Download Answers Chapters 1-3 bookwork - Dunmore High School

CHAPTER 1
CHEMISTRY: THE STUDY OF CHANGE
1.3
(a)
Quantitative. This statement clearly involves a measurable distance.
(b)
Qualitative. This is a value judgment. There is no numerical scale of measurement for artistic
excellence.
(c)
Qualitative. If the numerical values for the densities of ice and water were given, it would be a
quantitative statement.
(d)
Qualitative. Another value judgment.
(e)
Qualitative. Even though numbers are involved, they are not the result of measurement.
1.4
(a)
hypothesis
1.11
(a)
Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are
changed.
(b)
Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter
(different composition).
(c)
Physical property. The measurement of the boiling point of water does not change its identity or
composition.
(d)
Physical property. The measurement of the densities of lead and aluminum does not change their
composition.
(e)
Chemical property. When uranium undergoes nuclear decay, the products are chemically different
substances.
(a)
Physical change. The helium isn't changed in any way by leaking out of the balloon.
(b)
Chemical change in the battery.
(c)
Physical change. The orange juice concentrate can be regenerated by evaporation of the water.
(d)
Chemical change. Photosynthesis changes water, carbon dioxide, etc., into complex organic matter.
(e)
Physical change. The salt can be recovered unchanged by evaporation.
1.12
(b)
law
(c)
theory
1.13
Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum;
Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon.
1.14
(a)
(f)
K
Pu
1.15
(a)
element
1.16
(a)
(d)
(g)
homogeneous mixture
homogeneous mixture
heterogeneous mixture
(b)
(g)
Sn
S
(b)
(c)
(h)
compound
(b)
(e)
Cr
Ar
(d)
(i)
(c)
B
Hg
element
element
heterogeneous mixture
(e)
(d)
(c)
(f)
Ba
compound
compound
homogeneous mixture
2
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.21
density 
1.22
Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid.
Rearrange the density equation, Equation (1.1) of the text, to solve for mass.
mass
586 g

 3.12 g/mL
volume
188 mL
density 
mass
volume
Solution:
mass  density  volume
mass of ethanol 
1.23
? C = (F  32F) 
0.798 g
 17.4 mL  13.9 g
1 mL
5C
9F
5C
 35C
9F
5C
  11C
(12F  32F) 
9 F
5C
(102F  32F) 
 39C
9F
5C
 1011C
(1852F  32F) 
9F
9F 

 C  5C   32F


(a)
? C = (95F  32F) 
(b)
? C =
(c)
? C =
(d)
? C =
(e)
? F 
9F 

? F   273.15 C 
 32F   459.67F
C 
5

1.24
Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between
Celsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in the
problem into the appropriate equation.
(a)
Conversion from Fahrenheit to Celsius.
? C = (F  32F) 
5C
9F
? C = (105F  32F) 
(b)
5C
 41C
9F
Conversion from Celsius to Fahrenheit.
9F 

? F   C 
 32F
5C 

9F 

? F   11.5 C 
 32F  11.3 F
5
C 

CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(c)
Conversion from Celsius to Fahrenheit.
9F 

? F   C 
 32F
5
C 

9F 

? F   6.3  103 C 
 32F  1.1  104 F
5C 

(d)
Conversion from Fahrenheit to Celsius.
? C = (F  32F) 
5C
9F
? C = (451F  32F) 
1.25
1.26
K  (C  273C)
1K
1C
(a)
K  113C  273C  386 K
(b)
K  37C  273C  3.10  10 K
(c)
K  357C  273C  6.30  10 K
(a)
2
2
1K
1C
C  K  273  77 K  273  196C
K  (C  273C)
(b)
C  4.2 K  273  269C
(c)
C  601 K  273  328C
1.29
(a)
2.7  10
1.30
(a)
10
2
8
3.56  10
(b)
2
(c)
10
8
(a)
(b)
4
(d)
2
 0.0152
indicates that the decimal point must be moved 8 places to the left.
7.78  10
1.31
4.7764  10
indicates that the decimal point must be moved two places to the left.
1.52  10
(b)
5C
 233C
9F
8
1
 0.0000000778
145.75  (2.3  10 )  145.75  0.23  1.4598  10
79500
2.5  102
=
7.95  104
2.5  102
3
2
 3.2  102
4
3
3
(c)
(7.0  10 )  (8.0  10 )  (7.0  10 )  (0.80  10 )  6.2  10
(d)
(1.0  10 )  (9.9  10 )  9.9  10
4
6
10
3
9.6  10
2
3
4
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.32
(a)
Addition using scientific notation.
n
Strategy: Let's express scientific notation as N  10 . When adding numbers using scientific notation, we
must write each quantity with the same exponent, n. We can then add the N parts of the numbers, keeping
the exponent, n, the same.
Solution: Write each quantity with the same exponent, n.
3
n
3
Let’s write 0.0095 in such a way that n  3. We have decreased 10 by 10 , so we must increase N by 10 .
Move the decimal point 3 places to the right.
0.0095  9.5  10
3
Add the N parts of the numbers, keeping the exponent, n, the same.
3
9.5  10
3
 8.5  10
18.0  10
3
The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of
n
10 to express N between 1 and 10 (1.8), we must increase 10 by a factor of 10. The exponent, n, is
increased by 1 from 3 to 2.
18.0  10
(b)
3
 1.8  10
2
Division using scientific notation.
n
Strategy: Let's express scientific notation as N  10 . When dividing numbers using scientific notation,
divide the N parts of the numbers in the usual way. To come up with the correct exponent, n, we subtract the
exponents.
Solution: Make sure that all numbers are expressed in scientific notation.
653  6.53  10
2
Divide the N parts of the numbers in the usual way.
6.53  5.75  1.14
Subtract the exponents, n.
1.14  10
(c)
2  (8)
 1.14  10
2  8
 1.14  10
10
Subtraction using scientific notation.
n
Strategy: Let's express scientific notation as N  10 . When subtracting numbers using scientific notation,
we must write each quantity with the same exponent, n. We can then subtract the N parts of the numbers,
keeping the exponent, n, the same.
Solution: Write each quantity with the same exponent, n.
Let’s write 850,000 in such a way that n  5. This means to move the decimal point five places to the left.
850,000  8.5  10
5
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
5
Subtract the N parts of the numbers, keeping the exponent, n, the same.
5
8.5  10
5
 9.0  10
0.5  10
5
The usual practice is to express N as a number between 1 and 10. Since we must increase N by a factor of 10
n
to express N between 1 and 10 (5), we must decrease 10 by a factor of 10. The exponent, n, is decreased by
1 from 5 to 4.
5
0.5  10  5  10
(d)
4
Multiplication using scientific notation.
n
Strategy: Let's express scientific notation as N  10 . When multiplying numbers using scientific notation,
multiply the N parts of the numbers in the usual way. To come up with the correct exponent, n, we add the
exponents.
Solution: Multiply the N parts of the numbers in the usual way.
3.6  3.6  13
Add the exponents, n.
13  10
4  (6)
 13  10
2
The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of
n
10 to express N between 1 and 10 (1.3), we must increase 10 by a factor of 10. The exponent, n, is
increased by 1 from 2 to 3.
2
3
13  10  1.3  10
1.33
(a)
(e)
four
three
1.34
(a)
(e)
one
two or three
1.35
(a)
10.6 m
1.36
(a)
Division
(b)
(f)
two
one
(b)
(f)
(b)
three
one
0.79 g
(c)
(c)
(g)
five
one
(c)
(g)
three
one or two
(d)
(h)
two, three, or four
two
(d)
four
2
16.5 cm
Strategy: The number of significant figures in the answer is determined by the original number having the
smallest number of significant figures.
Solution:
7.310 km
 1.283
5.70 km
The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits.
Therefore, the answer has only three significant digits.
The correct answer rounded off to the correct number of significant figures is:
1.28
(Why are there no units?)
6
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(b)
Subtraction
Strategy: The number of significant figures to the right of the decimal point in the answer is determined by
the lowest number of digits to the right of the decimal point in any of the original numbers.
Solution: Writing both numbers in decimal notation, we have
0.00326 mg
 0.0000788 mg
0.0031812 mg
The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the
decimal point. Therefore, we carry five digits to the right of the decimal point in our answer.
The correct answer rounded off to the correct number of significant figures is:
0.00318 mg  3.18  10
(c)
3
mg
Addition
Strategy: The number of significant figures to the right of the decimal point in the answer is determined by
the lowest number of digits to the right of the decimal point in any of the original numbers.
Solution: Writing both numbers with exponents  7, we have
7
7
7
(0.402  10 dm)  (7.74  10 dm)  8.14  10 dm
7
Since 7.74  10 has only two digits to the right of the decimal point, two digits are carried to the right of the
decimal point in the final answer.
1.37
1 dm
 226 dm
0.1 m
(a)
? dm  22.6 m 
(b)
? kg  25.4 mg 
(c)
? L  556 mL 
(d)
?
0.001 g
1 kg

 2.54  105 kg
1 mg
1000 g
1  103 L
 0.556 L
1 mL
3
1.38
g
cm 3

10.6 kg
1 m3

1000 g  1  102 m 

  0.0106 g/cm 3
 1 cm 
1 kg


(a)
Strategy: The problem may be stated as
? mg  242 lb
A relationship between pounds and grams is given on the end sheet of your text (1 lb  453.6 g). This
relationship will allow conversion from pounds to grams. A metric conversion is then needed to convert
3
grams to milligrams (1 mg  1  10 g). Arrange the appropriate conversion factors so that pounds and
grams cancel, and the unit milligrams is obtained in your answer.
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
7
Solution: The sequence of conversions is
lb  grams  mg
Using the following conversion factors,
1 mg
453.6 g
1 lb
1  103 g
we obtain the answer in one step:
? mg  242 lb 
453.6 g
1 mg

 1.10  108 mg
1 lb
1  103 g
Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many
mg are in 1 lb? There are 453,600 mg in 1 lb.
(b)
Strategy: The problem may be stated as
3
3
? m  68.3 cm
Recall that 1 cm  1  10
2
3
3
m. We need to set up a conversion factor to convert from cm to m .
Solution: We need the following conversion factor so that centimeters cancel and we end up with meters.
1  102 m
1 cm
Since this conversion factor deals with length and we want volume, it must therefore be cubed to give
 1  102 m 
1  102 m 1  102 m 1  102 m


 

 1 cm 
1 cm
1 cm
1 cm


3
We can write
3
 1  102 m 
? m  68.3 cm  
  6.83  105 m 3
 1 cm 


3
3
3
Check: We know that 1 cm  1  10
6
5
1  10 gives 6.83  10 .
6
3
1
3
m . We started with 6.83  10 cm . Multiplying this quantity by
(c)
Strategy: The problem may be stated as
3
? L  7.2 m
3
3
In Chapter 1 of the text, a conversion is given between liters and cm (1 L  1000 cm ). If we can convert
3
3
2
m to cm , we can then convert to liters. Recall that 1 cm  1  10 m. We need to set up two conversion
3
3
3
factors to convert from m to L. Arrange the appropriate conversion factors so that m and cm cancel, and
the unit liters is obtained in your answer.
8
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
Solution: The sequence of conversions is
3
3
m  cm  L
Using the following conversion factors,
 1 cm 


 1  102 m 


3
1L
1000 cm3
the answer is obtained in one step:
3
 1 cm 
1L
 7.2  103 L
? L  7.2 m3  
 
 1  102 m  1000 cm3


3
3
3
Check: From the above conversion factors you can show that 1 m  1  10 L. Therefore, 7 m would
3
equal 7  10 L, which is close to the answer.
(d)
Strategy: The problem may be stated as
? lb  28.3 g
A relationship between pounds and grams is given on the end sheet of your text (1 lb  453.6 g). This
relationship will allow conversion from grams to pounds. If we can convert from g to grams, we can then
6
convert from grams to pounds. Recall that 1 g  1  10 g. Arrange the appropriate conversion factors so
that g and grams cancel, and the unit pounds is obtained in your answer.
Solution: The sequence of conversions is
g  g  lb
Using the following conversion factors,
1  106 g
1 g
1 lb
453.6 g
we can write
? lb  28.3 g 
1  106 g
1 lb

 6.24  108 lb
1 g
453.6 g
Check: Does the answer seem reasonable? What number does the prefix  represent? Should 28.3 g be a
very small mass?
1.39
1255 m
1 mi
3600 s


 2808 mi/h
1s
1609 m
1h
1.40
Strategy: The problem may be stated as
? s  365.24 days
You should know conversion factors that will allow you to convert between days and hours, between hours
and minutes, and between minutes and seconds. Make sure to arrange the conversion factors so that days,
hours, and minutes cancel, leaving units of seconds for the answer.
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
Solution: The sequence of conversions is
days  hours  minutes  seconds
Using the following conversion factors,
24 h
1 day
60 min
1h
60 s
1 min
we can write
? s = 365.24 day 
24 h 60 min
60 s


 3.1557  107 s
1 day
1h
1 min
Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year?
1.609 km 1000 m
1s
1 min



 8.3 min
8
1 mi
1 km
60 s
3.00  10 m
1.41
(93  106 mi) 
1.42
(a)
? in/s 
(b)
? m/min 
(c)
? km/h 
1.43
6.0 ft 
168 lb 
1 mi
5280 ft 12 in 1 min



 81 in/s
13 min
1 mi
1 ft
60 s
1 mi
1609 m

 1.2  102 m/min
13 min
1 mi
1 mi
1609 m
1 km
60 min



 7.4 km/h
13 min
1 mi
1000 m
1h
1m
 1.8 m
3.28 ft
453.6 g
1 kg

 76.2 kg
1 lb
1000 g
55 mi 1.609 km

 88 km/h
1h
1 mi
1.44
? km/h 
1.45
62 m
1 mi
3600 s


 1.4  102 mph
1s
1609 m
1h
1.46
0.62 ppm Pb 
1.47
0.62 g Pb
1  106 g blood
0.62 g Pb
6.0  103 g of blood 
 3.7  103 g Pb
6
1  10 g blood
(a)
1.42 yr 
365 day 24 h 3600 s 3.00  108 m
1 mi




 8.35  1012 mi
1 yr
1 day
1h
1s
1609 m
9
10
1.48
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
36 in 2.54 cm

 2.96  103 cm
1 yd
1 in
(b)
32.4 yd 
(c)
3.0  1010 cm
1 in
1 ft


 9.8  108 ft/s
1s
2.54 cm 12 in
(a)
? m  185 nm 
(b)
? s  (4.5  109 yr) 
(c)
 0.01 m 
5 3
? m  71.2 cm3  
  7.12  10 m
1
cm


(d)
 1 cm 
1L
? L  88.6 m3  
 8.86  104 L
 
 1  102 m  1000 cm3


1  109 m
 1.85  107 m
1 nm
365 day 24 h 3600 s


 1.4  1017 s
1 yr
1 day
1h
3
3
3
3
 1 cm 
1 kg
3
3

  2.70  10 kg/m
1000 g  0.01 m 
1.49
density 
2.70 g
1.50
density 
0.625 g
1L
1 mL


 6.25  104 g/cm 3
1L
1000 mL 1 cm3
1.51
Substance
(a) water
(b) carbon
(c) iron
(d) hydrogen gas
(e) sucrose
(f) table salt
(g) mercury
(h) gold
(i) air
1.52
See Section 1.6 of your text for a discussion of these terms.
1 cm3

Qualitative Statement
colorless liquid
black solid (graphite)
rusts easily
colorless gas
tastes sweet
tastes salty
liquid at room temperature
a precious metal
a mixture of gases
Quantitative Statement
freezes at 0C
3
density  2.26 g/cm
3
density  7.86 g/cm
melts at 255.3C
at 0C, 179 g of sucrose dissolves in 100 g of H2O
melts at 801C
boils at 357C
3
density  19.3 g/cm
contains 20% oxygen by volume
(a)
Chemical property. Iron has changed its composition and identity by chemically combining with
oxygen and water.
(b)
Chemical property. The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids,
thus changing the composition and identity of the water.
(c)
Physical property. The color of the hemoglobin can be observed and measured without changing its
composition or identity.
(d)
Physical property. The evaporation of water does not change its chemical properties. Evaporation is a
change in matter from the liquid state to the gaseous state.
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(e)
Chemical property. The carbon dioxide is chemically converted into other molecules.
1 ton
 4.75  107 tons of sulfuric acid
1.53
(95.0  109 lb of sulfuric acid) 
1.54
Volume of rectangular bar  length  width  height
density 
1.55
11
2.0  103 lb
m
52.7064 g
=
 2.6 g/cm 3
V
(8.53 cm)(2.4 cm)(1.0 cm)
mass  density  volume
3
4
3
4
(10.0 cm) ]  8.08  10 g
3
(a)
mass  (19.3 g/cm )  [
(b)

1 cm 
6
mass  (21.4 g/cm )   0.040 mm 
  1.4  10 g
10 mm 

(c)
mass  (0.798 g/mL)(50.0 mL)  39.9 g
3
1.56
3
You are asked to solve for the inner diameter of the tube. If you can calculate the volume that the mercury
2
occupies, you can calculate the radius of the cylinder, Vcylinder  r h (r is the inner radius of the cylinder,
and h is the height of the cylinder). The cylinder diameter is 2r.
volume of Hg filling cylinder 
volume of Hg filling cylinder 
mass of Hg
density of Hg
105.5 g
13.6 g/cm3
 7.76 cm3
Next, solve for the radius of the cylinder.
2
Volume of cylinder  r h
r 
volume
h
r 
7.76 cm3
 0.441 cm
  12.7 cm
The cylinder diameter equals 2r.
Cylinder diameter  2r  2(0.441 cm)  0.882 cm
1.57
From the mass of the water and its density, we can calculate the volume that the water occupies. The volume
that the water occupies is equal to the volume of the flask.
volume 
mass
density
Mass of water  87.39 g  56.12 g  31.27 g
12
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
Volume of the flask 
mass
31.27 g
=
 31.35 cm 3
density
0.9976 g/cm3
1.58
343 m
1 mi
3600 s


 767 mph
1s
1609 m
1h
1.59
The volume of silver is equal to the volume of water it displaces.
3
Volume of silver  260.5 mL  242.0 mL  18.5 mL  18.5 cm
194.3 g
density 
18.5 cm
1.60
3
 10.5 g/cm 3
In order to work this problem, you need to understand the physical principles involved in the experiment in
Problem 1.59. The volume of the water displaced must equal the volume of the piece of silver. If the silver
did not sink, would you have been able to determine the volume of the piece of silver?
The liquid must be less dense than the ice in order for the ice to sink. The temperature of the experiment must
be maintained at or below 0°C to prevent the ice from melting.
1.61
density 
mass
1.20  104 g

 11.4 g/cm 3
volume
1.05  103 cm3
1.62
Volume 
mass
density
Volume occupied by Li 
1.63
1.20  103 g
0.53 g / cm3
 2.3  103 cm 3
For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C.
5C
? C = 0.1F 
 0.056C
9F
The percent error is the amount of uncertainty in a measurement divided by the value of the measurement,
converted to percent by multiplication by 100.
Percent error 
known error in a measurement
 100%
value of the measurement
For the Fahrenheit thermometer,
percent error 
0.056C
 100%  0.14%
38.9C
For the Celsius thermometer,
percent error 
0.1C
 100%  0.26%
38.9C
Which thermometer is more accurate?
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.64
To work this problem, we need to convert from cubic feet to L. Some tables will have a conversion factor of
3
28.3 L  1 ft , but we can also calculate it using the factor-label method described in Section 1.9 of the text.
First, converting from cubic feet to liters:
 12 in   2.54 cm 
1 mL 1  103 L
(5.0  10 ft )  

 1.4  109 L
 
 
3
1 mL
 1 ft   1 in  1 cm
3
7
3
3
The mass of vanillin (in g) is:
2.0  1011 g vanillin
 (1.4  109 L)  2.8  102 g vanillin
1L
The cost is:
(2.8  102 g vanillin) 
1.65
$112
 $0.063  6.3¢
50 g vanillin
9F 

? F =  C 
+ 32F
5C 

Let temperature  t
9
t + 32F
5
9
t  t = 32F
5
4
 t = 32F
5
t =
t  40F  40C
1.66
There are 78.3  117.3  195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor.
 195.6oC 


 100oS 


Set up the equation like a Celsius to Fahrenheit conversion. We need to subtract 117.3C, because the zero
point on the new scale is 117.3C lower than the zero point on the Celsius scale.
 195.6C 
? C = 
 (? S )  117.3C
 100S 
1.67
13
Solving for ? °S gives:
 100S 
? S = (? C + 117.3C) 

 195.6C 
For 25°C we have:
 100S 
? S  (25C + 117.3C) 
  73S
 195.6C 
The key to solving this problem is to realize that all the oxygen needed must come from the 4% difference
(20% - 16%) between inhaled and exhaled air.
The 240 mL of pure oxygen/min requirement comes from the 4% of inhaled air that is oxygen.
14
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
240 mL of pure oxygen/min  (0.04)(volume of inhaled air/min)
Volume of inhaled air/min 
240 mL of oxygen/min
 6000 mL of inhaled air/min
0.04
Since there are 12 breaths per min,
volume of air/breath 
1.68
1.69
6000 mL of inhaled air
1 min

 5  102 mL/breath
1 min
12 breaths
(a)
6000 mL of inhaled air 0.001 L 60 min 24 h



 8.6  103 L of air/day
1 min
1 mL
1h
1 day
(b)
8.6  103 L of air 2.1  106 L CO

 0.018 L CO/day
1 day
1 L of air
The mass of the seawater is:
(1.5  1021 L) 
1 mL
1.03 g

 1.5  1024 g  1.5  1021 kg seawater
0.001 L 1 mL
Seawater is 3.1% NaCl by mass. The total mass of NaCl in kilograms is:
mass NaCl (kg)  (1.5  1021 kg seawater) 
mass NaCl (tons)  (4.7  1019 kg) 
1.70
3.1% NaCl
 4.7  1019 kg NaCl
100% seawater
2.205 lb
1 ton

 5.2  1016 tons NaCl
1 kg
2000 lb
First, calculate the volume of 1 kg of seawater from the density and the mass. We chose 1 kg of seawater,
because the problem gives the amount of Mg in every kg of seawater. The density of seawater is given in
Problem 1.69.
volume 
mass
density
volume of 1 kg of seawater 
1000 g
 971 mL  0.971 L
1.03 g/mL
In other words, there are 1.3 g of Mg in every 0.971 L of seawater.
Next, let’s convert tons of Mg to grams of Mg.
(8.0  104 tons Mg) 
2000 lb 453.6 g

 7.3  1010 g Mg
1 ton
1 lb
4
Volume of seawater needed to extract 8.0  10 ton Mg 
(7.3  1010 g Mg) 
0.971 L seawater
 5.5  1010 L of seawater
1.3 g Mg
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.71
15
Assume that the crucible is platinum. Let’s calculate the volume of the crucible and then compare that to the
volume of water that the crucible displaces.
volume 
mass
density
Volume of crucible 
860.2 g
 40.10 cm 3
21.45 g/cm3
Volume of water displaced 
(860.2  820.2)g
0.9986 g/cm
3
 40.1 cm 3
The volumes are the same (within experimental error), so the crucible is made of platinum.
1.72
Volume  surface area  depth
3
2
Recall that 1 L  1 dm . Let’s convert the surface area to units of dm and the depth to units of dm.
2
2
 1000 m   1 dm 
16
2
surface area  (1.8  108 km2 )  
 
  1.8  10 dm
1
km
0.1
m

 

depth  (3.9  103 m) 
1 dm
 3.9  104 dm
0.1 m
Volume  surface area  depth  (1.8  10
1.73
16
2
4
dm )(3.9  10 dm)  7.0  10
31.103 g Au
 75.0 g Au
1 troy oz Au
(a)
2.41 troy oz Au 
(b)
1 troy oz  31.103 g
? g in 1 oz  1 oz 
1 lb
453.6 g

 28.35 g
16 oz
1 lb
A troy ounce is heavier than an ounce.
1.74
Volume of sphere 
4 3
r
3
3
Volume 
4  15 cm 

 1.8  103 cm3
3  2 
mass  volume  density  (1.8  103 cm3 ) 
22.57 g Os
1 cm
41 kg Os 
1.75
(a)
2.205 lb
 9.0  101 lb Os
1 kg
|0.798 g/mL  0.802 g/mL|
 100%  0.5%
0.798 g/mL
3

1 kg
 41 kg Os
1000 g
20
3
dm  7.0  10
20
L
16
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(b)
1.76
4
62 kg  6.2  10 g
O:
C:
H:
1.77
|0.864 g  0.837 g|
 100%  3.1%
0.864 g
4
4
4
(6.2  10 g)(0.65)  4.0  10 g O
4
4
(6.2  10 g)(0.18)  1.1  10 g C
4
3
(6.2  10 g)(0.10)  6.2  10 g H
3
N: (6.2  10 g)(0.03)  2  10 g N
4
2
Ca: (6.2  10 g)(0.016)  9.9  10 g Ca
4
2
P: (6.2  10 g)(0.012)  7.4  10 g P
3 minutes 44.39 seconds  224.39 seconds
Time to run 1500 meters is:
1500 m 
1.78
1 mi
224.39 s

 209.19 s  3 min 29.19 s
1609 m
1 mi
2
2
? C  (7.3  10  273) K  4.6  10 C

? F   (4.6  102 C) 


9o F 
  32F  8.6  102 F
o 
5 C
1.79
? g Cu  (5.11  103 kg ore) 
1.80
(8.0  104 tons Au) 
1.81
? g Au 
34.63% Cu 1000 g

 1.77  106 g Cu
100% ore
1 kg
2000 lb Au 16 oz Au
$350


 $9.0  1011 or 900 billion dollars
1 ton Au
1 lb Au
1 oz Au
4.0  1012 g Au
1 mL

 (1.5  1021 L seawater)  6.0  1012 g Au
1 mL seawater
0.001 L
value of gold  (6.0  1012 g Au) 
1 lb
16 oz $350


 $7.4  1013
453.6 g
1 lb
1 oz
No one has become rich mining gold from the ocean, because the cost of recovering the gold would outweigh
the price of the gold.
1.1  1022 Fe atoms
 5.4  1022 Fe atoms
1.0 g Fe
1.82
? Fe atoms  4.9 g Fe 
1.83
mass of Earth's crust  (5.9  1021 tons) 
0.50% crust
 3.0  1019 tons
100% Earth
mass of silicon in crust  (3.0  1019 tons crust) 
27.2% Si
2000 lb
1 kg


 7.4  1021 kg Si
100% crust
1 ton
2.205 lb
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.84
17
10 cm  0.1 m. We need to find the number of times the 0.1 m wire must be cut in half until the piece left is
10
1.3  10 m long. Let n be the number of times we can cut the Cu wire in half. We can write:
n
1
10
m
 2   0.1 m  1.3  10
 
n
1
9
 2   1.3  10 m
 
Taking the log of both sides of the equation:
1
n log    log(1.3  109 )
2
n  30 times
5000 mi 1 gal gas 9.5 kg CO2


 9.5  1010 kg CO 2
1 car
20 mi
1 gal gas
1.85
(40  106 cars) 
1.86
Volume  area  thickness.
From the density, we can calculate the volume of the Al foil.
Volume 
mass
3.636 g

 1.347 cm3
density
2.699 g / cm3
2
2
Convert the unit of area from ft to cm .
2
2
 12 in   2.54 cm 
2
1.000 ft  
 
  929.0 cm
 1 ft   1 in 
2
thickness 
volume
1.347 cm3

 1.450  103 cm  1.450  102 mm
2
area
929.0 cm
1.87
(a)
(b)
1.88
First, let’s calculate the mass (in g) of water in the pool. We perform this conversion because we know there
is 1 g of chlorine needed per million grams of water.
homogeneous
heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous.
(2.0  104 gallons H 2 O) 
3.79 L
1 mL
1g


 7.6  107 g H 2 O
1 gallon 0.001 L 1 mL
Next, let’s calculate the mass of chlorine that needs to be added to the pool.
(7.6  107 g H 2 O) 
1 g chlorine
1  106 g H 2 O
 76 g chlorine
18
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
The chlorine solution is only 6 percent chlorine by mass. We can now calculate the volume of chlorine
solution that must be added to the pool.
76 g chlorine 
100% soln
1 mL soln

 1.3  103 mL of chlorine solution
6% chlorine
1 g soln
1 yr
1.89
(2.0  1022 J) 
1.90
We assume that the thickness of the oil layer is equivalent to the length of one oil molecule. We can
calculate the thickness of the oil layer from the volume and surface area.
1.8  1020 J
 1.1  102 yr
2
 1 cm 
5
2
40 m 2  
  4.0  10 cm
0.01
m


3
0.10 mL  0.10 cm
Volume  surface area  thickness
thickness 
volume
0.10 cm3

 2.5  107 cm
5
2
surface area
4.0  10 cm
Converting to nm:
(2.5  107 cm) 
1.91
0.01 m
1 nm

 2.5 nm
1 cm
1  109 m
The mass of water used by 50,000 people in 1 year is:
50, 000 people 
1 g H2O
150 gal water
3.79 L 1000 mL
365 days




 1.04  1013 g H 2 O/yr
1 person each day
1 gal.
1L
1 mL H 2 O
1 yr
A concentration of 1 ppm of fluorine is needed. In other words, 1 g of fluorine is needed per million grams
of water. NaF is 45.0% fluorine by mass. The amount of NaF needed per year in kg is:
(1.04  1013 g H 2 O) 
1g F
6
10 g H 2 O

100% NaF
1 kg

 2.3  104 kg NaF
45% F
1000 g
An average person uses 150 gallons of water per day. This is equal to 569 L of water. If only 6 L of water is
used for drinking and cooking, 563 L of water is used for purposes in which NaF is not necessary. Therefore
the amount of NaF wasted is:
563 L
 100%  99%
569 L
1.92
3
3
(a)
 1 ft   1 in  1 cm3
1 mL


 $3.06  103 /L
 
 
3
12
in
2.54
cm
1
mL
0.001
L
15.0 ft

 

(b)
2.1 L water 
$1.30
0.304 ft 3 gas
$1.30

 $0.055  5.5¢
1 L water
15.0 ft 3
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.93
19
To calculate the density of the pheromone, you need the mass of the pheromone, and the volume that it
occupies. The mass is given in the problem. First, let’s calculate the volume of the cylinder. Converting the
radius and height to cm gives:
0.50 mi 
40 ft 
1609 m
1 cm

 8.0  104 cm
1 mi
0.01 m
12 in 2.54 cm

 1.2  103 cm
1 ft
1 in
2
volume of a cylinder  area  height  r  h
4
2
3
volume  (8.0  10 cm)  (1.2  10 cm)  2.4  10
13
Density of gases is usually expressed in g/L. Let’s convert the volume to liters.
(2.4  1013 cm3 ) 
1 mL
1 cm
density 
3

1L
 2.4  1010 L
1000 mL
mass
1.0  108 g

 4.2  1019 g/L
10
volume
2.4  10 L
3
cm
CHAPTER 2
ATOMS, MOLECULES, AND IONS
2.7
First, convert 1 cm to picometers.
1 cm 
0.01 m
1 pm

 1  1010 pm
1 cm
1  1012 m
? He atoms  (1  1010 pm) 
2.8
1 He atom
2
1  10 pm
Note that you are given information to set up the unit factor relating meters and miles.
ratom  104 rnucleus  104  2.0 cm 
2.13
 1  108 He atoms
1m
1 mi

 0.12 mi
100 cm 1609 m
For iron, the atomic number Z is 26. Therefore the mass number A is:
A  26  28  54
2.14
Strategy: The 239 in Pu-239 is the mass number. The mass number (A) is the total number of neutrons
and protons present in the nucleus of an atom of an element. You can look up the atomic number (number of
protons) on the periodic table.
Solution:
mass number  number of protons  number of neutrons
number of neutrons  mass number  number of protons  239  94  145
2.15
2.16
Isotope
No. Protons
No. Neutrons
3
2 He
4
2 He
24
12 Mg
25
12 Mg
48
22Ti
79
35 Br
195
78 Pt
2
1
2
2
12
12
12
13
22
26
35
44
78
117
Isotope
No. Protons
No. Neutrons
No. Electrons
15
7N
33
16 S
63
29 Cu
84
38 Sr
130
56 Ba
186
74W
202
80 Hg
7
8
7
16
17
16
29
34
29
38
46
38
56
74
56
74
112
74
80
122
80
23
11 Na
(b)
64
28 Ni
2.17
(a)
2.18
The accepted way to denote the atomic number and mass number of an element X is as follows:
A
ZX
where,
A  mass number
Z  atomic number
CHAPTER 2: ATOMS, MOLECULES, AND IONS
(a)
186
74W
21
201
80 Hg
(b)
2.23
Helium and Selenium are nonmetals whose name ends with ium. (Tellerium is a metalloid whose name ends
in ium.)
2.24
(a)
Metallic character increases as you progress down a group of the periodic table. For example, moving
down Group 4A, the nonmetal carbon is at the top and the metal lead is at the bottom of the group.
(b)
Metallic character decreases from the left side of the table (where the metals are located) to the right
side of the table (where the nonmetals are located).
2.25
The following data were measured at 20C.
3
3
H2O (0.98 g/cm )
3
Hg (13.6 g/cm )
(a)
Li (0.53 g/cm )
K (0.86 g/cm )
(b)
Au (19.3 g/cm )
3
Pt (21.4 g/cm )
(c)
Os (22.6 g/cm )
(d)
Te (6.24 g/cm )
3
3
3
3
2.26
F and Cl are Group 7A elements; they should have similar chemical properties. Na and K are both Group 1A
elements; they should have similar chemical properties. P and N are both Group 5A elements; they should
have similar chemical properties.
2.31
(a)
(b)
(c)
This is a polyatomic molecule that is an elemental form of the substance. It is not a compound.
This is a polyatomic molecule that is a compound.
This is a diatomic molecule that is a compound.
2.32
(a)
(b)
(c)
This is a diatomic molecule that is a compound.
This is a polyatomic molecule that is a compound.
This is a polyatomic molecule that is the elemental form of the substance. It is not a compound.
2.33
Elements:
Compounds:
2.34
There are more than two correct answers for each part of the problem.
(a)
(d)
N 2 , S8 , H 2
NH3, NO, CO, CO2, SO2
H2 and F2
(b)
H2O and C12H22O11 (sucrose)

Na
11
10
2
Ca
20
18
HCl and CO
3
Al
13
10
2
Fe
26
24
(c)

I
53
54

F
9
10
S8 and P4
2
S
16
18
2
O
8
10
3
2.35
Ion
No. protons
No. electrons
N
7
10
2.36
The atomic number (Z) is the number of protons in the nucleus of each atom of an element. You can find
this on a periodic table. The number of electrons in an ion is equal to the number of protons minus the
charge on the ion.
number of electrons (ion)  number of protons  charge on the ion
22
CHAPTER 2: ATOMS, MOLECULES, AND IONS

Ion
No. protons
No. electrons
CN
K
19
18
(b)
CH
Mg
12
10
(c)
2
3
Fe
26
23
C9H20

Br
35
36
(d)
Mn
25
23
2
P2 O 5
4
C
6
10
Cu
29
27
(e)
2
BH3
2.43
(a)
2.44
Strategy: An empirical formula tells us which elements are present and the simplest whole-number ratio of
their atoms. Can you divide the subscripts in the formula by some factor to end up with smaller wholenumber subscripts?
Solution:
(a)
(b)
(c)
(d)
Dividing both subscripts by 2, the simplest whole number ratio of the atoms in Al2Br6 is AlBr3.
Dividing all subscripts by 2, the simplest whole number ratio of the atoms in Na2S2O4 is NaSO2.
The molecular formula as written, N2O5, contains the simplest whole number ratio of the atoms
present. In this case, the molecular formula and the empirical formula are the same.
The molecular formula as written, K2Cr2O7, contains the simplest whole number ratio of the atoms
present. In this case, the molecular formula and the empirical formula are the same.
2.45
The molecular formula of glycine is C2H5NO2.
2.46
The molecular formula of ethanol is C2H6O.
2.47
Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually
molecular.
Ionic:
Molecular:
2.48
LiF, BaCl2, KCl
SiCl4, B2H6, C2H4
Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually
molecular.
Ionic:
Molecular:
NaBr, BaF2, CsCl.
CH4, CCl4, ICl, NF3
potassium dihydrogen phosphate
potassium hydrogen phosphate
hydrogen bromide (molecular compound)
hydrobromic acid
lithium carbonate
potassium dichromate
ammonium nitrite
(h)
(i)
(j)
(k)
(l)
(m)
iodic acid
phosphorus pentafluoride
tetraphosphorus hexoxide
cadmium iodide
strontium sulfate
aluminum hydroxide
2.55
(a)
(b)
(c)
(d)
(e)
(f)
(g)
2.56
Strategy: When naming ionic compounds, our reference for the names of cations and anions is Table 2.3
of the text. Keep in mind that if a metal can form cations of different charges, we need to use the Stock
system. In the Stock system, Roman numerals are used to specify the charge of the cation. The metals that

have only one charge in ionic compounds are the alkali metals (1), the alkaline earth metals (2), Ag ,
2
2
3
Zn , Cd , and Al .
CHAPTER 2: ATOMS, MOLECULES, AND IONS
23
When naming acids, binary acids are named differently than oxoacids. For binary acids, the name is based
on the nonmetal. For oxoacids, the name is based on the polyatomic anion. For more detail, see Section 2.7
of the text.
Solution:

(a)
This is an ionic compound in which the metal cation (K ) has only one charge. The correct name is
potassium hypochlorite. Hypochlorite is a polyatomic ion with one less O atom than the chlorite ion,

ClO2
(b)
silver carbonate
(c)
This is an oxoacid that contains the nitrite ion, NO2 . The “-ite” suffix is changed to “-ous”. The
correct name is nitrous acid.
(d)
potassium permanganate
sulfate
(g)
This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to
specify the charge of the Fe ion. Since the oxide ion has a 2 charge, the Fe ion has a 2 charge. The
correct name is iron(II) oxide.
(h)
iron(III) oxide
(i)
This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to
specify the charge of the Ti ion. Since each of the four chloride ions has a 1 charge (total of 4), the
Ti ion has a 4 charge. The correct name is titanium(IV) chloride.
(j)
sodium hydride

(e)
(k)
cesium chlorate
(f)
lithium nitride
(l)
potassium ammonium
sodium oxide

2
(m) This is an ionic compound in which the metal cation (Na ) has only one charge. The O2 ion is called
the peroxide ion. Each oxygen has a 1 charge. You can determine that each oxygen only has a 1
charge, because each of the two Na ions has a 1 charge. Compare this to sodium oxide in part (l).
The correct name is sodium peroxide.
RbNO2
KH2PO4
(b)
(g)
K2S
IF7
(c)
(h)
NaHS
(NH4)2SO4
(d)
(i)
Mg3(PO4)2
AgClO4
(e)
(j)
CaHPO4
BCl3
2.57
(a)
(f)
2.58
Strategy: When writing formulas of molecular compounds, the prefixes specify the number of each type of
atom in the compound.
When writing formulas of ionic compounds, the subscript of the cation is numerically equal to the charge of
the anion, and the subscript of the anion is numerically equal to the charge on the cation. If the charges of
the cation and anion are numerically equal, then no subscripts are necessary. Charges of common cations
and anions are listed in Table 2.3 of the text. Keep in mind that Roman numerals specify the charge of the
cation, not the number of metal atoms. Remember that a Roman numeral is not needed for some metal
cations, because the charge is known. These metals are the alkali metals (1), the alkaline earth metals (2),

2
2
3
Ag , Zn , Cd , and Al .
When writing formulas of oxoacids, you must know the names and formulas of polyatomic anions (see Table
2.3 of the text).
Solution:
(a)
The Roman numeral I tells you that the Cu cation has a 1 charge. Cyanide has a 1 charge. Since, the
charges are numerically equal, no subscripts are necessary in the formula. The correct formula is
CuCN.
24
CHAPTER 2: ATOMS, MOLECULES, AND IONS
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)

Strontium is an alkaline earth metal. It only forms a 2 cation. The polyatomic ion chlorite, ClO2 , has
a 1 charge. Since the charges on the cation and anion are numerically different, the subscript of the
cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically
equal to the charge on the cation. The correct formula is Sr(ClO2)2.

Perbromic tells you that the anion of this oxoacid is perbromate, BrO4 . The correct formula is
HBrO4(aq). Remember that (aq) means that the substance is dissolved in water.

Hydroiodic tells you that the anion of this binary acid is iodide, I . The correct formula is HI(aq).

Na is an alkali metal. It only forms a 1 cation. The polyatomic ion ammonium, NH4 , has a 1
3
charge and the polyatomic ion phosphate, PO4 , has a 3 charge. To balance the charge, you need 2

Na cations. The correct formula is Na2(NH4)PO4.
The Roman numeral II tells you that the Pb cation has a 2 charge. The polyatomic ion carbonate,
2
CO3 , has a 2 charge. Since, the charges are numerically equal, no subscripts are necessary in the
formula. The correct formula is PbCO3.
The Roman numeral II tells you that the Sn cation has a 2 charge. Fluoride has a 1 charge. Since the
charges on the cation and anion are numerically different, the subscript of the cation is numerically
equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the
cation. The correct formula is SnF2.
This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the
molecule. The correct formula is P4S10.
The Roman numeral II tells you that the Hg cation has a 2 charge. Oxide has a 2 charge. Since, the
charges are numerically equal, no subscripts are necessary in the formula. The correct formula is HgO.
The Roman numeral I tells you that the Hg cation has a 1 charge. However, this cation exists as
2
2
Hg2 . Iodide has a 1 charge. You need two iodide ion to balance the 2 charge of Hg2 . The
correct formula is Hg2I2.
This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the
molecule. The correct formula is SeF6.
2.59
Uranium is radioactive. It loses mass because it constantly emits alpha () particles.
2.60
Changing the electrical charge of an atom usually has a major effect on its chemical properties. The two
electrically neutral carbon isotopes should have nearly identical chemical properties.
2.61
The number of protons  65  35  30. The element that contains 30 protons is zinc, Zn. There are two
2
fewer electrons than protons, so the charge of the cation is 2. The symbol for this cation is Zn .
2.62
Atomic number  127  74  53. This anion has 53 protons, so it is an iodide ion. Since there is one more

electron than protons, the ion has a 1 charge. The correct symbol is I .
2.63
(a)
(b)
(c)
Species with the same number of protons and electrons will be neutral. A, F, G.
Species with more electrons than protons will have a negative charge. B, E.
Species with more protons than electrons will have a positive charge. C, D.
(d)
A:
2.64
10
5B
B:
14 3
7N
C:
39 +
19 K
D:
NaCl is an ionic compound; it doesn’t form molecules.
66
2+
30 Zn
E:
81 
35 Br
F:
11
5B
G:
19
9F
CHAPTER 2: ATOMS, MOLECULES, AND IONS
25
2.65
Yes. The law of multiple proportions requires that the masses of sulfur combining with phosphorus must be
in the ratios of small whole numbers. For the three compounds shown, four phosphorus atoms combine with
three, seven, and ten sulfur atoms, respectively. If the atom ratios are in small whole number ratios, then the
mass ratios must also be in small whole number ratios.
2.66
The species and their identification are as follows:
(a)
(b)
(c)
(d)
(e)
(f)
SO2
S8
Cs
N2O5
O
O2
molecule and compound
element and molecule
element
molecule and compound
element
element and molecule
2.67
(a)
(b)
(c)
(d)
This is an ionic compound. Prefixes are not used. The correct name is barium chloride.
Iron has a 3 charge in this compound. The correct name is iron(III) oxide.

NO2 is the nitrite ion. The correct name is cesium nitrite.
Magnesium is an alkaline earth metal, which always has a 2 charge in ionic compounds. The roman
numeral is not necessary. The correct name is magnesium bicarbonate.
2.68
(a)
(b)
(c)
(d)
Ammonium is NH4 , not NH3 . The formula should be (NH4)2CO3.
Calcium has a 2 charge and hydroxide has a 1 charge. The formula should be Ca(OH)2.
2
2−
Sulfide is S , not SO3 . The correct formula is CdS.
2
2
Dichromate is Cr2O7 , not Cr2O4 . The correct formula is ZnCr2O7.
2.69
Symbol
Protons
Neutrons
Electrons
Net Charge

(g)
(h)
(i)
(j)
(k)
(l)
O3
CH4
KBr
S
P4
LiF
element and molecule
molecule and compound
compound
element
element and molecule
compound

11
5B
54 2+
26 Fe
31 3
15 P
196
79 Au
222
86 Rn
5
6
5
0
26
28
24
2
15
16
18
3
79
117
79
0
86
136
86
0
2.70
(a)
(b)
Ionic compounds are typically formed between metallic and nonmetallic elements.
In general the transition metals, the actinides and lanthanides have variable charges.
2.71
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Li , alkali metals always have a 1 charge in ionic compounds
2
S

I , halogens have a 1 charge in ionic compounds
3
N
3
Al , aluminum always has a 3 charge in ionic compounds

Cs , alkali metals always have a 1 charge in ionic compounds
2
Mg , alkaline earth metals always have a 2 charge in ionic compounds.
2.72
The symbol Na provides more information than 11Na. The mass number plus the chemical symbol
identifies a specific isotope of Na (sodium) while combining the atomic number with the chemical symbol
tells you nothing new. Can other isotopes of sodium have different atomic numbers?

23
26
CHAPTER 2: ATOMS, MOLECULES, AND IONS
2.73
The binary Group 7A element acids are: HF, hydrofluoric acid; HCl, hydrochloric acid; HBr, hydrobromic
acid; HI, hydroiodic acid. Oxoacids containing Group 7A elements (using the specific examples for
chlorine) are: HClO4, perchloric acid; HClO3, chloric acid; HClO2, chlorous acid: HClO, hypochlorous acid.
Examples of oxoacids containing other Group A-block elements are: H3BO3, boric acid (Group 3A);
H2CO3, carbonic acid (Group 4A); HNO3, nitric acid and H3PO4, phosphoric acid (Group 5A); and H2SO4,
sulfuric acid (Group 6A). Hydrosulfuric acid, H2S, is an example of a binary Group 6A acid while HCN,
hydrocyanic acid, contains both a Group 4A and 5A element.
2.74
Mercury (Hg) and bromine (Br2)
2.75
(a)
(b)
Isotope
No. Protons
No. Neutrons
4
2 He
20
10 Ne
40
18 Ar
84
36 Kr
132
54 Xe
2
2
10
10
18
22
36
48
54
78
neutron/proton ratio
1.00
1.00
1.22
1.33
1.44
The neutron/proton ratio increases with increasing atomic number.
2.76
H2, N2, O2, F2, Cl2, He, Ne, Ar, Kr, Xe, Rn
2.77
Cu, Ag, and Au are fairly chemically unreactive. This makes them specially suitable for making coins and
jewelry, that you want to last a very long time.
2.78
They do not have a strong tendency to form compounds. Helium, neon, and argon are chemically inert.
2.79
Magnesium and strontium are also alkaline earth metals. You should expect the charge of the metal to be the
same (2). MgO and SrO.
2.80
All isotopes of radium are radioactive. It is a radioactive decay product of uranium-238. Radium itself does
not occur naturally on Earth.
2.81
(a)
(b)
(c)
Berkelium (Berkeley, CA); Europium (Europe); Francium (France); Scandium (Scandinavia);
Ytterbium (Ytterby, Sweden); Yttrium (Ytterby, Sweden).
Einsteinium (Albert Einstein); Fermium (Enrico Fermi); Curium (Marie and Pierre Curie);
Mendelevium (Dmitri Mendeleev); Lawrencium (Ernest Lawrence).
Arsenic, Cesium, Chlorine, Chromium, Iodine.
2.82
Argentina is named after silver (argentum, Ag).
2.83
The mass of fluorine reacting with hydrogen and deuterium would be the same. The ratio of F atom to
hydrogen (or deuterium) is 1:1 in both compounds. This does not violate the law of definite proportions.
When the law of definite proportions was formulated, scientists did not know of the existence of isotopes.
2.84
(a)
(d)
NaH, sodium hydride
AlF3, aluminum fluoride
2.85
(a)
Br
(b)
Rn
(b)
(e)
(c)
Se
B2O3, diboron trioxide
OF2, oxygen difluoride
(d)
Rb
(c)
(f)
(e)
Pb
Na2S, sodium sulfide
SrCl2, strontium chloride
CHAPTER 2: ATOMS, MOLECULES, AND IONS
2.86
2.87
27
All of these are molecular compounds. We use prefixes to express the number of each atom in the molecule.
The names are nitrogen trifluoride (NF3), phosphorus pentabromide (PBr5), and sulfur dichloride (SCl2).
alkali
metals
alkaline
earth metals
noble gases
halogens
1A
8A
2A
3A 4A 5A 6A 7A
The metalloids are shown in gray.
2.88
Cation
2
Mg
Anion

HCO3
Sr
Cl
2
3
Fe
Mn
Sn
2
4
Co
2
Hg2
Cu
2


Li
3
Al

NO2

ClO3


Br
PO4
I
3

CO3
N
S
3
2
2
Formula
Mg(HCO3)2
SrCl2
Name
Magnesium bicarbonate
Strontium chloride
Fe(NO2)3
Iron(III) nitrite
Mn(ClO3)2
SnBr4
Manganese(II) chlorate
Tin(IV) bromide
Co3(PO4)2
Cobalt(II) phosphate
Hg2I2
Mercury(I) iodide
Cu2CO3
Li3N
Copper(I) carbonate
Lithium nitride
Al2S3
Aluminum sulfide
2.89
(a)
(b)
(c)
(d)
(e)
CO2(s), solid carbon dioxide
NaCl, sodium chloride
N2O, nitrous oxide
CaCO3, calcium carbonate
CaO, calcium oxide
2.90
(a)
Rutherford’s experiment is described in detail in Section 2.2 of the text. From the average magnitude
of scattering, Rutherford estimated the number of protons (based on electrostatic interactions) in the
nucleus.
(b)
Assuming that the nucleus is spherical, the volume of the nucleus is:
V 
(f)
(g)
(h)
(i)
(j)
Ca(OH)2, calcium hydroxide
NaHCO3, sodium bicarbonate
Na2CO310H2O, sodium carbonate decahydrate
CaSO42H2O, calcium sulfate dihydrate
Mg(OH)2, magnesium hydroxide
4 3
4
r  (3.04  1013 cm)3  1.18  1037 cm3
3
3
The density of the nucleus can now be calculated.
d 
3.82  1023 g
m

 3.24  1014 g/cm 3
37
3
V
1.18  10 cm
28
CHAPTER 2: ATOMS, MOLECULES, AND IONS
To calculate the density of the space occupied by the electrons, we need both the mass of 11 electrons,
and the volume occupied by these electrons.
The mass of 11 electrons is:
11 electrons 
9.1095  1028 g
 1.0020  1026 g
1 electron
The volume occupied by the electrons will be the difference between the volume of the atom and the
volume of the nucleus. The volume of the nucleus was calculated above. The volume of the atom is
calculated as follows:
186 pm 
Vatom 
1  1012 m
1 cm

 1.86  108 cm

2
1 pm
1  10 m
4 3
4
r  (1.86  108 cm)3  2.70  1023 cm3
3
3
Velectrons  Vatom  Vnucleus  (2.70  10
23
3
cm )  (1.18  10
37
3
cm )  2.70  10
23
3
cm
As you can see, the volume occupied by the nucleus is insignificant compared to the space occupied by
the electrons.
The density of the space occupied by the electrons can now be calculated.
d 
m
1.0020  1026 g

 3.71  104 g/cm 3
23
3
V
2.70  10 cm
The above results do support Rutherford's model. Comparing the space occupied by the electrons to the
volume of the nucleus, it is clear that most of the atom is empty space. Rutherford also proposed that
the nucleus was a dense central core with most of the mass of the atom concentrated in it. Comparing
the density of the nucleus with the density of the space occupied by the electrons also supports
Rutherford's model.
2.91
S
N
B
I
CHAPTER 3
MASS RELATIONSHIPS IN
CHEMICAL REACTIONS
3.5
(34.968 amu)(0.7553)  (36.956 amu)(0.2447)  35.45 amu
3.6
Strategy: Each isotope contributes to the average atomic mass based on its relative abundance.
Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the
average atomic mass of that particular isotope.
6
It would seem that there are two unknowns in this problem, the fractional abundance of Li and the fractional
7
abundance of Li. However, these two quantities are not independent of each other; they are related by the
6
fact that they must sum to 1. Start by letting x be the fractional abundance of Li. Since the sum of the two
abundance’s must be 1, we can write
7
Abundance Li  (1  x)
Solution:
Average atomic mass of Li  6.941 amu  x(6.0151 amu)  (1  x)(7.0160 amu)
6.941  1.0009x  7.0160
1.0009x  0.075
x  0.075
6
7
x  0.075 corresponds to a natural abundance of Li of 7.5 percent. The natural abundance of Li is
(1  x)  0.925 or 92.5 percent.
3.7
 6.022  1023 amu 
The unit factor required is 



1g


? g  13.2 amu 
3.8
6.022  1023 amu
 2.19  1023 g
 6.022  1023 amu 
The unit factor required is 



1g


? amu  8.4 g 
3.11
1g
6.022  1023 amu
= 5.1  1024 amu
1g
In one year:
(6.5  109 people) 
Total time 
365 days 24 hr 3600 s 2 particles



 4.1  1017 particles/yr
1 yr
1 day
1 hr
1 person
6.022  1023 particles
4.1  10 particles/yr
17
 1.5  106 yr
30
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.12
The thickness of the book in miles would be:
0.0036 in
1 ft
1 mi


 (6.022  1023 pages) = 3.4  1016 mi
1 page
12 in 5280 ft
The distance, in miles, traveled by light in one year is:
1.00 yr 
365 day 24 h 3600 s 3.00  108 m
1 mi




 5.88  1012 mi
1 yr
1 day
1h
1s
1609 m
The thickness of the book in light-years is:
(3.4  1016 mi) 
1 light-yr
 5.8  103 light - yr
5.88  10 mi
12
3
It will take light 5.8  10 years to travel from the first page to the last one!
6.022  1023 S atoms
 3.07  1024 S atoms
1 mol S
3.13
5.10 mol S 
3.14
(6.00  109 Co atoms) 
1 mol Co
6.022  10
23
= 9.96  1015 mol Co
Co atoms
1 mol Ca
 1.93 mol Ca
40.08 g Ca
3.15
77.4 g of Ca 
3.16
Strategy: We are given moles of gold and asked to solve for grams of gold. What conversion factor do we
need to convert between moles and grams? Arrange the appropriate conversion factor so moles cancel, and
the unit grams is obtained for the answer.
Solution: The conversion factor needed to covert between moles and grams is the molar mass. In the
periodic table (see inside front cover of the text), we see that the molar mass of Au is 197.0 g. This can be
expressed as
1 mol Au  197.0 g Au
From this equality, we can write two conversion factors.
1 mol Au
197.0 g Au
and
197.0 g Au
1 mol Au
The conversion factor on the right is the correct one. Moles will cancel, leaving the unit grams for the answer.
We write
? g Au = 15.3 mol Au 
197.0 g Au
= 3.01  103 g Au
1 mol Au
Check: Does a mass of 3010 g for 15.3 moles of Au seem reasonable? What is the mass of 1 mole of Au?
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.17
3.18
(a)
200.6 g Hg
1 mol Hg

 3.331  1022 g/Hg atom
1 mol Hg
6.022  1023 Hg atoms
(b)
20.18 g Ne
1 mol Ne

 3.351  1023 g/Ne atom
1 mol Ne
6.022  1023 Ne atoms
31
(a)
Strategy: We can look up the molar mass of arsenic (As) on the periodic table (74.92 g/mol). We want to
find the mass of a single atom of arsenic (unit of g/atom). Therefore, we need to convert from the unit mole
in the denominator to the unit atom in the denominator. What conversion factor is needed to convert between
moles and atoms? Arrange the appropriate conversion factor so mole in the denominator cancels, and the
unit atom is obtained in the denominator.
Solution: The conversion factor needed is Avogadro's number. We have
1 mol  6.022  10
23
particles (atoms)
From this equality, we can write two conversion factors.
1 mol As
and
6.022  1023 As atoms
6.022  1023 As atoms
1 mol As
The conversion factor on the left is the correct one. Moles will cancel, leaving the unit atoms in the
denominator of the answer.
We write
? g/As atom 
(b)
74.92 g As
1 mol As

 1.244  1022 g/As atom
23
1 mol As
6.022  10 As atoms
Follow same method as part (a).
? g/Ni atom 
58.69 g Ni
1 mol Ni

 9.746  1023 g/Ni atom
1 mol Ni
6.022  1023 Ni atoms
Check: Should the mass of a single atom of As or Ni be a very small mass?
3.19
3.20
1.00  1012 Pb atoms 
1 mol Pb
6.022  10
23
Pb atoms

207.2 g Pb
 3.44  1010 g Pb
1 mol Pb
Strategy: The question asks for atoms of Cu. We cannot convert directly from grams to atoms of copper.
What unit do we need to convert grams of Cu to in order to convert to atoms? What does Avogadro's
number represent?
Solution: To calculate the number of Cu atoms, we first must convert grams of Cu to moles of Cu. We use
the molar mass of copper as a conversion factor. Once moles of Cu are obtained, we can use Avogadro's
number to convert from moles of copper to atoms of copper.
1 mol Cu  63.55 g Cu
The conversion factor needed is
1 mol Cu
63.55 g Cu
32
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
Avogadro's number is the key to the second conversion. We have
1 mol  6.022  10
23
particles (atoms)
From this equality, we can write two conversion factors.
1 mol Cu
and
6.022  1023 Cu atoms
6.022  1023 Cu atoms
1 mol Cu
The conversion factor on the right is the one we need because it has number of Cu atoms in the numerator,
which is the unit we want for the answer.
Let's complete the two conversions in one step.
grams of Cu  moles of Cu  number of Cu atoms
? atoms of Cu  3.14 g Cu 
1 mol Cu
6.022  1023 Cu atoms

 2.98  1022 Cu atoms
63.55 g Cu
1 mol Cu
Check: Should 3.14 g of Cu contain fewer than Avogadro's number of atoms? What mass of Cu would
contain Avogadro's number of atoms?
3.21
1 mol H
6.022  1023 H atoms

 6.57  1023 H atoms
1.008 g H
1 mol H
For hydrogen:
1.10 g H 
For chromium:
14.7 g Cr 
1 mol Cr
6.022  1023 Cr atoms

 1.70  1023 Cr atoms
52.00 g Cr
1 mol Cr
There are more hydrogen atoms than chromium atoms.
3.22
2 Pb atoms 
1 mol Pb
6.022  10
(5.1  1023 mol He) 
23
Pb atoms

207.2 g Pb
= 6.881  1022 g Pb
1 mol Pb
4.003 g He
= 2.0  1022 g He
1 mol He
2 atoms of lead have a greater mass than 5.1  10
3.23
23
mol of helium.
Using the appropriate atomic masses,
(a)
(b)
(c)
(d)
(e)
(f)
(g)
CH4
NO2
SO3
C6H6
NaI
K2SO4
Ca3(PO4)2
12.01 amu  4(1.008 amu)  16.04 amu
14.01 amu  2(16.00 amu)  46.01 amu
32.07 amu  3(16.00 amu)  80.07 amu
6(12.01 amu)  6(1.008 amu)  78.11 amu
22.99 amu  126.9 amu  149.9 amu
2(39.10 amu)  32.07 amu  4(16.00 amu)  174.27 amu
3(40.08 amu)  2(30.97 amu)  8(16.00 amu)  310.18 amu
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.24
33
Strategy: How do molar masses of different elements combine to give the molar mass of a compound?
Solution: To calculate the molar mass of a compound, we need to sum all the molar masses of the elements
in the molecule. For each element, we multiply its molar mass by the number of moles of that element in one
mole of the compound. We find molar masses for the elements in the periodic table (inside front cover of the
text).
3.25
(a)
molar mass Li2CO3  2(6.941 g)  12.01 g  3(16.00 g)  73.89 g
(b)
molar mass CS2  12.01 g  2(32.07 g)  76.15 g
(c)
molar mass CHCl3  12.01 g  1.008 g  3(35.45 g)  119.37 g
(d)
molar mass C6H8O6  6(12.01 g)  8(1.008 g)  6(16.00 g)  176.12 g
(e)
molar mass KNO3  39.10 g  14.01 g  3(16.00 g)  101.11 g
(f)
molar mass Mg3N2  3(24.31 g)  2(14.01 g)  100.95 g
To find the molar mass (g/mol), we simply divide the mass (in g) by the number of moles.
152 g
 409 g/mol
0.372 mol
3.26
Strategy: We are given grams of ethane and asked to solve for molecules of ethane. We cannot convert
directly from grams ethane to molecules of ethane. What unit do we need to obtain first before we can
convert to molecules? How should Avogadro's number be used here?
Solution: To calculate number of ethane molecules, we first must convert grams of ethane to moles of
ethane. We use the molar mass of ethane as a conversion factor. Once moles of ethane are obtained, we can
use Avogadro's number to convert from moles of ethane to molecules of ethane.
molar mass of C2H6  2(12.01 g)  6(1.008 g)  30.07 g
The conversion factor needed is
1 mol C2 H 6
30.07 g C2 H 6
Avogadro's number is the key to the second conversion. We have
1 mol  6.022  10
23
particles (molecules)
From this equality, we can write the conversion factor:
6.022  10 23 ethane molecules
1 mol ethane
Let's complete the two conversions in one step.
grams of ethane  moles of ethane  number of ethane molecules
? molecules of C2 H 6  0.334 g C2 H 6 
 6.69  10
21
1 mol C2 H 6
6.022  1023 C2 H 6 molecules

30.07 g C2 H 6
1 mol C2 H 6
C2H6 molecules
34
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
Check: Should 0.334 g of ethane contain fewer than Avogadro's number of molecules? What mass of
ethane would contain Avogadro's number of molecules?
3.27
1.50 g glucose 
 3.01  10
1 mol glucose
6.022  1023 molecules glucose
6 C atoms


180.2 g glucose
1 mol glucose
1 molecule glucose
22
C atoms
The ratio of O atoms to C atoms in glucose is 1:1. Therefore, there are the same number of O atoms in
22
glucose as C atoms, so the number of O atoms  3.01  10 O atoms.
The ratio of H atoms to C atoms in glucose is 2:1. Therefore, there are twice as many H atoms in glucose as
22
22
C atoms, so the number of H atoms  2(3.01  10 atoms)  6.02  10 H atoms.
3.28
4
Strategy: We are asked to solve for the number of N, C, O, and H atoms in 1.68  10 g of urea. We
cannot convert directly from grams urea to atoms. What unit do we need to obtain first before we can
convert to atoms? How should Avogadro's number be used here? How many atoms of N, C, O, or H are in
1 molecule of urea?
4
Solution: Let's first calculate the number of N atoms in 1.68  10 g of urea. First, we must convert grams
of urea to number of molecules of urea. This calculation is similar to Problem 3.26. The molecular formula
of urea shows there are two N atoms in one urea molecule, which will allow us to convert to atoms of N. We
need to perform three conversions:
grams of urea  moles of urea  molecules of urea  atoms of N
The conversion factors needed for each step are: 1) the molar mass of urea, 2) Avogadro's number, and 3) the
number of N atoms in 1 molecule of urea.
We complete the three conversions in one calculation.
? atoms of N = (1.68  104 g urea) 
 3.37  10
26
1 mol urea
6.022  10 23 urea molecules
2 N atoms


60.06 g urea
1 mol urea
1 molecule urea
N atoms
The above method utilizes the ratio of molecules (urea) to atoms (nitrogen). We can also solve the problem
by reading the formula as the ratio of moles of urea to moles of nitrogen by using the following conversions:
grams of urea  moles of urea  moles of N  atoms of N
Try it.
4
Check: Does the answer seem reasonable? We have 1.68  10 g urea. How many atoms of N would
60.06 g of urea contain?
We could calculate the number of atoms of the remaining elements in the same manner, or we can use the
atom ratios from the molecular formula. The carbon atom to nitrogen atom ratio in a urea molecule is 1:2,
the oxygen atom to nitrogen atom ratio is 1:2, and the hydrogen atom to nitrogen atom ration is 4:2.
? atoms of C  (3.37  1026 N atoms) 
1 C atom
 1.69  1026 C atoms
2 N atoms
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.29
? atoms of O  (3.37  1026 N atoms) 
1 O atom
 1.69  1026 O atoms
2 N atoms
? atoms of H  (3.37  1026 N atoms) 
4 H atoms
 6.74  1026 H atoms
2 N atoms
The molar mass of C19H38O is 282.5 g.
1.0  1012 g 
1 mol
6.022  1023 molecules

 2.1  109 molecules
282.5 g
1 mol
Notice that even though 1.0  10
pheromone molecules!
3.30
35
Mass of water = 2.56 mL 
12
g is an extremely small mass, it still is comprised of over a billion
1.00 g
= 2.56 g
1.00 mL
Molar mass of H2O  (16.00 g)  2(1.008 g)  18.02 g/mol
? H 2 O molecules = 2.56 g H 2 O 
 8.56  10
3.33
22
1 mol H 2 O
6.022  1023 molecules H 2 O

18.02 g H 2 O
1 mol H 2 O
molecules

Since there are only two isotopes of carbon, there are only two possibilities for CF4 .
12 19 
6 C 9 F4
(molecular mass 88 amu) and
13 19 
6 C 9 F4
(molecular mass 89 amu)
There would be two peaks in the mass spectrum.
3.34
1
1
1
2
2
2
Since there are two hydrogen isotopes, they can be paired in three ways: H- H, H- H, and H- H. There
will then be three choices for each sulfur isotope. We can make a table showing all the possibilities (masses
in amu):
32
33
34
36
S
S
S
S
1
H2
1 2
2
H H
H2
34
35
36
35
36
37
36
37
38
38
39
40
There will be seven peaks of the following mass numbers: 34, 35, 36, 37, 38, 39, and 40.
1
Very accurate (and expensive!) mass spectrometers can detect the mass difference between two H and one
2
H. How many peaks would be detected in such a “high resolution” mass spectrum?
3.39
Molar mass of SnO2  (118.7 g)  2(16.00 g)  150.7 g
%Sn 
118.7 g/mol
 100%  78.77%
150.7 g/mol
%O 
(2)(16.00 g/mol)
 100%  21.23%
150.7 g/mol
36
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.40
Strategy: Recall the procedure for calculating a percentage. Assume that we have 1 mole of CHCl3. The
percent by mass of each element (C, H, and Cl) is given by the mass of that element in 1 mole of CHCl3
divided by the molar mass of CHCl3, then multiplied by 100 to convert from a fractional number to a
percentage.
Solution: The molar mass of CHCl3  12.01 g/mol  1.008 g/mol  3(35.45 g/mol)  119.4 g/mol. The
percent by mass of each of the elements in CHCl3 is calculated as follows:
%C 
12.01 g/mol
 100%  10.06%
119.4 g/mol
%H 
1.008 g/mol
 100%  0.8442%
119.4 g/mol
%Cl 
3(35.45) g/mol
 100%  89.07%
119.4 g/mol
Check: Do the percentages add to 100%? The sum of the percentages is (10.06%  0.8442%  89.07%) 
99.97%. The small discrepancy from 100% is due to the way we rounded off.
3.41
The molar mass of cinnamic alcohol is 134.17 g/mol.
(a)
(b)
%C 
(9)(12.01 g/mol)
 100%  80.56%
134.17 g/mol
%H 
(10)(1.008 g/mol)
 100%  7.51%
134.17 g/mol
%O 
16.00 g/mol
 100%  11.93%
134.17 g/mol
0.469 g C9 H10 O 
 2.11  10
3.42
21
1 mol C9 H10 O
6.022  1023 molecules C9 H10 O

134.17 g C9 H10 O
1 mol C9 H10 O
molecules C9H10O
Compound
Molar mass (g)
N% by mass
2(14.01 g)
 100% = 46.65%
60.06 g
(a)
(NH2)2CO
60.06
(b)
NH4NO3
80.05
2(14.01 g)
 100% = 35.00%
80.05 g
(c)
HNC(NH2)2
59.08
3(14.01 g)
 100% = 71.14%
59.08 g
(d)
NH3
17.03
14.01 g
 100% = 82.27%
17.03 g
Ammonia, NH3, is the richest source of nitrogen on a mass percentage basis.
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.43
37
Assume you have exactly 100 g of substance.
nC  44.4 g C 
1 mol C
 3.70 mol C
12.01 g C
nH  6.21 g H 
1 mol H
 6.16 mol H
1.008 g H
nS  39.5 g S 
1 mol S
 1.23 mol S
32.07 g S
nO  9.86 g O 
1 mol O
 0.616 mol O
16.00 g O
Thus, we arrive at the formula C3.70H6.16S1.23O0.616. Dividing by the smallest number of moles (0.616 mole)
gives the empirical formula, C6H10S2O.
To determine the molecular formula, divide the molar mass by the empirical mass.
molar mass
162 g

 1
empirical molar mass
162.3 g
Hence, the molecular formula and the empirical formula are the same, C6H10S2O.
3.44
METHOD 1:
Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the
percentages sum to 100%. The percentage of oxygen is found by difference:
100%  (19.8%  2.50%  11.6%)  66.1%
In 100 g of PAN there will be 19.8 g C, 2.50 g H, 11.6 g N, and 66.1 g O.
Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula
tells us which elements are present and the simplest whole-number ratio of their atoms. This ratio is
also a mole ratio. Use the molar masses of these elements as conversion factors to convert to moles.
nC = 19.8 g C 
1 mol C
= 1.65 mol C
12.01 g C
nH = 2.50 g H 
1 mol H
= 2.48 mol H
1.008 g H
nN = 11.6 g N 
1 mol N
= 0.828 mol N
14.01 g N
nO = 66.1 g O 
1 mol O
= 4.13 mol O
16.00 g O
Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript. The
formula is C1.65H2.48N0.828O4.13. Dividing the subscripts by 0.828 gives the empirical formula,
C2H3NO5.
To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater
than or equal to one.
molar mass
 1 (integer values)
empirical molar mass
38
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
In this case,
molar mass
120 g

 1
empirical molar mass
121.05 g
Hence, the molecular formula and the empirical formula are the same, C2H3NO5.
METHOD 2:
Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to
grams of each element. Then, use the molar mass to convert to moles of each element.
nC  (0.198)  (120 g) 
1 mol C
 1.98 mol C  2 mol C
12.01 g C
nH  (0.0250)  (120 g) 
1 mol H
 2.98 mol H  3 mol H
1.008 g H
nN  (0.116)  (120 g) 
1 mol N
 0.994 mol N  1 mol N
14.01 g N
nO  (0.661)  (120 g) 
1 mol O
 4.96 mol O  5 mol O
16.00 g O
Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this
method directly gives the molecular formula. The formula is C2H3NO5.
Step 3: Try to reduce the molecular formula to a simpler whole number ratio to determine the empirical
formula. The formula is already in its simplest whole number ratio. The molecular and empirical
formulas are the same. The empirical formula is C2H3NO5.
1 mol Fe 2 O3
2 mol Fe

 0.308 mol Fe
159.7 g Fe2 O3 1 mol Fe 2 O3
3.45
24.6 g Fe2 O3 
3.46
Using unit factors we convert:
g of Hg  mol Hg  mol S  g S
? g S  246 g Hg 
3.47
1 mol Hg
1 mol S
32.07 g S


 39.3 g S
200.6 g Hg 1 mol Hg
1 mol S
 2AlI3(s)
The balanced equation is: 2Al(s)  3I2(s) 
Using unit factors, we convert: g of Al  mol of Al  mol of I2  g of I2
20.4 g Al 
3.48
3 mol I 2 253.8 g I 2
1 mol Al


 288 g I 2
26.98 g Al 2 mol Al
1 mol I2
Strategy: Tin(II) fluoride is composed of Sn and F. The mass due to F is based on its percentage by mass
in the compound. How do we calculate mass percent of an element?
Solution: First, we must find the mass % of fluorine in SnF2. Then, we convert this percentage to a fraction
and multiply by the mass of the compound (24.6 g), to find the mass of fluorine in 24.6 g of SnF2.
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
39
The percent by mass of fluorine in tin(II) fluoride, is calculated as follows:
mass of F in 1 mol SnF2
 100%
molar mass of SnF2
mass % F 

2(19.00 g)
 100% = 24.25% F
156.7 g
Converting this percentage to a fraction, we obtain 24.25/100  0.2425.
Next, multiply the fraction by the total mass of the compound.
? g F in 24.6 g SnF2  (0.2425)(24.6 g)  5.97 g F
Check: As a ball-park estimate, note that the mass percent of F is roughly 25 percent, so that a quarter of
the mass should be F. One quarter of approximately 24 g is 6 g, which is close to the answer.
Note: This problem could have been worked in a manner similar to Problem 3.46. You could
complete the following conversions:
g of SnF2  mol of SnF2  mol of F  g of F
3.49
In each case, assume 100 g of compound.
(a)
2.1 g H 
1 mol H
 2.1 mol H
1.008 g H
65.3 g O 
1 mol O
 4.08 mol O
16.00 g O
32.6 g S 
1 mol S
 1.02 mol S
32.07 g S
This gives the formula H2.1S1.02O4.08. Dividing by 1.02 gives the empirical formula, H2SO4.
(b)
20.2 g Al 
1 mol Al
 0.749 mol Al
26.98 g Al
79.8 g Cl 
1 mol Cl
 2.25 mol Cl
35.45 g Cl
This gives the formula, Al0.749Cl2.25. Dividing by 0.749 gives the empirical formula, AlCl3.
3.50
(a)
Strategy: In a chemical formula, the subscripts represent the ratio of the number of moles of each element
that combine to form the compound. Therefore, we need to convert from mass percent to moles in order to
determine the empirical formula. If we assume an exactly 100 g sample of the compound, do we know the
mass of each element in the compound? How do we then convert from grams to moles?
Solution: If we have 100 g of the compound, then each percentage can be converted directly to grams. In
this sample, there will be 40.1 g of C, 6.6 g of H, and 53.3 g of O. Because the subscripts in the formula
represent a mole ratio, we need to convert the grams of each element to moles. The conversion factor needed
is the molar mass of each element. Let n represent the number of moles of each element so that
40
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
nC  40.1 g C 
1 mol C
 3.34 mol C
12.01 g C
nH  6.6 g H 
1 mol H
 6.5 mol H
1.008 g H
nO  53.3 g O 
1 mol O
 3.33 mol O
16.00 g O
Thus, we arrive at the formula C3.34H6.5O3.33, which gives the identity and the mole ratios of atoms present.
However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing
all the subscripts by the smallest subscript (3.33).
C:
3.34
 1
3.33
6.5
 2
3.33
H:
O:
3.33
 1
3.33
This gives the empirical formula, CH2O.
Check: Are the subscripts in CH2O reduced to the smallest whole numbers?
(b)
Following the same procedure as part (a), we find:
nC  18.4 g C 
1 mol C
 1.53 mol C
12.01 g C
nN  21.5 g N 
1 mol N
 1.53 mol N
14.01 g N
nK  60.1 g K 
1 mol K
 1.54 mol K
39.10 g K
Dividing by the smallest number of moles (1.53 mol) gives the empirical formula, KCN.
3.51
The molar mass of CaSiO3 is 116.17 g/mol.
%Ca 
40.08 g
 34.50%
116.17 g
%Si 
28.09 g
 24.18%
116.17 g
%O 
(3)(16.00 g)
 41.32%
116.17 g
Check to see that the percentages sum to 100%. (34.50%  24.18%  41.32%)  100.00%
3.52
The empirical molar mass of CH is approximately 13.02 g. Let's compare this to the molar mass to determine
the molecular formula.
Recall that the molar mass divided by the empirical mass will be an integer greater than or equal to one.
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
41
molar mass
 1 (integer values)
empirical molar mass
In this case,
molar mass
78 g

 6
empirical molar mass
13.02 g
Thus, there are six CH units in each molecule of the compound, so the molecular formula is (CH)6, or C6H6.
3.53
Find the molar mass corresponding to each formula.
For C4H5N2O:
4(12.01 g)  5(1.008 g)  2(14.01 g)  (16.00 g)  97.10 g
For C8H10N4O2:
8(12.01 g)  10(1.008 g)  4(14.01 g)  2(16.00 g)  194.20 g
The molecular formula is C8H10N4O2.
3.54
METHOD 1:
Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the
percentages sum to 100%. In 100 g of MSG there will be 35.51 g C, 4.77 g H, 37.85 g O, 8.29 g N,
and 13.60 g Na.
Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula
tells us which elements are present and the simplest whole-number ratio of their atoms. This ratio is
also a mole ratio. Let nC, nH, nO, nN, and nNa be the number of moles of elements present. Use the
molar masses of these elements as conversion factors to convert to moles.
nC  35.51 g C 
1 mol C
 2.957 mol C
12.01 g C
nH  4.77 g H 
1 mol H
 4.73 mol H
1.008 g H
nO  37.85 g O 
nN  8.29 g N 
1 mol O
 2.366 mol O
16.00 g O
1 mol N
 0.592 mol N
14.01 g N
nNa  13.60 g Na 
1 mol Na
 0.5916 mol Na
22.99 g Na
Thus, we arrive at the formula C2.957H4.73O2.366N0.592Na0.5916, which gives the identity and the ratios of
atoms present. However, chemical formulas are written with whole numbers.
Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript.
2.957
= 4.998  5
0.5916
0.592
N:
= 1.00
0.5916
C:
4.73
= 8.00
0.5916
0.5916
Na :
= 1
0.5916
H:
This gives us the empirical formula for MSG, C5H8O4NNa.
O:
2.366
= 3.999  4
0.5916
42
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater
than or equal to one.
molar mass
 1 (integer values)
empirical molar mass
In this case,
molar mass
169 g

 1
empirical molar mass
169.11 g
Hence, the molecular formula and the empirical formula are the same, C5H8O4NNa. It should come as no
surprise that the empirical and molecular formulas are the same since MSG stands for monosodiumglutamate.
METHOD 2:
Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to
grams of each element. Then, use the molar mass to convert to moles of each element.
nC  (0.3551)  (169 g) 
1 mol C
 5.00 mol C
12.01 g C
nH  (0.0477)  (169 g) 
1 mol H
 8.00 mol H
1.008 g H
nO  (0.3785)  (169 g) 
1 mol O
 4.00 mol O
16.00 g O
nN  (0.0829)  (169 g) 
1 mol N
 1.00 mol N
14.01 g N
nNa  (0.1360)  (169 g) 
1 mol Na
 1.00 mol Na
22.99 g Na
Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this
method directly gives the molecular formula. The formula is C5H8O4NNa.
3.59
3.60
The balanced equations are as follows:
(a)
2C  O2  2CO
(h)
N2  3H2  2NH3
(b)
2CO  O2  2CO2
(i)
Zn  2AgCl  ZnCl2  2Ag
(c)
H2  Br2  2HBr
(j)
S8  8O2  8SO2
(d)
2K  2H2O  2KOH  H2
(k)
2NaOH  H2SO4  Na2SO4  2H2O
(e)
2Mg  O2  2MgO
(l)
Cl2  2NaI  2NaCl  I2
(f)
2O3  3O2
(m) 3KOH  H3PO4  K3PO4  3H2O
(g)
2H2O2  2H2O  O2
(n)
CH4  4Br2  CBr4  4HBr
The balanced equations are as follows:
(a)
2N2O5  2N2O4  O2
(b)
2KNO3  2KNO2  O2
(c)
NH4NO3  N2O  2H2O
(d)
NH4NO2  N2  2H2O
(e)
2NaHCO3  Na2CO3  H2O  CO2
(f)
P4O10  6H2O  4H3PO4
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.63
(g)
2HCl  CaCO3  CaCl2  H2O  CO2
(h)
2Al  3H2SO4  Al2(SO4)3  3H2
(i)
CO2  2KOH  K2CO3  H2O
(j)
CH4  2O2  CO2  2H2O
(k)
Be2C  4H2O  2Be(OH)2  CH4
(l)
3Cu  8HNO3  3Cu(NO3)2  2NO  4H2O
(m) S  6HNO3  H2SO4  6NO2  2H2O
(n)
2NH3  3CuO  3Cu  N2  3H2O
43
On the reactants side there are 8 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 4 D
atoms. Writing an equation,
8A  4B  4C  4D
Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the
equation by four gives,
2A  B  C  D
The correct answer is choice (c).
3.64
On the reactants side there are 6 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 2 D
atoms. Writing an equation,
6A  4B  4C  2D
Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the
equation by two gives,
3A  2B  2C  D
The correct answer is choice (d).
3.65
The mole ratio from the balanced equation is 2 moles CO2 : 2 moles CO.
3.60 mol CO 
3.66
2 mol CO2
 3.60 mol CO 2
2 mol CO
Si(s)  2Cl2(g) 
 SiCl4(l)
Strategy: Looking at the balanced equation, how do we compare the amounts of Cl2 and SiCl4? We can
compare them based on the mole ratio from the balanced equation.
Solution: Because the balanced equation is given in the problem, the mole ratio between Cl2 and SiCl4 is
known: 2 moles Cl2  1 mole SiCl4. From this relationship, we have two conversion factors.
2 mol Cl2
1 mol SiCl4
and
1 mol SiCl4
2 mol Cl2
Which conversion factor is needed to convert from moles of SiCl4 to moles of Cl2? The conversion factor on
the left is the correct one. Moles of SiCl4 will cancel, leaving units of "mol Cl2" for the answer. We
calculate moles of Cl2 reacted as follows:
? mol Cl 2 reacted  0.507 mol SiCl4 
2 mol Cl2
 1.01 mol Cl 2
1 mol SiCl4
44
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
Check: Does the answer seem reasonable? Should the moles of Cl2 reacted be double the moles of SiCl4
produced?
3.67
Starting with the amount of ammonia produced (6.0 moles), we can use the mole ratio from the balanced
equation to calculate the moles of H2 and N2 that reacted to produce 6.0 moles of NH3.
3H2(g)  N2(g)  2NH3(g)
3.68
? mol H 2  6.0 mol NH3 
3 mol H 2
 9.0 mol H 2
2 mol NH3
? mol N 2  6.0 mol NH3 
1 mol N 2
 3.0 mol N 2
2 mol NH3
Starting with the 5.0 moles of C4H10, we can use the mole ratio from the balanced equation to calculate the
moles of CO2 formed.
2C4H10(g)  13O2(g)  8CO2(g)  10H2O(l)
? mol CO2  5.0 mol C4 H10 
3.69
It is convenient to use the unit ton-mol in this problem. We normally use a g-mol. 1 g-mol SO2 has a mass
of 64.07 g. In a similar manner, 1 ton-mol of SO2 has a mass of 64.07 tons. We need to complete the
following conversions: tons SO2  ton-mol SO2  ton-mol S  ton S.
(2.6  107 tons SO 2 ) 
3.70
8 mol CO 2
 20 mol CO2  2.0  101 mol CO 2
2 mol C4 H10
1 ton-mol SO 2
1 ton-mol S
32.07 ton S


 1.3  107 tons S
64.07 ton SO 2 1 ton-mol SO 2 1 ton-mol S
(a)
 Na2CO3  H2O  CO2
2NaHCO3 
(b)
Molar mass NaHCO3  22.99 g  1.008 g  12.01 g  3(16.00 g)  84.01 g
Molar mass CO2  12.01 g  2(16.00 g)  44.01 g
The balanced equation shows one mole of CO2 formed from two moles of NaHCO3.
mass NaHCO 3 = 20.5 g CO 2 
2 mol NaHCO3 84.01 g NaHCO3
1 mol CO 2


44.01 g CO2
1 mol CO 2
1 mol NaHCO3
 78.3 g NaHCO3
3.71
The balanced equation shows a mole ratio of 1 mole HCN : 1 mole KCN.
0.140 g KCN 
1 mol KCN
1 mol HCN 27.03 g HCN


 0.0581 g HCN
65.12 g KCN 1 mol KCN
1 mol HCN
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.72
45
C6H12O6 
 2C2H5OH  2CO2
glucose
ethanol
Strategy: We compare glucose and ethanol based on the mole ratio in the balanced equation. Before we
can determine moles of ethanol produced, we need to convert to moles of glucose. What conversion factor is
needed to convert from grams of glucose to moles of glucose? Once moles of ethanol are obtained, another
conversion factor is needed to convert from moles of ethanol to grams of ethanol.
Solution: The molar mass of glucose will allow us to convert from grams of glucose to moles of glucose.
The molar mass of glucose  6(12.01 g)  12(1.008 g)  6(16.00 g)  180.16 g. The balanced equation is
given, so the mole ratio between glucose and ethanol is known; that is 1 mole glucose  2 moles ethanol.
Finally, the molar mass of ethanol will convert moles of ethanol to grams of ethanol. This sequence of three
conversions is summarized as follows:
grams of glucose  moles of glucose  moles of ethanol  grams of ethanol
? g C2 H 5OH  500.4 g C6 H12 O6 
1 mol C6 H12 O6
2 mol C2 H5 OH
46.07 g C2 H5 OH


180.16 g C6 H12 O6
1 mol C6 H12 O6
1 mol C2 H5OH
 255.9 g C2H5OH
Check: Does the answer seem reasonable? Should the mass of ethanol produced be approximately half the
mass of glucose reacted? Twice as many moles of ethanol are produced compared to the moles of glucose
reacted, but the molar mass of ethanol is about one-fourth that of glucose.
The liters of ethanol can be calculated from the density and the mass of ethanol.
volume 
mass
density
Volume of ethanol obtained =
3.73
255.9 g
= 324 mL = 0.324 L
0.789 g/mL
The mass of water lost is just the difference between the initial and final masses.
Mass H2O lost  15.01 g  9.60 g  5.41 g
moles of H 2 O  5.41 g H 2 O 
3.74
The balanced equation shows that eight moles of KCN are needed to combine with four moles of Au.
? mol KCN  29.0 g Au 
3.75
1 mol H 2 O
 0.300 mol H 2 O
18.02 g H 2 O
The balanced equation is:
1.0 kg CaCO3 
1 mol Au
8 mol KCN
= 0.294 mol KCN

197.0 g Au
4 mol Au
CaCO3(s) 
 CaO(s)  CO2(g)
1 mol CaCO3
1000 g
1 mol CaO
56.08 g CaO



 5.6  102 g CaO
1 kg
100.1 g CaCO3 1 mol CaCO3
1 mol CaO
46
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.76
(a)
NH4NO3(s) 
 N2O(g)  2H2O(g)
(b)
Starting with moles of NH4NO3, we can use the mole ratio from the balanced equation to find moles of
N2O. Once we have moles of N2O, we can use the molar mass of N2O to convert to grams of N2O.
Combining the two conversions into one calculation, we have:
mol NH4NO3  mol N2O  g N2O
? g N 2 O  0.46 mol NH 4 NO3 
3.77
1 mol N 2 O
44.02 g N 2 O

 2.0  101 g N 2O
1 mol NH 4 NO3
1 mol N 2 O
The quantity of ammonia needed is:
1.00  108 g (NH 4 )2 SO4 
2 mol NH3
17.03 g NH3
1 mol (NH 4 )2 SO 4
1 kg



132.2 g (NH 4 ) 2 SO 4
1 mol (NH 4 ) 2 SO 4
1 mol NH3
1000 g
4
 2.58  10 kg NH3
3.78
The balanced equation for the decomposition is :
2KClO3(s) 
 2KCl(s)  3O2(g)
? g O 2  46.0 g KClO3 
3.81
1 mol KClO3
3 mol O 2
32.00 g O 2


 18.0 g O 2
122.6 g KClO3 2 mol KClO3
1 mol O 2
2A  B  C
(a)
The number of B atoms shown in the diagram is 5. The balanced equation shows 2 moles A  1 mole B.
Therefore, we need 10 atoms of A to react completely with 5 atoms of B. There are only 8 atoms of A
present in the diagram. There are not enough atoms of A to react completely with B.
A is the limiting reagent.
(b)
There are 8 atoms of A. Since the mole ratio between A and B is 2:1, 4 atoms of B will react with 8
atoms of A, leaving 1 atom of B in excess. The mole ratio between A and C is also 2:1. When 8 atoms
of A react, 4 molecules of C will be produced.
B
C
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.82
47
N2  3H2  2NH3
(a)
The number of N2 molecules shown in the diagram is 3. The balanced equation shows
3 moles H2  1 mole N2. Therefore, we need 9 molecules of H2 to react completely with 3 molecules
of N2. There are 10 molecules of H2 present in the diagram. H2 is in excess.
N2 is the limiting reagent.
(b)
9 molecules of H2 will react with 3 molecules of N2, leaving 1 molecule of H2 in excess. The mole
ratio between N2 and NH3 is 1:2. When 3 molecules of N2 react, 6 molecules of NH3 will be produced.
H2
NH3
3.83
This is a limiting reagent problem. Let's calculate the moles of NO2 produced assuming complete reaction
for each reactant.
2NO(g)  O2(g)  2NO2(g)
0.886 mol NO 
0.503 mol O2 
2 mol NO2
 0.886 mol NO 2
2 mol NO
2 mol NO2
 1.01 mol NO 2
1 mol O2
NO is the limiting reagent; it limits the amount of product produced. The amount of product produced is
0.886 mole NO2.
3.84
Strategy: Note that this reaction gives the amounts of both reactants, so it is likely to be a limiting reagent
problem. The reactant that produces fewer moles of product is the limiting reagent because it limits the
amount of product that can be produced. How do we convert from the amount of reactant to amount of
product? Perform this calculation for each reactant, then compare the moles of product, NO2, formed by the
given amounts of O3 and NO to determine which reactant is the limiting reagent.
Solution: We carry out two separate calculations. First, starting with 0.740 g O3, we calculate the number
of moles of NO2 that could be produced if all the O3 reacted. We complete the following conversions.
grams of O3  moles of O3  moles of NO2
Combining these two conversions into one calculation, we write
? mol NO 2  0.740 g O3 
1 mol O3
1 mol NO 2

 0.0154 mol NO2
48.00 g O3
1 mol O3
48
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
Second, starting with 0.670 g of NO, we complete similar conversions.
grams of NO  moles of NO  moles of NO2
Combining these two conversions into one calculation, we write
? mol NO 2  0.670 g NO 
1 mol NO 2
1 mol NO

 0.0223 mol NO 2
30.01 g NO 1 mol NO
The initial amount of O3 limits the amount of product that can be formed; therefore, it is the limiting
reagent.
The problem asks for grams of NO2 produced. We already know the moles of NO2 produced, 0.0154 mole.
Use the molar mass of NO2 as a conversion factor to convert to grams (Molar mass NO2  46.01 g).
? g NO 2  0.0154 mol NO2 
46.01 g NO 2
 0.709 g NO 2
1 mol NO 2
Check: Does your answer seem reasonable? 0.0154 mole of product is formed. What is the mass of 1
mole of NO2?
Strategy: Working backwards, we can determine the amount of NO that reacted to produce 0.0154 mole of
NO2. The amount of NO left over is the difference between the initial amount and the amount reacted.
Solution: Starting with 0.0154 mole of NO2, we can determine the moles of NO that reacted using the
mole ratio from the balanced equation. We can calculate the initial moles of NO starting with 0.670 g and
using molar mass of NO as a conversion factor.
mol NO reacted  0.0154 mol NO2 
mol NO initial  0.670 g NO 
1 mol NO
 0.0154 mol NO
1 mol NO2
1 mol NO
 0.0223 mol NO
30.01 g NO
mol NO remaining  mol NO initial  mol NO reacted.
mol NO remaining  0.0223 mol NO  0.0154 mol NO  0.0069 mol NO
3.85
3.86
C3H8(g)  5O2(g) 
 3CO2(g)  4H2O(l)
(a)
The balanced equation is:
(b)
The balanced equation shows a mole ratio of 3 moles CO2 : 1 mole C3H8. The mass of CO2 produced
is:
3 mol CO 2
44.01 g CO 2
3.65 mol C3H8 

 482 g CO 2
1 mol C3 H8
1 mol CO 2
This is a limiting reagent problem. Let's calculate the moles of Cl2 produced assuming complete reaction for
each reactant.
0.86 mol MnO2 
48.2 g HCl 
1 mol Cl2
= 0.86 mol Cl2
1 mol MnO2
1 mol Cl2
1 mol HCl
= 0.330 mol Cl2

36.46 g HCl 4 mol HCl
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
49
HCl is the limiting reagent; it limits the amount of product produced. It will be used up first. The amount of
product produced is 0.330 mole Cl2. Let's convert this to grams.
? g Cl 2  0.330 mol Cl2 
3.89
The balanced equation is given:
70.90 g Cl2
= 23.4 g Cl 2
1 mol Cl2
CaF2  H2SO4 
 CaSO4  2HF
The balanced equation shows a mole ratio of 2 moles HF : 1 mole CaF2. The theoretical yield of HF is:
(6.00  103 g CaF2 ) 
1 mol CaF2
2 mol HF
20.01 g HF
1 kg



 3.08 kg HF
78.08 g CaF2 1 mol CaF2
1 mol HF
1000 g
The actual yield is given in the problem (2.86 kg HF).
3.90
(a)
% yield 
actual yield
 100%
theoretical yield
% yield 
2.86 kg
 100%  92.9%
3.08 kg
Start with a balanced chemical equation. It’s given in the problem. We use NG as an abbreviation for
nitroglycerin. The molar mass of NG  227.1 g/mol.
4C3H5N3O9 
 6N2  12CO2  10H2O  O2
Map out the following strategy to solve this problem.
g NG  mol NG  mol O2  g O2
Calculate the grams of O2 using the strategy above.
? g O 2  2.00  102 g NG 
(b)
3.91
1 mol O2
32.00 g O 2
1 mol NG


 7.05 g O 2
227.1 g NG 4 mol NG
1 mol O 2
The theoretical yield was calculated in part (a), and the actual yield is given in the problem (6.55 g).
The percent yield is:
% yield 
actual yield
 100%
theoretical yield
% yield 
6.55 g O 2
 100% = 92.9%
7.05 g O 2
The balanced equation shows a mole ratio of 1 mole TiO2 : 1 mole FeTiO3. The molar mass of FeTiO3 is
151.7 g/mol, and the molar mass of TiO2 is 79.88 g/mol. The theoretical yield of TiO2 is:
8.00  106 g FeTiO3 
3
1 mol FeTiO3
1 mol TiO 2
79.88 g TiO2
1 kg



151.7 g FeTiO3 1 mol FeTiO3
1 mol TiO2
1000 g
 4.21  10 kg TiO2
50
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3
The actual yield is given in the problem (3.67  10 kg TiO2).
% yield 
3.92
actual yield
3.67  103 kg
 100% 
 100%  87.2%
theoretical yield
4.21  103 kg
The actual yield of ethylene is 481 g. Let’s calculate the yield of ethylene if the reaction is 100 percent
efficient. We can calculate this from the definition of percent yield. We can then calculate the mass of
hexane that must be reacted.
% yield 
actual yield
 100%
theoretical yield
42.5% yield 
481 g C2 H 4
 100%
theoretical yield
3
theoretical yield C2H4  1.13  10 g C2H4
The mass of hexane that must be reacted is:
(1.13  103 g C2 H 4 ) 
3.93
1 mol C6 H14 86.15 g C6 H14
1 mol C2 H 4


 3.47  103 g C6 H14
28.05 g C2 H 4 1 mol C2 H 4
1 mol C6 H14
This is a limiting reagent problem. Let's calculate the moles of Li3N produced assuming complete reaction
for each reactant.
6Li(s)  N2(g)  2Li3N(s)
12.3 g Li 
2 mol Li3 N
1 mol Li

 0.591 mol Li3 N
6.941 g Li
6 mol Li
33.6 g N 2 
2 mol Li3 N
1 mol N 2

 2.40 mol Li3 N
28.02 g N 2
1 mol N 2
Li is the limiting reagent; it limits the amount of product produced. The amount of product produced is
0.591 mole Li3N. Let's convert this to grams.
? g Li3 N  0.591 mol Li3 N 
34.83 g Li3 N
 20.6 g Li 3 N
1 mol Li3 N
This is the theoretical yield of Li3N. The actual yield is given in the problem (5.89 g). The percent yield is:
% yield 
3.94
actual yield
5.89 g
 100% 
 100%  28.6%
theoretical yield
20.6 g
This is a limiting reagent problem. Let's calculate the moles of S2Cl2 produced assuming complete reaction
for each reactant.
S8(l)  4Cl2(g)  4S2Cl2(l)
4.06 g S8 
1 mol S8
4 mol S2 Cl2

 0.0633 mol S2 Cl2
256.6 g S8
1 mol S8
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
6.24 g Cl2 
51
1 mol Cl2
4 mol S2 Cl2

 0.0880 mol S2 Cl2
70.90 g Cl2
4 mol Cl2
S8 is the limiting reagent; it limits the amount of product produced. The amount of product produced is
0.0633 mole S2Cl2. Let's convert this to grams.
? g S2 Cl2  0.0633 mol S2 Cl2 
135.04 g S2 Cl2
 8.55 g S 2Cl 2
1 mol S2 Cl2
This is the theoretical yield of S2Cl2. The actual yield is given in the problem (6.55 g). The percent yield is:
% yield 
3.95
actual yield
6.55 g
 100% 
 100%  76.6%
theoretical yield
8.55 g
All the carbon from the hydrocarbon reactant ends up in CO2, and all the hydrogen from the hydrocarbon
reactant ends up in water. In the diagram, we find 4 CO2 molecules and 6 H2O molecules. This gives a ratio
between carbon and hydrogen of 4:12. We write the formula C4H12, which reduces to the empirical formula
CH3. The empirical molar mass equals approximately 15 g, which is half the molar mass of the hydrocarbon.
Thus, the molecular formula is double the empirical formula or C2H6. Since this is a combustion reaction,
the other reactant is O2. We write:
C2H6  O2  CO2  H2O
Balancing the equation,
2C2H6  7O2  4CO2  6H2O
3.96
2H2(g)  O2(g)  2H2O(g)
We start with 8 molecules of H2 and 3 molecules of O2. The balanced equation shows 2 moles H2  1 mole
O2. If 3 molecules of O2 react, 6 molecules of H2 will react, leaving 2 molecules of H2 in excess. The
balanced equation also shows 1 mole O2  2 moles H2O. If 3 molecules of O2 react, 6 molecules of H2O
will be produced.
After complete reaction, there will be 2 molecules of H2 and 6 molecules of H2O. The correct diagram is
choice (b).
3.97
First, let's convert to moles of HNO3 produced.
1.00 ton HNO3 
1 mol HNO3
2000 lb 453.6 g


 1.44  104 mol HNO3
1 ton
1 1b
63.02 g HNO3
Now, we will work in the reverse direction to calculate the amount of reactant needed to produce 1.44  10
mol of HNO3. Realize that since the problem says to assume an 80% yield for each step, the amount of
100%
, compared to a standard stoichiometry
reactant needed in each step will be larger by a factor of
80%
calculation where a 100% yield is assumed.
3
52
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
Referring to the balanced equation in the last step, we calculate the moles of NO2.
(1.44  104 mol HNO3 ) 
2 mol NO2
100%

 3.60  104 mol NO 2
1 mol HNO3
80%
4
Now, let's calculate the amount of NO needed to produce 3.60  10 mol NO2. Following the same
procedure as above, and referring to the balanced equation in the middle step, we calculate the moles of NO.
(3.60  104 mol NO 2 ) 
1 mol NO 100%

 4.50  104 mol NO
1 mol NO 2
80%
4
Now, let's calculate the amount of NH3 needed to produce 4.5  10 mol NO. Referring to the balanced
equation in the first step, the moles of NH3 is:
(4.50  104 mol NO) 
4 mol NH3 100%

 5.63  104 mol NH3
4 mol NO
80%
Finally, converting to grams of NH3:
5.63  104 mol NH3 
3.98
17.03 g NH3
 9.59  105 g NH 3
1 mol NH3
We assume that all the Cl in the compound ends up as HCl and all the O ends up as H2O. Therefore, we
need to find the number of moles of Cl in HCl and the number of moles of O in H2O.
mol Cl  0.233 g HCl 
1 mol HCl
1 mol Cl

 0.00639 mol Cl
36.46 g HCl 1 mol HCl
mol O  0.403 g H 2 O 
1 mol H 2 O
1 mol O

 0.0224 mol O
18.02 g H 2 O 1 mol H 2 O
Dividing by the smallest number of moles (0.00639 mole) gives the formula, ClO3.5. Multiplying both
subscripts by two gives the empirical formula, Cl2O7.
3.99
The number of moles of Y in 84.10 g of Y is:
27.22 g X 
1 mol X
1 mol Y

 0.8145 mol Y
33.42 g X 1 mol X
The molar mass of Y is:
molar mass Y 
84.10 g Y
 103.3 g/mol
0.8145 mol Y
The atomic mass of Y is 103.3 amu.
3.100
The symbol “O” refers to moles of oxygen atoms, not oxygen molecule (O2). Look at the molecular
formulas given in parts (a) and (b). What do they tell you about the relative amounts of carbon and oxygen?
(a)
0.212 mol C 
1 mol O
= 0.212 mol O
1 mol C
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
(b)
0.212 mol C 
53
2 mol O
= 0.424 mol O
1 mol C
3.101
The observations mean either that the amount of the more abundant isotope was increasing or the amount of
the less abundant isotope was decreasing. One possible explanation is that the less abundant isotope was
undergoing radioactive decay, and thus its mass would decrease with time.
3.102
This is a calculation involving percent composition. Remember,
percent by mass of each element 
mass of element in 1 mol of compound
 100%
molar mass of compound
The molar masses are: Al, 26.98 g/mol; Al2(SO4)3, 342.2 g/mol; H2O, 18.02 g/mol. Thus, using x as the
number of H2O molecules,


2(molar mass of Al)
mass % Al  
  100%
 molar mass of Al2 (SO4 )3  x (molar mass of H 2 O) 


2(26.98 g)
8.20%  
  100%
 342.2 g  x(18.02 g) 
x  17.53
Rounding off to a whole number of water molecules, x  18. Therefore, the formula is Al2(SO4)318 H2O.
3.103
Molar mass of C4H8Cl2S  4(12.01 g)  8(1.008 g)  2(35.45 g)  32.07 g  159.1 g
%C 
4(12.01 g/mol)
 100%  30.19%
159.1 g/mol
%H 
8(1.008 g/mol)
 100%  5.069%
159.1 g/mol
%Cl 
2(35.45 g/mol)
 100%  44.56%
159.1 g/mol
%S 
3.104
The number of carbon atoms in a 24-carat diamond is:
24 carat 
3.105
32.07 g/mol
 100%  20.16%
159.1 g/mol
200 mg C 0.001 g C
1 mol C
6.022  1023 atoms C
= 2.4  1023 atoms C



1 carat
1 mg C
12.01 g C
1 mol C
The amount of Fe that reacted is:
1
 664 g  83.0 g reacted
8
The amount of Fe remaining is:
664 g  83.0 g  581 g remaining
Thus, 83.0 g of Fe reacts to form the compound Fe2O3, which has two moles of Fe atoms per 1 mole of
compound. The mass of Fe2O3 produced is:
54
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
83.0 g Fe 
1 mol Fe2 O3 159.7 g Fe2 O3
1 mol Fe


 119 g Fe2O 3
55.85 g Fe
2 mol Fe
1 mol Fe2 O3
The final mass of the iron bar and rust is:
3.106
581 g Fe  119 g Fe2O3  700 g
The mass of oxygen in MO is 39.46 g  31.70 g  7.76 g O. Therefore, for every 31.70 g of M, there is
7.76 g of O in the compound MO. The molecular formula shows a mole ratio of 1 mole M : 1 mole O. First,
calculate moles of M that react with 7.76 g O.
mol M  7.76 g O 
molar mass M 
1 mol O
1 mol M

 0.485 mol M
16.00 g O 1 mol O
31.70 g M
 65.4 g/mol
0.485 mol M
Thus, the atomic mass of M is 65.4 amu. The metal is most likely Zn.
3.107
(a)
Zn(s)  H2SO4(aq) 
 ZnSO4(aq)  H2(g)
(b)
We assume that a pure sample would produce the theoretical yield of H2. The balanced equation shows
a mole ratio of 1 mole H2 : 1 mole Zn. The theoretical yield of H2 is:
3.86 g Zn 
1 mol H 2 2.016 g H 2
1 mol Zn


 0.119 g H 2
65.39 g Zn 1 mol Zn
1 mol H 2
percent purity 
(c)
3.108
0.0764 g H 2
 100%  64.2%
0.119 g H 2
We assume that the impurities are inert and do not react with the sulfuric acid to produce hydrogen.
The wording of the problem suggests that the actual yield is less than the theoretical yield. The percent yield
will be equal to the percent purity of the iron(III) oxide. We find the theoretical yield :
(2.62  103 kg Fe 2 O3 ) 
1000 g Fe2 O3
1 mol Fe2 O3
2 mol Fe
55.85 g Fe
1 kg Fe




1 kg Fe2 O3
159.7 g Fe2 O3 1 mol Fe 2 O3
1 mol Fe
1000 g Fe
3
 1.83  10 kg Fe
percent yield 
percent yield =
3.109
actual yield
 100%
theoretical yield
1.64  103 kg Fe
1.83  103 kg Fe
 100% = 89.6%  purity of Fe 2O 3
The balanced equation is: C6H12O6  6O2 
 6CO2  6H2O
6 mol CO 2
44.01 g CO 2 365 days
5.0  102 g glucose
1 mol glucose




 (6.5  109 people)
1 day
180.2 g glucose 1 mol glucose
1 mol CO2
1 yr
 1.7  10
15
g CO2/yr
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.110
The carbohydrate contains 40 percent carbon; therefore, the remaining 60 percent is hydrogen and oxygen.
The problem states that the hydrogen to oxygen ratio is 2:1. We can write this 2:1 ratio as H2O.
Assume 100 g of compound.
40.0 g C 
1 mol C
 3.33 mol C
12.01 g C
60.0 g H 2 O 
1 mol H 2 O
 3.33 mol H 2 O
18.02 g H 2 O
Dividing by 3.33 gives CH2O for the empirical formula.
To find the molecular formula, divide the molar mass by the empirical mass.
molar mass
178 g
=
 6
empirical mass
30.03 g
Thus, there are six CH2O units in each molecule of the compound, so the molecular formula is (CH2O)6, or
C6H12O6.
3.111
The molar mass of chlorophyll is 893.5 g/mol. Finding the mass of a 0.0011-mol sample:
0.0011 mol chlorophyll 
893.5 g chlorophyll
 0.98 g chlorophyll
1 mol chlorophyll
The chlorophyll sample has the greater mass.
3.112
If we assume 100 g of compound, the masses of Cl and X are 67.2 g and 32.8 g, respectively. We can
calculate the moles of Cl.
1 mol Cl
67.2 g Cl 
 1.90 mol Cl
35.45 g Cl
Then, using the mole ratio from the chemical formula (XCl3), we can calculate the moles of X contained in
32.8 g.
1 mol X
1.90 mol Cl 
 0.633 mol X
3 mol Cl
0.633 mole of X has a mass of 32.8 g. Calculating the molar mass of X:
32.8 g X
 51.8 g/mol
0.633 mol X
The element is most likely chromium (molar mass  52.00 g/mol).
3.113
(a)
The molar mass of hemoglobin is:
2952(12.01 g)  4664(1.008 g)  812(14.01 g)  832(16.00 g)  8(32.07 g)  4(55.85 g)
4
 6.532  10 g
(b)
To solve this problem, the following conversions need to be completed:
L  mL  red blood cells  hemoglobin molecules  mol hemoglobin  mass hemoglobin
55
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
56
We will use the following abbreviations: RBC  red blood cells, HG  hemoglobin
5.00 L 
1 mL
5.0  109 RBC 2.8  108 HG molecules
1 mol HG
6.532  104 g HG




0.001 L
1 mL
1 RBC
1 mol HG
6.022  1023 molecules HG
2
 7.6  10 g HG
3.114
A 100 g sample of myoglobin contains 0.34 g of iron (0.34% Fe). The number of moles of Fe is:
0.34 g Fe 
1 mol Fe
 6.1  103 mol Fe
55.85 g Fe
Since there is one Fe atom in a molecule of myoglobin, the moles of myoglobin also equal 6.1  10
The molar mass of myoglobin can be calculated.
molar mass myoglobin 
3.115
(a)
8.38 g KBr 
100 g myoglobin
6.1  10
3
mol myoglobin
1 mol KBr
6.022  1023 KBr
1 K  ion


 4.24  1022 K + ions
119.0 g KBr
1 mol KBr
1 KBr



5.40 g Na 2SO 4 

7.45 g Ca 3 (PO 4 )2 
 4.34  10

Br ions
2
2
ions  2.29  10
22
SO4
1 mol Ca 3 (PO 4 )2
6.022  1023 Ca 3 (PO 4 )2
3 Ca 2 ions


310.2 g Ca 3 (PO 4 )2
1 mol Ca 3 (PO 4 )2
1 Ca 3 (PO 4 )2
22
Ca
Since there are three Ca
2
4.34  1022 Ca 2 ions 
3.116
22
1 mol Na 2SO 4
6.022  1023 Na 2SO4 2 Na  ions


 4.58  1022 Na + ions
142.05 g Na 2SO 4
1 mol Na 2SO 4
1 Na 2SO4
Since there are two Na for every one SO4 , the number of SO4
(c)
mole.
 1.6  104 g/mol
Since there is one Br for every one K , the number of Br ions  4.24  10
(b)
3
2
ions
3
for every two PO4 , the number of PO4
2 PO34 ions
3
ions is:
 2.89  1022 PO43 ions
3 Ca 2 ions
Assume 100 g of sample. Then,
mol Na  32.08 g Na 
mol O  36.01 g O 
1 mol Na
 1.395 mol Na
22.99 g Na
1 mol O
 2.251 mol O
16.00 g O
mol Cl  19.51 g Cl 
1 mol Cl
 0.5504 mol Cl
35.45 g Cl
Since Cl is only contained in NaCl, the moles of Cl equals the moles of Na contained in NaCl.
mol Na (in NaCl)  0.5504 mol
2
ions
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
57
The number of moles of Na in the remaining two compounds is: 1.395 mol  0.5504 mol  0.8446 mol Na.
To solve for moles of the remaining two compounds, let
x  moles of Na2SO4
y  moles of NaNO3
Then, from the mole ratio of Na and O in each compound, we can write
2x  y  mol Na  0.8446 mol
4x  3y  mol O  2.251 mol
Solving two equations with two unknowns gives
x  0.1414  mol Na2SO4
and
y  0.5618  mol NaNO3
Finally, we convert to mass of each compound to calculate the mass percent of each compound in the sample.
Remember, the sample size is 100 g.
mass % NaCl  0.5504 mol NaCl 
58.44 g NaCl
1

 100%  32.17% NaCl
1 mol NaCl
100 g sample
mass % Na 2 SO4  0.1414 mol Na 2SO4 
mass % NaNO 3  0.5618 mol NaNO3 
3.117
142.1 g Na 2SO 4
1

 100%  20.09% Na 2SO4
1 mol Na 2SO 4
100 g sample
85.00 g NaNO3
1

 100%  47.75% NaNO 3
1 mol NaNO3
100 g sample
(a)
16 amu, CH4
17 amu, NH3
18 amu, H2O
64 amu, SO2
(b)
The formula C3H8 can also be written as CH3CH2CH3. A CH3 fragment could break off from this
molecule giving a peak at 15 amu. No fragment of CO2 can have a mass of 15 amu. Therefore, the
substance responsible for the mass spectrum is most likely C3H8.
(c)
First, let’s calculate the masses of CO2 and C3H8.
molecular mass CO2  12.00000 amu  2(15.99491 amu)  43.98982 amu
molecular mass C3H8  3(12.00000 amu)  8(1.00797 amu)  44.06376 amu
These masses differ by only 0.07394 amu. The measurements must be precise to 0.030 amu.
43.98982  0.030 amu  44.02 amu
44.06376  0.030 amu  44.03 amu
3.118
The mass percent of an element in a compound can be calculated as follows:
percent by mass of each element 
mass of element in 1 mol of compound
 100%
molar mass of compound
The molar mass of Ca3(PO4)2  310.18 g/mol
% Ca 
(3)(40.08 g)
 100%  38.76% Ca
310.18 g
58
3.119
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
(a)
%P 
(2)(30.97 g)
 100%  19.97% P
310.18 g
%O 
(8)(16.00 g)
 100%  41.27% O
310.18 g
First, calculate the mass of C in CO2, the mass of H in H2O, and the mass of N in NH3. For now, we
will carry more than 3 significant figures and then round to the correct number at the end of the problem.
? g C  3.94 g CO 2 
1 mol CO 2
1 mol C
12.01 g C


 1.075 g C
44.01 g CO 2 1 mol CO2
1 mol C
? g H  1.89 g H 2 O 
1 mol H 2 O
2 mol H
1.008 g H


 0.2114 g H
18.02 g H 2 O 1 mol H 2 O
1 mol H
? g N  0.436 g NH3 
1 mol NH3
1 mol N
14.01 g N


 0.3587 g N
17.03 g NH3 1 mol NH3
1 mol N
Next, we can calculate the %C, %H, and the %N in each sample, then we can calculate the %O by
difference.
1.075 g C
%C 
 100%  49.43% C
2.175 g sample
%H 
0.2114 g H
 100%  9.720% H
2.175 g sample
%N 
0.3587 g N
 100%  19.15% N
1.873 g sample
The % O  100%  (49.43%  9.720%  19.15%)  21.70% O
Assuming 100 g of compound, calculate the moles of each element.
? mol C  49.43 g C 
1 mol C
 4.116 mol C
12.01 g C
? mol H  9.720 g H 
1 mol H
 9.643 mol H
1.008 g H
? mol N  19.15 g N 
1 mol N
 1.367 mol N
14.01 g N
? mol O  21.70 g O 
1 mol O
 1.356 mol O
16.00 g O
Thus, we arrive at the formula C4.116H9.643N1.367O1.356. Dividing by 1.356 gives the empirical
formula, C3H7NO.
(b)
The empirical molar mass is 73.10 g. Since the approximate molar mass of lysine is 150 g, we have:
150 g
 2
73.10 g
Therefore, the molecular formula is (C3H7NO)2 or C6H14N2O2.
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.120
59
Yes. The number of hydrogen atoms in one gram of hydrogen molecules is the same as the number in one
gram of hydrogen atoms. There is no difference in mass, only in the way that the particles are arranged.
Would the mass of 100 dimes be the same if they were stuck together in pairs instead of separated?
3.121
The mass of one fluorine atom is 19.00 amu. The mass of one mole of fluorine atoms is 19.00 g. Multiplying
the mass of one atom by Avogadro’s number gives the mass of one mole of atoms. We can write:
19.00 amu
 (6.022  1023 F atoms)  19.00 g F
1 F atom
or,
6.022  10
23
amu  1 g
This is why Avogadro’s numbers has sometimes been described as a conversion factor between amu and
grams.
3.122
Since we assume that water exists as either H2O or D2O, the natural abundances are 99.985 percent and
0.015 percent, respectively. If we convert to molecules of water (both H2O or D2O), we can calculate the
molecules that are D2O from the natural abundance (0.015%).
The necessary conversions are:
mL water  g water  mol water  molecules water  molecules D2O
400 mL water 
1 g water
1 mol water
6.022  1023 molecules 0.015% molecules D 2 O



1 mL water 18.02 g water
1 mol water
100% molecules water
 2.01  10
3.123
21
molecules D2O
There can only be one chlorine per molecule, since two chlorines have a combined mass in excess of 70 amu.
35
Since the Cl isotope is more abundant, let’s subtract 35 amu from the mass corresponding to the more
intense peak.
50 amu  35 amu  15 amu
15 amu equals the mass of one
12
1
C and three H. To explain the two peaks, we have:
12 1
35
molecular mass C H3 Cl  12 amu  3(1 amu)  35 amu  50 amu
12 1 37
molecular mass C H3 Cl  12 amu  3(1 amu)  37 amu  52 amu
35
Cl is three times more abundant than
the 52 amu peak.
3.124
37
Cl; therefore, the 50 amu peak will be three times more intense than
First, we can calculate the moles of oxygen.
2.445 g C 
1 mol C
1 mol O

 0.2036 mol O
12.01 g C 1 mol C
Next, we can calculate the molar mass of oxygen.
molar mass O 
3.257 g O
 16.00 g/mol
0.2036 mol O
If 1 mole of oxygen atoms has a mass of 16.00 g, then 1 atom of oxygen has an atomic mass of 16.00 amu.
60
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.125
The molecular formula for Cl2O7 means that there are 2 Cl atoms for every 7 O atoms or 2 moles of Cl atoms
for every 7 moles of O atoms. We can write:
mole ratio 
3.126
1 mol Cl 2
2 mol Cl

7 mol O
3.5 mol O 2
(a)
The mass of chlorine is 5.0 g.
(b)
From the percent by mass of Cl, we can calculate the mass of chlorine in 60.0 g of NaClO3.
mass % Cl 
35.45 g Cl
 100%  33.31% Cl
106.44 g compound
mass Cl  60.0 g  0.3331  20.0 g Cl
(c)
0.10 mol of KCl contains 0.10 mol of Cl.
0.10 mol Cl 
(d)
35.45 g Cl
 3.5 g Cl
1 mol Cl
From the percent by mass of Cl, we can calculate the mass of chlorine in 30.0 g of MgCl2.
mass % Cl 
(2)(35.45 g Cl)
 100%  74.47% Cl l
95.21 g compound
mass Cl  30.0 g  0.7447  22.3 g Cl
(e)
The mass of Cl can be calculated from the molar mass of Cl2.
0.50 mol Cl2 
70.90 g Cl
 35.45 g Cl
1 mol Cl2
Thus, (e) 0.50 mol Cl2 contains the greatest mass of chlorine.
3.127
(a)
We need to compare the mass % of K in both KCl and K2SO4.
%K in KCl 
39.10 g
 100%  52.45% K
74.55 g
%K in K 2SO4 
2(39.10 g)
 100%  44.87% K
174.27 g
The price is dependent on the %K.
Price of K 2SO 4
%K in K 2SO 4

Price of KCl
%K in KCl
Price of K 2SO4 = Price of KCl ×
Price of K 2SO4 
%K in K 2SO4
%K in KCl
$0.55 44.87%

 $0.47 /kg
kg
52.45%
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
(b)
61
First, calculate the number of moles of K in 1.00 kg of KCl.
(1.00  103 g KCl) 
1 mol KCl
1 mol K

 13.4 mol K
74.55 g KCl 1 mol KCl
Next, calculate the amount of K2O needed to supply 13.4 mol K.
13.4 mol K 
3.128
1 mol K 2 O 94.20 g K 2 O
1 kg


 0.631 kg K 2O
2 mol K
1 mol K 2 O
1000 g
Both compounds contain only Pt and Cl. The percent by mass of Pt can be calculated by subtracting the
percent Cl from 100 percent.
Compound A: Assume 100 g of compound.
26.7 g Cl 
1 mol Cl
 0.753 mol Cl
35.45 g Cl
73.3 g Pt 
1 mol Pt
 0.376 mol Pt
195.1 g Pt
Dividing by the smallest number of moles (0.376 mole) gives the empirical formula, PtCl2.
Compound B: Assume 100 g of compound.
42.1 g Cl 
1 mol Cl
 1.19 mol Cl
35.45 g Cl
57.9 g Pt 
1 mol Pt
 0.297 mol Pt
195.1 g Pt
Dividing by the smallest number of moles (0.297 mole) gives the empirical formula, PtCl4.
3.129
The mass of the metal (X) in the metal oxide is 1.68 g. The mass of oxygen in the metal oxide is
2.40 g  1.68 g  0.72 g oxygen. Next, find the number of moles of the metal and of the oxygen.
moles X  1.68 g 
1 mol X
 0.0301 mol X
55.9 g X
moles O  0.72 g 
1 mol O
 0.045 mol O
16.00 g O
This gives the formula X0.0301O0.045. Dividing by the smallest number of moles (0.0301 moles) gives the
formula X1.00O1.5. Multiplying by two gives the empirical formula, X2O3.
The balanced equation is:
3.130
X2O3(s)  3CO(g) 
 2X(s)  3CO2(g)
Both compounds contain only Mn and O. When the first compound is heated, oxygen gas is evolved. Let’s
calculate the empirical formulas for the two compounds, then we can write a balanced equation.
(a) Compound X: Assume 100 g of compound.
63.3 g Mn 
1 mol Mn
 1.15 mol Mn
54.94 g Mn
62
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
36.7 g O 
1 mol O
 2.29 mol O
16.00 g O
Dividing by the smallest number of moles (1.15 moles) gives the empirical formula, MnO2.
Compound Y: Assume 100 g of compound.
72.0 g Mn 
28.0 g O 
1 mol Mn
 1.31 mol Mn
54.94 g Mn
1 mol O
 1.75 mol O
16.00 g O
Dividing by the smallest number of moles gives MnO1.33. Recall that an empirical formula must have whole
number coefficients. Multiplying by a factor of 3 gives the empirical formula Mn3O4.
(b) The unbalanced equation is:
Balancing by inspection gives:
3.131
MnO2 
 Mn3O4  O2
3MnO2 
 Mn3O4  O2
The mass of the water is the difference between 1.936 g of the hydrate and the mass of water-free
(anhydrous) BaCl2. First, we need to start with a balanced equation for the reaction. Upon treatment with
sulfuric acid, BaCl2 dissolves, losing its waters of hydration.
 BaSO4(s)  2HCl(aq)
BaCl2(aq)  H2SO4(aq) 
Next, calculate the mass of anhydrous BaCl2 based on the amount of BaSO4 produced.
1.864 g BaSO 4 
1 mol BaSO 4
1 mol BaCl2
208.2 g BaCl2


 1.663 g BaCl2
233.4 g BaSO4 1 mol BaSO 4
1 mol BaCl2
The mass of water is (1.936 g  1.663 g)  0.273 g H2O. Next, we convert the mass of H2O and the mass
of BaCl2 to moles to determine the formula of the hydrate.
0.273 g H 2 O 
1 mol H 2 O
 0.0151 mol H 2 O
18.02 g H 2 O
1.663 g BaCl2 
1 mol BaCl2
 0.00799 mol BaCl2
208.2 g BaCl2
The ratio of the number of moles of H2O to the number of moles of BaCl2 is 0.0151/0.00799  1.89. We
round this number to 2, which is the value of x. The formula of the hydrate is BaCl2  2H2O.
3.132
SO2 is converted to H2SO4 by reaction with water. The mole ratio between SO2 and H2SO4 is 1:1.
This is a unit conversion problem. You should come up with the following strategy to solve the problem.
tons SO2  ton-mol SO2  ton-mol H2SO4  tons H2SO4
? tons H 2SO4  (4.0  105 tons SO 2 ) 
5
 6.1  10 tons H2SO4
1 ton-mol SO2
1 ton-mol H 2SO 4 98.09 tons H 2SO4


64.07 tons SO 2
1 ton-mol SO 2
1 ton-mol H 2SO4
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
63
Tip: You probably won’t come across a ton-mol that often in chemistry. However, it was
convenient to use in this problem. We normally use a g-mol. 1 g-mol SO2 has a mass of
64.07 g. In a similar manner, 1 ton-mol of SO2 has a mass of 64.07 tons.
3.133
The mass of water lost upon heating the mixture is (5.020 g  2.988 g)  2.032 g water. Next, if we let
x  mass of CuSO4  5H2O, then the mass of MgSO4  7H2O is (5.020  x)g. We can calculate the amount of
water lost by each salt based on the mass % of water in each hydrate. We can write:
(mass CuSO4  5H2O)(% H2O)  (mass MgSO4  7H2O)(% H2O)  total mass H2O  2.032 g H2O
Calculate the % H2O in each hydrate.
% H 2 O (CuSO 4  5H 2 O) 
(5)(18.02 g)
 100%  36.08% H 2 O
249.7 g
% H 2 O (MgSO 4  7H 2 O) 
(7)(18.02 g)
 100%  51.17% H 2 O
246.5 g
Substituting into the equation above gives:
(x)(0.3608)  (5.020  x)(0.5117)  2.032 g
0.1509x  0.5367
x  3.557 g  mass of CuSO4  5H2O
Finally, the percent by mass of CuSO4  5H2O in the mixture is:
3.557 g
 100%  70.86%
5.020 g
3.134
We assume that the increase in mass results from the element nitrogen. The mass of nitrogen is:
0.378 g  0.273 g  0.105 g N
The empirical formula can now be calculated. Convert to moles of each element.
0.273 g Mg 
0.105 g N 
1 mol Mg
 0.0112 mol Mg
24.31 g Mg
1 mol N
 0.00749 mol N
14.01 g N
Dividing by the smallest number of moles gives Mg1.5N. Recall that an empirical formula must have whole
number coefficients. Multiplying by a factor of 2 gives the empirical formula Mg3N2. The name of this
compound is magnesium nitride.
3.135
The balanced equations are:
CH4  2O2 
 CO2  2H2O
2C2H6  7O2 
 4CO2  6H2O
If we let x  mass of CH4, then the mass of C2H6 is (13.43  x) g.
64
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
Next, we need to calculate the mass of CO2 and the mass of H2O produced by both CH4 and C2H6. The sum
of the masses of CO2 and H2O will add up to 64.84 g.
? g CO 2 (from CH 4 )  x g CH 4 
1 mol CH 4
1 mol CO2 44.01 g CO 2


 2.744 x g CO2
16.04 g CH 4 1 mol CH 4
1 mol CO 2
? g H 2 O (from CH 4 )  x g CH 4 
1 mol CH 4
2 mol H 2 O 18.02 g H 2 O


 2.247 x g H 2 O
16.04 g CH 4 1 mol CH 4
1 mol H 2 O
? g CO 2 (from C2 H 6 )  (13.43  x ) g C2 H 6 
1 mol C2 H 6
4 mol CO2
44.01 g CO 2


30.07 g C2 H 6 2 mol C2 H 6
1 mol CO2
 2.927(13.43  x) g CO2
? g H 2 O (from C2 H6 )  (13.43  x) g C2 H 6 
1 mol C2 H 6
6 mol H 2 O 18.02 g H 2 O


30.07 g C2 H 6 2 mol C2 H 6
1 mol H 2 O
 1.798(13.43  x) g H2O
Summing the masses of CO2 and H2O:
2.744x g  2.247x g  2.927(13.43  x) g  1.798(13.43  x) g  64.84 g
0.266x  1.383
x  5.20 g
The fraction of CH4 in the mixture is
3.136
5.20 g
 0.387
13.43 g
Step 1: Calculate the mass of C in 55.90 g CO2, and the mass of H in 28.61 g H2O. This is a factor-label
problem. To calculate the mass of each component, you need the molar masses and the correct mole
ratio.
You should come up with the following strategy:
g CO2  mol CO2  mol C  g C
Step 2:
? g C  55.90 g CO2 
1 mol CO2
1 mol C
12.01 g C


 15.25 g C
44.01 g CO 2 1 mol CO2
1 mol C
? g H  28.61 g H 2 O 
1 mol H 2 O
2 mol H
1.008 g H


 3.201 g H
18.02 g H 2 O 1 mol H 2 O
1 mol H
Similarly,
Since the compound contains C, H, and Pb, we can calculate the mass of Pb by difference.
51.36 g  mass C  mass H  mass Pb
51.36 g  15.25 g  3.201 g  mass Pb
mass Pb  32.91 g Pb
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
65
Step 3: Calculate the number of moles of each element present in the sample. Use molar mass as a
conversion factor.
? mol C  15.25 g C 
1 mol C
 1.270 mol C
12.01 g C
? mol H = 3.201 g H 
1 mol H
 3.176 mol H
1.008 g H
Similarly,
? mol Pb  32.91 g Pb 
1 mol Pb
 0.1588 mol Pb
207.2 g Pb
Thus, we arrive at the formula Pb0.1588C1.270H3.176, which gives the identity and the ratios of atoms present.
However, chemical formulas are written with whole numbers.
Step 4: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript.
Pb:
0.1588
 1.00
0.1588
C:
1.270
 8
0.1588
H:
3.176
 20
0.1588
This gives the empirical formula, PbC8H20.
3.137
First, calculate the mass of C in CO2 and the mass of H in H2O.
? g C  30.2 g CO 2 
1 mol CO 2
1 mol C
12.01 g C


 8.24 g C
44.01 g CO2 1 mol CO2
1 mol C
? g H  14.8 g H 2 O 
1 mol H 2 O
2 mol H
1.008 g H


 1.66 g H
18.02 g H 2 O 1 mol H 2 O
1 mol H
Since the compound contains C, H, and O, we can calculate the mass of O by difference.
12.1 g  mass C  mass H  mass O
12.1 g  8.24 g  1.66 g  mass O
mass O  2.2 g O
Next, calculate the moles of each element.
? mol C = 8.24 g C 
1 mol C
 0.686 mol C
12.01 g C
? mol H = 1.66 g H 
1 mol H
 1.65 mol H
1.008 g H
? mol O = 2.2 g O 
1 mol O
 0.14 mol O
16.00 g O
Thus, we arrive at the formula C0.686H1.65O0.14. Dividing by 0.14 gives the empirical formula, C5H12O.
3.138
(a)
The following strategy can be used to convert from the volume of the Mg cube to the number of Mg
atoms.
3
cm  grams  moles  atoms
66
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
1.0 cm3 
(b)
1.74 g Mg
1 cm3

1 mol Mg
6.022  1023 Mg atoms

 4.3  1022 Mg atoms
24.31 g Mg
1 mol Mg
Since 74 percent of the available space is taken up by Mg atoms, 4.3  10
volume:
3
22
atoms occupy the following
3
0.74  1.0 cm  0.74 cm
We are trying to calculate the radius of a single Mg atom, so we need the volume occupied by a single
Mg atom.
0.74 cm3
 1.7  1023 cm3 /Mg atom
volume Mg atom 
22
4.3  10 Mg atoms
4 3
r . Solving for the radius:
3
4
cm3  r 3
3
The volume of a sphere is
V  1.7  1023
3
r  4.1  10
r  1.6  10
24
8
3
cm
cm
Converting to picometers:
radius Mg atom  (1.6  108 cm) 
3.139
0.01 m
1 pm

 1.6  102 pm
1 cm
1  1012 m
The balanced equations for the combustion of sulfur and the reaction of SO2 with CaO are:
S(s)  O2(g) 
 SO2(g)
SO2(g)  CaO(s) 
 CaSO3(s)
First, find the amount of sulfur present in the daily coal consumption.
(6.60  106 kg coal) 
1.6% S
 1.06  105 kg S  1.06  108 g S
100%
The daily amount of CaO needed is:
(1.06  108 g S) 
3.140
1 mol SO2 1 mol CaO 56.08 g CaO
1 mol S
1 kg




 1.85  105 kg CaO
32.07 g S
1 mol S
1 mol SO 2
1 mol CaO
1000 g
The molar mass of air can be calculated by multiplying the mass of each component by its abundance and
adding them together. Recall that nitrogen gas and oxygen gas are diatomic.
molar mass air  (0.7808)(28.02 g/mol)  (0.2095)(32.00 g/mol)  (0.0097)(39.95 g/mol)  28.97 g/mol
3.141
(a)
Assuming the die pack with no empty space between die, the volume of one mole of die is:
3
23
(1.5 cm)  (6.022  10 )  2.0  10
24
3
cm
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
(b)
8
6371 km  6.371  10 cm
Volume  area  height (h)
h 
3.142
volume
2.0  1024 cm3

 3.9  105 cm  3.9 × 103 m
area
4(6.371  108 cm)2
The surface area of the water can be calculated assuming that the dish is circular.
2
2
2
2
surface area of water  r  (10 cm)  3.1  10 cm
2
The cross-sectional area of one stearic acid molecule in cm is:
2
 1  109 m   1 cm 2
0.21 nm  
 2.1  1015 cm 2 /molecule
 
 1 nm   0.01 m 


2
Assuming that there is no empty space between molecules, we can calculate the number of stearic acid
2
2
molecules that will fit in an area of 3.1  10 cm .
(3.1  102 cm 2 ) 
1 molecule
2.1  10
15
cm
2
 1.5  1017 molecules
Next, we can calculate the moles of stearic acid in the 1.4  10
Avogadro’s number (the number of molecules per mole).
1.4  104 g stearic acid 
g sample. Then, we can calculate
1 mol stearic acid
 4.9  107 mol stearic acid
284.5 g stearic acid
Avogadro's number ( N A ) 
3.143
4
1.5  1017 molecules
4.9  10
7
 3.1  1023 molecules/mol
mol
The balanced equations for the combustion of octane are:
2C8H18  25O2 
 16CO2  18H2O
 16CO  18H2O
2C8H18  17O2 
The quantity of octane burned is 2650 g (1 gallon with a density of 2.650 kg/gallon). Let x be the mass of
octane converted to CO2; therefore, (2650  x) g is the mass of octane converted to CO.
The amounts of CO2 and H2O produced by x g of octane are:
x g C8 H18 
1 mol C8 H18
16 mol CO 2
44.01 g CO2


 3.083 x g CO 2
114.2 g C8 H18 2 mol C8 H18
1 mol CO 2
x g C8 H18 
1 mol C8 H18
18 mol H 2 O 18.02 g H 2 O


 1.420 x g H 2 O
114.2 g C8 H18 2 mol C8 H18
1 mol H 2 O
The amounts of CO and H2O produced by (2650  x) g of octane are:
(2650  x) g C8 H18 
1 mol C8 H18
16 mol CO
28.01 g CO


 (5200  1.962 x) g CO
114.2 g C8 H18 2 mol C8 H18
1 mol CO
67
68
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
(2650  x) g C8 H18 
1 mol C8 H18
18 mol H 2 O 18.02 g H 2 O


 (3763  1.420 x) g H 2 O
114.2 g C8 H18 2 mol C8 H18
1 mol H 2 O
The total mass of CO2  CO  H2O produced is 11530 g. We can write:
11530 g  3.083x  1.420x  5200  1.962x  3763  1.420x
x  2290 g
Since x is the amount of octane converted to CO2, we can now calculate the efficiency of the process.
efficiency 
3.144
(a)
g octane converted
2290 g
 100% 
 100%  86.49%
g octane total
2650 g
The balanced chemical equation is:
C3H8(g)  3H2O(g) 
 3CO(g)  7H2(g)
(b)
You should come up with the following strategy to solve this problem. In this problem, we use kg-mol
to save a couple of steps.
kg C3H8  mol C3H8  mol H2  kg H2
? kg H 2  (2.84  103 kg C3 H8 ) 
1 kg-mol C3 H8
7 kg-mol H 2
2.016 kg H 2


44.09 kg C3 H8 1 kg-mol C3 H8 1 kg-mol H 2
2
 9.09  10 kg H2
3.145
For the first step of the synthesis, the yield is 90% or 0.9. For the second step, the yield will be 90% of 0.9 or
(0.9  0.9)  0.81. For the third step, the yield will be 90% of 0.81 or (0.9  0.9  0.9)  0.73. We see that
the yield will be:
n
Yield  (0.9)
where n  number of steps in the reaction. For 30 steps,
Yield  (0.9)
30
 0.04  4%