542 = 3.8 (maximum of 2 significant figures allowed since one of the original 4.8 g numbers in the calculation has only 2 significant figures) b)

7.87 mL 16.1 mL  8.44 mL

= 1.0274 = 1.0 (After the subtraction, the denominator has 2 significant figures; only one

digit is allowed to the right of the decimal in the value in the denominator since 16.1 has only one digit to the right of the decimal.) 2 3 c) V = (6.23 cm) (4.630 cm) = 564.556 = 565 cm (maximum of 3 significant figures allowed since one of the original numbers in the calculation has only 3 significant figures) 1.56

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1-20

Solution: 5 a) 1.310000 × 10 –4 b) 4.7 × 10 5 c) 2.10006 × 10 3 d) 2.1605 × 10

(Note that all zeros are significant.) (No zeros are significant.)

1.57

Plan: Review the procedure for changing a number to scientific notation. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Moving the decimal point to the left results in a positive exponent while moving the decimal point to the right results in a negative exponent. Solution: 2 a) 2.820 × 10 (Note that the zero is significant.) –2 b) 3.80 × 10 (Note the one significant zero.) 3 c) 4.2708 × 10 4 d) 5.82009 × 10

1.58

Plan: Review the examples for changing a number from scientific notation to standard notation. If the exponent is positive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left. Solution: a) 5550 (Do not use terminal decimal point since the zero is not significant.) b) 10070. (Use terminal decimal point since final zero is significant.) c) 0.000000885 d) 0.003004

1.59

Plan: Review the examples for changing a number from scientific notation to standard notation. If the exponent is positive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left. Solution: a) 6500. (Use terminal decimal point since the final zero is significant.) b) 0.0000346 c) 750 (Do not use terminal decimal point since the zero is not significant.) d) 188.56

1.60

Plan: In most cases, this involves a simple addition or subtraction of values from the exponents. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Solution: 4 2 2 4 a) 8.025 × 10 (The decimal point must be moved an additional 2 places to the left: 10 + 10 = 10 ) –3 3 –6 –3 b) 1.0098 × 10 (The decimal point must be moved an additional 3 places to the left: 10 + 10 = 10 ) –11 –2 –9 –11 c) 7.7 × 10 (The decimal point must be moved an additional 2 places to the right: 10 + 10 = 10 )

1.61

(The decimal point must be moved an additional 1 place to the left: 10 + 10 = 10 ) 2 –2 0 (The decimal point must be moved an additional 2 places to the left: 10 + 10 = 10 ) 3 –3 0 (The decimal point must be moved an additional 3 places to the left: 10 + 10 = 10 )

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1-21

1.62

6.626 10

34



J/s 2.9979 10 8 m/s 9

48910 m

= 4.06 × 10

–19

 = 4.062185 × 10

–19

–9

J (489 × 10 m limits the answer to 3 significant figures; units of m and s cancel)

6.0221023 molecules/mol1.2310 2 g 46.07 g/mol 24

= 1.6078 × 10

= 1.61 × 10 molecules (1.23 × 10 g limits answer to 3 significant figures; units of mol and g cancel) 1 1 5 c) 6.0221023 atoms/mol2.181018 J/atom 2  2  = 1.82333 × 10 2 3  5 –18 = 1.82 × 10 J/mol (2.18 × 10 J/atom limits answer to 3 significant figures; unit of atoms cancels) 1.63

Plan: Calculate a temporary answer by simply entering the numbers into a calculator. Then you will need to round the value to the appropriate number of significant figures. Cancel units as you would cancel numbers, and place the remaining units after your numerical answer. Solution: 4.3210 7 g 3 a) = 1.3909 = 1.39 g/cm 3 4 2 3.1416 1.9510 cm 3 7 (4.32 × 10 g limits the answer to 3 significant figures) 1.84 102 g44.7 m/s2 5 5 2 2 b) = 1.8382 × 10 = 1.84 × 10 g·m /s 2 2 (1.84 × 10 g limits the answer to 3 significant figures)

1.07104 mol / L 3.8103 mol / L c) = 0.16072 = 0.16 L/mol 3 8.35105 mol / L 1.48102 mol / L  2

–3

(3.8 × 10 mol/L limits the answer to 2 significant figures; mol /L in the numerator cancels mol /L in the denominator to leave mol/L in the denominator or units of L/mol) 1.64

Plan: Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a group are examples of exact numbers. Solution: a) The height of Angel Falls is a measured quantity. This is not an exact number. b) The number of planets in the solar system is a number count. This is an exact number. c) The number of grams in a pound is not a unit definition. This is not an exact number. d) The number of millimeters in a meter is a definition of the prefix “milli–.” This is an exact number.

1.65

Plan: Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a group are examples of exact numbers. Solution: a) The speed of light is a measured quantity. It is not an exact number. b) The density of mercury is a measured quantity. It is not an exact number. c) The number of seconds in an hour is based on the definitions of minutes and hours. This is an exact number. d) The number of states is a counted value. This is an exact number.

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1-22

1.66

Plan: Observe the figure, and estimate a reading the best you can. Solution: The scale markings are 0.2 cm apart. The end of the metal strip falls between the mark for 7.4 cm and 7.6 cm. If we assume that one can divide the space between markings into fourths, the uncertainty is one-fourth the separation between the marks. Thus, since the end of the metal strip falls between 7.45 and 7.55 we can report its length as 7.50 ± 0.05 cm. (Note: If the assumption is that one can divide the space between markings into halves only, then the result is 7.5 ± 0.1 cm.)

1.67

Plan: You are given the density values for five solvents. Use the mass and volume given to calculate the density of the solvent in the cleaner and compare that value to the density values given to identify the solvent. Use the uncertainties in the mass and volume to recalculate the density. Solution: mass 11.775 g  a) Density (g/mL)  = 0.7850 g/mL. The closest value is isopropanol. volume 15.00 mL b) Ethanol is denser than isopropanol. Recalculating the density using the maximum mass = (11.775 + 0.003) g with the minimum volume = (15.00 – 0.02) mL, gives mass 11.778 g Density (g/mL)   = 0.7862 g/mL. This result is still clearly not ethanol. volume 14.98 mL Yes, the equipment is precise enough.

1.68

Plan: Calculate the average of each data set. Remember that accuracy refers to how close a measurement is to the actual or true value while precision refers to how close multiple measurements are to each other. Solution: 8.72 g  8.74 g  8.70 g a) Iavg = = 8.7200 = 8.72 g 3 8.56 g  8.77 g  8.83 g IIavg = = 8.7200 = 8.72 g 3 8.50 g  8.48 g  8.51 g IIIavg = = 8.4967 = 8.50 g 3 8.41 g  8.72 g  8.55 g IVavg = = 8.5600 = 8.56 g 3 Sets I and II are most accurate since their average value, 8.72 g, is closest to the true value, 8.72 g. b) To get an idea of precision, calculate the range of each set of values: largest value – smallest value. A small range is an indication of good precision since the values are close to each other. Irange = 8.74 g – 8.70 g = 0.04 g IIrange = 8.83 g – 8.56 g = 0.27 g IIIrange = 8.51 g – 8.48 g = 0.03 g IVrange = 8.72 g – 8.41 g = 0.31 g Set III is the most precise (smallest range), but is the least accurate (the average is the farthest from the actual value). c) Set I has the best combination of high accuracy (average value = actual value) and high precision (relatively small range). d) Set IV has both low accuracy (average value differs from actual value) and low precision (has the largest range). Plan: Remember that accuracy refers to how close a measurement is to the actual or true value; since the bull’seye represents the actual value, the darts that are closest to the bull’s-eye are the most accurate. Precision refers to how close multiple measurements are to each other; darts that are positioned close to each other on the target have high precision.

1.69

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1-23

Solution: a) Experiments II and IV — the averages appear to be near each other. b) Experiments III and IV — the darts are closely grouped. c) Experiment IV and perhaps Experiment II — the average is in or near the bull’s-eye. d) Experiment III — the darts are close together, but not near the bull’s-eye. 1.70

Plan: If it is necessary to force something to happen, the potential energy will be higher. Solution: a) b)

a) The balls on the relaxed spring have a lower potential energy and are more stable. The balls on the compressed spring have a higher potential energy, because the balls will move once the spring is released. This configuration is less stable. b) The two + charges apart from each other have a lower potential energy and are more stable. The two + charges near each other have a higher potential energy, because they repel one another. This arrangement is less stable. 1.71

Plan: A physical change is one in which the physical form (or state) of a substance, but not its composition, is altered. A chemical change is one in which a substance is converted into a different substance with different composition and properties. Solution: a) Bonds have been broken in three yellow diatomic molecules. Bonds have been broken in three red diatomic molecules. The six resulting yellow atoms have reacted with three of the red atoms to form three molecules of a new substance. The remaining three red atoms have reacted with three blue atoms to form a new diatomic substance. b) There has been one physical change as the blue atoms at 273 K in the liquid phase are now in the gas phase at 473 K.

1.72

Plan: Use the concentrations of bromine given. Solution: Mass bromine in Dead Sea 0.50 g/L = = 7.7/1 Mass bromine in seawater 0.065 g/L

1.73

Plan: The swimming pool is a rectangle so the volume of the water can be calculated by multiplying the three dimensions of length, width, and the depth of the water in the pool. The depth in ft must be converted to units of 3 m before calculating the volume. The volume in m is then converted to volume in gal. The density of water is used to find the mass of this volume of water. Solution: 2 12 in   2.54 cm 10 m   a) Depth of water (m) = 4.8 ft   1 in  1 cm  = 1.46304 m  1 ft    3 3 Volume (m ) = length × width × depth = 50.0 m 25.0 m 1.46304 m = 1828.8 m 3 10 L  1.057 qt  1 gal   5 5 Volume (gal) = 1828.8 m 3   3   1 L  4 qt  = 4.8326 × 10 = 4.8 × 10 gal  1 m   











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1-24

b) Using the density of water = 1.0 g/mL.  4 qt  1000 mL 1.0 g  1 kg   6 6 Mass (kg) = 4.8326 10 5 gal     = 1.8288 × 10 = 1.8 × 10 kg  1 gal     1.057 qt  mL 1000 g 





1.74

Plan: In each case, calculate the overall density of the ball and contents and compare to the density of air. The 3 volume of the ball in cm is converted to units of L to find the density of the ball itself in g/L. The densities of the ball and the gas in the ball are additive because the volume of the ball and the volume of the gas are the same. Solution: a) Density of evacuated ball: the mass is only that of the sphere itself: 3  1 mL  10 L   Volume of ball (L) = 560 cm 3    = 0.560 = 0.56 L 3   1 mL  1 cm  mass 0.12 g   0.21 g/L Density of evacuated ball = volume 0.560 The evacuated ball will float because its density is less than that of air. b) Because the density of CO2 is greater than that of air, a ball filled with CO2 will sink. c) Density of ball + density of hydrogen = 0.0899 + 0.21 g/L = 0.30 g/L The ball will float because the density of the ball filled with hydrogen is less than the density of air. d) Because the density of O2 is greater than that of air, a ball filled with O2 will sink. e) Density of ball + density of nitrogen = 0.21 g/L + 1.165 g/L = 1.38 g/L The ball will sink because the density of the ball filled with nitrogen is greater than the density of air. 1.189 g   0.66584 g f) To sink, the total mass of the ball and gas must weigh 0.560 L  1 L  For the evacuated ball: 0.66584 – 0.12 g = 0.54585 = 0.55 g. More than 0.55 g would have to be added to make the ball sink. For ball filled with hydrogen:  0.0899 g   0.0503 g Mass of hydrogen in the ball = 0.56 L  1 L  Mass of hydrogen and ball = 0.0503 g + 0.12 g = 0.17 g 0.66584 – 0.17 g = 0.4958 = 0.50 g. More than 0.50 g would have to be added to make the ball sink.

1.75

Plan: Convert the crossto mm and then use the tensile strength of grunerite to find the mass that can be held up by a strand of grunerite with that cross-sectional area. Calculate the area of aluminum and steel that can match that mass. Solution:  1106 m 2   1 mm 2  2 2    = 1.0 × 10–6 mm2 Cross-sectional area (mm ) = 1.0  2 2    1  1103 m  

–6

Calculate the mass that can be held up by grunerite with a cross-sectional area of 1.0 × 10 mm :  3.5102 kg    3.5104 kg 1106 mm 2   1 mm 2  –4

Calculate the area of aluminum required to match a mass of 3.5 × 10 kg: 2 2  2.205 lb  2.54 cm2  10 mm   1.9916105 = 2.0 × 10–5 mm2  1 in    3.5104 kg   2 2    1 kg  2.510 4 lb  1 in   1 cm  –4

Calculate the area of steel required to match a mass of 3.5 × 10 kg: 2 2  2.205 lb  2.54 cm 2  10 mm   9.9580106 = 1.0 × 10–5 mm2  1 in    3.5104 kg  2 2    1 kg  5.0 10 4 lb  1 in  1 cm  Copyright

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1-25

1.76

Plan: Convert the surface area to m and then use the surface area and the depth to determine the volume of the 3 oceans (area × depth = volume) in m . The volume is then converted to liters, and finally to the mass of gold using the density of gold in g/L. Once the mass of the gold is known, its density is used to find the volume of that amount of gold. The mass of gold is converted to troy oz and the price of gold per troy oz gives the total price. Solution: 1000 m2  2  = 3.63 × 1014 m2 a) Area of ocean (m ) = 3.63108 km 2   1 km2  3 14 2 18 3 Volume of ocean (m ) = (area)(depth) = (3.63 × 10 m )(3800 m) = 1.3794 × 10 m  1 L  5.8109 g   = 8.00052 × 1012 = 8.0 × 1012 g Mass of gold (g) = 1.3794 1018 m 3  3 3  10 m  L  b) Use the density of gold to convert mass of gold to volume of gold:  1 cm3 0.01 m3  3  = 4.14535 × 105 = 4.1 × 105 m3  Volume of gold (m ) = 8.00052 1012 g  1 cm3  19.3 g  c) Value of gold = 

1.77





  

  

  



$4.11014

Plan: The mass of zinc in the sample of yellow zinc in part (a) is found from the percent of zinc in the sample. The mass of copper is found by subtracting the mass of zinc from the total mass of yellow zinc. In part (b), subtract the mass percent of zinc from 100 to find the mass percent of copper. Solution:   34% zinc  = 62.9 g Zn a) Mass of zinc in the 34% zinc sample = 185 g yellow zinc 100% yellow zinc 

  37% zinc  = 68.45 g Zn Mass of zinc in the 37% zinc sample = 185 g yellow zinc 100% yellow zinc  Mass copper = total mass – mass zinc Mass copper (34% zinc sample) = 185 g – 62.9 g = 122.1 = 122 g Mass copper (37% zinc sample) = 185 g – 68.45 g = 116.55 = 117 g 117 to 122 g copper b) The 34% zinc sample contains 100 – 34 = 66% copper. The 37% zinc sample contains 100 – 37 = 63% copper.  34% zinc   = 23.95 = 24 g Mass of zinc = 46.5 g copper   66% copper   37% zinc   = 27.31 = 27 g Mass of zinc = 46.5 g copper   63% copper  24 to 27 g zinc 1.78

Plan: Use the equations for temperature conversion given in the chapter. The mass of nitrogen is conserved when the gas is liquefied; the mass of the nitrogen gas equals the mass of the liquid nitrogen. Use the density of nitrogen gas to find the mass of the nitrogen; then use the density of liquid nitrogen to find the volume of that mass of liquid nitrogen. Solution: a) T (in °C) = T (in K) – 273.15 = 77.36 K – 273.15 = –195.79°C 9 9 b) T (in °F) = T(in °C) + 32 = (–195.79°C) + 32 = –320.422 = –320.42°F 5 5

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1-26

 4.566 g  c) Mass of liquid nitrogen = mass of gaseous nitrogen = 895.0 L   = 4086.57 g N2  1 L 

1.79

 1 L   = 5.0514 = 5.05 L Volume of liquid N2 = 4086.57 g  809 g  Plan: For part (a), convert mi to m and h to s. For part (b), time is converted from h to min and length from mi to km. For part (c), convert the distance in ft to mi and use the average speed in mi/h to find the time necessary to cover the given distance. Solution:  5.9 mi  1000 m   1 h   a) Speed (m/s) =    = 2.643 = 2.6 m/s  h  0.62 mi   3600 s   1 h  5.9 mi  1 km   b) Distance (km) = 98 min   = 15.543 = 16 km   60 min   h  0.62 mi 





 1 mi  1 h  c) Time (h) = 4.7510 4 ft    = 1.5248 = 1.5 h  5280 ft  5.9 mi 

If she starts running at 11:15 am, 1.5 hours later the time is 12:45 pm. 1.80

Plan: A physical change is one in which the physical form (or state) of a substance, but not its composition, is altered. A chemical change is one in which a substance is converted into a different substance with different composition and properties. A physical property is a characteristic shown by a substance itself, without interacting with or changing into other substances. A chemical property is a characteristic of a substance that appears as it interacts with, or transforms into, other substances. Solution: a) Scene A shows a physical change. The substance changes from a solid to a gas but a new substance is not formed. b) Scene B shows a chemical change. Two diatomic elements form from a diatomic compound. c) Both Scenes A and B result in different physical properties. Physical and chemical changes result in different physical properties. d) Scene B is a chemical change; therefore, it results in different chemical properties. e) Scene A results in a change in state. The substance changes from a solid to a gas.

1.81

Plan: In visualizing the problem, the two scales can be set next to each other. Solution: There are 50 divisions between the freezing point and boiling point of benzene on the °X scale and 74.6 divisions  50X  o o (80.1 C – 5.5 C) on the °C scale. So °X =  °C  74.6C  This does not account for the offset of 5.5 divisions in the °C scale from the zero point on the °X scale.  50X  So °X =  (°C – 5.5°C)  74.6C  Check: Plug in 80.1°C and see if result agrees with expected value of 50°X.  50X  So °X =  (80.1°C – 5.5°C) = 50°X  74.6C  Use this formula to find the freezing and boiling points of water on the °X scale.  50X  fpwater °X =  (0.00°C – 5.5°C) = 3.68°X = –3.7°X  74.6C 

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1-27

 50X  bpwater °X =  (100.0°C – 5.5°C) = 63.3°X  74.6C 

1.82

Plan: Determine the total mass of Earth’s crust in metric tons (t) by finding the volume of crust 3 3 (surface area × depth) in km and then in cm and then using the density to find the mass of this volume, using conversions from the inside back cover. The mass of each individual element comes from the concentration of that element multiplied by the mass of the crust. Solution: 3 10 3 Volume of crust (km ) = area × depth = 35 km 5.10 108 km 2  = 1.785 × 10 km 3 1000 m3  3  1 cm  = 1.785 × 1025 cm3 Volume of crust (cm ) = 1.7851010 km 3  3   1 km  0.01 m3   2.8 g  1 kg   1 t  = 4.998 × 1019 t  Mass of crust (t) = 1.7851025 cm 3   1000 kg  3 1 cm 1000 g  

 4.55105 g oxygen   = 2.2741 × 1025 = 2.3 × 1025 g oxygen Mass of oxygen (g) = 4.9981019 t   1t   2.72 105 g silicon   = 1.3595 × 1025 = 1.4 × 1025 g silicon Mass of silicon (g) = 4.9981019 t    1t  1104 g element   Mass of ruthenium = mass of rhodium = 4.9981019 t  1t   15

= 4.998 × 10 = 5 × 10 g each of ruthenium and rhodium

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1-28

CHAPTER 2 THE COMPONENTS OF MATTER FOLLOW---UP PROBLEMS 2.1A

Plan: An element has only one kind of atom; a compound is composed of at least two kinds of atoms. A mixture consists of two or more substances mixed together in the same container. Solution: (a) There is only one type of atom (blue) present, so this is an element. (b) Two different atoms (brown and green) appear in a fixed ratio of 1/1, so this is a compound. (c) These molecules consist of one type of atom (orange), so this is an element.

2.1B

Plan: An element has only one kind of atom; a compound is composed of at least two kinds of atoms. Solution: The circle on the left contains molecules with either only orange atoms or only blue atoms. This is a mixture of two different elements. In the circle on the right, the molecules are composed of one orange atom and one blue atom so this is a compound.

2.2A

Plan: Use the mass fraction of iron in fool’s gold to find the mass of fool’s gold that contains 86.2 g of iron. Subtract the amount of iron in the mass fraction from the amount of fool’s gold in the mass fraction to obtain the mass of sulfur in that amount of fool’s gold. Find the mass fraction of sulfur in fool’s gold and multiply the amount of fool’s gold by the mass fraction of sulfur to determine the mass of sulfur in the sample. Solution:   Mass (g) of fool’s gold = 86.2 g iron ×   = 185.195 g fool’s gold   Mass (g) of sulfur in 110.0 g of fool’s gold = 110.0 g fool’s gold --- 51.2 g iron = 58.8 g sulfur    = 98.995 = 99.0 g sulfur Mass (g) of sulfur = 185.195 g fool’s gold ×   

2.2B

Plan: Subtract the amount of silver from the amount of silver bromide to find the mass of bromine in 26.8 g of silver bromide. Use the mass fraction of silver in silver bromide to find the mass of silver in 3.57 g of silver bromide. Use the mass fraction of bromine in silver bromide to find the mass of bromine in 3.57 g of silver bromide. Solution: Mass (g) of bromine in 26.8 g silver bromide = 26.8 g silver bromide --- 15.4 g silver = 11.4 g bromine    = 2.05 g silver Mass (g) of silver in 3.57 g silver bromide = 3.57 g silver bromide   

 Mass (g) of bromine in 3.57 g silver bromide = 3.57 g silver bromide   2.3A

  = 1.52 g bromine 

Plan: The law of multiple proportions states that when two elements react to form two compounds, the different masses of element B that react with a fixed mass of element A is a ratio of small whole numbers. The law of definite composition states that the elements in a compound are present in fixed parts by mass. The law of mass conversation states that the total mass before and after a reaction is the same. Solution:

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2-1

The law of mass conservation is illustrated because the number of atoms does not change as the reaction proceeds (there are 14 red spheres and 12 black spheres before and after the reaction occurs). The law of multiple proportions is illustrated because two compounds are formed as a result of the reaction. One of the compounds has a ratio of 2 red spheres to 1 black sphere. The other has a ratio of 1 red sphere to 1 black sphere. The law of definite proportions is illustrated because each compound has a fixed ratio of red-to-black atoms. 2.3B

Plan: The law of multiple proportions states that when two elements react to form two compounds, the different masses of element B that react with a fixed mass of element A is a ratio of small whole numbers. Solution: Only Sample B shows two different bromine-fluorine compounds. In one compound there are three fluorine atoms for every one bromine atom; in the other compound, there is one fluorine atom for every bromine atom.

2.4A

Plan: The subscript (atomic number = Z ) gives the number of protons, and for an atom, the number of electrons. The atomic number identifies the element. The superscript gives the mass number (A) which is the total of the protons plus neutrons. The number of neutrons is simply the mass number minus the atomic number (A --- Z ). Solution: 46 + --0 Ti Z = 22 and A = 46, there are 22 p and 22 e and 46 --- 22 = 24 n 47 + --0 Ti Z = 22 and A = 47, there are 22 p and 22 e and 46 --- 22 = 25 n 48 + --0 Ti Z = 22 and A = 48, there are 22 p and 22 e and 46 --- 22 = 26 n 49 + --0 Ti Z = 22 and A = 49, there are 22 p and 22 e and 46 --- 22 = 27 n 50 + --0 Ti Z = 22 and A = 50, there are 22 p and 22 e and 46 --- 22 = 28 n

2.4B

Plan: The subscript (atomic number = Z ) gives the number of protons, and for an atom, the number of electrons. The atomic number identifies the element. The superscript gives the mass number (A) which is the total of the protons plus neutrons. The number of neutrons is simply the mass number minus the atomic number (A --- Z ). Solution: (a) Mass number --- number of neutrons = 88 --- 50 = 38 = atomic number. The element is strontium. (b) Mass number --- atomic number = 86 --- 38 = 48 = number of neutrons.

2.5A

Plan: First, divide the percent abundance value by 100 to obtain the fractional value for each isotope. Multiply each isotopic mass by the fractional value, and add the resulting masses to obtain silicon’s atomic mass. Solution: 28 28 29 29 Atomic Mass = ( Si mass) (fractional abundance of Si) + ( Si mass) (fractional abundance of Si) + 30 30 ( Si mass) (fractional abundance of Si) Atomic mass = (27.97693 amu)(0.9223) + 28.976495 amu)(0.0467) + (29.973770 amu)(0.0310) = 28.09 amu

2.5B

Plan: To find the percent abundance of each B isotope, let x equal the fractional abundance of B and (1 --- x) 11 10 equal the fractional abundance of B. Remember that atomic mass = isotopic mass of B x fractional abundance) 11 + (isotopic mass of B x fractional abundance). Solution: 10 10 11 11 Atomic Mass = ( B mass) (fractional abundance of B) + ( B mass) (fractional abundance of B) 10 11 10 11 Amount of B + Amount B = 1 (setting B = x gives B = 1 --- x) 10.81 amu = (10.0129 amu)(x) + (11.0093 amu) (1 --- x) 10.81 amu = 11.0093 --- 11.0093x + 10.0129 x 10.81 amu = 11.0093 --- 0.9964 x ---0.1993 = ---0.9964x x = 0.20; 1 --- x = 0.80 (10.81 --- 11.0093 limits the answer to 2 significant figures) Fraction x 100% = percent abundance. 10 11 % abundance of B = 20.%; % abundance of B = 80.%

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2-2

2.6A

Plan: Use the provided atomic numbers (the Z numbers) to locate these elements on the periodic table. The name of the element is on the periodic table or on the list of elements inside the front cover of the textbook. Use the periodic table to find the group/column number (listed at the top of each column) and the period/row number (listed at the left of each row) in which the element is located. Classify the element from the color coding in the periodic table. Solution: (a) Z = 14: Silicon, Si; Group 4A(14) and Period 3; metalloid (b) Z = 55: Cesium, Cs; Group 1A(1) and Period 6; main-group metal (c) Z = 54: Xenon, Xe; Group 8A(18) and Period 5; nonmetal

2.6B

Plan: Use the provided atomic numbers (the Z numbers) to locate these elements on the periodic table. The name of the element is on the periodic table or on the list of elements inside the front cover of the textbook. Use the periodic table to find the group/column number (listed at the top of each column) and the period/row number (listed at the left of each row) in which the element is located. Classify the element from the color coding in the periodic table. Solution: (a) Z = 12: Magnesium, Mg; Group 2A(2) and Period 3; main-group metal (b) Z = 7: Nitrogen, N; Group 5A(15) and Period 2; nonmetal (c) Z = 30: Zinc, Zn; Group 2B(12) and Period 4; transition metal

2.7A

Plan: Locate these elements on the periodic table and predict what ions they will form. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number ---8. Or, relate the element’s position to the nearest noble gas. Elements after a noble gas lose electrons to become positive ions, while those before a noble gas gain electrons to become negative ions. Solution: 2--a) 16S [Group 6A(16); 6 --- 8 = ---2]; sulfur needs to gain 2 electrons to match the number of electrons in 18Ar. + b) 37Rb [Group 1A(1)]; rubidium needs to lose 1 electron to match the number of electrons in 36Kr. 2+ c) 56Ba [Group 2A(2)]; barium needs to lose 2 electrons to match the number of electrons in 54Xe.

2.7B

Plan: Locate these elements on the periodic table and predict what ions they will form. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number ---8. Or, relate the element’s position to the nearest noble gas. Elements after a noble gas lose electrons to become positive ions, while those before a noble gas gain electrons to become negative ions. Solution: 2+ a) 38Sr [Group 2A(2)]; strontium needs to lose 2 electrons to match the number of electrons in 36Kr. 2--b) 8O [Group 6A(16); 6 --- 8 = ---2]; oxygen needs to gain 2 electrons to match the number of electrons in 10Ne. + c) 55Cs [Group 1A(1)]; cesium needs to lose 1 electron to match the number of electrons in 54Xe.

2.8A

Plan: When dealing with ionic binary compounds, the first name is that of the metal and the second name is that of the nonmetal. If there is any doubt, refer to the periodic table. The metal name is unchanged, while the nonmetal has an -ide suffix added to the nonmetal root. Solution: a) Zinc is the metal and oxygen is the nonmetal: zinc oxide. b) Silver is the metal and bromine is the nonmetal: silver bromide. c) Lithium is the metal and chlorine is the nonmetal: lithium chloride. d) Aluminum is the metal and sulfur is the nonmetal: aluminum sulfide.

2.8B

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2-3

Solution: a) Potassium is the metal and sulfur is the nonmetal: potassium sulfide. b) Barium is the metal and iodine is the nonmetal: barium iodide. c) Cesium is the metal and nitrogen is the nonmetal: cesium nitride. d) Sodium is the metal and hydrogen is the nonmetal: sodium hydride. 2.9A

Plan: Use the charges of the ions to predict the lowest ratio leading to a neutral compound. The sum of the total charges must be 0. Solution: 2+ 2--a) Zinc should form Zn and oxygen should form O ; these will combine to give ZnO. The charges cancel [+2 + ---2 = 0], so this is an acceptable formula. + --b) Silver should form Ag and bromine should form Br ; these will combine to give AgBr. The charges cancel [+1 + ---1 = 0], so this is an acceptable formula. + --c) Lithium should form Li and chlorine should form Cl ; these will combine to give LiCl. The charges cancel [+1 + ---1 = 0], so this is an acceptable formula. 3+ 2--d) Aluminum should form Al and sulfur should form S ; to produce a neutral combination the formula is Al2S3. This way the charges will cancel [2(+3) + 3(---2) = 0], so this is an acceptable formula.

2.9B

Plan: Use the charges of the ions to predict the lowest ratio leading to a neutral compound. The sum of the total charges must be 0. Solution: + 2--a) Potassium should form K and sulfur should form S ; these will combine to give K2S. The charges cancel [2(+1) + ---2 = 0], so this is an acceptable formula. 2+ – b) Barium should form Ba and iodine should form I , these will combine to give BaI2. The charges cancel [+2 + 2(–1) = 0], so this is an acceptable formula. + 3--c) Cesium should form Cs and nitrogen should form N ; these will combine to give Cs 3N. The charges cancel [3(+1) + ---3 = 0], so this is an acceptable formula. + --d) Sodium should form Na and hydrogen should form H ; to produce a neutral combination the formula is NaH. This way the charges will cancel (+1 + ---1 = 0), so this is an acceptable formula.

2.10A

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions are written first. Review the rules for nomenclature covered in the chapter. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: 4+ 2--a) The Roman numeral means that the lead is Pb ; oxygen produces the usual O . The neutral combination is [+4 + 2(---2) = 0], so the formula is PbO2. b) Sulfide (Group 6A(16)), like oxide, is ---2 (6 --- 8 = ---2). This is split between two copper ions, each of which must be +1. This is one of the two common charges for copper ions. The +1 charge on the copper is indicated with a Roman numeral. This gives the name copper(I) sulfide (common name = cuprous sulfide). c) Bromine (Group 7A(17)), like other elements in the same column of the periodic table, forms a ---1 ion. Two of these ions require a total of +2 to cancel them out. Thus, the iron must be +2 (indicated with a Roman numeral). This is one of the two common charges on iron ions. This gives the name iron(II) bromide (or ferrous bromide). 2+ --d) The mercuric ion is Hg , and two ---1 ions (Cl ) are needed to cancel the charge. This gives the formula HgCl2.

2.10B

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2-4

Solution: 2+ 3--a) The Roman numeral means that the copper ion is Cu ; nitride produces the usual N . The neutral combination 2+ 3--is [3(+2) + 2(---3) = 0], so the formula is Cu 3N2. Three Cu ions balance two N ions. 2+ b) The anion is iodide, I , and the formula shows two I . Therefore the cation must be Pb , lead(II) ion: PbI2 is lead(II) iodide. (The common name is plumbous iodide.) 3+ 2--c) Chromic is the common name for chromium(III) ion, Cr ; sulfide ion is S . To balance the charges, the formula is Cr2S3. [The systematic name is chromium(III) sulfide.] 2--2+ d) The anion is oxide, O , which requires that the cation be Fe . The name is iron(II) oxide. (The common name is ferrous oxide.) 2.11A

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Solution: 2+ --a) The cupric ion, Cu , requires two nitrate ions, NO3 , to cancel the charges. Trihydrate means three water molecules. These combine to give Cu(NO3)2 3H2O. 2+ --b) The zinc ion, Zn , requires two hydroxide ions, OH , to cancel the charges. These combine to give Zn(OH) 2. + --c) Lithium only forms the Li ion, so Roman numerals are unnecessary. The cyanide ion, CN , has the appropriate charge. These combine to give lithium cyanide.

2.11B

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Solution: + 2--a) Two ammonium ions, NH4 , are needed to balance the charge on one sulfate ion, SO4 . These combine to give (NH4)2SO4. --2+ b) The nickel ion is combined with two nitrate ions, NO3 , so the charge on the nickel ion is 2+, Ni . There are 6 water molecules (hexahydrate). Therefore, the name is nickel(II) nitrate hexahydrate. + --c) Potassium forms the K ion. The bicarbonate ion, HCO3 , has the appropriate charge to balance out one potassium ion. Therefore, the formula of this compound is KHCO3.

2.12A

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Make corrections accordingly. Solution: + 3--a) The ammonium ion is NH4 and the phosphate ion is PO4 . To give a neutral compound they should combine [3(+1) + (---3) = 0] to give the correct formula (NH4)3PO4. 3+ --b) Aluminum gives Al and the hydroxide ion is OH . To give a neutral compound they should combine [+3 + 3(---1) = 0] to give the correct formula Al(OH) 3. Parentheses are required around the polyatomic ion. 2+ c) Manganese is Mn, and Mg, in the formula, is magnesium. Magnesium only forms the Mg ion, so Roman --numerals are unnecessary. The other ion is HCO3 , which is called the hydrogen carbonate (or bicarbonate) ion. The correct name is magnesium hydrogen carbonate or magnesium bicarbonate.

2.12B

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Make corrections accordingly. Solution: 3----a) Either use the ‘‘-ic’’ suffix or the ‘‘(III)’’ but not both. Nitride is N , and nitrate is NO3 . This gives the correct name: chromium(III) nitrate (the common name is chromic nitrate). ----b) Cadmium is Cd, and Ca, in the formula, is calcium. Nitrate is NO3 , and nitrite is NO2 . The correct name is calcium nitrite. ----c) Potassium is K, and P, in the formula, is phosphorus. Perchlorate is ClO4 , and chlorate is ClO3 . Additionally, parentheses are not needed when there is only one of a given polyatomic ion. The correct formula is KClO3.

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2-5

2.13A

Plan: Use the name of the acid to determine the name of the anion of the acid. The name hydro______ic acid indicates that the anion is a monatomic nonmetal. The name ______ic acid indicates that the anion is an oxoanion with an --- ate ending. The name ______ous acid indicates that the anion is an oxoanion with an ---ite ending. Solution: --a) Chloric acid is derived from the chlorate ion, ClO3 . The ---1 charge on the ion requires one hydrogen. These combine to give the formula HClO3. --b) Hydrofluoric acid is derived from the fluoride ion, F . The ---1 charge on the ion requires one hydrogen. These combine to give the formula HF. ----c) Acetic acid is derived from the acetate ion, which may be written as CH3COO or as C 2H3O2 . The ---1 charge means that one H is needed. These combine to give the formula CH3COOH or HC2H3O2. --d) Nitrous acid is derived from the nitrite ion, NO2 . The ---1 charge on the ion requires one hydrogen. These combine to give the formula HNO2.

2.13B

Plan: Remove a hydrogen ion to determine the formula of the anion. Identify the corresponding name of the anion and use the name of the anion to name the acid. For the oxoanions, the -ate suffix changes to -ic acid and the -ite suffix changes to -ous acid. For the monatomic nonmetal anions, the name of the acid includes a hydro- prefix and the ---ide suffix changes to ---ic acid. Solution: --a) Removing a hydrogen ion from the formula H2SO3 gives the oxoanion HSO3 , hydrogen sulfite; removing two 2--hydrogen ions gives the oxoanion SO3 , sulfite. To name the acid, the ‘‘-ite of ‘‘sulfite’’ must be replaced with ‘‘-ous.’’ The corresponding name is sulfurous acid. --b) HBrO is an oxoacid containing the BrO ion (hypobromite ion). To name the acid, the ‘‘-ite’’ must be replaced with ‘‘-ous.’’ This gives the name: hypobromous acid. --c) HClO2 is an oxoacid containing the ClO2 ion (chlorite ion). To name the acid, the ‘‘-ite’’ must be replaced with ‘‘-ous.’’ This gives the name: chlorous acid. --d) HI is a binary acid containing the I ion (iodide ion). To name the acid, a ‘‘hydro-’’ prefix is used, and the ‘‘-ide’’ must be replaced with ‘‘-ic.’’ This gives the name: hydroiodic acid.

2.14A

Plan: Determine the names or symbols of each of the species present. Since these are binary compounds consisting of two nonmetals, the number of each type of atom is indicated with a Greek prefix. Solution: a) Sulfur trioxide ----- one sulfur and three (tri) oxygens, as oxide, are present. b) Silicon dioxide ----- one silicon and two (di) oxygens, as oxide, are present. c) N2O Nitrogen has the prefix ‘‘di’’ = 2, and oxygen has the prefix ‘‘mono’’ = 1 (understood in the formula). d) SeF6 Selenium has no prefix (understood as = 1), and the fluoride has the prefix ‘‘hexa’’ = 6.

2.14B

Plan: Determine the names or symbols of each of the species present. Since these are binary compounds consisting of two nonmetals, the number of each type of atom is indicated with a Greek prefix. Solution: a) Sulfur dichloride ----- one sulfur and two (di) chlorines, as chloride, are present. b) Dinitrogen pentoxide ----- two (di) nitrogen and five (penta) oxygens, as oxide, are present. Note that the ‘‘a’’ in ‘‘penta’’ is dropped when this prefix is combined with ‘‘oxide’’. c) BF3 Boron doesn’t have a prefix, so there is one boron atom present. Fluoride has the prefix ‘‘tri’’ = 3. d) IBr3 Iodine has no prefix (understood as = 1), and the bromide has the prefix ‘‘tri’’ = 3.

2.15A

Plan: Determine the names or symbols of each of the species present. For compounds between nonmetals, the number of atoms of each type is indicated by a Greek prefix. If both elements in the compound are in the same group, the one with the higher period number is named first. Solution: a) Suffixes are not used in the common names of the nonmetal listed first in the formula. Sulfur does not qualify for the use of a suffix. Chlorine correctly has an ‘‘ide’’ suffix. There are two of each nonmetal atom, so both names require a ‘‘di’’ prefix. This gives the name disulfur dichloride.

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2-6

b) Both elements are nonmetals, and there is just one nitrogen and one oxygen. These combine to give the formula NO. c) Br has a higher period number than Cl and should be named first. The three chlorides are correctly named. The correct name is bromine trichloride. 2.15B

Plan: Determine the names or symbols of each of the species present. For compounds between nonmetals, the number of atoms of each type is indicated by a Greek prefix. If both elements in the compound are in the same group, the one with the higher period number is named first. Solution: a) The name of the element phosphorus ends in ---us, not ---ous. Additionally, the prefix hexa- is shortened to hexbefore oxide. The correct name is tetraphosphorus hexoxide. b) Because sulfur is listed first in the formula (and has a lower group number), it should be named first. The fluorine should come second in the name, modified with an ---ide ending. The correct name is sulfur hexafluoride. c) Nitrogen’s symbol is N, not Ni. Additionally, the second letter of an element symbol should be lowercase (Br, not BR). The correct formula is NBr3.

2.16A

Plan: First, write a formula to match the name. Next, multiply the number of each type of atom by the atomic mass of that atom. Sum all the masses to get an overall mass. Solution: 2--a) The peroxide ion is O2 , which requires two hydrogen atoms to cancel the charge: H2O2. Molecular mass = (2 × 1.008 amu) + (2 × 16.00 amu) = 34.016 = 34.02 amu. +1 2--b) Two Cs ions are required to balance the charge on one CO3 ion: Cs2CO3; formula mass = (2 × 132.9 amu) + (1 × 12.01 amu) + (3 × 16.00 amu) = 325.81 = 325.8 amu.

2.16B

Plan: First, write a formula to match the name. Next, multiply the number of each type of atom by the atomic mass of that atom. Sum all the masses to get an overall mass. Solution: 2--a) Sulfuric acid contains the sulfate ion, SO4 , which requires two hydrogen atoms to cancel the charge: H2SO4; molecular mass = (2 × 1.008 amu) + 32.06 amu + (4 × 16.00 amu) = 98.076 = 98.08 amu. 2--+ b) The sulfate ion, SO4 , requires two +1 potassium ions, K , to give K2SO4; formula mass = (2 × 39.10 amu) + 32.06 amu + (4 × 16.00 amu) = 174.26 amu.

2.17A

Plan: Since the compounds only contain two elements, finding the formulas by counting each type of atom and developing a ratio. Name the compounds. Multiply the number of each type of atom by the atomic mass of that atom. Sum all the masses to get an overall mass. Solution: a) There are two brown atoms (sodium) for every red (oxygen). The compound contains a metal with a nonmetal. Thus, the compound is sodium oxide, with the formula Na2O. The formula mass is twice the mass of sodium plus the mass of oxygen: 2 (22.99 amu) + (16.00 amu) = 61.98 amu b) There is one blue (nitrogen) and two reds (oxygen) in each molecule. The compound only contains nonmetals. Thus, the compound is nitrogen dioxide, with the formula NO2. The molecular mass is the mass of nitrogen plus twice the mass of oxygen: (14.01 amu) + 2 (16.00 amu) = 46.01 amu.

2.17B

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2-7

b) There is one green (chlorine) and three golds (fluorine) in each molecule. The compound only contains nonmetals. Thus, the compound is chlorine trifluoride, with the formula ClF3. The molecular mass is the mass of chlorine plus three times the mass of fluorine: (35.45 amu) +3 (19.00 amu) = 92.45 amu. END---OF---CHAPTER PROBLEMS 2.1

Plan: Refer to the definitions of an element and a compound. Solution: Unlike compounds, elements cannot be broken down by chemical changes into simpler materials. Compounds contain different types of atoms; there is only one type of atom in an element.

2.2

Plan: Refer to the definitions of a compound and a mixture. Solution: 1) A compound has constant composition but a mixture has variable composition. 2) A compound has distinctly different properties than its component elements; the components in a mixture retain their individual properties. Plan: Recall that a substance has a fixed composition. Solution: a) The fixed mass ratio means it has constant composition, thus, it is a pure substance (compound). b) All the atoms are identical, thus, it is a pure substance (element). c) The composition can vary, thus, this is an impure substance (a mixture). d) The specific arrangement of different atoms means it has constant composition, thus, it is a pure substance (compound).

2.3

2.4

Plan: Remember that an element contains only one kind of atom while a compound contains at least two different elements (two kinds of atoms) in a fixed ratio. A mixture contains at least two different substances in a composition that can vary. Solution: a) The presence of more than one element (calcium and chlorine) makes this pure substance a compound. b) There are only atoms from one element, sulfur, so this pure substance is an element. c) This is a combination of two compounds and has a varying composition, so this is a mixture. d) The presence of more than one type of atom means it cannot be an element. The specific, not variable, arrangement means it is a compound.

2.5

Some elements, such as the noble gases (He, Ne, Ar, etc.) occur as individual atoms. Many other elements, such as most other nonmetals (O2, N2, S8, P4, etc.) occur as molecules.

2.6

Compounds contain atoms from two or more elements, thus the smallest unit must contain at least a pair of atoms in a molecule.

2.7

Mixtures have variable composition; therefore, the amounts may vary. Compounds, as pure substances, have constant composition so their composition cannot vary.

2.8

The tap water must be a mixture, since it consists of some unknown (and almost certainly variable) amount of dissolved substance in solution in the water.

2.9

Plan: Recall that an element contains only one kind of atom; the atoms in an element may occur as molecules. A compound contains two kinds of atoms (different elements). Solution: a) This scene has 3 atoms of an element, 2 molecules of one compound (with one atom each of two different elements), and 2 molecules of a second compound (with 2 atoms of one element and one atom of a second element).

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2-8

b) This scene has 2 atoms of one element, 2 molecules of a diatomic element, and 2 molecules of a compound (with one atom each of two different elements). c) This scene has 2 molecules composed of 3 atoms of one element and 3 diatomic molecules of the same element. 2.10

Plan: Recall that a mixture is composed of two or more substances physically mixed, with a composition that can vary. Solution: The street sample is a mixture. The mass of vitamin C per gram of drug sample can vary. Therefore, if several samples of the drug have the same mass of vitamin C per gram of sample, this is an indication that the samples all have a common source. Samples of the street drugs with varying amounts of vitamin C per gram of sample have different sources. The constant mass ratio of the components indicates mixtures that have the same composition by accident, not of necessity.

2.11

Separation techniques allow mixtures (with varying composition) to be separated into the pure substance components which can then be analyzed by some method. Only when there is a reliable way of determining the composition of a sample, can you determine if the composition is constant.

2.12

Plan: Restate the three laws in your own words. Solution: a) The law of mass conservation applies to all substances ----- elements, compounds, and mixtures. Matter can neither be created nor destroyed, whether it is an element, compound, or mixture. b) The law of definite composition applies to compounds only, because it refers to a constant, or definite, composition of elements within a compound. c) The law of multiple proportions applies to compounds only, because it refers to the combination of elements to form compounds.

2.13

In ordinary chemical reactions (i.e., those that do not involve nuclear transformations), mass is conserved and the law of mass conservation is still valid.

2.14

Plan: Review the three laws: law of mass conservation, law of definite composition, and law of multiple proportions. Solution: a) Law of Definite Composition ----- The compound potassium chloride, KCl, is composed of the same elements and same fraction by mass, regardless of its source (Chile or Poland). b) Law of Mass Conservation ----- The mass of the substances inside the flashbulb did not change during the chemical reaction (formation of magnesium oxide from magnesium and oxygen). c) Law of Multiple Proportions ----- Two elements, O and As, can combine to form two different compounds that have different proportions of As present.

2.15

Plan: The law of multiple proportions states that two elements can form two different compounds in which the proportions of the elements are different. Solution: Scene B illustrates the law of multiple proportions for compounds of chlorine and oxygen. The law of multiple proportions refers to the different compounds that two elements can form that have different proportions of the elements. Scene B shows that chlorine and oxygen can form both Cl2O, dichlorine monoxide, and ClO2, chlorine dioxide.

2.16

Plan: Review the definition of percent by mass. Solution: a) No, the mass percent of each element in a compound is fixed. The percentage of Na in the compound NaCl is 39.34% (22.99 amu/58.44 amu), whether the sample is 0.5000 g or 50.00 g.

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2-9

b) Yes, the mass of each element in a compound depends on the mass of the compound. A 0.5000 g sample of NaCl contains 0.1967 g of Na (39.34% of 0.5000 g), whereas a 50.00 g sample of NaCl contains 19.67 g of Na (39.34% of 50.00 g). 2.17

Generally no, the composition of a compound is determined by the elements used, not their amounts. If too much of one element is used, the excess will remain as unreacted element when the reaction is over.

2.18

Plan: Review the mass laws: law of mass conservation, law of definite composition, and law of multiple proportions. For each experiment, compare the mass values before and after each reaction and examine the ratios of the mass of white compound to the mass of colorless gas. Solution: Experiment 1: mass before reaction = 1.00 g; mass after reaction = 0.64 g + 0.36 g = 1.00 g Experiment 2: mass before reaction = 3.25 g; mass after reaction = 2.08 g + 1.17 g = 3.25 g Both experiments demonstrate the law of mass conservation since the total mass before reaction equals the total mass after reaction. Experiment 1: mass white compound/mass colorless gas = 0.64 g/0.36 g = 1.78 Experiment 2: mass white compound/mass colorless gas = 2.08 g/1.17 g = 1.78 Both Experiments 1 and 2 demonstrate the law of definite composition since the compound has the same composition by mass in each experiment.

2.19

Plan: Review the mass laws: law of mass conservation, law of definite composition, and law of multiple proportions. For each experiment, compare the mass values before and after each reaction and examine the ratios of the mass of reacted copper to the mass of reacted iodine. Solution: Experiment 1: mass before reaction = 1.27 g + 3.50 g = 4.77 g; mass after reaction = 3.81 g + 0.96 g = 4.77 g Experiment 2: mass before reaction = 2.55 g + 3.50 g = 6.05 g; mass after reaction = 5.25 g + 0.80 g = 6.05 g Both experiments demonstrate the law of mass conversation since the total mass before reaction equals the total mass after reaction. Experiment 1: mass of reacted copper = 1.27 g; mass of reacted iodine = 3.50 g --- 0.96 g = 2.54 g Mass reacted copper/mass reacted iodine = 1.27 g/2.54 g = 0.50 Experiment 2: mass of reacted copper = 2.55 g --- 0.80 g = 1.75 g; mass of reacted iodine = 3.50 g Mass reacted copper/mass reacted iodine = 1.75 g/3.50 g = 0.50 Both Experiments 1 and 2 demonstrate the law of definite composition since the compound has the same composition by mass in each experiment.

2.20

Plan: Fluorite is a mineral containing only calcium and fluorine. The difference between the mass of fluorite and the mass of calcium gives the mass of fluorine. Mass fraction is calculated by dividing the mass of element by the mass of compound (fluorite) and mass percent is obtained by multiplying the mass fraction by 100. Solution: a) Mass (g) of fluorine = mass of fluorite --- mass of calcium = 2.76 g --- 1.42 g = 1.34 g fluorine mass Ca 1.42 g Ca = b) Mass fraction of Ca = = 0.51449 = 0.514 mass fluorite 2.76 g fluorite mass F 1.34 g F Mass fraction of F = = = 0.48551 = 0.486 mass fluorite 2.76 g fluorite c) Mass percent of Ca = 0.51449 × 100 = 51.449 = 51.4% Mass percent of F = 0.48551 × 100 = 48.551 = 48.6%

2.21

Plan: Galena is a mineral containing only lead and sulfur. The difference between the mass of galena and the mass of lead gives the mass of sulfur. Mass fraction is calculated by dividing the mass of element by the mass of compound (galena) and mass percent is obtained by multiplying the mass fraction by 100.

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2-10

Solution: a) Mass (g) of sulfur = mass of galena --- mass of sulfur = 2.34 g --- 2.03 g = 0.31 g sulfur mass Pb 2.03 g Pb = b) Mass fraction of Pb = = 0.8675214 = 0.868 mass galena 2.34 g galena mass S 0.31 g S Mass fraction of S = = = 0.1324786 = 0.13 mass galena 2.34 g galena c) Mass percent of Pb = (0.8675214)(100) = 86.752 = 86.8% Mass percent of S = (0.1324786)(100) = 13.248 = 13% 2.22

Plan: Dividing the mass of magnesium by the mass of the oxide gives the ratio. Multiply the mass of the second sample of magnesium oxide by this ratio to determine the mass of magnesium. Solution: a) If 1.25 g of MgO contains 0.754 g of Mg, then the mass ratio (or fraction) of magnesium in the oxide mass Mg 0.754 g Mg = compound is = 0.6032 = 0.603. mass MgO 1.25 g MgO

 0.6032 g Mg    = 322.109 = 322 g magnesium b) Mass (g) of magnesium = 534 g MgO   1 g MgO 



2.23



Plan: Dividing the mass of zinc by the mass of the sulfide gives the ratio. Multiply the mass of the second sample of zinc sulfide by this ratio to determine the mass of zinc. Solution: a) If 2.54 g of ZnS contains 1.70 g of Zn, then the mass ratio (or fraction) of zinc in the sulfide compound is mass Zn 1.70 g Zn = = 0.66929 = 0.669. mass ZnS 2.54 g ZnS

 0.66929 kg Zn    = 2.5567 = 2.56 kg zinc b) Mass (g) of zinc = 3.82 kg ZnS   1 kg ZnS 



2.24



Plan: Since copper is a metal and sulfur is a nonmetal, the sample contains 88.39 g Cu and 44.61 g S. Calculate the mass fraction of each element in the sample by dividing the mass of element by the total mass of compound. Multiply the mass of the second sample of compound in grams by the mass fraction of each element to find the mass of each element in that sample. Solution: Mass (g) of compound = 88.39 g copper + 44.61 g sulfur = 133.00 g compound  88.39 g copper   = 0.664586 Mass fraction of copper =  133.00 g compound  103 g compound   0.664586 g copper  Mass (g) of copper = 5264 kg compound  1 g compound   1 kg compound   6

= 3.49838 × 10 = 3.498 × 10 g copper  44.61 g sulfur   = 0.335414 Mass fraction of sulfur =  133.00 g compound 

103 g compound   0.335414 g sulfur  Mass (g) of sulfur = 5264 kg compound   1 g compound   1 kg compound  6 6 = 1.76562 × 10 = 1.766 × 10 g sulfur

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2-11

2.25

Plan: Since cesium is a metal and iodine is a nonmetal, the sample contains 63.94 g Cs and 61.06 g I. Calculate the mass fraction of each element in the sample by dividing the mass of element by the total mass of compound. Multiply the mass of the second sample of compound by the mass fraction of each element to find the mass of each element in that sample. Solution: Mass of compound = 63.94 g cesium + 61.06 g iodine = 125.00 g compound  63.94 g cesium   = 0.51152 Mass fraction of cesium =  125.00 g compound 

 0.51152 g cesium    = 19.83163 = 19.83 g cesium Mass (g) of cesium = 38.77 g compound   1 g compound   61.06 g iodine   = 0.48848 Mass fraction of iodine =  125.00 g compound 





 0.48848 g iodine    = 18.9384 = 18.94 g iodine Mass (g) of iodine = 38.77 g compound   1 g compound 



2.26



Plan: The law of multiple proportions states that if two elements form two different compounds, the relative amounts of the elements in the two compounds form a whole-number ratio. To illustrate the law we must calculate the mass of one element to one gram of the other element for each compound and then compare this mass for the two compounds. The law states that the ratio of the two masses should be a small whole-number ratio such as 1:2, 3:2, 4:3, etc. Solution: 47.5 mass % S Compound 1: = 0.90476 = 0.905 52.5 mass % Cl Compound 2:

31.1 mass % S 68.9 mass % Cl

= 0.451379 = 0.451

0.90476 = 2.0044 = 2.00:1.00 0.451379 Thus, the ratio of the mass of sulfur per gram of chlorine in the two compounds is a small whole-number ratio of 2:1, which agrees with the law of multiple proportions. Ratio:

2.27

Plan: The law of multiple proportions states that if two elements form two different compounds, the relative amounts of the elements in the two compounds form a whole-number ratio. To illustrate the law we must calculate the mass of one element to one gram of the other element for each compound and then compare this mass for the two compounds. The law states that the ratio of the two masses should be a small whole-number ratio such as 1:2, 3:2, 4:3, etc. Solution: 77.6 mass % Xe Compound 1: = 3.4643 = 3.46 22.4 mass % F Compound 2:

63.3 mass % Xe 36.7 mass % F

= 1.7248 = 1.72

3.4643 = 2.0085 = 2.01:1.00 1.7248 Thus, the ratio of the mass of xenon per gram of fluorine in the two compounds is a small whole-number ratio of 2:1, which agrees with the law of multiple proportions. Ratio:

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2-12

2.28

Plan: Calculate the mass percent of calcium in dolomite by dividing the mass of calcium by the mass of the sample and multiply by 100. Compare this mass percent to that in fluorite. The compound with the larger mass percent of calcium is the richer source of calcium. Solution: 1.70 g calcium 100% = 21.767 = 21.8% Ca Mass percent calcium = 7.81 g dolomite Fluorite (51.4%) is the richer source of calcium.

2.29

Plan: Determine the mass percent of sulfur in each sample by dividing the grams of sulfur in the sample by the total mass of the sample and multiplying by 100. The coal type with the smallest mass percent of sulfur has the smallest environmental impact. Solution:  11.3 g sulfur  100% = 2.9894 = 2.99% S (by mass) Mass % in Coal A =   378 g sample 

 19.0 g sulfur  100% = 3.8384 = 3.84% S (by mass) Mass % in Coal B =   495 g sample   20.6 g sulfur  100% = 3.0519 = 3.05% S (by mass) Mass % in Coal C =   675 g sample  Coal A has the smallest environmental impact. 2.30

We now know that atoms of one element may change into atoms of another element. We also know that atoms of an element can have different masses (isotopes). Finally, we know that atoms are divisible into smaller particles. Based on the best available information in 1805, Dalton was correct. This model is still useful, since its essence (even if not its exact details) remains true today.

2.31

Plan: This question is based on the law of definite composition. If the compound contains the same types of atoms, they should combine in the same way to give the same mass percentages of each of the elements. Solution: Potassium nitrate is a compound composed of three elements ----- potassium, nitrogen, and oxygen ----- in a specific ratio. If the ratio of these elements changed, then the compound would be changed to a different compound, for example, to potassium nitrite, with different physical and chemical properties. Dalton postulated that atoms of an element are identical, regardless of whether that element is found in India or Italy. Dalton also postulated that compounds result from the chemical combination of specific ratios of different elements. Thus, Dalton’s theory explains why potassium nitrate, a compound comprised of three different elements in a specific ratio, has the same chemical composition regardless of where it is mined or how it is synthesized.

2.32

Plan: Review the discussion of the experiments in this chapter. Solution: Millikan determined the minimum charge on an oil drop and that the minimum charge was equal to the charge on one electron. Using Thomson’s value for the mass/charge ratio of the electron and the determined value for the ---28 charge on one electron, Millikan calculated the mass of an electron (charge/(charge/mass)) to be 9.109 × 10 g.

2.33

Plan: The charges on the oil droplets should be whole-number multiples of a minimum charge. Determine that minimum charge by dividing the charges by small integers to find the common factor. Solution: ---19 ---19 ---3.204 × 10 C/2 = ---1.602 × 10 C ---19 ---19 ---4.806 × 10 C/3 = ---1.602 × 10 C ---19 ---19 ---8.010 × 10 C/5 = ---1.602 × 10 C ---18 ---19 ---1.442 × 10 C/4 = ---1.602 × 10 C

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2-13

The value ---1.602 × 10

---19

C is the common factor and is the charge for the electron.

2.34

Thomson’s ‘‘plum pudding’’ model described the atom as a ‘‘blob’’ of positive charge with tiny electrons embedded in it. The electrons could be easily removed from the atoms when a current was applied and ejected as a stream of ‘‘cathode rays.’’

2.35

Rutherford and co-workers expected that the alpha particles would pass through the foil essentially unaffected, or perhaps slightly deflected or slowed down. The observed results (most passing through straight, a few deflected, a very few at large angles) were partially consistent with expectations, but the large-angle scattering could not be explained by Thomson’s model. The change was that Rutherford envisioned a small (but massive) positively charged nucleus in the atom, capable of deflecting the alpha particles as observed.

2.36

Plan: Re-examine the definitions of atomic number and the mass number. Solution: The atomic number is the number of protons in the nucleus of an atom. When the atomic number changes, the identity of the element also changes. The mass number is the total number of protons and neutrons in the nucleus of an atom. Since the identity of an element is based on the number of protons and not the number of neutrons, the mass number can vary (by a change in number of neutrons) without changing the identity of the element.

2.37

Plan: Recall that the mass number is the sum of protons and neutrons while the atomic number is the number of protons. Solution: Mass number (protons plus neutrons) --- atomic number (protons) = number of neutrons (c).

2.38

The actual masses of the protons, neutrons, and electrons are not whole numbers so their sum is not a whole number.

2.39

Plan: There is one peak for each type of Cl atom and peaks for the Cl2 molecule. The m/e ratio equals the mass divided by 1+. Solution: 35 37 a) There is one peak for the Cl atom and another peak for the Cl atom. There are three peaks for the three 35 35 37 37 35 37 possible Cl2 molecules: Cl Cl (both atoms are mass 35), Cl Cl (both atoms are mass 37), and Cl Cl (one atom is mass 35 and one is mass 37). So the mass of chlorine will have 5 peaks. b) Peak m/e ratio 35 Cl 35 lightest particle 37 Cl 37 35 35 Cl Cl 70 (35 + 35) 35 37 Cl Cl 72 (35 + 37) 37 37 Cl Cl 74 (35 + 37) heaviest particle

2.40

Plan: The superscript is the mass number, the sum of the number of protons and neutrons. Consult the periodic table to get the atomic number (the number of protons). The mass number --- the number of protons = the number of neutrons. For atoms, the number of protons and electrons are equal. Solution: Isotope Mass Number # of Protons # of Neutrons # of Electrons 36 Ar 36 18 18 18 38 Ar 38 18 20 18 40 Ar 40 18 22 18

2.41

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2-14

Solution: Isotope 35 Cl 37 Cl 2.42

Mass Number 35 37

# of Protons 17 17

# of Neutrons 18 20

# of Electrons 17 17

Plan: The superscript is the mass number (A), the sum of the number of protons and neutrons; the subscript is the atomic number (Z, number of protons). The mass number --- the number of protons = the number of neutrons. For atoms, the number of protons = the number of electrons. Solution: 16 17 a) 8 O and 8 O have the same number of protons and electrons (8), but different numbers of neutrons. 16 8

O and 178 O are isotopes of oxygen, and 168 O has 16 --- 8 = 8 neutrons whereas 178 O has 17 --- 8 = 9 neutrons. Same

Z value 40

b) 18 Ar and 19 K have the same number of neutrons (Ar: 40 --- 18 = 22; K: 41 --- 19 = 22) but different numbers of protons and electrons (Ar = 18 protons and 18 electrons; K = 19 protons and 19 electrons). Same N value 60

c) 27 Co and 28 Ni have different numbers of protons, neutrons, and electrons. Co: 27 protons, 27 electrons, and 60 --- 27 = 33 neutrons; Ni: 28 protons, 28 electrons and 60 --- 28 = 32 neutrons. However, both have a mass number of 60. Same A value 2.43

a) 1 H and 2 He have different numbers of protons, neutrons, and electrons. H: 1 proton, 1 electron, and 3 --- 1 = 2 neutrons; He: 2 protons, 2 electrons, and 3 --- 2 = 1 neutron. However, both have a mass number of 3. Same A value 14 15 b) 6 C and 7 N have the same number of neutrons (C: 14 --- 6 = 8; N: 15 --- 7 = 8) but different numbers of protons and electrons (C = 6 protons and 6 electrons; N = 7 protons and 7 electrons). Same N value c)

19 9

F and 189 F have the same number of protons and electrons (9), but different numbers of neutrons.

19 9

F and 189 F are isotopes of oxygen, and 199 F has 19 --- 9 = 10 neutrons whereas 189 F has 18 --- 9 = 9 neutrons. Same Z value 2.44

Plan: Combine the particles in the nucleus (protons + neutrons) to give the mass number (superscript, A). The number of protons gives the atomic number (subscript, Z) and identifies the element. Solution: 38 a) A = 18 + 20 = 38; Z = 18; 18 Ar 55

b) A = 25 + 30 = 55; Z = 25; 25 Mn 109

c) A = 47 + 62 = 109; Z = 47; 47 Ag 2.45

b) A = 40 + 50 = 90; Z = 40; 40 Zr 61

c) A = 28 + 33 = 61; Z = 28; 28 Ni Copyright

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2-15

2.46

Plan: Determine the number of each type of particle. The superscript is the mass number (A) and the subscript is the atomic number (Z, number of protons). The mass number --- the number of protons = the number of neutrons. For atoms, the number of protons = the number of electrons. The protons and neutrons are in the nucleus of the atom. Solution: 48 79 11 a) 22Ti b) 34 Se c) 5 B 22 protons

34 protons

5 protons

22 electrons

34 electrons

5 electrons

48 --- 22 = 26 neutrons

79 --- 34 = 45 neutrons

11 --- 5 = 6 neutrons

2.47

a) 82 Pb 82 protons 82 electrons 207 --- 82 = 125 neutrons

b) 4 Be 4 protons 4 electrons 9 --- 4 = 5 neutrons

c) 33 As 33 protons 33 electrons 75 --- 33 = 42 neutrons

2.48

Plan: To calculate the atomic mass of an element, take a weighted average based on the natural abundance of the isotopes: (isotopic mass of isotope 1 × fractional abundance) + (isotopic mass of isotope 2 × fractional abundance). Solution:  60.11%   39.89%  Atomic mass of gallium = 68.9256 amu   70.9247 amu  = 69.7230 = 69.72 amu  100%   100% 

2.49

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2-16

Solution:

 78.99%  10.00%  11.01%  Atomic mass of Mg = 23.9850 amu    24.9858 amu    25.9826 amu    100%   100%   100% 













= 24.3050 = 24.31 amu 35

2.50

Plan: To find the percent abundance of each Cl isotope, let x equal the fractional abundance of Cl and (1 --- x) 37 equal the fractional abundance of Cl since the sum of the fractional abundances must equal 1. Remember that 35 37 atomic mass = (isotopic mass of Cl × fractional abundance) + (isotopic mass of Cl × fractional abundance). Solution: 35 37 Atomic mass = (isotopic mass of Cl × fractional abundance) + (isotopic mass of Cl × fractional abundance) 35.4527 amu = 34.9689 amu(x) + 36.9659 amu(1 --- x) 35.4527 amu = 34.9689 amu(x) + 36.9659 amu --- 36.9659 amu(x) 35.4527 amu = 36.9659 amu --- 1.9970 amu(x) 1.9970 amu(x) = 1.5132 amu x = 0.75774 and 1 --- x = 1 --- 0.75774 = 0.24226 35 37 % abundance Cl = 75.774% % abundance Cl = 24.226%

2.51

Plan: To find the percent abundance of each Cu isotope, let x equal the fractional abundance of Cu and (1 --- x) 65 equal the fractional abundance of Cu since the sum of the fractional abundances must equal 1. Remember that 63 65 atomic mass = (isotopic mass of Cu × fractional abundance) + (isotopic mass of Cu × fractional abundance). Solution: 63 65 Atomic mass = (isotopic mass of Cu × fractional abundance) + (isotopic mass of Cu × fractional abundance) 63.546 amu = 62.9396 amu(x) + 64.9278 amu(1 --- x) 63.546 amu = 62.9396 amu(x) + 64.9278 amu --- 64.9278 amu(x) 63.546 amu = 64.9278 amu --- 1.9882 amu(x) 1.9882 amu(x) = 1.3818 amu x = 0.69500 and 1 --- x = 1 --- 0.69500 = 0.30500 63 65 % abundance Cu = 69.50% % abundance Cu = 30.50%

2.52

Iodine has more protons in its nucleus (higher Z), but iodine atoms must have, on average, fewer neutrons than Te atoms and thus a lower atomic mass.

2.53

Plan: Review the section in the chapter on the periodic table. Solution: a) In the modern periodic table, the elements are arranged in order of increasing atomic number. b) Elements in a column or group (or family) have similar chemical properties, not those in the same period or row. c) Elements can be classified as metals, metalloids, or nonmetals.

2.54

The metalloids lie along the ‘‘staircase’’ line, with properties intermediate between metals and nonmetals.

2.55

Plan: Review the section on the classification of elements as metals, nonmetals, or metalloids. Solution: To the left of the “staircase” are the metals, which are generally hard, shiny, malleable, ductile, good conductors of heat and electricity, and form positive ions by losing electrons. To the right of the “staircase” are the nonmetals, which are generally soft or gaseous, brittle, dull, poor conductors of heat and electricity, and form negative ions by gaining electrons.

2.56

Plan: Review the properties of these two columns in the periodic table. Solution: The alkali metals (Group 1A(1)) are metals and readily lose one electron to form cations whereas the halogens (Group 7A(17)) are nonmetals and readily gain one electron to form anions.

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2-17

2.57

Plan: Locate each element on the periodic table. The Z value is the atomic number of the element. Metals are to the left of the ‘‘staircase,’’ nonmetals are to the right of the ‘‘staircase,’’ and the metalloids are the elements that lie along the ‘‘staircase’’ line. Solution: a) Germanium Ge 4A(14) metalloid b) Phosphorus P 5A(15) nonmetal c) Helium He 8A(18) nonmetal d) Lithium Li 1A(1) metal e) Molybdenum Mo 6B(6) metal

2.58

Plan: Locate each element on the periodic table. The Z value is the atomic number of the element. Metals are to the left of the ‘‘staircase,’’ nonmetals are to the right of the ‘‘staircase,’’ and the metalloids are the elements that lie along the ‘‘staircase’’ line. Solution: a) Arsenic As 5A(15) metalloid b) Calcium Ca 2A(2) metal c) Bromine Br 7A(17) nonmetal d) Potassium K 1A(1) metal e) Aluminum Al 3A(13) metal

2.59

Plan: Review the section in the chapter on the periodic table. Remember that alkaline earth metals are in Group 2A(2), the halogens are in Group 7A(17), and the metalloids are the elements that lie along the ‘‘staircase’’ line; periods are horizontal rows. Solution: a) The symbol and atomic number of the heaviest alkaline earth metal are Ra and 88. b) The symbol and atomic number of the lightest metalloid in Group 4A(14) are Si and 14. c) The symbol and atomic mass of the coinage metal whose atoms have the fewest electrons are Cu and 63.55 amu. d) The symbol and atomic mass of the halogen in Period 4 are Br and 79.90 amu.

2.60

Plan: Review the section in the chapter on the periodic table. Remember that the noble gases are in Group 8A(18), the alkali metals are in Group 1A(1), and the transition elements are the groups of elements located between Groups 2A(s) and 3A(13); periods are horizontal rows and metals are located to the left of the ‘‘staircase’’ line. Solution: a) The symbol and atomic number of the heaviest nonradioactive noble gas are Xe and 54, respectively. b) The symbol and group number of the Period 5 transition element whose atoms have the fewest protons are Y and 3B(3). c) The symbol and atomic number of the only metallic chalcogen are Po and 84. d) The symbol and number of protons of the Period 4 alkali metal atom are K and 19.

2.61

Plan: Review the section of the chapter on the formation of ionic compounds. Solution: Reactive metals and nometals will form ionic bonds, in which one or more electrons are transferred from the metal atom to the nonmetal atom to form a cation and an anion, respectively. The oppositely charged ions attract, forming the ionic bond.

2.62

Plan: Review the section of the chapter on the formation of covalent compounds. Solution: Two nonmetals will form covalent bonds, in which the atoms share two or more electrons.

2.63

The total positive charge of the cations is balanced by the total negative charge of the anions.

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2-18

2.64

Plan: Assign charges to each of the ions. Since the sizes are similar, there are no differences due to the sizes. Solution: Coulomb’s law states the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of charges in MgO (+2 × ---2 = ---4) is greater than the product of charges in LiF (+1 × ---1 = ---1). Thus, MgO has stronger ionic bonding.

2.65

There are no molecules; BaF2 is an ionic compound consisting of Ba and F ions.

2.66

There are no ions present; P and O are both nonmetals, and they will bond covalently to form P 4O6 molecules.

2.67

Plan: Locate these groups on the periodic table and assign charges to the ions that would form. Solution: + + + The monatomic ions of Group 1A(1) have a +1 charge (e.g., Li , Na , and K ) whereas the monatomic ions of ------Group 7A(17) have a ---1 charge (e.g., F , Cl , and Br ). Elements gain or lose electrons to form ions with the same number of electrons as the nearest noble gas. For example, Na loses one electron to form a cation with the same number of electrons as Ne. The halogen F gains one electron to form an anion with the same number of electrons as Ne.

2.68

Plan: A metal and a nonmetal will form an ionic compound. Locate these elements on the periodic table and predict their charges. Solution: Magnesium chloride (MgCl2) is an ionic compound formed from a metal (magnesium) and a nonmetal (chlorine). 2+ Magnesium atoms transfer electrons to chlorine atoms. Each magnesium atom loses two electrons to form a Mg ion and the same number of electrons (10) as the noble gas neon. Each chlorine atom gains one electron to form a --2+ --Cl ion and the same number of electrons (18) as the noble gas argon. The Mg and Cl ions attract each other to 2+ --form an ionic compound with the ratio of one Mg ion to two Cl ions. The total number of electrons lost by the magnesium atoms equals the total number of electrons gained by the chlorine atoms.

2.69

Plan: A metal and a nonmetal will form an ionic compound. Locate these elements on the periodic table and predict their charges. Solution: Potassium sulfide (K2S) is an ionic compound formed from a metal (potassium) and a nonmetal (sulfur). Potassium atoms transfer electrons to sulfur atoms. Each potassium atom loses one electron to form an ion with +1 charge and the same number of electrons (18) as the noble gas argon. Each sulfur atom gains two electrons to form an ion with a ---2 charge and the same number of electrons (18) as the noble gas argon. The oppositely + 2--+ 2--charged ions, K and S , attract each other to form an ionic compound with the ratio of two K ions to one S ion. The total number of electrons lost by the potassium atoms equals the total number of electrons gained by the sulfur atoms.

2.70

Plan: Recall that ionic bonds occur between metals and nonmetals, whereas covalent bonds occur between nonmetals. Solution: --KNO3 shows both ionic and covalent bonding, covalent bonding between the N and O in NO 3 and ionic bonding --+ between the NO3 and the K .

2.71

Plan: Locate these elements on the periodic table and predict what ions they will form. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Solution: + --Potassium (K) is in Group 1A(1) and forms the K ion. Bromine (Br) is in Group 7A(17) and forms the Br ion (7 --- 8 = ---1).

2.72

---

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2-19

Solution: 2+ 2--Radium in Group 2A(2) forms a +2 ion: Ra . Selenium in Group 6A(16) forms a ---2 ion: Se (6 --- 8 = ---2). 2.73

Plan: Use the number of protons (atomic number) to identify the element. Add the number of protons and neutrons together to get the mass number. Locate the element on the periodic table and assign its group and period number. Solution: a) Oxygen (atomic number = 8) mass number = 8p + 9n = 17 Group 6A(16) Period 2 b) Fluorine (atomic number = 9) mass number = 9p + 10n = 19 Group 7A(17) Period 2 c) Calcium (atomic number = 20) mass number = 20p + 20n = 40 Group 2A(2) Period 4

2.74

Plan: Use the number of protons (atomic number) to identify the element. Add the number of protons and neutrons together to get the mass number. Locate the element on the periodic table and assign its group and period number. Solution: a) Bromine (atomic number = 35) mass number = 35p + 44n = 79 Group 7A(17) Period 4 b) Nitrogen (atomic number = 7) mass number = 7p + 8n = 15 Group 5A(15) Period 2 c) Rubidium (atomic number = 37) mass number = 37p + 48n = 85 Group 1A(1) Period 5

2.75

Plan: Determine the charges of the ions based on their position on the periodic table. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Next, determine the ratio of the charges to get the ratio of the ions. Solution: + 2--Lithium [Group 1A(1)] forms the Li ion; oxygen [Group 6A(16)] forms the O ion (6 --- 8 = ---2). The ionic + compound that forms from the combination of these two ions must be electrically neutral, so two Li ions 2--+ 2--combine with one O ion to form the compound Li2O. There are twice as many Li ions as O ions in a sample of Li2O.  1 O2 ion  2-- = 4.2 × 10 21 O2--- ions Number of O ions = (8.4 1021 Li ions)   2 Li ions 

2.76

Plan: Determine the charges of the ions based on their position on the periodic table. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Next, determine the ratio of the charges to get the ratio of the ions. Solution: 2+ --Ca [Group 2A(2)] forms Ca and I [Group 7A(17)] forms I ions (7 --- 8 = ---1). The ionic compound that forms 2+ --from the combination of these two ions must be electrically neutral, so one Ca ion combines with two I ions to --2+ form the compound CaI2. There are twice as many I ions as Ca ions in a sample of CaI2.  2 I ions  -- = 1.48 × 1022 = 1.5 × 10 22 I--- ions Number of I ions = (7.4 1021 Ca 2 ions)  2 1 Ca ion 

2.77

Plan: The key is the size of the two alkali metal ions. The charges on the sodium and potassium ions are the same as both are in Group 1A(1), so there will be no difference due to the charge. The chloride ions are the same in size and charge, so there will be no difference due to the chloride ion. Solution: Coulomb’s law states that the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of the charges is the same in both compounds because both sodium and potassium ions have a +1 charge. Attraction increases as distance decreases, + so the ion with the smaller radius, Na , will form a stronger ionic interaction (NaCl).

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2-20

2.78

Plan: The key is the charge of the two metal ions. The sizes of the lithium and magnesium ions are about the same (magnesium is slightly smaller), so there will be little difference due to ion size. The oxide ions are the same in size and charge, so there will be no difference due to the oxide ion. Solution: Coulomb’s law states the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of charges in MgO (+2 × ---2 = ---4) is greater than the product of charges in Li2O (+1 × ---2 = ---2). Thus, MgO has stronger ionic bonding.

2.79

Plan: Review the definition of molecular formula. Solution: The subscripts in the formula, MgF2, give the number of ions in a formula unit of the ionic compound. The --2+ subscripts indicate that there are two F ions for every one Mg ion. Using this information and the mass of each element, we could calculate the percent mass of each element.

2.80

Plan: Review the definitions of molecular and structural formulas. Solution: Both the structural and molecular formulas show the actual numbers of the atoms of the molecule; in addition, the structural formula shows the arrangement of the atoms (i.e., how the atoms are connected to each other).

2.81

Plan: Review the concepts of atoms and molecules. Solution: The mixture is similar to the sample of hydrogen peroxide in that both contain 20 billion oxygen atoms and 20 billion hydrogen atoms since both O2 and H2O2 contain 2 oxygen atoms per molecule and both H2 and H2O2 contain 2 hydrogen atoms per molecule. They differ in that they contain different types of molecules: H2O2 molecules in the hydrogen peroxide sample and H2 and O2 molecules in the mixture. In addition, the mixture contains 20 billion molecules (10 billion H2 molecules + 10 billion O2 molecules) while the hydrogen peroxide sample contains 10 billion molecules.

2.82

Plan: Review the rules for naming compounds. Solution: Roman numerals are used when naming ionic compounds that contain a metal that can form more than one ion. This is generally true for the transition metals, but it can be true for some non-transition metals as well (e.g., Sn).

2.83

Plan: Review the rules for naming compounds. Solution: Greek prefixes are used only in naming covalent compounds.

2.84

Molecular formulas cannot be written for ionic compounds since they only have ions and there are no molecules.

2.85

Plan: Locate each of the individual elements on the periodic table, and assign charges to each of the ions. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: a) Sodium is a metal that forms a +1 (Group 1A) ion and nitrogen is a nonmetal that forms a ---3 ion (Group 5A, 5 --- 8 = ---3). +3 ---3 +1 ---3 +1 Na N Na3N The compound is Na3N, sodium nitride. b) Oxygen is a nonmetal that forms a ---2 ion (Group 6A, 6 --- 8 = ---2) and strontium is a metal that forms a +2 ion (Group 2A). +2 ---2 Sr O The compound is SrO, strontium oxide.

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2-21

c) Aluminum is a metal that forms a +3 ion (Group 3A) and chlorine is a nonmetal that forms a ---1 ion (Group 7A, 7--- 8 = ---1). +3 ---3 +3 ---1 +3 ---1 Al Cl AlCl3 The compound is AlCl3, aluminum chloride. 2.86

Plan: Locate each of the individual elements on the periodic table, and assign charges to each of the ions. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: a) Cesium is a metal that forms a +1 (Group 1A) ion and bromine is a nonmetal that forms a ---1 ion (Group 7A, 7 --- 8 = ---1). +1 ---1 Cs Br The compound is CsBr, cesium bromide. b) Sulfur is a nonmetal that forms a ---2 ion (Group 6A, 6 --- 8 = ---2) and barium is a metal that forms a +2 ion (Group 2A). +2 ---2 Ba S The compound is BaS, barium sulfide. c) Fluorine is a nonmetal that forms a ---1 ion (Group 7A, 7 --- 8 = ---1) and calcium is a metal that forms a +2 ion (Group 2A). +2 ---2 +2 ---1 +2 ---1 Ca F CaF2 The compound is CaF2, calcium fluoride.

2.87

Plan: Based on the atomic numbers (the subscripts) locate the elements on the periodic table. Once the atomic numbers are located, identify the element and based on its position, assign a charge. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: 2+ a) 12L is the element Mg (Z = 12). Magnesium [Group 2A(2)] forms the Mg ion. 9M is the element F (Z = 9). --Fluorine [Group 7A(17)] forms the F ion (7 --- 8 = ---1). The compound formed by the combination of these two elements is MgF2, magnesium fluoride. 2+ b) 30L is the element Zn (Z = 30). Zinc forms the Zn ion (see Table 2.3). 16M is the element S (Z = 16). 2--Sulfur [Group 6A(16)] will form the S ion (6 --- 8 = ---2). The compound formed by the combination of these two elements is ZnS, zinc sulfide. --c) 17L is the element Cl (Z = 17). Chlorine [Group 7A(17)] forms the Cl ion (7 --- 8 = ---1). 38M is the element Sr 2+ (Z = 38). Strontium [Group 2A(2)] forms the Sr ion. The compound formed by the combination of these two elements is SrCl2, strontium chloride.

2.88

Plan: Based on the atomic numbers (the subscripts) locate the elements on the periodic table. Once the atomic numbers are located, identify the element and based on its position, assign a charge. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: + a) 37Q is the element Rb (Z = 37). Rubidium [Group 1A(1)] forms the Rb ion. 35R is the element Br (Z = 35). --Bromine [Group 7A(17)] forms the Br ion (7 --- 8 = ---1). The compound formed by the combination of these two elements is RbBr, rubidium bromide. 2--b) 8Q is the O (Z = 8). Oxygen [Group 6A(16)] will form the O ion (6 --- 8 = ---2). 13R is the element Al (Z = 13). 3+ Aluminum [Group 3A(13)] forms the Al ion. The compound formed by the combination of these two elements is Al2O3, aluminum oxide. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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2-22

c) 20Q is the element Ca (Z = 20). Calcium [Group 2A(2)] forms the Ca ion. 53R is the element I (Z = 53). Iodine --[Group 7A(17)] forms the I ion (7 --- 8 = ---1). The compound formed by the combination of these two elements is CaI2, calcium iodide. 2.89

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds, name the metal, followed by the nonmetal name with an -ide suffix. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: 4+ --a) tin(IV) chloride = SnCl4 The (IV) indicates that the metal ion is Sn which requires 4 Cl ions for a neutral compound. b) FeBr3 = iron(III) bromide (common name is ferric bromide); the charge on the iron ion is +3 to match the ---3 --charge of 3 Br ions. The +3 charge of the Fe is indicated by (III). +6 ---6 c) cuprous bromide = CuBr (cuprous is +1 copper ion, cupric is +2 copper ion). +3 ---2 d) Mn2O3 = manganese(III) oxide Use (III) to indicate the +3 ionic charge of Mn: Mn2O3

2.90

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds, name the metal, followed by the nonmetal name with an -ide suffix. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: a) CoO = cobalt(II) oxide Cobalt forms more than one monatomic ion so the ionic charge must be indicated with 2--a Roman numeral. Since the Co is paired with one O ion, the charge of Co is +2. b) mercury(I) chloride = Hg 2Cl2 The Roman numeral I indicates that mercury has an ionic charge of +1; mercury 2+ + is an unusual case in which the +1 ion formed is Hg2 , not Hg . 3+ c) chromic oxide = Cr2O3 ‘‘chromic’’ denotes a +3 charge (see Table 2.4), oxide has a ---2 charge. Two Cr ions 2--are required for every three O ions. (d) CuBr2 = copper(II) bromide Copper forms more than one monatomic ion so the ionic charge must be --indicated with a Roman numeral. Since the Cu is paired with two Br ions, the charge of Cu is +2.

2.91

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. Hydrates, compounds with a specific number of water molecules associated with them, are named with a prefix before the word hydrate to indicate the number of water molecules. Solution: + 2--a) Na2HPO4 = sodium hydrogen phosphate Sodium [Group 1A(1)] forms the Na ion; HPO4 is the hydrogen phosphate ion. + b) ammonium perchlorate = NH4ClO4 Ammonium is the polyatomic ion NH4 and perchlorate is the polyatomic --+ --ion ClO4 . One NH4 is required for every one ClO4 ion. --c) Pb(C2H3O2)2 O = lead(II) acetate trihydrate The C2H3O2 ion has a ---1 charge (see Table 2.5); since there 2 are two of these ions, the lead ion has a +2 charge which must be indicated with the Roman numeral II. The O indicates a hydrate in which the number of H2O molecules is indicated by the prefix tri-. 2 --d) NaNO2 = sodium nitrite NO2 is the nitrite polyatomic ion.

2.92

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: a) Sn(SO3)2 = tin(IV) sulfite Tin forms more than one monatomic ion so the ionic charge must be indicated with 2--a Roman numeral. Each SO3 polyatomic ion has a charge of ---2, so the ionic charge of tin is +4.

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2-23

2---

b) potassium dichromate = K2Cr2O7 Dichromate is the polyatomic ion Cr2O7 ; two K ions are required for a neutral compound. c) FeCO3 = iron(II) carbonate Iron forms more than one monatomic ion so the ionic charge must be indicated 2--with a Roman numeral. The CO3 polyatomic ion has a charge of ---2, so the ionic charge of iron is +2. + 2--d) potassium carbonate dihydrate = K2CO3 2H2O Potassium [Group 1A(1)] forms the K ion; carbonate is CO3 . + 2--Two K ions are required for every one CO3 ion. Dihydrate indicates the presence of two H2O molecules in the formula unit, indicated with O. 2 2.93

Plan: Review the rules for nomenclature covered in the chapter. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Compounds must be neutral. Solution: 2+ 2--a) Barium [Group 2A(2)] forms Ba and oxygen [Group 6A(16)] forms O (6 --- 8 = ---2) so the neutral compound 2+ 2--forms from one Ba ion and one O ion. Correct formula is BaO. 2+ --b) Iron(II) indicates Fe and nitrate is NO3 so the neutral compound forms from one iron(II) ion and two nitrate ions. Correct formula is Fe(NO3)2. c) Mn is the symbol for manganese. Mg is the correct symbol for magnesium. Correct formula is MgS. 2--2--Sulfide is the S ion and sulfite is the SO3 ion. + d) P is the symbol for phosphorus. K is the correct symbol for potassium. Potassium [Group 1A(1)] forms K and --+ --iodine [Group 7A(17)] forms I , so the neutral compound forms from one K ion and one I ion. Correct formula is KI.

2.94

Plan: Review the rules for nomenclature covered in the chapter. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Compounds must be neutral. Solution: --a) copper(I) iodide Cu is copper, not cobalt; since iodide is I , this must be copper(I). --b) iron(III) hydrogen sulfate HSO4 is hydrogen sulfate, and this must be iron(III) to be neutral. 2+ 2--c) magnesium dichromate Mg forms Mg and Cr2O7 is named dichromate ion. 2+ d) calcium chloride Ca [Group 2A(2)] forms a Ca ion only, so no Roman numeral is needed in the name; it also --means that there must be two chloride ions (Cl ) for the compound to be neutral, so a prefix like ‘‘di-’’ is unnecessary.

2.95

Plan: Acids donate H ion to the solution, so the acid is a combination of H and a negatively charged ion. Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. Oxoacids (H + an oxoanion) are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Solution: --a) Hydrogen carbonate is HCO3 , so its source acid is H2CO3. The name of the acid is carbonic acid (-ate becomes ---ic acid). --b) HIO4, periodic acid. IO4 is the periodate ion: -ate becomes ---ic acid. --c) Cyanide is CN ; its source acid is HCN hydrocyanic acid (binary acid). d) H2S, hydrosulfuric acid (binary acid).

2.96

Plan: Acids donate H ion to the solution, so the acid is a combination of H and a negatively charged ion. Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. Oxoacids (H + an oxoanion) are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Solution: --a) Perchlorate is ClO4 , so the source acid is HClO4. Name of acid is perchloric acid (-ate becomes -ic acid). --b) nitric acid, HNO 3 NO3 is the nitrate ion: -ate becomes -ic acid. --c) Bromite is BrO2 , so the source acid is HBrO2. Name of acid is bromous acid (-ite becomes -ous acid). --d) H2PO4 is dihydrogen phosphate, so its source acid is H3PO4. The name of the acid is phosphoric acid (-ate becomes ---ic acid).

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2-24

2.97

Plan: Use the formulas of the polyatomic ions. Recall that oxoacids are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Compounds must be neutral. Solution: + a) ammonium ion = NH4 ammonia = NH3 b) magnesium sulfide = MgS magnesium sulfite = MgSO3 magnesium sulfate = MgSO4 2--2--2--Sulfide = S ; sulfite = SO3 ; sulfate = SO4 . c) hydrochloric acid = HCl chloric acid = HClO3 chlorous acid = HClO2 Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. Chloric indicates the ----polyatomic ion ClO3 while chlorous indicates the polyatomic ion ClO2 . d) cuprous bromide = CuBr cupric bromide = CuBr2 The suffix -ous indicates the lower charge, +1, while the suffix -ic indicates the higher charge, +2.

2.98

Plan: Use the formulas of the polyatomic ions. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Compounds must be neutral. Solution: a) lead(II) oxide = PbO lead(IV) oxide = PbO2 2+ 4+ Lead(II) indicates Pb while lead(IV) indicates Pb . b) lithium nitride = Li3N lithium nitrite = LiNO2 lithium nitrate = LiNO3 3------Nitride = N ; nitrite = NO2 ; nitrate = NO3 . c) strontium hydride = SrH2 strontium hydroxide = Sr(OH) 2 ----Hydride = H ; hydroxide = OH . d) magnesium oxide = MgO manganese(II) oxide = MnO

2.99

Plan: This compound is composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: disulfur tetrafluoride S 2F 4 Di- indicates two S atoms and tetra- indicates four F atoms.

2.100

Plan: This compound is composed of two nonmetals. When a compound contains oxygen and a halogen, the halogen is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: dichlorine monoxide Cl2O Di- indicates two Cl atoms and mono- indicates one O atom.

2.101

Plan: These compounds are composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: a) Tetraphosphorus decoxide is P4O10. Tetra- indicates four P atoms and deca- indicates ten O atoms. b) Diboron trioxide is B2O3. Di-indicates two B atoms and triindicates three O atoms. c) Phosphorus trifluoride is PF3. P has a lower group number than F and is written first.

2.102

Plan: These compounds are composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: a) CBr4 is carbon tetrabromide. Tetra- is used to indicate that there are 4 Br atoms in the compound. b) IF7 is iodine heptafluoride. Hepta- is used to indicate that there are 7 I atoms in the compound. c) NO is nitrogen monoxide. Roman numerals are not used when naming molecular compounds. Mono- is used to indicate that there is only one O atom in the compound. Mono- is generally not used with the first element.

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2-25

2.103

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: + 2--a) (NH4)2SO4 ammonium is NH4 and sulfate is SO4 N = 2(14.01 amu) = 28.02 amu H = 8(1.008 amu) = 8.064 amu S = 1(32.06 amu) = 32.06 amu O = 4(16.00 amu) = 64.00 amu 132.14 amu + --b) NaH2PO4 sodium is Na and dihydrogen phosphate is H2PO4 Na = 1(22.99 amu) = 22.99 amu H = 2(1.008 amu) = 2.016 amu P = 1(30.97 amu) = 30.97 amu O = 4(16.00 amu) = 64.00 amu 119.98 amu + --c) KHCO3 potassium is K and bicarbonate is HCO3 K = 1(39.10 amu) = 39.10 amu H = 1(1.008 amu) = 1.008 amu C = 1(12.01 amu) = 12.01 amu O = 3(16.00 amu) = 48.00 amu 100.12 amu

2.104

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: + 2--a) Na2Cr2O7 sodium is Na and dichromate is Cr2O7 Na = 2(22.99 amu) = 45.98 amu Cr = 2(52.00 amu) = 104.00 amu O = 7(16.00 amu) = 112.00 amu 261.98 amu + --b) NH4ClO4 ammonium is NH4 and perchlorate is ClO4 N = 1(14.01 amu) = 14.01 amu H = 4(1.008 amu) = 4.032 amu Cl = 1(35.45 amu) = 35.45 amu O = 4(16.00 amu) = 64.00 amu 117.49 amu 2+ --c) Mg(NO2)2 3H2O magnesium is Mg , nitrite is NO2 , and trihydrate is 3H2O Mg = 1(24.31 amu) = 24.31 amu N = 2(14.01 amu) = 28.02 amu H = 6(1.008 amu) = 6.048 amu O = 7(16.00 amu) = 112.00 amu 170.38 amu

2.105

Plan: Convert the names to the appropriate chemical formulas. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) dinitrogen pentoxide N2O5 (di- = 2 and penta- = 5) N = 2(14.01 amu) = 28.02 amu O = 5(16.00 amu) = 80.00 amu 108.02 amu

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2-26

---

b) lead(II) nitrate Pb = N = O =

Pb(NO3)2 (lead(II) is Pb and nitrate is NO3 ) 1(207.2 amu) = 207.2 amu 2(14.01 amu) = 28.02 amu 6(16.00 amu) = 96.00 amu 331.2 amu

c) calcium peroxide Ca = O =

CaO2 (calcium is Ca and peroxide is O2 ) 1(40.08 amu) = 40.08 amu 2(16.00 amu) = 32.00 amu 72.08 amu

2---

2.106

Plan: Convert the names to the appropriate chemical formulas. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: 2+ --a) iron(II) acetate tetrahydrate Fe(C2H3O2)2 4H2O (iron(II) is Fe , acetate is C2H3O2 , and tetrahydrate is 4H2O) Fe = 1(55.85 amu) = 55.85 amu C = 4(12.01 amu) = 48.04 amu H = 14(1.008 amu) = 14.112 amu O = 8(16.00 amu) = 128.00 amu 246.00 amu b) sulfur tetrachloride SCl4 (tetra- = 4) S = 1 (32.06 amu) = 32.06 amu Cl = 4(35.45 amu) = 141.80 amu 173.86 amu + --c) potassium permanganate KMnO4 (potassium is K and permanganate is MnO4 ) K = 1(39.10 amu) = 39.10 amu Mn = 1(54.94 amu) = 54.94 amu O = 4(16.00 amu) = 64.00 amu 158.04 amu

2.107

Plan: Break down each formula to the individual elements and count the number of atoms of each element by observing the subscripts. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: a) There are 12 atoms of oxygen in Al2(SO4)3. The molecular mass is: Al = 2(26.98 amu) = 53.96 amu S = 3(32.06 amu) = 96.18 amu O = 12(16.00 amu) = 192.00 amu 342.14 amu b) There are 9 atoms of hydrogen in (NH4)2HPO4. The molecular mass is: N = 2(14.01 amu) = 28.02 amu H = 9(1.008 amu) = 9.072 amu P = 1(30.97 amu) = 30.97 amu O = 4(16.00 amu) = 64.00 amu 132.06 amu c) There are 8 atoms of oxygen in Cu3(OH)2(CO3)2. The molecular mass is: Cu = 3(63.55 amu) = 190.65 amu O = 8(16.00 amu) = 128.00 amu H = 2(1.008 amu) = 2.016 amu C = 2(12.01 amu) = 24.02 amu 344.69 amu

2.108

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2-27

Solution: a) There are 9 atoms of hydrogen in C6H5COONH4. The molecular mass is: C = 7(12.01 amu) = 84.07 amu H = 9(1.008 amu) = 9.072 amu O = 2(16.00 amu) = 32.00 amu N = 1(14.01 amu) = 14.01 amu 139.15 amu b) There are 2 atoms of nitrogen in N2H6SO4. The molecular mass is: N = 2(14.01 amu) = 28.02 amu H = 6(1.008 amu) = 6.048 amu S = 1(32.06 amu) = 32.06 amu O = 4(16.00 amu) = 64.00 amu 130.13 amu c) There are 12 atoms of oxygen in Pb4SO4(CO3)2(OH)2. The molecular mass is: Pb = 4(207.2 amu) = 828.8 amu S = 1(32.06 amu) = 32.06 amu O = 12(16.00 amu) = 192.00 amu C = 2(12.01 amu) = 24.02 amu H = 2(1.008 amu) = 2.016 amu 1078.9 amu 2.109

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Formula is SO3. Name is sulfur trioxide (the prefix triindicates 3 oxygen atoms). S = 1(32.06 amu) = 32.06 amu O = 3(16.00 amu) = 48.00 amu 80.06 amu b) Formula is C 3H8. Since it contains only carbon and hydrogen it is a hydrocarbon and with three carbons its name is propane. C = 3(12.01 amu) = 36.03 amu H = 8(1.008 amu) = 8.064 amu 44.09 amu

2.110

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Formula is N2O. Name is dinitrogen monoxide (the prefix diindicates 2 nitrogen atoms and mono- indicates 1 oxygen atom). N = 2(14.01 amu) = 28.02 amu O = 1(16.00 amu) = 16.00 amu 44.02 amu b) Formula is C 2H6. Since it contains only carbon and hydrogen it is a hydrocarbon and with three carbons its name is ethane. C = 2(12.01 amu) = 24.02 amu H = 6(1.008 amu) = 6.048 amu 30.07 amu

2.111

Plan: Review the nomenclature rules in the chapter. For ionic compounds, name the metal, followed by the nonmetal name with an -ide suffix. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. For metals, like many transition metals, that can form more than one ion each

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2-28

with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Oxoacids (H + an oxoanion) are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Greek numerical prefixes are used to indicate the number of atoms of each element in a compound composed of two nonmetals. Solution: a) blue vitriol CuSO4 5H2O copper(II) sulfate pentahydrate 2--SO4 = sulfate; II is used to indicate the 2+ charge of Cu; penta- is used to indicate the 5 waters of hydration. b) slaked lime Ca(OH)2 calcium hydroxide --The anion OH is hydroxide. c) oil of vitriol H2SO4 sulfuric acid 2--SO4 is the sulfate ion; since this is an acid, -ate becomes -ic acid. d) washing soda Na2CO3 sodium carbonate 2--CO3 is the carbonate ion. e) muriatic acid HCl hydrochloric acid Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. f) Epsom salts MgSO4 7H2O magnesium sulfate heptahydrate 2--SO4 = sulfate; hepta- is used to indicate the 7 waters of hydration. g) chalk CaCO3 calcium carbonate 2--CO3 is the carbonate ion. h) dry ice CO2 carbon dioxide The prefix diindicates 2 oxygen atoms; since there is only one carbon atom, no prefix is used. i) baking soda NaHCO3 sodium hydrogen carbonate --HCO3 is the hydrogen carbonate ion. j) lye NaOH sodium hydroxide --The anion OH is hydroxide. 2.112

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Each molecule has 2 blue spheres and 1 red sphere so the molecular formula is N2O. This compound is composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The prefix diindicates 2 nitrogen atoms and mono- indicates 1 oxygen atom. The name is dinitrogen monoxide. N = 2(14.01 amu) = 28.02 amu O = 1(16.00 amu) = 16.00 amu 44.02 amu b) Each molecule has 2 green spheres and 1 red sphere so the molecular formula is Cl2O. This compound is composed of two nonmetals. When a compound contains oxygen and a halogen, the halogen is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The prefix diindicates 2 chlorine atoms and mono- indicates 1 oxygen atom. The name is dichlorine monoxide. Cl = 2(35.45 amu) = 70.90 amu O = 1(16.00 amu) = 16.00 amu 86.90 amu

2.113

Plan: Review the discussion on separations. Solution: Separating the components of a mixture requires physical methods only; that is, no chemical changes (no changes in composition) take place and the components maintain their chemical identities and properties throughout. Separating the components of a compound requires a chemical change (change in composition).

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2-29

2.114

Plan: Review the definitions of homogeneous and heterogeneous. Solution: A homogeneous mixture is uniform in its macroscopic, observable properties; a heterogeneous mixture shows obvious differences in properties (density, color, state, etc.) from one part of the mixture to another.

2.115

A solution (such as salt or sugar dissolved in water) is a homogeneous mixture.

2.116

Plan: Review the definitions of homogeneous and heterogeneous. The key is that a homogeneous mixture has a uniform composition while a heterogeneous mixture does not. A mixture consists of two or more substances physically mixed together while a compound is a pure substance. Solution: a) Distilled water is a compound that consists of H2O molecules only. b) Gasoline is a homogeneous mixture of hydrocarbon compounds of uniform composition that can be separated by physical means (distillation). c) Beach sand is a heterogeneous mixture of different size particles of minerals and broken bits of shells. d) Wine is a homogeneous mixture of water, alcohol, and other compounds that can be separated by physical means (distillation). e) Air is a homogeneous mixture of different gases, mainly N2, O2, and Ar.

2.117

Plan: Review the definitions of homogeneous and heterogeneous. The key is that a homogeneous mixture has a uniform composition while a heterogeneous mixture does not. A mixture consists of two or more substances physically mixed together while a compound is a pure substance. Solution: a) Orange juice is a heterogeneous mixture of water, juice, and bits of orange pulp. b) Vegetable soup is a heterogeneous mixture of water, broth, and vegetables. c) Cement is a heterogeneous mixture of various substances. d) Calcium sulfate is a compound of calcium, sulfur, and oxygen in a fixed proportion. e) Tea is a homogeneous mixture.

2.118

Plan: Review the discussion on separations. Solution: a) Salt dissolves in water and pepper does not. Procedure: add water to mixture and filter to remove solid pepper. Evaporate water to recover solid salt. b) The water/soot mixture can be filtered; the water will flow through the filter paper, leaving the soot collected on the filter paper. c) Allow the mixture to warm up, and then pour off the melted ice (water); or, add water, and the glass will sink and the ice will float. d) Heat the mixture; the alcohol will boil off (distill), while the sugar will remain behind. e) The spinach leaves can be extracted with a solvent that dissolves the pigments. Chromatography can be used to separate one pigment from the other.

2.119

Plan: Review the discussion on separations. Solution: a) Filtration ----- separating the mixture on the basis of differences in particle size. The water moves through the holes in the colander but the larger pasta cannot. b) Extraction ----- The colored impurities are extracted into a solvent that is rinsed away from the raw sugar (or chromatography). A sugar solution is passed through a column in which the impurities stick to the stationary phase and the sugar moves through the column in the mobile phase.

2.120

Analysis time can be shortened by operating the column at a higher temperature or by increasing the rate of flow of the gaseous mobile phase.

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2-30

2.121

Plan: Use the equation for the volume of a sphere in part (a) to find the volume of the nucleus and the volume of the atom. Calculate the fraction of the atom volume that is occupied by the nucleus. For part (b), calculate the total mass of the two electrons; subtract the electron mass from the mass of the atom to find the mass of the nucleus. Then calculate the fraction of the atom’s mass contributed by the mass of the nucleus. Solution: 3 4 4 3 15 3 ---44 3 a) Volume (m ) of nucleus = r = 2.510 m = 6.54498 × 10 m 3 3 3 4 4 3 11 3 ---31 3 Volume (m ) of atom = r = 3.110 m = 1.24788 × 10 m 3 3







Fraction of volume =



volume of Nucleus volume of Atom

6.544981044 m 3 ---13 ---13 = 5.2449 × 10 = 5.2 × 10 1.247881031 m 3

b) Mass of nucleus = mass of atom --- mass of electrons = 6.64648 × 10 Fraction of mass =

---24

g --- 2(9.10939 × 10

massof Nucleus mass of Atom

---28

g) = 6.64466 × 10

6.64466 10

6.6464810

24 24

g g

---24

 = 0.99972617 = 0.999726



As expected, the volume of the nucleus relative to the volume of the atom is small while its relative mass is large. 2.122

Plan: Use Coulomb’s law which states that the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. Choose the largest ionic charges and smallest radii for the strongest ionic bonding and the smallest ionic charges and largest radii for the weakest ionic bonding. Solution: 2+ 2+ 2– Strongest ionic bonding: MgO. Mg , Ba , and O have the largest charges. Attraction increases as distance 2+ 2+ decreases, so the positive ion with the smaller radius, Mg , will form a stronger ionic bond than the larger ion Ba . + + ----Weakest ionic bonding: RbI. K , Rb , Cl , and I have the smallest charges. Attraction decreases as distance + --increases, so the ions with the larger radii, Rb and I , will form the weakest ionic bond.

2.123

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. These compounds are composed of two nonmetals. Greek numerical prefixes are used to indicate the number of atoms of each element in each compound. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Formula is BrF3. When a compound is composed of two elements from the same group, the element with the higher period number is named first. The prefix triindicates 3 fluorine atoms. A prefix is used with the first word in the name only when more than one atom of that element is present. The name is bromine trifluoride. Br = 1(79.90 amu) = 79.90 amu F = 3(19.00 amu) = 57.00 amu 136.90 amu b) The formula is SCl2. The element with the lower group number is the first word in the name. The prefix diindicates 2 chlorine atoms. A prefix is used with the first word in the name only when more than one atom of that element is present. The name is sulfur dichloride. S = 1(32.06 amu) = 32.06 amu Cl = 2(35.45 amu) = 70.90 amu 102.96 amu

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2-31

c) The formula is PCl3. The element with the lower group number is the first word in the name. The prefix triindicates 3 chlorine atoms. A prefix is used with the first word in the name only when more than one atom of that element is present. The name is phosphorus trichloride. P = 1(30.97 amu) = 30.97 amu Cl = 3(35.45 amu) = 106.35 amu 137.32 amu d) The formula is N2O5. The element with the lower group number is the first word in the name. The prefix diindicates 2 nitrogen atoms and the prefix penta- indicates 5 oxygen atoms. Only the second element is named with the suffix -ide. The name is dinitrogen pentoxide. N = 2(14.01 amu) = 28.02 amu O = 5(16.00 amu) = 80.00 amu 108.02 amu 2.124

Plan: These polyatomic ions are oxoanions composed of oxygen and another nonmetal. Oxoanions with the same number of oxygen atoms and nonmetals in the same group will have the same suffix ending. Only the nonmetal root name will change. Solution: 2--2--a) SeO4 selenate ion from SO4 = sulfate ion 3--3--b) AsO4 arsenate ion from PO4 = phosphate ion ----c) BrO2 bromite ion from ClO2 = chlorite ion ----d) HSeO4 hydrogen selenate ion from HSO4 = hydrogen sulfate ion 2--2--e) TeO3 tellurite ion from SO3 = sulfite ion

2.125

Plan: Write the formula of the compound and find the molecular mass. Determine the mass percent of nitrogen or phosphorus by dividing the mass of nitrogen or phosphorus in the compound by the molecular mass and multiplying by 100. For part (b), multiply the 100. g sample of compound by the mass ratio of ammonia to compound. Solution: + --a) Ammonium is NH4 and dihydrogen phosphate is H2PO4 . The formula is NH4H2PO4. N = 1(14.01 amu) = 14.01 amu H = 6(1.008 amu) = 6.048 amu P = 1(30.97 amu) = 30.97 amu O = 4(16.00 amu) = 64.00 amu 115.03 amu 14.01 amu N 100 = 12.18% N Mass percent of N = 115.03 amu compound Mass percent of P =

30.97 amu P 100 = 26.92% P 115.03 amu compound

  17.03 amu NH3  = 14.80 g NH3 b) Mass (g) of ammonia (NH3) = 100. g NH4 H2 PO4  115.03 amu NH4 H2 PO4  2.126

Plan: Determine the percent oxygen in each oxide by subtracting the percent nitrogen from 100%. Express the percentage in amu and divide by the atomic mass of the appropriate elements. Then divide each amount by the smaller number and convert to the simplest whole-number ratio. To find the mass of oxygen per 1.00 g of nitrogen, divide the mass percentage of oxygen by the mass percentage of nitrogen.

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2-32

Solution: a) I

(100.00 --- 46.69 N)% = 53.31% O  46.69 amu N    = 3.3326 N   14.01 amu N 

3.3326 N

2.6303

= 1.0000 mol N

III

= 1.0000 O

 63.15 amu O    = 3.9469 O  16.00 amu O 

3.9469 O = 1.5001 O 2.6303

The simplest whole-number ratio is 1:1.5 N:O = 2:3 N:O. (100.00 --- 25.94 N)% = 74.06% O  25.94 amu N   74.06 amu O     = 1.8515 N  = 4.6288 O    14.01amu N   16.00 amu O 

1.8515 N = 1.0000 N 1.8515

3.3319 O

3.3319 3.3319 The simplest whole-number ratio is 1:1 N:O. (100.00 --- 36.85 N)% = 63.15% O  36.85 amu N    = 2.6303 N   14.01 amu N  2.6303N

III

= 1.0002 N

 53.31 amu O    = 3.3319 O  16.00 amu O 

4.6288 O = 2.5000 O 1.8515

The simplest whole-number ratio is 1:2.5 N:O = 2:5 N:O.  53.31 amu O    = 1.1418 = 1.14 g O   46.69 amu N   63.15 amu O    = 1.7137 = 1.71 g O   36.85 amu N   74.06 amu O    = 2.8550 = 2.86 g O   25.94 amu N 

2.127

Plan: Recall that density = mass/volume. Solution: The mass of an atom of Pb is several times that of one of Al. Thus, the density of Pb would be expected to be several times that of Al if approximately equal numbers of each atom were occupying the same volume.

2.128

Plan: Review the law of mass conservation and law of definite composition. For each experiment, compare the mass values before and after each reaction and examine the ratios of the mass of reacted sodium to the mass of reacted chlorine. Solution: In each case, the mass of the starting materials (reactants) equals the mass of the ending materials (products), so the law of mass conservation is observed. Case 1: 39.34 g + 60.66 g = 100.00 g Case 2: 39.34 g + 70.00 g = 100.00 g + 9.34 g Case 3: 50.00 g + 50.00 g = 82.43 g + 17.57 g

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2-33

Each reaction yields the product NaCl, not Na2Cl or NaCl2 or some other variation, so the law of definite composition is observed. In each case, the ratio of the mass of sodium to the mass of chlorine in the compound is the same. Case 1: Mass Na/mass Cl2 = 39.34 g/60.66 g = 0.6485 Case 2: Mass of reacted Cl2 = initial mass --- excess mass = 70.00 g --- 9.34 g = 60.66 g Cl2 Mass Na/mass Cl2 = 39.34 g/60.66 g = 0.6485 Case 3: Mass of reacted Na = initial mass --- excess mass = 50.00 g --- 17.57 g = 32.43 g Na Mass Na/mass Cl2 = 32.43 g/50.00 g = 0.6486 2.129

Plan: Recall the definitions of solid, liquid, gas (from Chapter 1), element, compound, and homogeneous and heterogeneous mixtures. Solution: a) Gas is the phase of matter that fills its container. A mixture must contain at least two different substances. B, F, G, and I each contain only one gas. D and E each contain a mixture; E is a mixture of two different gases while D is a mixture of a gas and a liquid of a second substance. b) An element is a substance that cannot be broken down into simpler substances. A, C, G, and I are elements. c) The solid phase has a very high resistance to flow since it has a fixed shape. A shows a solid element. d) A homogeneous mixture contains two or more substances and has only one phase. E and H are examples of this. E is a homogeneous mixture of two gases and H is a homogeneous mixture of two liquid substances. e) A liquid conforms to the container shape and forms a surface. C shows one element in the liquid phase. f) A diatomic particle is a molecule composed of two atoms. B and G contain diatomic molecules of gas. g) A compound can be broken down into simpler substances. B and F show molecules of a compound in the gas phase. h) The compound shown in F has molecules composed of two white atoms and one blue atom for a 2:1 atom ratio. i) Mixtures can be separated into the individual components by physical means. D, E, and H are each a mixture of two different substances. j) A heterogeneous mixture like D contains at least two different substances with a visible boundary between those substances. k) Compounds obey the law of definite composition. B and F depict compounds.

2.130

Plan: To find the mass percent divide the mass of each substance in mg by the amount of seawater in mg and multiply by 100. The percent of an ion is the mass of that ion divided by the total mass of ions. Solution: 1000 g  1000 mg  = 1106 mg a) Mass (mg) of seawater = 1 kg  1 g   1 kg   mass of substance   100% Mass % =   mass of seawater   18,980 mg Cl   --100% = 1.898% Cl--Mass % Cl =   110 6 mg seawater     10.560 mg Na    + 100% = 1.056% Na+ Mass % Na =   1106 mg seawater     2650 mg SO 2   2--4 100% = 0.265% SO 2--Mass % SO4 =  4  1106 mg seawater     1270 mg Mg 2   2+ 2+ Mass % Mg =  100% = 0.127% Mg  1106 mg seawater   

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2-34

 400 mg Ca 2    2+ 100% = 0.04% Ca2+ Mass % Ca =   1106 mg seawater        380 mg K  + + Mass % K =  100% = 0.038% K  1106 mg seawater     140 mg HCO    --3 100% = 0.014% HCO --Mass % HCO3 =  3  1106 mg seawater    The mass percents do not add to 100% since the majority of seawater is H2O. b) Total mass of ions in 1 kg of seawater = 18,980 mg + 10,560 mg + 2650 mg + 1270 mg + 400 mg + 380 mg + 140 mg = 34,380 mg  10,560 mg Na +  100 = 30.71553 = 30.72% % Na + =   34,380 mg total ions  2+

c) Alkaline earth metal ions are Mg and Ca (Group 2 ions). 2+ 2+ Total mass % = 0.127% Mg + 0.04% Ca = 0.167% + + + + Alkali metal ions are Na and K (Group 1 ions). Total mass % = 1.056% Na + 0.038% K = 1.094% Mass % of alkali metal ions 1.094% = = 6.6 Mass % of alkaline earth metal ions 0.167% Total mass percent for alkali metal ions is 6.6 times greater than the total mass percent for alkaline earth metal ions. Sodium ions (alkali metal ions) are dominant in seawater. --2----d) Anions are Cl , SO4 , and HCO3 . --2----Total mass % = 1.898% Cl + 0.265% SO4 + 0.014% HCO3 = 2.177% anions + 2+ 2+ + Cations are Na , Mg , Ca , and K . + 2+ 2+ + Total mass % = 1.056% Na + 0.127% Mg + 0.04% Ca + 0.038% K = 1.2610 = 1.26% cations The mass fraction of anions is larger than the mass fraction of cations. Is the solution neutral since the mass of anions exceeds the mass of cations? Yes, although the mass is larger, the number of positive charges equals the number of negative charges. 2.131

Plan: Review the mass laws in the chapter. Solution: The law of mass conservation is illustrated in this change. The first flask has six oxygen atoms and six nitrogen atoms. The same number of each type of atom is found in both of the subsequent flasks. The mass of the substances did not change. The law of definite composition is also illustrated. During both temperature changes, the same compound, N2O, was formed with the same composition.

2.132

Plan: Use the density values to convert volume of each element to mass. Find the mass ratio of Ba to S in the compound and compare that to the mass ratio present. Solution: For barium sulfide the barium to sulfur mass ratio is (137.3 g Ba/32.06 g S) = 4.283 g Ba/g S  3.51 g Ba  Mass (g) of barium = 2.50 cm 3 Ba   = 8.775 g Ba  1 cm3 Ba 





 2.07 g S  Mass (g) of sulfur = 1.75 cm 3 S  = 3.6225 g S  1 cm 3 S  8.775 g Ba Barium to sulfur mass ratio = = 2.4224 = 2.42 g Ba/g S 3.6225 g S No, the ratio is too low; there is insufficient barium.





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2-35

2.133

Plan: First, count each type of atom present to produce a molecular formula. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Divide the mass of each element in the compound by the molecular mass and multiply by 100 to obtain the mass percent of each element. Solution: The molecular formula of succinic acid is C 4H6O4. C = 4(12.01 amu) = 48.04 amu H = 6(1.008 amu) = 6.048 amu O = 4 (16.00 amu) = 64.00 amu 118.09 amu  48.04 amu C   100% = 40.6815 = 40.68% C % C =   118.088 amu   6.048 amu H   100% = 5.1216 = 5.122% H % H =   118.088 amu   64.00 amu O   100% = 54.1969 = 54.20% O % O =   118.088 amu  Check: Total = (40.68 + 5.122 + 54.20)% = 100.00% The answer checks.

2.134

Plan: The toxic level of fluoride ion for a 70-kg person is 0.2 g. Convert this mass to mg and use the concentration of fluoride ion in drinking water to find the volume of water that contains the toxic amount. Convert the volume of the reservoir to liters and use the concentration of 1mg of fluoride ion per liter of water to find the mass of sodium fluoride required. Solution: --A 70-kg person would have to consume 0.2 mg of F to reach the toxic level.  1 mg F     = 200 mg F--Mass (mg) of fluoride for a toxic level = 0.2 g F   0.001 g F  1 L water    = 200 = 2 × 10 2 L water Volume (L) of water = 200 mg   1 mg F 





 4 qt   1 L   8 Volume (L) of reservoir = 8.50 10 7 gal    = 3.216651 × 10 L 1 gal  1.057 qt  ---

The molecular mass of NaF = 22.99 amu Na + 19.00 amu F = 41.99 amu. There are 19.00 mg of F in every 41.99 mg of NaF. 3  1 mg F    41.99 mg NaF  10 g  1 kg NaF   8 Mass (kg) of NaF = 3.21665110 L      1 L H 2 O  19.00 mg F   1 mg 10 3 g NaF       = 710.88 = 711 kg NaF 2.135

Plan: Z = the atomic number of the element. A is the mass number. To find the percent abundance of each Sb isotope, let x equal the fractional abundance of one isotope and (1 --- x) equal the fractional abundance of the second isotope since the sum of the fractional abundances must equal 1. Remember that atomic mass = (isotopic mass of the first isotope × fractional abundance) + (isotopic mass of the second isotope × fractional abundance). Solution: 121 a) Antimony is element 51so Z = 51. Isotope of mass 120.904 amu has a mass number of 121: 51 Sb 123 Isotope of mass 122.904 amu has a mass number of 123: 51 Sb b) Let x = fractional abundance of antimony-121. This makes the fractional abundance of antimony-123 = 1 --- x × (120.904 amu) + (1 --- x) (122.904 amu) = 121.8 amu McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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2-36

120.904 amu(x) + 122.904 amu --- 122.904 amu(x) = 121.8 amu 2x = 1.104 x = 0.552 = 0.55 fraction of antimony-121 1 --- x = 1 --- 0.552 = 0.45 fraction of antimony-123 2.136

Plan: List all possible combinations of the isotopes. Determine the masses of each isotopic composition. The molecule consisting of the lower abundance isotopes (N-15 and O-18) is the least common, and the one containing only the more abundant isotopes (N-14 and O-16) will be the most common. Solution: a) b) Formula Mass (amu) 15 18 N2 O 2(15 amu N) + 18 amu O = 48 least common 15 16 N2 O 2(15 amu N) + 16 amu O = 46 14 18 N2 O 2(14 amu N) + 18 amu O = 46 14 16 N2 O 2(14 amu N) + 16 amu O = 44 most common 15 14 18 N N O 1(15 amu N) + 1(14 amu N) + 18 amu O = 47 15 14 16 N N O 1(15 amu N) + 1(14 amu N) + 16 amu O = 45

2.137

Plan: Review the information about the periodic table in the chapter. Solution: a) Nonmetals are located in the upper-right portion of the periodic table: Black, red, green, and purple b) Metals are located in the large left portion of the periodic table: Brown and blue c) Some nonmetals, such as oxygen, chlorine, and argon, are gases: Red, green, and purple d) Most metals, such as sodium and barium are solids; carbon is a solid: Brown, blue, and black e) Nonmetals form covalent compounds; most noble gases do not form compounds: Black and red or black and green or red and green f) Nonmetals form covalent compounds; most noble gases do not form compounds: Black and red or black and green or red and green g) Metals react with nonmetals to form ionic compounds. For a compound with a formula of MX, the + --2+ 2--ionic charges of the metals and nonmetal must be equal in magnitude like Na and Cl or Ba and O : Brown and green or blue and red h) Metals react with nonmetals to form ionic compounds. For a compound with a formula of MX, the + --2+ 2--ionic charges of the metals and nonmetal must be equal in magnitude like Na and Cl or Ba and O : Brown and green or blue and red i) Metals react with nonmetals to form ionic compounds. For a compound with a formula of M2X, the + 2--2+ 4--ionic charge of the nonmetal must be twice as large as that of the metal like Na and O or Ba and C : Brown and red or blue and black j) Metals react with nonmetals to form ionic compounds. For a compound with a formula of MX2, the 2+ --ionic charge of the metal must be twice as large as that of the nonmetal like Ba and Cl : Blue and green k) Most Group 8A(18) elements are unreactive: Purple 2--2--l) Different compounds often exist between the same two nonmetal elements. Since oxygen exists as O or O2 , metals can sometimes form more than one compound with oxygen: Black and red or red and green or black and green or brown and red or blue and red

2.138

Plan: To find the formula mass of potassium fluoride, add the atomic masses of potassium and fluorine. Fluorine has only one naturally occurring isotope, so the mass of this isotope equals the atomic mass of fluorine. The atomic mass of potassium is the weighted average of the two isotopic masses: (isotopic mass of isotope 1 × fractional abundance) + (isotopic mass of isotope 2 × fractional abundance). Solution:

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2-37

Average atomic mass of K = 39 41 (isotopic mass of K × fractional abundance) + (isotopic mass of K × fractional abundance)  93.258%   6.730% Average atomic mass of K = (38.9637 amu)   (40.9618 amu)  = 39.093 amu  100%   100%  The formula for potassium fluoride is KF, so its molecular mass is (39.093 + 18.9984) = 58.091 amu 10

2.139

Plan: List all possible combinations of the isotopes. BF3 contains either B or B. Determine the masses of each isotopic composition and also the masses of each molecule missing one, two, or all three F atoms. Solution: 10 19 B F3 = 10 amu B + 3(19 amu F) = 67 amu 10 19 B F2 = 10 amu B + 2(19 amu F) = 48 amu 10 19 B F = 10 amu B + 19 amu F = 29 amu 10 B = 10 amu B = 10. amu 11 19 B F3 = 11 amu B + 3(19 amu F) = 68 amu 11 19 B F2 = 11 amu B + 2(19 amu F) = 49 amu 11 19 B F = 11 amu B + 19 amu F = 30. amu 11 B = 11 amu B = 11 amu

2.140

Plan: One molecule of NO is released per atom of N in the medicine. Divide the total mass of NO released by the molecular mass of the medicine and multiply by 100 for mass percent. Solution: NO = (14.01 + 16.00) amu = 30.01 amu Nitroglycerin: C3H5N3O9 = 3(12.01 amu C) + 5(1.008 amu H) + 3(14.01 amu N) + 9(16.00 amu O) = 227.10 amu In C3H5N3O9 (molecular mass = 227.10 amu), there are 3 atoms of N; since 1 molecule of NO is released per atom of N, this medicine would release 3 molecules of NO. The molecular mass of NO = 30.01 amu. total mass of NO 3(30.01 amu) 100  100 = 39.6433 = 39.64% Mass percent of NO = mass of compound 227.10 amu Isoamyl nitrate: C5H11NO3 = 5(12.01 amu C) + 11(1.008 amu H) + 1(14.01 amu N) + 3(16.00 amu O) = 133.15 amu In (CH3)2CHCH2CH2ONO2 (molecular mass = 133.15 amu), there is one atom of N; since 1 molecule of NO is released per atom of N, this medicine would release 1 molecule of NO. total mass of NO 1(30.01 amu) 100  100 = 22.5385 = 22.54% Mass percent of NO = mass of compound 133.15 amu

2.141

Plan: Each peak in the mass spectrum of carbon represents a different isotope of carbon. The heights of the peaks correspond to the natural abundances of the isotopes. Solution: 12 13 14 12 Carbon has three naturally occurring isotopes: C, C, and C. C has an abundance of 98.89% and would have the tallest peak in the mass spectrum as the most abundant isotope. 13 C has an abundance of 1.11% and thus would have a significantly shorter peak; the shortest 14 peak in the mass spectrum would correspond to the least abundant isotope, C, the abundance of 12 which is less than 0.01%. Peak Y, as the tallest peak, has a m/e ratio of 12 ( C); X, the shortest 14 13 peak, has a m/e ratio of 14( C). Peak Z corresponds to C with a m/e ratio of 13.

2.142

Plan: First, count each type of atom present to produce a molecular formula. Determine the mass fraction of each total mass of the element element. Mass fraction = . The mass of TNT multiplied by the mass fraction of each molecular mass of TNT element gives the mass of that element.

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2-38

Solution: The molecular formula for TNT is C7H5O6N3. The molecular mass of TNT is: C = 7(12.01 amu) = 84.07 amu H = 5(1.008 amu) = 5.040 amu O = 6(16.00 amu) = 96.00 amu N = 3(14.01 amu) = 42.03 amu 227.14 amu The mass fraction of each element is: 84.07 amu 5.040 amu C= = 0.3701 C H= = 0.02219 H 227.14 amu 227.14 amu

96.00 amu 42.03 amu = 0.4226 O N= = 0.1850 N 227.14 amu 227.14 amu Masses of each element in 1.00 lb of TNT = mass fraction of element × 1.00 lb. Mass (lb) C = 0.3701 × 1.00 lb = 0.370 lb C Mass (lb) H = 0.02219 × 1.00 lb = 0.0222 lb H Mass (lb) O = 0.4226 × 1.00 lb = 0.423 lb O Mass (lb) N = 0.1850 × 1.00 lb = 0.185 lb N O=

2.143

Plan: The superscript is the mass number, the sum of the number of protons and neutrons. Consult the periodic table to get the atomic number (the number of protons). The mass number --- the number of protons = the number of neutrons. Divide the number of neutrons by the number of protons to obtain the N/Z ratio. For atoms, the number of protons and electrons are equal. Solution: neutrons (N) protons (Z) N/Z 144 a) 62 Sm 144 --- 62 = 82 62 82/62 = 1.3 56

b) 26 Fe

56 --- 26 = 30

30/26 = 1.2

20 10

20 --- 10 = 10

10/10 = 1.0

107 --- 47 = 60

60/47 = 1.3

neutrons

protons

electrons

107 d) 47

2.144

238 92

238 --- 92 = 146

234 92

234 --- 92 = 142

214 82

214 --- 82 = 132

210 82

210 --- 82 = 128

206 82

206 --- 82 = 124

Plan: Determine the mass percent of platinum by dividing the mass of Pt in the compound by the molecular mass of the compound and multiplying by 100. For part (b), divide the total amount of money available by the cost of Pt per gram to find the mass of Pt that can be purchased. Use the mass percent of Pt to convert from mass of Pt to mass of compound. Solution: a) The molecular formula for platinol is Pt(NH3)2Cl2. Its molecular mass is: Pt = 1(195.1 amu) = 195.1 amu N = 2 (14.01 amu) = 28.02 amu H = 6(1.008 amu) = 6.048 amu Cl = 2(35.45 amu) = 70.90 amu 300.1 amu McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

2-39

Mass % Pt =

195.1 amu mass of Pt 100 = 65.012 = 65.01% Pt 100 = 300.1 amu molecular mass of compound

   = 19,608 g Pt b) Mass (g) of Pt = $1.00106   $51  Mass (g) of platinol = 

  65.01 g Pt



  = 3.0162 × 104 = 3.0 × 104 g platinol 

2.145

Plan: Obtain the information from the periodic table. The period number of an element is its row number while the group number is its column number. Solution: a) Building-block elements: Name Symbol Atomic number Atomic mass Period number Group number Hydrogen H 1 1.008 amu 1 1A(1) Carbon C 6 12.01 amu 2 4A(14) Nitrogen N 7 14.01 amu 2 5A(15) Oxygen O 8 16.00 amu 2 6A(16) b) Macronutrients: Sodium Na 11 22.99 amu 3 1A(1) Magnesium Mg 12 24.31 amu 3 2A(2) Potassium K 19 39.10 amu 4 1A(1) Calcium Ca 20 40.08 amu 4 2A(2) Phosphorus P 15 30.97 amu 3 5A(15) Sulfur S 16 32.06 amu 3 6A(16) Chlorine Cl 17 35.45 amu 3 7A(17)

2.146

Plan: Review the definitions of pure substance, element, compound, homogeneous mixture, and heterogeneous mixture. Solution: Matter is divided into two categories: pure substances and mixtures. Pure substances are divided into elements and compounds. Mixtures are divided into solutions (homogeneous mixtures) and heterogeneous mixtures.

2.147

Plan: A change is physical when there has been a change in physical form but not a change in composition. In a chemical change, a substance is converted into a different substance. Solution: 1) Initially, all the molecules are present in blue-blue or red-red pairs. After the change, there are no red-red pairs, and there are now red-blue pairs. Changing some of the pairs means there has been a chemical change. 2) There are two blue-blue pairs and four red-blue pairs both before and after the change, thus no chemical change occurred. The different types of molecules are separated into different boxes. This is a physical change. 3) The identity of the box contents has changed from pairs to individuals. This requires a chemical change. 4) The contents have changed from all pairs to all triplets. This is a change in the identity of the particles, thus, this is a chemical change. 5) There are four red-blue pairs both before and after, thus there has been no change in the identity of the individual units. There has been a physical change.

McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

2-40

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS FOLLOW–UP PROBLEMS 3.1A

Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles. Solution: 103 g   1 mol C   –2 –2 Moles of carbon = 315 mg C    = 2.6228 × 10 = 2.62 × 10 mol C  1 mg   12.01 g C  Road map: Mass (mg) of C 3

10 mg = 1 g Mass (g) of C Divide by

(g/mol)

Amount (moles) of C 3.1B

Plan: The number of moles of aluminum must be changed to g. Then the mass of aluminum per can can be used to calculate the number of soda cans that can be made from 52 mol of Al. Solution:     = 100.21 = 100 soda cans Number of soda cans = 52 mol Al        Road map: Amount (mol) of Al Multiply by (g/mol) (1 mol Al = 26.98 g Al) Mass (g) of Al 14 g Al = 1 soda can Number of cans

3.2A

Plan: Avogadro’s number is needed to convert the number of nitrogen molecules to moles. Since nitrogen molecules are diatomic (composed of two N atoms), the moles of molecules must be multiplied by 2 to obtain moles of atoms.

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3-1

Solution:   1 mol N 2  2 N atoms   Moles of N atoms = 9.72 10 21 N 2 molecules     6.022 10 23 N 2 molecules   1 mol N 2 





–2

= 3.2281634 × 10 = 3.23 × 10 mol N Road map: No. of N2 molecules Divide by Avogadro’s number (molecules/mol) Amount (moles) of N2 Use chemical formula (1 mol N2 = 2 mol N) Amount (moles) of N 3.2B

Plan: Avogadro’s number is needed to convert the number of moles of He to atoms. Solution:    23 26 26 Number of He atoms = 325 mol He   = 1.9572 × 10 = 1.96 × 10 He atoms   Road map: Amount (mol) of He Multiply by Avogadro’s number 23 (1 mol He = 6.022 × 10 He atoms) Number of He atoms

3.3A

Plan: Avogadro’s number is needed to convert the number of atoms to moles. The molar mass of manganese can then be used to determine the number of grams. Solution:   1 mol Mn 54.94 g Mn     Mass (g) of Mn = 3.22  10 20 Mn atoms  23  1 mol Mn   6.022  10 Mn atoms  





–2

= 2.9377 × 10 = 2.94 × 10 g Mn Road map: No. of Mn atoms Divide by Avogadro’s number (molecules/mol) Amount (moles) of Mn

3-2

Multiply by

(g/mol)

Mass (g) of Mn 3.3B

Plan: Use the molar mass of copper to calculate the number of moles of copper present in a penny. Avogadro’s number is then needed to convert the number of moles of Cu to Cu atoms. Solution:     23  Number of Cu atoms = 0.0625 g Cu        20

= 5.9225 × 10 = 5.92 × 10 Cu atoms Road map: Mass (g) of Cu Divide by (g/mol) (1 mol Cu = 63.55 g Cu) Amount (moles) of Cu Multiply by Avogadro’s number 23 (1 mol Cu = 6.022 × 10 Cu atoms) No. of Cu atoms

3.4A

Plan: Avogadro’s number is used to change the number of molecules to moles. Moles may be changed to mass by multiplying by the molar mass. The molar mass of tetraphosphorus decoxide is obtained from its chemical formula. Each molecule has four phosphorus atoms, so the total number of atoms is four times the number of molecules. Solution: a) Tetra = 4, and deca = 10 to give P4O10. The molar mass, , is the sum of the atomic weights, expressed in g/mol: P = 4(30.97) = 123.88 g/mol O = 10(16.00) = 160.00 g/mol = 283.88 g/mol of P4O10   1 mol 283.88 g     Mass (g) of P4O10 = 4.6510 22 molecules P4 O10   1 mol   6.022 10 23 molecules   = 21.9203 = 21.9 g P4O10  4 atoms P   23 b) Number of P atoms = 4.6510 22 molecules P4 O10   = 1.86 × 10 P atoms 1 P4 O10 molecule 



3.4B



Plan: The mass of calcium phosphate is converted to moles of calcium phosphate by dividing by the molar mass. Avogadro’s number is used to change the number of moles to formula units. Each formula unit has two phosphate ions, so the total number of phosphate ions is two times the number of formula units.

3-3

Solution: a) The formula of calcium phosphate is Ca3(PO4)2. The molar mass, , is the sum of the atomic weights, expressed in g/mol: = (3 × of Ca) + (2 × of P) + (8 × of O) = (3 × 40.08 g/mol Ca) + (2 × 30.97 g/mol P) + (8 × 16.00 g/mol O) = 310.18 g/mol Ca3(PO4)2  No. of formula units Ca3(PO4)2 = 75.5 g Ca3(PO4)2   

   23  3 4 2      3 4 2 3 4 2 23 23 = 1.4658 × 10 = 1.47 × 10 formula units Ca3(PO4)2 3

4 2

 3 23 b) No. of phosphate (PO4 ) ions = 1.47 × 10 formula units Ca3(PO4)2  

3– 4 3

   4 2

= 2.94 × 10 phosphate ions 3.5A

Plan: Calculate the molar mass of glucose. The total mass of carbon in the compound divided by the molar mass of the compound, multiplied by 100% gives the mass percent of C. Solution: The formula for glucose is C6H12O6. There are 6 atoms of C per each formula. Molar mass of C6H12O6 = (6 × of C) + (12 × of H) + (6 × of O) = (6 × 12.01 g/mol) + (12 × 1.008 g/mol) + (6 × 16.00 g/mol) = 180.16 g/mol



Mass % of C = 6

3.5B

(100) = 39.9978 = 40.00% C

Plan: Calculate the molar mass of CCl3F. The total mass of chlorine in the compound divided by the molar mass of the compound, multiplied by 100% gives the mass percent of Cl. Solution: The formula is CCl3F. There are 3 atoms of Cl per each formula. Molar mass of CCl3F = (1 × of C) + (3 × of Cl) + (1 × of F) = (1 × 12.01 g/mol) + (3 × 35.45 g/mol) + (1 × 19.00 g/mol) = 137.36 g/mol Mass % of Cl =

(100) =



(100) = 77.4243 = 77.42% Cl

3.6A

Plan: Multiply the mass of the sample by the mass fraction of C found in the preceding problem. Solution:    = 6.6196 = 6.620 g C Mass (g) of C = 16.55 g C6H12O6    6 12 6 

3.6B

Plan: Multiply the mass of the sample by the mass fraction of Cl found in the preceding problem. Solution:     = 1819.47 = 1820 g Cl Mass (g) of Cl = 2.35 kg CCl3F       3 

3.7A

Plan: Calculate the number of moles of each element in the sample by dividing by the molar mass of the corresponding element. The calculated numbers of moles are the fractional amounts of the elements and can be used as subscripts in a chemical formula. Convert the fractional amounts to whole numbers by dividing each number by the smallest subscripted number.

3-4

Solution:  Moles of H = 1.23 g H  

  = 1.2202 mol H 

 Moles of P = 12.64 g P  

  = 0.40814 mol P 

 Moles of O = 26.12 g O  

  = 1.6325 mol O 

Divide each subscript by the smaller value, 0.408: H 1.2202 P0.40814 O 1.6325 = H3PO4, this is phosphoric acid. 3.7B

Plan: The moles of sulfur may be calculated by dividing the mass of sulfur by the molar mass of sulfur. The moles of sulfur and the chemical formula will give the moles of M. The mass of M divided by the moles of M will give the molar mass of M. The molar mass of M can identify the element. Solution:  1 mol S    = 0.089832 mol S Moles of S = 2.88 g S   32.06 g S   2 mol M    = 0.059888 mol M Moles of M = 0.089832 mol S   3 mol S  3.12 g M Molar mass of M = = 52.0972 = 52.1 g/mol 0.059888 mol M The element is Cr (52.00 g/mol); M is Chromium and M2S3 is chromium(III) sulfide.







3.8A



Plan: If we assume there are 100 grams of this compound, then the masses of carbon and hydrogen, in grams, are numerically equivalent to the percentages. Divide the atomic mass of each element by its molar mass to obtain the moles of each element. Dividing each of the moles by the smaller value gives the simplest ratio of C and H. The smallest multiplier to convert the ratios to whole numbers gives the empirical formula. To obtain the molecular formula, divide the given molar mass of the compound by the molar mass of the empirical formula to find the whole-number by which the empirical formula is multiplied. Solution: Assuming 100 g of compound gives 95.21 g C and 4.79 g H:  1 mol C    = 7.92756 mol C Moles of C = 95.21 g C  12.01 g C   1 mol H    = 4.75198 mol H Mole of H = 4.79 g H  1.008 g H  Divide each of the moles by 4.75198, the smaller value: C 7.92756 H 4.75198 = C1.6683H1 4.75198

4.75198

The value 1.668 is 5/3, so the moles of C and H must each be multiplied by 3. If it is not obvious that the value is near 5/3, use a trial and error procedure whereby the value is multiplied by the successively larger integer until a value near an integer results. This gives C5H3 as the empirical formula. The molar mass of this formula is: (5 × 12.01 g/mol) + (3 × 1.008 g/mol) = 63.074 g/mol

3-5

Whole-number multiple =

molar mass of compound 252.30 g/mol = =4 molar mass of empirical formula 63.074 g/mol

Thus, the empirical formula must be multiplied by 4 to give 4(C5H3) = C20H12 as the molecular formula of benzo[a]pyrene. 3.8B

Plan: If we assume there are 100 grams of this compound, then the masses of carbon, hydrogen, nitrogen, and oxygen, in grams, are numerically equivalent to the percentages. Divide the atomic mass of each element by its molar mass to obtain the moles of each element. Dividing each of the moles by the smaller value gives the simplest ratio of C, H, N, and O. To obtain the molecular formula, divide the given molar mass of the compound by the molar mass of the empirical formula to find the whole-number by which the empirical formula is multiplied. Solution: Assuming 100 g of compound gives 49.47 g C, 5.19 g H, 28.86 g N, and 16.48 g O:    = 4.1191 mol C Moles of C = 49.47 g C     Moles of H = 5.19 g H  

  = 5.1488 mol H 

 Moles of N = 28.86 g N  

  = 2.0600 mol N 

 Moles of O = 16.48 g O  

  = 1.0300 mol O 

Divide each subscript by the smaller value, 1.030: C 4.1191 H 5.1488 N 2.0600 O 1.0300 = C4H5N2O This gives C4H5N2O as the empirical formula. The molar mass of this formula is: (4 × 12.01 g/mol) + (5 × 1.008 g/mol) + (2 × 14.01 g/mol) + (1 × 16.00 g/mol) = 97.10 g/mol The molar mass of caffeine is 194.2 g/mol, which is larger than the empirical formula mass of 97.10 g/mol, so the molecular formula must be a whole-number multiple of the empirical formula. Whole-number multiple =

=2 ol Thus, the empirical formula must be multiplied by 2 to give 2(C4H5N2O) = C8H10N4O2 as the molecular formula of caffeine. 3.9A

Plan: The carbon in the sample is converted to carbon dioxide, the hydrogen is converted to water, and the remaining material is chlorine. The grams of carbon dioxide and the grams of water are both converted to moles. One mole of carbon dioxide gives one mole of carbon, while one mole of water gives two moles of hydrogen. Using the molar masses of carbon and hydrogen, the grams of each of these elements in the original sample may be determined. The original mass of sample minus the masses of carbon and hydrogen gives the mass of chlorine. The mass of chlorine and the molar mass of chlorine will give the moles of chlorine. Once the moles of each of the elements have been calculated, divide by the smallest value, and, if necessary, multiply by the smallest number required to give a set of whole numbers for the empirical formula. Compare the molar mass of the empirical formula to the molar mass given in the problem to find the molecular formula. Solution: Determine the moles and the masses of carbon and hydrogen produced by combustion of the sample.  1 mol CO 2  12.01 g C   1 mol C     = 0.12307 g C 0.451 g CO 2     0.010248 mol C   44.01 g CO 2   1 mol C  1 mol CO 2 

3-6

 1 mol H 2 O  1.008 g H   2 mol H     = 0.006904 g H 0.0617 g H 2 O     0.0068495 mol H   18.016 g H 2 O   1 mol H 2 O   1 mol H 

The mass of chlorine is given by: 0.250 g sample – (0.12307 g C + 0.006904 g H) = 0.12003 g Cl The moles of chlorine are:  1 mol Cl   0.12003 g Cl   = 0.0033859 mol Cl. This is the smallest number of moles.  35.45 g Cl  Divide each mole value by the lowest value, 0.0033850: C 0.010248 H 0.0068495 Cl 0.0033859 = C3H2Cl 0.0033859

0.0033859

The empirical formula has the following molar mass: (3 × 12.01 g/mol) + (2 × 1.008 g/mol) + (35.45 g/mol) = 73.496 g/mol C3H2Cl molar mass of compound 146.99 g/mol = Whole-number multiple = =2 molar mass of empirical formula 73.496 g/mol Thus, the molecular formula is two times the empirical formula, 2(C3H2Cl) = C6H4Cl2. 3.9B

Plan: The carbon in the sample is converted to carbon dioxide, the hydrogen is converted to water, and the remaining material is oxygen. The grams of carbon dioxide and the grams of water are both converted to moles. One mole of carbon dioxide gives one mole of carbon, while one mole of water gives two moles of hydrogen. Using the molar masses of carbon and hydrogen, the grams of each of these elements in the original sample may be determined. The original mass of sample minus the masses of carbon and hydrogen gives the mass of oxygen. The mass of oxygen and the molar mass of oxygen will give the moles of oxygen. Once the moles of each of the elements have been calculated, divide by the smallest value, and, if necessary, multiply by the smallest number required to give a set of whole numbers for the empirical formula. Compare the molar mass of the empirical formula to the molar mass given in the problem to find the molecular formula. Solution: Determine the moles and the masses of carbon and hydrogen produced by combustion of the sample.  3.516 g CO2    1.007 g H2O  

2 2

   

   2

   

  

    

 = 0.95949 g C   = 0.11266 g H 

The mass of oxygen is given by: 1.200 g sample – (0.95949 g C + 0.11266 g H) = 0.12785 g O The moles of C and H are calculated above. The moles of oxygen are:    = 0.0079906mol O. This is the smallest number of moles. 0.12785 g O    Divide each subscript by the smallest value, 0.00800: C 0.079891 H 0.11176 O 0.0079906 = C10H14O 0.0079906

0.0079906

This gives C10H14O as the empirical formula. The molar mass of this formula is: (10 × 12.01 g/mol) + (14 × 1.008 g/mol) + (1 × 16.00 g/mol) = 150.21 g/mol The molar mass of the steroid is 300.42 g/mol, which is larger than the empirical formula mass of 150.21 g/mol, so the molecular formula must be a whole-number multiple of the empirical formula. Whole-number multiple =

mol

3-7

Thus, the empirical formula must be multiplied by 2 to give 2(C10H14O) = C20H28O2 as the molecular formula of the steroid. 3.10A

Plan: In each part it is necessary to determine the chemical formulas, including the physical states, for both the reactants and products. The formulas are then placed on the appropriate sides of the reaction arrow. The equation is then balanced. Solution: a) Sodium is a metal (solid) that reacts with water (liquid) to produce hydrogen (gas) and a solution of sodium hydroxide (aqueous). Sodium is Na; water is H2O; hydrogen is H2; and sodium hydroxide is NaOH. Na(s) + H2O(l)  H2(g) + NaOH(aq) is the equation. Balancing will precede one element at a time. One way to balance hydrogen gives: Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq) Next, the sodium will be balanced: 2Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq) On inspection, we see that the oxygen is already balanced. b) Aqueous nitric acid reacts with calcium carbonate (solid) to produce carbon dioxide (gas), water (liquid), and aqueous calcium nitrate. Nitric acid is HNO3; calcium carbonate is CaCO3; carbon dioxide is CO2; water is H2O; and calcium nitrate is Ca(NO3)2. The starting equation is HNO3(aq) + CaCO3(s)  CO2(g) + H2O(l) + Ca(NO3)2(aq) Initially, Ca and C are balanced. Proceeding to another element, such as N, or better yet the group of elements in – NO3 gives the following partially balanced equation: 2HNO3(aq) + CaCO3(s)  CO2(g) + H2O(l) + Ca(NO3)2(aq) Now, all the elements are balanced. c) We are told all the substances involved are gases. The reactants are phosphorus trichloride and hydrogen fluoride, while the products are phosphorus trifluoride and hydrogen chloride. Phosphorus trifluoride is PF 3; phosphorus trichloride is PCl3; hydrogen fluoride is HF; and hydrogen chloride is HCl. The initial equation is: PCl3(g) + HF(g)  PF3(g) + HCl(g) Initially, P and H are balanced. Proceed to another element (either F or Cl); if we will choose Cl, it balances as: PCl3(g) + HF(g)  PF3(g) + 3HCl(g) The balancing of the Cl unbalances the H, this should be corrected by balancing the H as: PCl3(g) + 3HF(g)  PF3(g) + 3HCl(g) Now, all the elements are balanced.

3.10B

3-8

One element should be picked and balanced. Any element except oxygen will work. Oxygen will not work in this case because it appears more than once on one side of the reaction arrow. We will start with carbon. Balancing C gives: C3H5N3O9(l)  3CO2(g) + H2O(g) + N2(g) + O2(g) Now balancing the hydrogen gives: C3H5N3O9(l)  3CO2(g) + 5/2H2O(g) + N2(g) + O2(g) Similarly, if we balance N we get: C3H5N3O9(l)  3CO2(g) + 5/2H2O(g) + 3/2N2(g) + O2(g) Clear the fractions by multiplying everything except the unbalanced oxygen by 2: 2C3H5N3O9(l)  6CO2(g) + 5H2O(g) + 3N2(g) + O2(g) This leaves oxygen to balance. Balancing oxygen gives: 2C3H5N3O9(l)  6CO2(g) + 5H2O(g) + 3N2(g) + 1/2O2(g) Again clearing fractions by multiplying everything by 2 gives: 4C3H5N3O9(l)  12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) Now all the elements are balanced. b) Potassium superoxide (KO2) is a solid. Carbon dioxide (CO2) and oxygen (O2) are gases. Potassium carbonate (K2CO3) is a solid. The initial equation is: KO2(s) + CO2(g)  O2(g) + K2CO3(s) Counting the atoms indicates that the carbons are balanced, but none of the other atoms are balanced. One element should be picked and balanced. Any element except oxygen will work (oxygen will be more challenging to balance because it appears more than once on each side of the reaction arrow). Because the carbons are balanced, we will start with potassium. Balancing potassium gives: 2KO2(s) + CO2(g)  O2(g) + K2CO3(s) Now all elements except for oxygen are balanced. Balancing oxygen by adding a coefficient in front of the O2 gives: 2KO2(s) + CO2(g)  3/2O2(g) + K2CO3(s) Clearing the fractions by multiplying everything by 2 gives: 4KO2(s) + 2CO2(g)  3O2(g) + 2K2CO3(s) Now all the elements are balanced. c) Iron(III) oxide (Fe2O3) is a solid, as is iron metal (Fe). Carbon monoxide (CO) and carbon dioxide (CO 2) are gases. The initial equation is: Fe2O3(s) + CO(g)  Fe(s) + CO2(g) Counting the atoms indicates that the carbons are balanced, but none of the other atoms are balanced. One element should be picked and balanced. Because oxygen appears in more than one compound on one side of the reaction arrow, it is best not to start with that element. Because the carbons are balanced, we will start with iron. Balancing iron gives: Fe2O3(s) + CO(g)  2Fe(s) + CO2(g) Now all the atoms but oxygen are balanced. There are 4 oxygen atoms on the left-hand side of the reaction arrow and 2 oxygen atoms on the right-hand side of the reaction arrow. In order to balance the oxygen, we want to change the coefficients in front of the carbon-containing compounds (if we changed the coefficient in front of the iron(III) oxide, the iron atoms would no longer be balanced). To maintain the balance of carbons, the coefficients in front of the carbon monoxide and the carbon dioxide must be the same. On the left-hand side of the equation, there are 3 oxygens in Fe2O3 plus 1X oxygen atoms from the CO (where X is the coefficient in the balanced Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-9

equation). On the right-hand side of the equation, there are 2X oxygen atoms. The number of oxygen atoms on both sides of the equation should be the same: 3 + 1X = 2X 3=X Balancing oxygen by adding a coefficient of 3 in front of the CO and CO2 gives: Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) Now all the elements are balanced. 3.11A

Plan: Count the number of each type of atom in each molecule to write the formulas of the reactants and products. Solution: 6CO(g) + 3O2(g)  6CO2(g) or,

3.11B

Plan: Count the number of each type of atom in each molecule to write the formulas of the reactants and products. Solution: 6H2(g) + 2N2(g)  4NH3(g) or,

3.12A

2CO(g) + O2(g)  2CO2(g)

3H2(g) + N2(g)  2NH3(g)

Plan: The reaction, like all reactions, needs a balanced chemical equation. The balanced equation gives the molar ratio between the moles of iron and moles of iron(III) oxide. Solution: The names and formulas of the substances involved are: iron(III) oxide, Fe2O3, and aluminum, Al, as reactants, and aluminum oxide, Al2O3, and iron, Fe, as products. The iron is formed as a liquid; all other substances are solids. The equation begins as: Fe2O3(s) + Al(s)  Al2O3(s) + Fe(l) There are 2 Fe, 3 O, and 1 Al on the reactant side and 1 Fe, 3 O, and 2 Al on the product side. Balancing aluminum:

Fe2O3(s) + 2Al(s)  Al2O3(s) + Fe(l)

Fe2O3(s) + 2Al(s)  Al2O3(s) + 2Fe(l) 1 mol Fe2 O3  3 Moles of Fe2O3 = 3.60 103 mol Fe = 1.80 × 10 mol Fe2O3  2 mol Fe  Balancing iron:

Road map: Amount (moles) of Fe Molar ratio (2 mol Fe = 1 mol Fe2O3) Amount (moles) of Fe2O3 3.12B

Plan: Divide the mass of silver sulfide by its molar mass to obtain moles of the compound. The balanced equation gives the molar ratio between moles of silver sulfide and moles of silver. Solution:  Moles of Ag = 32.6 g Ag2S  

    2

  = 0.2630 = 0.263 mol Ag  2 

3-10

Road map: Mass (g) of Ag2S Divide by (g/mol) (247.9 g Ag2S = 1 mol Ag2S) Amount (moles) of Ag2S Molar ratio (3 mol Ag2S = 6 mol Ag) Amount (moles) of Ag 3.13A

Plan: The mass of aluminum oxide must be converted to moles by dividing by its molar mass. The balanced chemical equation (follow-up problem 3.14A) shows there are two moles of aluminum for every mole of aluminum oxide. Multiply the moles of aluminum by Avogadro’s number to obtain atoms of Al. Solution: 23  1 mol Al O      2 mol Al  6.022 10 atoms Al  2 3   Atoms of Al = 1.00 g Al2 O3     101.96 g Al2 O3  1 mol Al 1 mol Al2 O3   





= 1.18125 × 10 = 1.18 × 10 atoms Al Road map: Mass (g) of Al2O3 Divide by

(g/mol)

Amount (moles) of Al2O3 Molar ratio Amount (moles) of Al Multiply by Avogadro’s number Number of Al atoms 3.13B

Plan: The mass of aluminum sulfide must be converted to moles by dividing by its molar mass. The balanced chemical equation (follow-up problem 3.14B) shows there are two moles of aluminum for every mole of aluminum sulfide. Multiply the moles of aluminum by its molar mass to obtain the mass (g) of aluminum. Solution:  Mass (g) of Al = 12.1 g Al2S3  

3 2

   

  

 = 4.3487 = 4.35 g Al 

3-11

Road map: Mass (g) of Al2S3 Divide by (g/mol) (150.14 g Al2S3 = 1 mol Al2S3) Amount (moles) of Al2S3 Molar ratio (1 mol Al2S3 = 2 mol Al) Amount (moles) of Al Multiply by (g/mol) (1 mol Al = 26.98 g Al) Mass (g) of Al 3.14A

Plan: Write the balanced chemical equation for each step. Add the equations, canceling common substances. Solution: Step 1 2SO2(g) + O2(g)  2SO3(g) Step 2 SO3(g) + H2O(l)  H2SO4(aq) Adjust the coefficients since 2 moles of SO3 are produced in Step 1 but only 1 mole of SO3 is consumed in Step 2. We have to double all of the coefficients in Step 2 so that the amount of SO 3 formed in Step 1 is used in Step 2. Step 1 2SO2(g) + O2(g)  2SO3(g) Step 2 2SO3(g) + 2H2O(l)  2H2SO4(aq) Add the two equations and cancel common substances. Step 1 2SO2(g) + O2(g)  2SO3(g) Step 2 2SO3(g) + 2H2O(l)  2H2SO4(aq) 2SO2(g) + O2(g) + 2SO3(g) + 2H2O(l)  2SO3(g) + 2H2SO4(aq) Or

3.14B

2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq)

Plan: Write the balanced chemical equation for each step. Add the equations, canceling common substances. Solution: Step 1 N2(g) + O2(g)  2NO(g) Step 2 NO(g) + O3(g)  NO2(g) + O2(g) Adjust the coefficients since 2 moles of NO are produced in Step 1 but only 1 mole of NO is consumed in Step 2. We have to double all of the coefficients in Step 2 so that the amount of NO formed in Step 1 is used in Step 2. Step 1 N2(g) + O2(g)  2NO(g) Step 2 2NO(g) + 2O3(g)  2NO2(g) + 2O2(g)

3-12

Add the two equations and cancel common substances. Step 1 N2(g) + O2(g)  2NO(g) Step 2 2NO(g) + 2O3(g)  2NO2(g) + 2O2(g) N2(g) + O2(g) + 2NO(g) + 2O3(g)  2NO(g) + 2NO2(g) + 2O2(g) Or 3.15A

N2(g) + 2O3(g)  2NO2(g) + O2(g)

Plan: Count the molecules of each type and find the simplest ratio. The simplest ratio leads to a balanced chemical equation. The substance with no remaining particles is the limiting reagent. Solution: 4 AB molecules react with 3 B2 molecules to produce 4 molecules of AB2, with 1 B2 molecule remaining unreacted. The balanced chemical equation is 4AB(g) + 2B2(g)  4AB2(g) or 2AB(g) + B2(g)  2AB2(g) The limiting reagent is AB since there is a B2 molecule left over (excess).

3.15B

Plan: Write a balanced equation for the reaction. Use the molar ratios in the balanced equation to find the amount (molecules) of SO3 produced when each reactant is consumed. The reactant that gives the smaller amount of product is the limiting reagent. Solution: 5 SO2 molecules react with 2 O2 molecules to produce molecules of SO3. The balanced chemical equation is 2SO2(g) + O2(g)  2SO3(g)  Amount (molecules) of SO3 produced from the SO2 = 5 molecules SO2    Amount (molecules) of SO3 produced from the O2 = 2 molecules SO2  

 = 5 molecules SO3  2 3

3 2

  = 4 molecules SO3 

O2 is the limiting reagent since it produces less SO3 than the SO2 does. 3.16A

Plan: Use the molar ratios in the balanced equation to find the amount of AB2 produced when 1.5 moles of each reactant is consumed. The smaller amount of product formed is the actual amount. Solution:  2 mol AB2  Moles of AB2 from AB = 1.5 mol AB  = 1.5 mol AB2  2 mol AB   2 mol AB2   = 3.0 mol AB2 Moles of AB2 from B2 = 1.5 mol B2   1 mol B2  Thus AB is the limiting reagent and only 1.5 mol of AB2 will form.

3.16B

Plan: Use the molar ratios in the balanced equation to find the amount of SO3 produced when 4.2 moles of SO2 are consumed and, separately, the amount of SO3 produced when 3.6 moles of O2 are consumed. The smaller amount of product formed is the actual amount. Solution: The balanced chemical equation is 2SO2(g) + O2(g)  2SO3(g)

3-13

 Amount (mol) of SO3 produced from the SO2 = 4.2 mol SO2  

 = 4.2 mol SO3  2  3

 Amount (mol) of SO3 produced from the O2 = 3.6 mol SO2  

  = 7.2 mol SO3  2  3

4.2 mol of SO3 (the smaller amount) will be produced. 3.17A

Plan: First, determine the formulas of the materials in the reaction and write a balanced chemical equation. Using the molar mass of each reactant, determine the moles of each reactant. Use molar ratios from the balanced equation to determine the moles of aluminum sulfide that may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change the moles of aluminum sulfide from the limiting reactant to the grams of product using the molar mass of aluminum sulfide. To find the excess reactant amount, find the amount of excess reactant required to react with the limiting reagent and subtract that amount from the amount given in the problem. Solution: The balanced equation is 2Al(s) + 3S(s)  Al2S3(s) Determining the moles of product from each reactant:    Moles of Al2S3 from Al = (10.0 g Al)     Moles of Al2S3 from S = (15.0 g S)  

  

  = 0.18532 mol Al2S3 

  = 0.155958 mol Al2S3 

Sulfur produces less product so it is the limiting reactant.   2 3  = 23.4155 = 23.4 g Al2S3 Mass (g) of Al2S3 = (0.155958 mol Al2S3)    2 3  The mass of aluminum used in the reaction is now determined:       Mass (g) of Al = (15.0 g S)  = 8.4155 g Al used      Subtracting the mass of aluminum used from the initial aluminum gives the mass remaining. Excess Al = Initial mass of Al – mass of Al reacted = 10.0 g – 8.4155 g = 1.5845 = 1.6 g Al 3.17B

Plan: First, determine the formulas of the materials in the reaction and write a balanced chemical equation. Using the molar mass of each reactant, determine the moles of each reactant. Use molar ratios and the molar mass of carbon dioxide from the balanced equation to determine the mass of carbon dioxide that may be produced from each reactant. The reactant that generates the smaller mass of carbon dioxide is limiting. To find the excess reactant amount, find the amount of excess reactant required to react with the limiting reagent and subtract that amount from the amount given in the problem. Solution: The balanced equation is: 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g) Determining the mass of product formed from each reactant:  Mass (g) of CO2 from C4H10 = 4.65 g C4H10  

10 4

   

2 4

   

  = 14.0844 = 14.1 g CO2  2  2

3-14

 Mass (g) of CO2 from O2 = 10.0 g O2  

    2

  = 8.4635 = 8.46 g CO2  2 

Oxygen produces the smallest amount of product, so it is the limiting reagent, and 8.46 gof CO2 are produced. The mass of butane used in the reaction is now determined:    2  4 10    Mass (g) of C4H10=10.0 g O2       2

  = 2.7942 = 2.79 g C4H10 used  10  10

Subtracting the mass of butane used from the initial butane gives the mass remaining. Excess butane = Initial mass of butane – mass of butane reacted = 4.65 g – 2.79 g = 1.86 g butane 3.18A

Plan: Determine the formulas, and then balance the chemical equation. The mass of marble is converted to moles, the molar ratio (from the balanced equation) gives the moles of CO2, and finally the theoretical yield of CO2 is determined from the moles of CO2 and its molar mass. To calculate percent yield, divide the given actual yield of CO2 by the theoretical yield, and multiply by 100. Solution: The balanced equation: CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g) Find the theoretical yield of carbon dioxide.  1 mol CaCO3   1 mol CO2   44.01 g CO2   Mass (g) of CO2 = 10.0 g CaCO3     100.09 g CaCO3  1 mol CaCO3   1 mol CO2 





= 4.39704 g CO2 The percent yield:

 3.65 g CO2   actual yield   100% = 83.0104 = 83.0%   100% =   theoretical yield   4.39704 g CO2  3.18B

Plan: Determine the formulas, and then balance the chemical equation. The mass of potassium iodide is converted to moles, the molar ratio (from the balanced equation) gives the moles of lead(II) iodide, and finally the theoretical yield of lead(II) iodide is determined from the moles of lead(II) iodide and its molar mass. To calculate actual yield, multiply the theoretical yield of lead(II) iodide by the percent yield, and divide by 100. Solution: The balanced equation: 2KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2KNO3(aq) Find the theoretical yield of lead(II) iodide.  Mass (g) of PbI2 = 112 g KI  

   

  

  = 155.518 = 156 g PbI2 2  2

The actual yield:   Actual yield (g) of PbI2 =    100% 



100%

(156 g PbI2) = 140.40 = 140. g PbI2

3-15

END–OF–CHAPTER PROBLEMS 3.1

Plan: The atomic mass of an element expressed in amu is numerically the same as the mass of 1 mole of the element expressed in grams. We know the moles of each element and have to find the mass (in g). To convert moles of element to grams of element, multiply the number of moles by the molar mass of the element. Solution: Al 26.98 amu  26.98 g/mol Al  26.98 g Al  Mass Al (g) = 3 mol Al = 80.94 g Al  1 mol Al  Cl

35.45 amu  35.45 g/mol Cl

 35.45 g Cl  Mass Cl (g) = 2 mol Cl  = 70.90 g Cl  1 mol Cl  3.2

Plan: The molecular formula of sucrose tells us that 1 mole of sucrose contains 12 moles of carbon atoms. Multiply the moles of sucrose by 12 to obtain moles of carbon atoms; multiply the moles of carbon atoms by Avogadro’s number to convert from moles to atoms. Solution:  12 mol C    = 12 mol C a) Moles of C atoms = 1 mol C12 H 22 O11  1 mol C12 H 22 O11 





23  12 mol C   6.022 10 C atoms  25 b) C atoms = 2 mol C12 H 22 O11    = 1.445 × 10 C atoms 1 mol C H O   1 mol C  12 22 11 



3.3

3.4



Plan: Review the list of elements that exist as diatomic or polyatomic molecules. Solution: “1 mol of chlorine” could be interpreted as a mole of chlorine atoms or a mole of chlorine molecules, Cl2. Specify which to avoid confusion. The same problem is possible with other diatomic or polyatomic molecules, e.g., F2, Br2, I2, H2, O2, N2, S8, and P4. For these elements, as for chlorine, it is not clear if atoms or molecules are being discussed. The molecular mass is the sum of the atomic masses of the atoms or ions in a molecule. The molar mass is the mass of 1 mole of a chemical entity. Both will have the same numeric value for a given chemical substance but molecular mass will have the units of amu and molar mass will have the units of g/mol.

3.5

A mole of a particular substance represents a fixed number of chemical entities and has a fixed mass. Therefore the mole gives us an easy way to determine the number of particles (atoms, molecules, etc.) in a sample by weighing it. The mole maintains the same mass relationship between macroscopic samples as exist between individual chemical entities. It relates the number of chemical entities (atoms, molecules, ions, electrons) to the mass.

3.6

Plan: The mass of the compound is given. Divide the given mass by the molar mass of the compound to convert from mass of compound to number of moles of compound. The molecular formula of the compound tells us that 1 mole of compound contains 2 moles of phosphorus atoms. Use the ratio between P atoms and P4 molecules (4:1) to convert moles of phosphorus atoms to moles of phosphorus molecules. Finally, multiply moles of P4 molecules by Avogadro’s number to find the number of molecules.

3-16

Solution: Roadmap Mass (g) of Ca3(PO4)2 Divide by

(g/mol)

Amount (mol) of Ca3(PO4)2 Molar ratio between Ca3(PO4)2 and P atoms Amount (moles) of P atoms Molar ratio between P atoms and P4 molecules Amount (moles) of P4 molecules 23

Multiply by 6.022 × 10 formula units/mol Number of P4 molecules

3.7

Plan: The relative atomic masses of each element can be found by counting the number of atoms of each element and comparing the overall masses of the two samples. Solution: a) The element on the left (green) has the higher molar mass because only 5 green balls are necessary to counterbalance the mass of 6 yellow balls. Since the green ball is heavier, its atomic mass is larger, and therefore its molar mass is larger. b) The element on the left (red) has more atoms per gram. This figure requires more thought because the number of red and blue balls is unequal and their masses are unequal. If each pan contained 3 balls, then the red balls would be lighter. The presence of 6 red balls means that they are that much lighter. Because the red ball is lighter, more red atoms are required to make 1 g. c) The element on the left (orange) has fewer atoms per gram. The orange balls are heavier, and it takes fewer orange balls to make 1 g. d) Neither element has more atoms per mole. Both the left and right elements have the same number of 23 atoms per mole. The number of atoms per mole (6.022 × 10 ) is constant and so is the same for every element.

3.8

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (1 × of Sr) + (2 × of O) + (2 × of H) = (1 × 87.62 g/mol Sr) + (2 × 16.00 g/mol O) + (2 × 1.008 g/mol H) = 121.64 g/mol of Sr(OH)2 b) = (2 × of N) + (3 × of O) = (2 × 14.01 g/mol N) + (3 × 16.00 g/mol O) = 76.02 g/mol of N2O3 c) = (1 × of Na) + (1 × of Cl) + (3 × of O) = (1 × 22.99 g/mol Na) + (1 × 35.45 g/mol Cl) + (3 × 16.00 g/mol O) = 106.44 g/mol of NaClO3

3-17

= (2 × of Cr) + (3 × of O) = (2 × 52.00 g/mol Cr) + (3 × 16.00 g/mol O) = 152.00 g/mol of Cr2O3

3.9

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (3 × of N) + (12 × of H) + (1 × of P) + (4 × of O) = (3 × 14.01 g/mol N) + (12 × 1.008 g/mol H) + (1 × 30.97 g/mol P) + (4 × 16.00 g/mol O) = 149.10 g/mol of (NH4)3PO4 b) = (1 × of C) + (2 × of H) + (2 × of Cl) = (1 × 12.01 g/mol C) + (2 × 1.008 g/mol H) + (2 × 35.45 g/mol Cl) = 84.93 g/mol of CH2Cl2 c) = (1 × of Cu) + (1 × of S) + (9 × of O) + (10 × of H) = (1 × 63.55 g/mol Cu) + (1 × 32.06 g/mol S) + (9 × 16.00 g/mol O) + (10 × 1.008 g/mol H) = 249.69 g/mol of CuSO4 5H2O d) = (1 × of Br) + (3 × of F) = (1 × 79.90 g/mol Br) + (3 × 19.00 g/mol F) = 136.90 g/mol of BrF3

3.10

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (1 × of Sn) + (1 × of O) = (1 × 118.7 g/mol Sn) + (1 × 16.00 g/mol O) = 134.7 g/mol of SnO b) = (1 × of Ba) + (2 × of F) = (1 × 137.3 g/mol Ba) + (2 × 19.00 g/mol F) = 175.3 g/mol of BaF2 c) = (2 × of Al) + (3 × of S) + (12 × of O) = (2 × 26.98 g/mol Al) + (3 × 32.06 g/mol S) + (12 × 16.00 g/mol O) = 342.14 g/mol of Al2(SO4)3 d) = (1 × of Mn) + (2 × of Cl) = (1 × 54.94 g/mol Mn) + (2 × 35.45 g/mol Cl) = 125.84 g/mol of MnCl2

3.11

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (2 × of N) + (4 × of O) = (2 × 14.01 g/mol N) + (4 × 16.00 g/mol O) = 92.02 g/mol of N2O4 b) = (4 × of C) + (10 × of H) + (1 × of O) = (4 × 12.01 g/mol C) + (10 × 1.008 g/mol H) + (1 × 16.00 g/mol O) = 74.12 g/mol of C4H9OH c) = (1 × of Mg) + (1 × of S) + (11 × of O) + (14 × of H) = (1 × 24.31 g/mol Mg) + (1 × 32.06 g/mol S) + (11 × 16.00 g/mol O) + (14 × 1.008 g/mol H) = 246.48 g/mol of MgSO4 7H2O

3-18

3.12

= (1 × of Ca) + (4 × of C) + (6 × of H) + (4 × of O) = (1 × 40.08 g/mol Ca) + (4 × 12.01 g/mol C) + (6 × 1.008 g/mol H) + (4 × 16.00 g/mol O) = 158.17 g/mol of Ca(C2H3O2)2

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part (a), multiply the number of moles by the molar mass of Zn. In part (b), first multiply by Avogadro’s number to obtain the number of F2 molecules. The molecular formula tells us that there are 2 F atoms in each molecule of F2; use the 2:1 ratio to convert F2 molecules to F atoms. In part (c), convert mass of Ca to moles of Ca by dividing by the molar mass of Ca. Then multiply by Avogadro’s number to obtain the number of Ca atoms. Solution:

 a) (0.346 mol Zn)    b) (2.62 mol F2)  

 c) 28.5 g Ca  

3.13

3.14

  = 22.6 g Zn  

    

2 2

  



 24  = 3.16 × 10 F atoms 

 23  = 4.28 × 10 Ca atoms

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. In part (a), 3 convert mg units to g units by dividing by 10 ; then convert mass of Mn to moles of Mn by dividing by the molar mass of Mn. In part (b) convert number of Cu atoms to moles of Cu by dividing by Avogadro’s number. In part (c) divide by Avogadro’s number to convert number of Li atoms to moles of Li; then multiply by the molar mass of Li to find the mass. Solution:      = 1.13 × 103 mol Mn a) (62.0 mg Mn)     

 22 b) (1.36 × 10 Cu atoms)  



 24 c) (8.05 × 10 Li atoms)  



 = 0.0226 mol Cu 

   

 = 92.8 g Li 

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part (a), multiply the number of moles by the molar mass of the substance. In part (b), first convert mass of compound to moles of compound by dividing by the molar mass of the compound. The molecular formula of the compound tells us that 1 mole of compound contains 6 moles of oxygen atoms; use the 1:6 ratio to convert moles of compound to moles of oxygen atoms. In part (c), convert mass of compound to moles of compound by dividing by the molar mass of the compound. Since 1 mole of compound contains 6 moles of oxygen atoms, multiply the moles of compound by 6 to obtain moles of oxygen atoms; then multiply by Avogadro’s number to obtain the number of oxygen atoms. Solution: a) of KMnO4 = (1 × of K) + (1 × of Mn) + (4 × of O) = (1 × 39.10 g/mol K) + (1 × 54.94 g/mol Mn) + (4 × 16.00 g/mol O) = 158.04 g/mol of KMnO4 158.04 g KMnO 4    = 107.467 = 1.1 × 102 g KMnO Mass of KMnO4 = 0.68 mol KMnO 4  4  1 mol KMnO 4 





of Ba(NO3)2 = (1 × of Ba) + (2 × of N) + (6 × of O) = (1 × 137.3 g/mol Ba) + (2 × 14.01 g/mol N) + (6 × 16.00 g/mol O) = 261.3 g/mol Ba(NO3)2

3-19

 1 mol Ba(NO3 )2    = 0.031305 mol Ba(NO ) Moles of Ba(NO3)2 = 8.18 g Ba(NO3 )2  3 2  261.3 g Ba(NO3 )2 





 6 mol O atoms    = 0.18783 = 0.188 mol O atoms Moles of O atoms = 0.031305 mol Ba(NO3 )2  1 mol Ba(NO3 )2  of CaSO4 2H2O = (1 × of Ca) + (1 × of S) + (6 × of O) + (4 × of H) = (1 × 40.08 g/mol Ca) + (1 × 32.06 g/mol S) + (6 × 16.00 g/mol O) + (4 × 1.008 g/mol H) = 172.17 g/mol (Note that the waters of hydration are included in the molar mass.)





 1 mol CaSO  2H O   4 2 –5 Moles of CaSO4 2H2O = 7.3103 g CaSO 4  2H 2 O   = 4.239995 × 10 mol 172.17 g CaSO4  2H 2 O 





 6 mol O atoms    Moles of O atoms = 4.239995105 mol CaSO 4  2H 2 O  1 mol CaSO4  2H 2 O 





–5

= 2.543997 × 10 mol O atoms

 6.022 10 23 O atoms    Number of O atoms = 2.54399710 mol O atoms   1 mol O atoms  20 20 = 1.5320 × 10 = 1.5 × 10 O atoms



3.15



4

Plan: Determine the molar mass of each substance, then perform the appropriate molar conversions. To find the mass in part (a), divide the number of molecules by Avogadro’s number to find moles of compound and then multiply the mole amount by the molar mass in grams; convert from mass in g to mass in kg. In part (b), first convert mass of compound to moles of compound by dividing by the molar mass of the compound. The molecular formula of the compound tells us that 1 mole of compound contains 2 moles of chlorine atoms; use the 1:2 ratio to convert moles of compound to moles of chlorine atoms. In part (c), convert mass of compound to moles of compound by dividing by the – molar mass of the compound. Since 1 mole of compound contains 2 moles of H ions, multiply the moles of compound by – – 2 to obtain moles of H ions; then multiply by Avogadro’s number to obtain the number of H ions. Solution: a) of NO2 = (1 × of N) + (2 × of O) = (1 × 14.01 g/mol N) + (2 × 16.00 g/mol O) = 46.01g/mol of NO2   1 mol NO2   = 7.63866 × 10–3 mol NO Moles of NO2 = 4.610 21 molecules NO2  2 23  6.022 10 molecules NO2 





 46.01 g NO     1 kg  2 3 –4 –4 Mass (kg) of NO2 = 7.6386610 mol NO2   3  = 3.51455 × 10 = 3.5 × 10 kg NO2  1 mol NO2  10 g 





of C2H4Cl2 = (2 × of C) + (4 × of H) + (2 × of Cl) = (2 × 12.01g/mol C) + (4 × 1.008 g/mol H) + (2 × 35.45 g/mol Cl) = 98.95 g/mol of C2H4Cl2  1 mol C 2 H 4 Cl 2    = 6.21526 × 10–4 mol C H Cl Moles of C2H4Cl2 = 0.0615 g C 2 H 4 Cl2  2 4 2  98.95 g C 2 H 4 Cl2 





 2 mol Cl atoms   4  = 1.2431 × 10–3 Moles of Cl atoms = 6.2152610 mol C2 H 4 Cl2   1 mol C2 H 4 Cl 2 





–3

= 1.24 × 10 mol Cl atoms Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-20

c) of SrH2 = (1 × SrH2

of Sr) + (2 ×

of H) = (1 × 87.62 g/mol Sr) + (2 × 1.008 g/mol H) = 89.64 g/mol of

 1 mol SrH 2    = 0.0649264 mol SrH Moles of SrH2 = 5.82 g SrH 2  2  89.64 g SrH 2 





 2 mol H   –  = 0.1298528 mol H– ions Moles of H ions = 0.0649264 mol SrH 2  1 mol SrH 2 





 6.022 10 23 H ions  –  = 7.81974 × 1022 = 7.82 × 1022 H– ions Number of H ions = 0.1298528 mol H ions    1 mol H 



3.16



Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part (a), multiply the number of moles by the molar mass of the substance. In part (b), first convert the mass of compound in kg to mass in g and divide by the molar mass of the compound to find moles of compound. In part (c), convert mass of compound in mg to mass in g and divide by the molar mass of the compound to find moles of compound. Since 1 mole of compound contains 2 moles of nitrogen atoms, multiply the moles of compound by 2 to obtain moles of nitrogen atoms; then multiply by Avogadro’s number to obtain the number of nitrogen atoms. Solution: a) of MnSO4 = (1 × of Mn) + (1 × of S) + (4 × of O) = (1 × 54.94 g/mol Mn) + (1 × 32.06 g/mol S) + (4 × 16.00 g/mol O) = 151.00 g/mol of MnSO4 151.00 g MnSO  2  4 Mass (g) of MnSO4 = 6.44 10 mol MnSO 4   = 9.7244 = 9.72 g MnSO4  1 mol MnSO 4  b)

of Fe(ClO4)3 = (1 × of Fe) + (3 × of Cl) + (12 × of O) = (1 × 55.85 g/mol Fe) + (3 × 35.45 g/mol S) + (12 × 16.00 g/mol O) = 354.20 g/mol of Fe(ClO4)3 103 g   = 1.58 × 104 kg Fe(ClO4)3 Mass (g) of Fe(ClO4)3 = 15.8 kg Fe(ClO 4 )3   1 kg   1 mol Fe(ClO )  4  4 3  Moles of Fe(ClO4)3 = 1.5810 g Fe(ClO 4 )3   = 44.6076 = 44.6 mol Fe(ClO4)3  354.20 g Fe(ClO 4 )3 

of NH4NO2 = (2 × of N) + (4 × of H) + (2 × of O) = (2 × 14.01 g/mol N) + (4 × 1.008 g/mol H) + (2 × 16.00 g/mol O) = 64.05 g/mol NH4NO2 103 g   = 0.0926 g NH4NO2 Mass (g) of NH4NO2 = 92.6 mg NH 4 NO2   1 mg 





 1 mol NH 4 NO2   –3 Moles of NH4NO2 = 0.0926 g NH 4 NO2   = 1.44575 × 10 mol NH4NO2  64.05 g NH 4 NO2 





 2 mol N atoms    = 2.8915 × 10–3 mol N atoms Moles of N atoms = 1.44575103 mol NH 4 NO2  1 mol NH 4 NO2  Number of N atoms = 2.891510

3

 6.022 1023 N atoms    mol N atoms  1 mol N atoms  21

= 1.74126 × 10 = 1.74 × 10 N atoms

3-21

3.17

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. In part (a), divide the mass by the molar mass of the compound to find moles of compound. Since 1 mole of compound 2+ – contains 3 moles of ions (1 mole of Sr and 2 moles of F ), multiply the moles of compound by 3 to obtain moles of ions and then multiply by Avogadro’s number to obtain the number of ions. In part (b), multiply the number of moles by the molar mass of the substance to find the mass in g and then convert to kg. In part (c), divide the number of formula units by Avogadro’s number to find moles; multiply the number of moles by the molar mass to obtain the mass in g and then convert to mg. Solution: a) of SrF2 = (1 × of Sr) + (2 × of F) = (1 × 87.62 g/mol Sr) + (2 × 19.00 g/mol F) = 125.62 g/mol of SrF2  1 mol SrF2    = 0.303296 mol SrF Moles of SrF2 = 38.1 g SrF2  2 125.62 g SrF2 





 3 mol ions    = 0.909888 mol ions Moles of ions = 0.303296 mol SrF2  1 mol SrF2 





 6.022 1023 ions   23 23 Number of ions = 0.909888 mol ions   = 5.47935 × 10 = 5.48 × 10 ions  1 mol ions  b) of CuCl2 2H2O = (1 × of Cu) + (2 × of Cl) + (4 × of H) + (2 × of O) = (1 × 63.55 g/mol Cu) + (2 × 35.45 g/mol Cl) + (4 × 1.008 g/mol H) + (2 × 16.00 g/mol O) = 170.48 g/mol of CuCl2 2H2O (Note that the waters of hydration are included in the molar mass.) 170.48 g CuCl2  2H 2 O    = 610.32 g CuCl 2H O Mass (g) of CuCl2 2H2O = 3.58 mol CuCl2  2H 2 O  2 2  1 mol CuCl2  2H 2 O 









  Mass (kg) of CuCl2 2H2O = (610.32 g CuCl2 2H2O)  3  = 0.61032 = 0.610 kg CuCl2 2H2O   c) of Bi(NO3)3 5H2O = (1 × of Bi) + (3 × of N) + (10 × of H) + (14 × of O) = (1 × 209.0 g/mol Bi) + (3 × 14.01 g/mol N) + (10 × 1.008 g/mol H) + (14 × 16.00 g/mol H) = 485.11 g/mol of Bi(NO3)3 5H2O (Note that the waters of hydration are included in the molar mass.)  1 mol  Moles of Bi(NO3)3 5H2O = 2.8810 22 FU   = 0.047825 mol Bi(NO3)3 5H2O  6.022 10 23 FU 





 485.1 g Bi(NO3 )3  5H 2 O   = 23.1999 g Mass (g) of Bi(NO3)3 5H2O = (0.047825 mol Bi(NO3)3 5H2O)   1 mol Bi(NO3 )3  5H 2 O   1mg  Mass (mg) of Bi(NO3)3 5H2O = (23.1999 g Bi(NO3)3 5H2O)  3  10 g  4

= 23199.9 = 2.32 × 10 mg Bi(NO3)3 5H2O

3-22

3.18

a) Carbonate is a polyatomic anion with the formula, CO3 . Copper(I) indicates Cu . The correct formula for this ionic compound is Cu2CO3. of Cu2CO3 = (2 × of Cu) + (1 × of C) + (3 × of O) = (2 × 63.55 g/mol Cu) + (1 × 12.01 g/mol C) + (3 × 16.00 g/mol O) = 187.11 g/mol of Cu2CO3

187.11 g Cu2 CO3    = 1562.4 = 1.56 × 103 g Cu CO Mass (g) of Cu2CO3 = 8.35 mol Cu2 CO3  2 3  1 mol Cu2 CO3 





b) Dinitrogen pentaoxide has the formula N2O5. Di- indicates 2 N atoms and penta- indicates 5 O atoms. of N2O5 = (2 × of N) + (5 × of O) = (2 × 14.01 g/mol N) + (5 × 16.00 g/mol O) = 108.02 g/mol of N2O5

  1 mol N 2 O5   = 6.7087 × 10–4 mol N O Moles of N2O5 = 4.04 1020 N 2 O5 molecules  2 5  6.022 10 23 N 2 O5 molecules 





108.02 g N O   2 5 Mass (g) of N2O5 = 6.7087104 mol N 2 O5   = 0.072467 = 0.0725 g N2O5  1 mol N2 O5 





c) The correct formula for this ionic compound is NaClO4; Na has a charge of +1 (Group 1 ion) and the –

perchlorate ion is ClO4 . of NaClO4 = (1 × of Na) + (1 × of Cl) + (4 × of O) = (1 × 22.99 g/mol Na) + (1 × 35.45 g/mol Cl) + (4 × 16.00 g/mol O) = 122.44 g/mol of NaClO4

 1 mol NaClO 4    = 0.644397 = 0.644 mol NaClO Moles of NaClO4 = 78.9 g NaClO 4  4 122.44 g NaClO4 





FU = formula units

 6.022 10 23 FU NaClO   4 FU of NaClO4 = 0.644397 mol NaClO4   1 mol NaClO4  





= 3.88056 × 10 = 3.88 × 10 FU NaClO4

 1 Na  ion   +  = 3.88 × 1023 Na+ ions d) Number of Na ions = 3.88056 1023 FU NaClO4  1 FU NaClO4   1 ClO  ion   4 –  = 3.88 × 1023 ClO – ions Number of ClO4 ions = 3.88056 1023 FU NaClO4  4 1 FU NaClO4   1 Cl atom    = 3.88 × 1023 Cl atoms Number of Cl atoms = 3.88056 1023 FU NaClO4  1 FU NaClO4   4 O atoms    = 1.55 × 1024 O atoms Number of O atoms = 3.88056 1023 FU NaClO4  1 FU NaClO4  Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-23

3.19

Plan: The formula of each compound must be determined from its name. The molar mass for each formula comes from the formula and atomic masses from the periodic table. Determine the molar mass of each substance, then perform the appropriate molar conversions. In part (a), multiply the moles by the molar mass of the compound to find the mass of the sample. In part (b), divide the number of molecules by Avogadro’s number to find moles; multiply the number of moles by the molar mass to obtain the mass. In part (c), divide the mass by the molar mass to find moles of compound and multiply moles by Avogadro’s number to find the number of formula units. In part (d), use the fact that each formula unit contains 2 Li ions, 1 sulfate ion, 1 S atom, and 4 O atoms. Solution: 2– 3+ a) Sulfate is a polyatomic anion with the formula, SO4 . Chromium(III) indicates Cr . Decahydrate indicates 10 water molecules (“waters of hydration”). The correct formula for this ionic compound is Cr2(SO4)3 10H2O. of Cr2(SO4)3 10H2O = (2 × of Cr) + (3 × of S) + (22 × of O) + (20 × of H) = (2 × 52.00 g/mol Cr) + (3 × 32.06 g/mol S) + (22 × 16.00 g/mol O) + (20 × 1.008 g/mol H) =572.34 g/mol of Cr2(SO4)3 10H2O  572.34 g   Mass (g) of Cr2(SO4)3 10H2O = 8.42 mol Cr2 (SO 4 )3  10H 2 O   mol  3

= 4819.103 = 4.82 × 10 g Cr2(SO4)3 10H2O b) Dichlorine heptaoxide has the formula Cl2O7. Di- indicates 2 Cl atoms and hepta- indicates 7 O atoms. of Cl2O7 = (2 × of Cl) + (7 × of O) = (2 × 35.45 g/mol Cl) + (7 × 16.00 g/mol O) = 182.9 g/mol of Cl2O7

  1 mol   = 3.038858 mol Cl O Moles of Cl2O7 = 1.8310 24 molecules Cl2 O 7  2 7 23  6.022 10 molecules 





182.9 g Cl2 O7    = 555.807 = 5.56 × 102 g Cl O Mass (g) of Cl2O7 = 3.038858mol Cl2 O7  2 7   1 mol 





c) The correct formula for this ionic compound is Li2SO4; Li has a charge of +1 (Group 1 ion) and the 2–

sulfate ion is SO4 . of Li2SO4 = (2 × of Li) + (1 × of S) + (4 × of O) = (2 × 6.941 g/mol Li) + (1 × 32.06 g/mol S) + (4 × 16.00 g/mol O) = 109.94 g/mol of Li2SO4  1 mol Li 2 SO 4    = 0.056394 = 0.056 mol Li SO Moles of Li2SO4 = 6.2 g Li 2 SO4  2 4 109.94 g Li2 SO4 





 6.022 1023 FU   22 22 FU of Li2SO4 = 0.056394 mol Li2 SO4   = 3.3960 × 10 = 3.4 × 10 FU Li2SO4  1 mol Li 2 SO4 





 2 Li ions   +  = 6.7920 × 1022 = 6.8 × 1022 Li+ ions d) Number of Li ions = 3.39601022 FU Li2 SO4  1 FU Li 2 SO4   1 SO 2 ion   4 2–  = 3.3960 × 1022 = 3.4 × 1022 SO 2– ions Number of SO4 ions = 3.39601022 FU Li2 SO4  4 1 FU Li 2 SO4 

3-24

 1 S atom    = 3.3960 × 1022 = 3.4 × 1022 S atoms Number of S atoms = 3.39601022 FU Li2 SO4  1 FU Li 2 SO4   4 O atoms    = 1.3584 × 1023 = 1.4 × 1023 O atoms Number of O atoms = 3.39601022 FU Li2 SO4  1 FU Li 2 SO4  3.20

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative number of moles of each element present. Multiply the number of moles of each element by its molar mass to find the total total mass of element 100 . mass of element in 1 mole of compound. Mass percent = molar mass of compound Solution: + – a) Ammonium bicarbonate is an ionic compound consisting of ammonium ions, NH 4 and bicarbonate ions, HCO3 . The formula of the compound is NH4HCO3. of NH4HCO3 = (1 × of N) + (5 × of H) + (1 × of C) + (3 × of O) = (1 × 14.01 g/mol N) + (5 × 1.008 g/mol H) + (1 × 12.01 g/mol C) + (3 × 16.00 g/mol O) = 79.06 g/mol of NH4HCO3 There are 5 moles of H in 1 mole of NH4HCO3. 1.008 g H  Mass (g) of H = 5 mol H = 5.040 g H  1 mol H  total mass H 5.040 g H 100 = 100 = 6.374905 = 6.375% H Mass percent = molar mass of compound 79.06 g NH 4 HCO3 + b) Sodium dihydrogen phosphate heptahydrate is a salt that consists of sodium ions, Na , dihydrogen phosphate – ions, H2PO4 , and seven waters of hydration. The formula is NaH2PO4 7H2O. Note that the waters of hydration are included in the molar mass. of NaH2PO4 7H2O = (1 × of Na) + (16 × of H) + (1 × of P) + (11 × of O) = (1 × 22.99 g/mol Na) + (16 × 1.008 g/mol H) + (1 × 30.97 g/mol P) + (11 × 16.00 g/mol O) = 246.09 g/mol NaH2PO4 7H2O There are 11 moles of O in 1 mole of NaH2PO4 7H2O. 16.00 g O  Mass (g) of O = 11 mol O = 176.00 g O  1 mol O  total mass O 176.00 g O 100 = 100 Mass percent = molar mass of compound 246.09 g NaH2 PO 4  7H2 O = 71.51855 = 71.52% O

3.21

–

a) Strontium periodate is an ionic compound consisting of strontium ions, Sr and periodate ions, IO4 . The formula of the compound is Sr(IO4)2. of Sr(IO4)2 = (1 × of Sr) + (2 × of I) + (8 × of O) = (1 × 87.62 g/mol Sr) + (2 × 126.9 g/mol I) + (8 × 16.00 g/mol O) = 469.4 g/mol of Sr(IO4)2 There are 2 moles of I in 1 mole of Sr(IO4)2. Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-25

126.9 g I  Mass (g) of I = 2 mol I = 253.8 g I  1 mol I  total mass I 253.8 g I 100 = 100 = 54.0690 = 54.07% I Mass percent = molar mass of compound 469.4 g Sr(IO4 )2 +

–

b) Potassium permanganate is an ionic compound consisting of potassium ions, K and permanganate ions, MnO4 . The formula of the compound is KMnO4. of KMnO4 = (1 × of K) + (1 × of Mn) + (4 × of O) = (1 × 39.10 g/mol K) + (1 × 54.94 g/mol Mn) + (4 × 16.00 g/mol O) = 158.04 g/mol of KMnO4 There is 1 mole of Mn in 1 mole of KMnO4.

 54.94 g Mn  Mass (g) of Mn = 1 mol Mn  = 54.94 g Mn  1 mol Mn  Mass percent =

3.22

total mass Mn 54.94 g Mn 100 = 100 = 34.76335 = 34.76% Mn molar mass of compound 158.04 g KMnO4

–

a) Cesium acetate is an ionic compound consisting of Cs cations and C2H3O2 anions. (Note that the formula for –

–

acetate ions can be written as either C2H3O2 or CH3COO .) The formula of the compound is CsC2H3O2. of CsC2H3O2 = (1 × of Cs) + (2 × of C) + (3 × of H) + (2 × of O) = (1 × 132.9 g/mol Cs) + (2 × 12.01 g/mol C) + (3 × 1.008 g/mol H) + (2 × 16.00 g/mol O) = 191.9 g/mol of CsC2H3O2 There are 2 moles of C in 1 mole of CsC2H3O2. 12.01 g C  Mass (g) of C = 2 mol C = 24.02 g C  1 mol C  total mass C 24.02 g C = Mass fraction = = 0.125169 = 0.1252 mass fraction C molar mass of compound 191.9 g CsC 2 H 3 O 2 2+

2–

b) Uranyl sulfate trihydrate is a salt that consists of uranyl ions, UO2 , sulfate ions, SO4 , and three waters of hydration. The formula is UO2SO4 3H2O. Note that the waters of hydration are included in the molar mass. of UO2SO4 3H2O = (1 × of U) + (9 × of O) + (1 × of S) + (6 × of H) = (1 × 238.0 g/mol U) + (9 × 16.00 g/mol O) + (1 × 32.06 g/mol S) + (6 × 1.008 g/mol H) = 420.1 g/mol of UO2SO4 3H2O There are 9 moles of O in 1 mole of UO2SO4 3H2O. 16.00 g O  Mass (g) of O = 9 mol O = 144.0 g O  1 mol O  total mass O 144.0 g O = Mass fraction = = 0.3427755 = 0.3428 mass fraction O molar mass of compound 420.1 g UO2 SO4  3H 2 O

3-26

3.23

–

a) Calcium chlorate is an ionic compound consisting of Ca cations and ClO3 anions. The formula of the compound is Ca(ClO3)2. of Ca(ClO3)2 = (1 × of Ca) + (2 × of Cl) + (6 × of O) = (1 × 40.08 g/mol Ca) + (2 × 35.45 g/mol Cl) + (6 × 16.00 g/mol O) = 206.98 g/mol of Ca(ClO3)2 There are 2 moles of Cl in 1 mole of Ca(ClO3)2.

 35.45 g Cl  Mass (g) of Cl = 2 mol Cl  = 70.92 g Cl  1 mol Cl  Mass fraction =

total mass Cl 70.90 g Cl = = 0.342545 = 0.3425 mass fraction Cl molar mass of compound 206.98 g Ca(ClO3 )2

b) Dinitrogen trioxide has the formula N2O3. Di- indicates 2 N atoms and triindicates 3 O atoms. of N2O3 = (2 × of N) + (3 × of O) = (2 × 14.01 g/mol N) + (3 × 16.00 g/mol O) = 76.02 g/mol of N2O3 There are 2 moles of N in 1 mole of N2O3. 14.01 g N  Mass (g) of N = 2 mol N = 28.02 g N  1 mol N  total mass N 28.02 g N = Mass fraction = = 0.368587 = 0.3686 mass fraction N molar mass of compound 76.02 g N 2 O3 3.24

Plan: Divide the mass given by the molar mass of O2 to find moles. Since 1 mole of oxygen molecules contains 2 moles of oxygen atoms, multiply the moles by 2 to obtain moles of atoms and then multiply by Avogadro’s number to obtain the number of atoms. Solution:  1 mol O2   Moles of O2 = 38.0 g O2   = 1.1875 mol O2  32.00 g O2 





 2 mol O atoms    = 2.375 mol O atoms Moles of O atoms = 1.1875 mol O2   1 mol O2 





 6.022 1023 O atoms    = 1.430225 × 1024 = 1.43 × 1024 O atoms Number of O atoms = 2.375 mol O atoms   1 mol O atoms 



3.25



Plan: Determine the formula of cisplatin from the figure, and then calculate the molar mass from the formula. Divide the mass given by the molar mass to find moles of cisplatin. Since 1 mole of cisplatin contains 6 moles of hydrogen atoms, multiply the moles given by 6 to obtain moles of hydrogen and then multiply by Avogadro’s number to obtain the number of atoms. Solution: The formula for cisplatin is Pt(Cl)2(NH3) 2. of Pt(Cl)2(NH3) 2 = (1 × of Pt) + (2 × of Cl) + (2 × of N) + (6 × of H) = (1 × 195.1 g/mol Pt) + (2 × 35.45 g/mol Cl) + (2 × 14.01 g/mol N) + (6 × 1.008 g/mol H) = 300.1 g/mol of Pt(Cl)2(NH3)2

3-27

 1 mol cisplatin    = 0.9506831 = 0.9507 mol cisplatin a) Moles of cisplatin = 285.3 g cisplatin   300.1 g cisplatin 





 6 mol H    = 5.88 mol H atoms b) Moles of H atoms = 0.98 mol cisplatin  1 mol cisplatin 





 6.0221023 H atoms    = 3.540936 × 1024 = 3.5 × 1024 H atoms Number of H atoms = 5.88 mol H atoms   1 mol H atoms 



3.26



Plan: Determine the formula of allyl sulfide from the figure, and then calculate the molar mass from the formula. In part (a), multiply the given amount in moles by the molar mass to find the mass of the sample. In part (b), divide the given mass by the molar mass to find moles of compound. Since 1 mole of compound contains 6 moles of carbon atoms, multiply the moles of compound by 6 to obtain moles of carbon and then multiply by Avogadro’s number to obtain the number of atoms. Solution: The formula, from the figure, is (C3H5)2S. of (C3H5)2S = (6 × of C) + (10 × of H) + (1 × of S) = (6 × 12.01 g/mol C) + (10 × 1.008 g/mol H) + (1 × 32.06 g/mol S) = 114.20 g/mol of (C3H5)2S 114.20 g allyl sulfide    = 300.3460 = 300 g allyl sulfide a) Mass (g) of allyl sulfide = 2.63 mol allyl sulfide   1 mol allyl sulfide 





 1 mol (C3 H 5 )2 S    = 0.312609 mol allyl sulfide b) Moles of allyl sulfide = 35.7 g (C3 H 5 )2 S  114.20 g (C3 H 5 )2 S 





  6 mol C   = 1.8757 mol C atoms Moles of C atoms = 0.312609 mol (C3 H 5 )2 S  1 mol (C3 H 5 )2 S 





 6.022 1023 C atoms    = 1.129546 × 1024 = 1.13 × 1024 C atoms Number of C atoms = 1.8757 mol C atoms   1 mol C atoms 



3.27



Plan: Determine the molar mass of rust. Convert mass in kg to mass in g and divide by the molar mass to find the moles of rust. Since each mole of rust contains 1 mole of Fe2O3, multiply the moles of rust by 1 to obtain moles of Fe2O3. Multiply the moles of Fe2O3 by 2 to obtain moles of Fe (1:2 Fe2O3:Fe mole ratio) and multiply by the molar mass of Fe to convert to mass. Solution: a) of Fe2O3 4H2O = (2 × of Fe) + (7 × of O) + (8 × of H) = (2 × 55.85 g/mol Fe) + (7 × 16.00 g/mol O) + (8 × 1.008 g/mol H) = 231.76 g/mol 103 g   = 4.52 × 104 g Mass (g) of rust = 45.2 kg rust   1 kg 





 1 mol rust    = 195.029 = 195 mol rust Moles of rust = 4.52 10 4 g rust   231.76 g rust 





b) The formula shows that there is 1 mole of Fe2O3 for every mole of rust, so there are also 195 mol of Fe2O3.

3-28

 2 mol Fe    = 390.058 mol Fe c) Moles of iron = 195.029 mol Fe2 O3  1 mol Fe 2 O3 





 55.85 g Fe    = 21784.74 = 2.18 × 104 g Fe Mass (g) of iron = 390.058 mol Fe   1 mol Fe 



3.28



Plan: Determine the molar mass of propane. Divide the given mass by the molar mass to find the moles. Since each mole of propane contains 3 moles of carbon, multiply the moles of propane by 3 to obtain moles of C atoms. Multiply the moles of C by its molar mass to obtain mass of carbon. Solution: a) The formula of propane is C3H8. of C3H8 = (3 × of C) + (8 × of H) = (3 × 12.01 g/mol C) + (8 × 1.008 g/mol H) = 44.09 g/mol  1 mol C3 H8   Moles of C3H8 = 85.5 g C3 H8   = 1.939215 = 1.94 mol C3H8  44.09 g C3 H8 





 3 mol C    = 5.817645 mol C b) Moles of C = 1.939215 mol C3 H8  1 mol C3 H8 





12.01 g C    = 69.86992 = 69.9 g C Mass (g) of C = 5.817645 mol C   1 mol C 



3.29



Plan: Determine the formula and the molar mass of each compound. The formula gives the relative number of moles of nitrogen present. Multiply the number of moles of nitrogen by its molar mass to find the total mass of nitrogen in 1 mole of compound. Divide the total mass of nitrogen by the molar mass of compound and multiply mol Nmolar mass N 100 . Then rank the values in by 100 to determine mass percent. Mass percent = molar mass of compound order of decreasing mass percent N. Solution: Name

Formula

Molar Mass (g/mol)

Potassium nitrate

KNO3

101.11

Ammonium nitrate

NH4NO3

80.05

Ammonium sulfate

(NH4)2SO4

132.14

Urea

CO(NH2)2

60.06

Mass % N in potassium nitrate =

1 mol N 14.01 g/mol N100

Mass % N in ammonium nitrate = Mass % N in ammonium sulfate =

Mass % N in urea =

101.11 g/mol

= 13.856196 = 13.86% N

2 mol N14.01 g/mol N 100 80.05 g/mol

= 35.003123 = 35.00% N

2 mol N  14.01 g/mol N  100 = 21.20478 = 21.20% N 132.14 g/mol

2 mol N14.01 g/mol N 100 60.06 g/mol

= 46.6533 = 46.65% N

Rank is CO(NH2)2> NH4NO3> (NH4)2SO4> KNO3 Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-29

3.30

Plan: The volume must be converted from cubic feet to cubic centimeters. The volume and the density will give the mass of galena which is then divided by molar mass to obtain moles. Part (b) requires a conversion from cubic decimeters to cubic centimeters. The density allows a change from volume in cubic centimeters to mass which is then divided by the molar mass to obtain moles; the amount in moles is multiplied by Avogadro’s number to obtain formula units of PbS which is also the number of Pb atoms due to the 1:1 PbS:Pb mole ratio. Solution: 2+ 2– Lead(II) sulfide is composed of Pb and S ions and has a formula of PbS. of PbS = (1 × of Pb) + (1 × of S) = (1 × 207.2 g/mol Pb) + (1 × 32.06 g/mol S) = 239.3 g/mol 3 3  12 in    3  2.54 cm   = 28316.85 cm3 a) Volume (cm ) = 1.00 ft 3 PbS  3  1 in 3   1 ft  

 7.46 g PbS  Mass (g) of PbS = 28316.85 cm 3 PbS  = 211243.7 g PbS  1 cm 3   1 mol PbS   = 882.7568 = 883 mol PbS Moles of PbS = 211243.7 g PbS  239.3 g PbS  0.1 m3  1 cm3  3 3 3   b) Volume (cm ) = 1.00 dm 3 PbS  2 3  = 1.00 × 10 cm   1 dm3  10 m     

 7.46 g PbS  Mass (g) of PbS = 1.00 103 cm 3 PbS  = 7460 g PbS  1 cm 3   1 mol PbS   = 31.17426 mol PbS Moles of PbS = 7460 g PbS  239.3 g PbS 

 1 mol Pb  Moles of Pb = 31.17426 mol PbS  = 31.17426 mol Pb 1 mol PbS   6.022 10 23 Pb atoms  25 25 Number of lead atoms = 31.17426 mol Pb   = 1.87731 × 10 = 1.88 × 10 Pb atoms  1 mol Pb  3.31

Plan: If the molecular formula for hemoglobin (Hb) were known, the number of Fe ions in a molecule of hemoglobin could be calculated. It is possible to calculate the mass of iron from the percentage of iron and the molar mass of the compound. Assuming you have 1 mole of hemoglobin, take 0.33% of its molar mass as the mass of Fe in that 1 mole. Divide the mass of Fe by its molar mass to find moles of Fe in 1 mole of hemoglobin which is also the number of ions in 1 molecule. Solution:

 0.33% Fe  6.810 4 g    Mass of Fe =   = 224.4 g Fe   100% Hb  mol   1 mol Fe    = 4.0179 = 4.0 mol Fe2+/mol Hb Moles of Fe = 224.4 g Fe  55.85 g Fe  2+ Thus, there are 4 Fe /molecule Hb. 3.32

Plan: Review the definitions of empirical and molecular formulas. Solution: An empirical formula describes the type and simplest ratio of the atoms of each element present in a compound, whereas a molecular formula describes the type and actual number of atoms of each element in a molecule of the compound. The empirical formula and the molecular formula can be the same. For example, the compound

3-30

formaldehyde has the molecular formula, CH2O. The carbon, hydrogen, and oxygen atoms are present in the ratio of 1:2:1. The ratio of elements cannot be further reduced, so formaldehyde’s empirical formula and molecular formula are the same. Acetic acid has the molecular formula, C2H4O2. The carbon, hydrogen, and oxygen atoms are present in the ratio of 2:4:2, which can be reduced to 1:2:1. Therefore, acetic acid’s empirical formula is CH2O, which is different from its molecular formula. Note that the empirical formula does not uniquely identify a compound, because acetic acid and formaldehyde share the same empirical formula but are different compounds. 3.33

1. Compositional data may be given as the mass of each element present in a sample of compound. 2. Compositional data may be provided as mass percents of each element in the compound. 3. Compositional data obtained through combustion analysis provides the mass of C and H in a compound.

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Plan: Remember that the molecular formula tells the actual number of moles of each element in one mole of compound. Solution: a) No, this information does not allow you to obtain the molecular formula. You can obtain the empirical formula from the number of moles of each type of atom in a compound, but not the molecular formula. b) Yes, you can obtain the molecular formula from the mass percentages and the total number of atoms. Plan: 1) Assume a 100.0 g sample and convert masses (from the mass % of each element) to moles using molar mass. 2) Identify the element with the lowest number of moles and use this number to divide into the number of moles for each element. You now have at least one elemental mole ratio (the one with the smallest number of moles) equal to 1.00 and the remaining mole ratios that are larger than one. 3) Examine the numbers to determine if they are whole numbers. If not, multiply each number by a whole-number factor to get whole numbers for each element. You will have to use some judgment to decide when to round. Write the empirical formula using these whole numbers. 4) Check the total number of atoms in the empirical formula. If it equals the total number of atoms given then the empirical formula is also the molecular formula. If not, then divide the total number of atoms given by the total number of atoms in the empirical formula. This should give a whole number. Multiply the number of atoms of each element in the empirical formula by this whole number to get the molecular formula. If you do not get a whole number when you divide, return to step 3 and revise how you multiplied and rounded to get whole numbers for each element. Roadmap: Mass (g) of each element (express mass percent directly as grams) Divide by

(g/mol)

Amount (mol) of each element Use numbers of moles as subscripts Preliminary empirical formula Change to integer subscripts Empirical formula Divide total number of atoms in molecule by the number of atoms in the empirical formula and multiply the empirical formula by that factor Molecular formula

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c) Yes, you can determine the molecular formula from the mass percent and the number of atoms of one element in a compound. Plan: 1) Follow steps 1–3 in part (b). 2) Compare the number of atoms given for the one element to the number in the empirical formula. Determine the factor the number in the empirical formula must be multiplied by to obtain the given number of atoms for that element. Multiply the empirical formula by this number to get the molecular formula. Roadmap: (Same first three steps as in b). Empirical formula Divide the number of atoms of the one element in the molecule by the number of atoms of that element in the empirical formula and multiply the empirical formula by that factor Molecular formula d) No, the mass % will only lead to the empirical formula. e) Yes, a structural formula shows all the atoms in the compound. Plan: Count the number of atoms of each type of element and record as the number for the molecular formula. Roadmap: Structural formula Count the number of atoms of each element and use these numbers as subscripts Molecular formula 3.35

MgCl2 is an empirical formula, since ionic compounds such as MgCl2 do not contain molecules.

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Plan: Examine the number of atoms of each type in the compound. Divide all atom numbers by the common factor that results in the lowest whole-number values. Add the molar masses of the atoms to obtain the empirical formula mass. Solution: a) C2H4 has a ratio of 2 carbon atoms to 4 hydrogen atoms, or 2:4. This ratio can be reduced to 1:2, so that the empirical formula is CH2. The empirical formula mass is 12.01 g/mol C + 2(1.008 g/mol H) = 14.03 g/mol. b) The ratio of atoms is 2:6:2, or 1:3:1. The empirical formula is CH3O and its empirical formula mass is 12.01 g/mol C + 3(1.008 g/mol H) + 16.00 g/mol O = 31.03 g/mol. c) Since, the ratio of elements cannot be further reduced, the molecular formula and empirical formula are the same, N2O5. The formula mass is 2(14.01 g/mol N) + 5(16.00 g/mol O) = 108.02 g/mol. d) The ratio of elements is 3 atoms of barium to 2 atoms of phosphorus to 8 atoms of oxygen, or 3:2:8. This ratio cannot be further reduced, so the empirical formula is also Ba3(PO4)2, with a formula mass of 3(137.3 g/mol Ba) + 2(30.97 g/mol P) + 8(16.00 g/mol O) = 601.8 g/mol. e) The ratio of atoms is 4:16, or 1:4. The empirical formula is TeI4, and the formula mass is 127.6 g/mol Te + 4(126.9 g/mol I) = 635.2 g/mol.

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