Chemistry 10th Edition Student Solutions Manual (Raymong Chang) by Raymond Chang (z-lib.org).pdf

CHAPTER 1
CHEMISTRY: THE STUDY OF CHANGE
Problem Categories
Biological: 1.24, 1.48, 1.69, 1.70, 1.78, 1.84, 1.93, 1.95, 1.96, 1.97, 1.105.
Conceptual: 1.3, 1.4, 1.11, 1.12, 1.15, 1.16, 1.54, 1.62, 1.89, 1.101, 1.103.
Environmental: 1.70, 1.87, 1.89, 1.92, 1.98.
Industrial: 1.51, 1.55, 1.72, 1.81, 1.91.
Difficulty Level
Easy: 1.3, 1.11, 1.13, 1.14, 1.15, 1.21, 1.22, 1.23, 1.24, 1.25, 1.26, 1.29, 1.30, 1.31, 1.32, 1.33, 1.34, 1.54, 1.55, 1.63,
1.64, 1.77, 1.80, 1.84, 1.89, 1.91.
Medium: 1.4, 1.12, 1.16, 1.35, 1.36, 1.37, 1.38, 1.39, 1.40, 1.41, 1.42, 1.43, 1.44, 1.45, 1.46, 1.47, 1.48, 1.49, 1.50,
1.51, 1.52, 1.53, 1.56, 1.57, 1.59, 1.60, 1.61, 1.62, 1.70, 1.71, 1.72, 1.73, 1.74, 1.75, 1.76, 1.78, 1.79, 1.81, 1.82, 1.83,
1.85, 1.94, 1.95, 1.96, 1.97, 1.98.
Difficult: 1.58, 1.65, 1.66, 1.67, 1.68, 1.69, 1.86, 1.87, 1.88, 1.90, 1.92, 1.93, 1.99, 1.100, 1.101, 1.102, 1.103, 1.104,
1.105, 1.106.
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 (b) law (c) theory
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.
1.12 (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.

CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
2
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) K (b) Sn (c) Cr (d) B (e) Ba
(f) Pu (g) S (h) Ar (i) Hg
1.15 (a) element (b) compound (c) element (d) compound
1.16 (a) homogeneous mixture (b) element (c) compound
(d) homogeneous mixture (e) heterogeneous mixture (f) homogeneous mixture
(g) heterogeneous mixture
1.21
mass 586 g
volume 188 mL
= = =
density 3.12 g/mL
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
density
volume
=
Solution:
mass = density × volume
0.798 g
17.4 mL
1 mL
= × =
mass of ethanol 13.9 g
1.23
5 C
? C = ( F 32 F)
9 F
°
° ° − ° ×
°
(a)
5 C
? C = (95 32) F
9 F
°
° − ° × =
°
35 C
°
(b)
5 C
? C = (12 32) F
9 F
°
° − ° × =
°
11 C
− °
(c)
5 C
? C = (102 32) F
9 F
°
° − ° × =
°
39 C
°
(d)
5 C
? C = (1852 32) F
9 F
°
° − ° × =
°
1011 C
°
(e)
9 F
? F C 32 F
5 C
°
⎛ ⎞
° = ° × + °
⎜ ⎟
°
⎝ ⎠
9 F
? F 273.15 C 32 F
5 C
°
⎛ ⎞
° = − ° × + ° =
⎜ ⎟
°
⎝ ⎠
459.67 F
− °
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.
5 C
? C = ( F 32 F)
9 F
°
° ° − ° ×
°

CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE 3
5 C
= (105 32) F
9 F
°
− ° × =
°
? C 41 C
° °
(b) Conversion from Celsius to Fahrenheit.
9 F
? F C 32 F
5 C
°
⎛ ⎞
° = ° × + °
⎜ ⎟
°
⎝ ⎠
9 F
11.5 C 32 F
5 C
°
⎛ ⎞
= − ° × + ° =
⎜ ⎟
°
⎝ ⎠
? F 11.3 F
° °
(c) Conversion from Celsius to Fahrenheit.
9 F
? F C 32 F
5 C
°
⎛ ⎞
° = ° × + °
⎜ ⎟
°
⎝ ⎠
3 9 F
6.3 10 C 32 F
5 C
°
⎛ ⎞
= × ° × + ° =
⎜ ⎟
°
⎝ ⎠
4
? F 1.1 10 F
° × °
(d) Conversion from Fahrenheit to Celsius.
5 C
? C = ( F 32 F)
9 F
°
° ° − ° ×
°
5 C
= (451 32) F
9 F
°
− ° × =
°
? C 233 C
° °
1.25
1 K
K ( C 273 C)
1 C
= ° + °
°
(a) K = 113°C + 273°C = 386 K
(b) K = 37°C + 273°C = 3.10 × 10
2
K
(c) K = 357°C + 273°C = 6.30 × 10
2
K
1.26 (a)
1 K
K ( C 273 C)
1 C
= ° + °
°
°C = K − 273 = 77 K − 273 = −196°C
(b) °C = 4.2 K − 273 = −269°C
(c) °C = 601 K − 273 = 328°C
1.29 (a) 2.7 × 10
−8
(b) 3.56 × 10
2
(c) 4.7764 × 10
4
(d) 9.6 × 10
−2
1.30 (a) 10
−2
indicates that the decimal point must be moved two places to the left.
1.52 × 10
−2
= 0.0152
(b) 10
−8
indicates that the decimal point must be moved 8 places to the left.
7.78 × 10
−8
= 0.0000000778

4
1.31 (a) 145.75 + (2.3 × 10
−1
) = 145.75 + 0.23 = 1.4598 × 10
2
(b)
4
2 2
79500 7.95 10
=
2.5 10 2.5 10
×
=
× ×
2
3.2 10
×
(c) (7.0 × 10
−3
) − (8.0 × 10
−4
) = (7.0 × 10
−3
) − (0.80 × 10
−3
) = 6.2 × 10
−3
(d) (1.0 × 10
4
) × (9.9 × 10
6
) = 9.9 × 10
10
1.32 (a) Addition using scientific notation.
Strategy: Let's express scientific notation as N × 10
n
. 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.
Let’s write 0.0095 in such a way that n = −3. We have decreased 10
n
by 10
3
, so we must increase N by 10
3
.
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.
9.5 × 10
−3
+ 8.5 × 10
−3
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 10
to express N between 1 and 10 (1.8), we must increase 10
n
by a factor of 10. The exponent, n, is increased by
1 from −3 to −2.
18.0 × 10
−3
= 1.8 × 10
−2
(b) Division using scientific notation.
n
. 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
+2 − (−8)
= 1.14 × 10
+2 + 8
= 1.14 × 10
10
(c) Subtraction using scientific notation.
n
. 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
Subtract the N parts of the numbers, keeping the exponent, n, the same.
8.5 × 10
5
− 9.0 × 10
5
−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
to express N between 1 and 10 (5), we must decrease 10
n
by a factor of 10. The exponent, n, is decreased by
1 from 5 to 4.
−0.5 × 10
5
= −5 × 10
4
(d) Multiplication using scientific notation.
n
. 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 10
to express N between 1 and 10 (1.3), we must increase 10
n
by a factor of 10. The exponent, n, is increased by
1 from 2 to 3.
13 × 10
2
= 1.3 × 10
3
1.33 (a) four (b) two (c) five (d) two, three, or four
(e) three (f) one (g) one (h) two
1.34 (a) one (b) three (c) three (d) four
(e) two or three (f) one (g) one or two
1.35 (a) 10.6 m (b) 0.79 g (c) 16.5 cm
2
(d) 1 × 10
6
g/cm
3
1.36 (a) Division
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
1.28
5.70 km
= 3
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
(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
−3
mg
(c) 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
(0.402 × 10
7
dm) + (7.74 × 10
7
dm) = 8.14 × 10
7
dm
Since 7.74 × 10
7
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.
(d) Subtraction, addition, and division
Strategy: For subtraction and addition, the number of significant figures to the right of the decimal point in
that part of the calculation is determined by the lowest number of digits to the right of the decimal point in
any of the original numbers. For the division part of the calculation, the number of significant figures in the
answer is determined by the number having the smallest number of significant figures. First, perform the
subtraction and addition parts to the correct number of significant figures, and then perform the division.
Solution:
(7.8 m 0.34 m) 7.5 m
/
(1.15 s 0.82 s) 1.97 s
−
= =
+
3.8 m s
1.37 Calculating the mean for each set of date, we find:
Student A: 87.6 mL
Student B: 87.1 mL
Student C: 87.8 mL
From these calculations, we can conclude that the volume measurements made by Student B were the most
accurate of the three students. The precision in the measurements made by both students B and C are fairly
high, while the measurements made by student A are less precise. In summary:
Student A: neither accurate nor precise
Student B: both accurate and precise
Student C: precise, but not accurate

1.38 Calculating the mean for each set of date, we find:
Tailor X: 31.5 in
Tailor Y: 32.6 in
Tailor Z: 32.1 in
From these calculations, we can conclude that the seam measurements made by Tailor Z were the most
accurate of the three tailors. The precision in the measurements made by both tailors X and Z are fairly high,
while the measurements made by tailor Y are less precise. In summary:
Tailor X: most precise
Tailor Y: least accurate and least precise
Tailor Z: most accurate
1.39 (a)
1 dm
22.6 m
0.1 m
= × =
? dm 226 dm
(b)
0.001 g 1 kg
25.4 mg
1 mg 1000 g
= × × = 5
? kg 2.54 10 kg
−
×
(c)
3
1 10 L
556 mL
1 mL
−
×
= × =
? L 0.556 L
(d)
3
2
3
10.6 kg 1000 g 1 10 m
1 kg 1 cm
1 m
−
⎛ ⎞
×
= × × =
⎜ ⎟
⎜ ⎟
⎝ ⎠
3
3
g
? 0.0106 g/cm
cm
1.40 (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
grams to milligrams (1 mg = 1 × 10
−3
g). Arrange the appropriate conversion factors so that pounds and
grams cancel, and the unit milligrams is obtained in your answer.
Solution: The sequence of conversions is
lb → grams → mg
Using the following conversion factors,
453.6 g
1 lb 3
1 mg
1 10 g
−
×
we obtain the answer in one step:
3
453.6 g 1 mg
242 lb
1 lb 1 10 g
−
= × × =
×
8
? mg 1.10 10 mg
×
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.

8
(b)
? m
3
= 68.3 cm
3
Recall that 1 cm = 1 × 10
−2
m. We need to set up a conversion factor to convert from cm
3
to m
3
.
Solution: We need the following conversion factor so that centimeters cancel and we end up with meters.
2
1 10 m
1 cm
−
×
Since this conversion factor deals with length and we want volume, it must therefore be cubed to give
3
2 2 2 2
1 10 m 1 10 m 1 10 m 1 10 m
1 cm 1 cm 1 cm 1 cm
− − − −
⎛ ⎞
× × × ×
× × = ⎜ ⎟
⎜ ⎟
⎝ ⎠
We can write
3
2
3 1 10 m
68.3 cm
1 cm
−
⎛ ⎞
×
= × =
⎜ ⎟
⎜ ⎟
⎝ ⎠
3 5 3
? m 6.83 10 m
−
×
Check: We know that 1 cm
3
= 1 × 10
−6
m
3
. We started with 6.83 × 10
1
cm
3
. Multiplying this quantity by
1 × 10
−6
gives 6.83 × 10
−5
.
(c)
? L = 7.2 m
3
In Chapter 1 of the text, a conversion is given between liters and cm
3
(1 L = 1000 cm
3
). If we can convert m
3
to cm
3
, we can then convert to liters. Recall that 1 cm = 1 × 10
−2
m. We need to set up two conversion
factors to convert from m
3
to L. Arrange the appropriate conversion factors so that m
3
and cm
3
cancel, and
the unit liters is obtained in your answer.
m
3
→ cm
3
→ L
3
2
1 cm
1 10 m
−
⎛ ⎞
⎜ ⎟
⎜ ⎟
×
⎝ ⎠
3
1 L
1000 cm
the answer is obtained in one step:
3
3
2 3
1 cm 1 L
7.2 m
1 10 m 1000 cm
−
⎛ ⎞
= × × =
⎜ ⎟
⎜ ⎟
×
⎝ ⎠
3
? L 7.2 10 L
×
Check: From the above conversion factors you can show that 1 m
3
= 1 × 10
3
L. Therefore, 7 m
3
would
equal 7 × 10
3
L, which is close to the answer.

(d)
? lb = 28.3 μg
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 μg to grams, we can then
convert from grams to pounds. Recall that 1 μg = 1 × 10
−6
g. Arrange the appropriate conversion factors so
that μg and grams cancel, and the unit pounds is obtained in your answer.
μg → g → lb
6
1 10 g
1 g
−
×
μ
1 lb
453.6 g
we can write
6
1 10 g 1 lb
28.3 g
1 g 453.6 g
−
×
= μ × × =
μ
8
? lb 6.24 10 lb
−
×
Check: Does the answer seem reasonable? What number does the prefix μ represent? Should 28.3 μg be a
very small mass?
1.41
1255 m 1 mi 3600 s
1 s 1609 m 1 h
× × = 2808 mi/h
1.42 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.
days → hours → minutes → seconds
24 h
1 day
60 min
1 h
60 s
1 min
we can write
24 h 60 min 60 s
= 365.24 day
1 day 1 h 1 min
× × × = 7
? s 3.1557 10 s
×
Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year?
1.43 6
8
1.609 km 1000 m 1 s 1 min
(93 10 mi)
1 mi 1 km 60 s
3.00 10 m
× × × × × =
×
8.3 min

10
1.44 (a)
1 mi 5280 ft 12 in 1 min
13 min 1 mi 1 ft 60 s
= × × × =
? in/s 81 in/s
(b)
1 mi 1609 m
13 min 1 mi
= × = 2
? m/min 1.2 10 m/min
×
(c)
1 mi 1609 m 1 km 60 min
13 min 1 mi 1000 m 1 h
= × × × =
? km/h 7.4 km/h
1.45
1 m
6.0 ft
3.28 ft
× = 1.8 m
453.6 g 1 kg
168 lb
1 lb 1000 g
× × = 76.2 kg
1.46
55 mi 1.609 km
1 h 1 mi
= × =
? km/h 88 km/h
1.47
62 m 1 mi 3600 s
1 s 1609 m 1 h
× × = 2
1.4 10 mph
×
1.48
6
0.62 g Pb
0.62 ppm Pb
1 10 g blood
=
×
3
6
0.62 g Pb
6.0 10 g of blood
1 10 g blood
× × =
×
3
3.7 10 g Pb
−
×
1.49 (a)
8
365 day 24 h 3600 s 3.00 10 m 1 mi
1.42 yr
1 yr 1 day 1 h 1 s 1609 m
×
× × × × × = 12
8.35 10 mi
×
(b)
36 in 2.54 cm
32.4 yd
1 yd 1 in
× × = 3
2.96 10 cm
×
(c)
10
3.0 10 cm 1 in 1 ft
1 s 2.54 cm 12 in
×
× × = 8
9.8 10 ft/s
×
1.50 (a)
9
1 10 m
185 nm
1 nm
−
×
= × = 7
? m 1.85 10 m
−
×
(b) 9 365 day 24 h 3600 s
(4.5 10 yr)
1 yr 1 day 1 h
= × × × × = 17
? s 1.4 10 s
×
(c)
3
3 0.01 m
71.2 cm
1 cm
⎛ ⎞
= × =
⎜ ⎟
⎝ ⎠
3 5 3
? m 7.12 10 m
−
×
(d)
3
3
2 3
1 cm 1 L
88.6 m
1 10 m 1000 cm
−
⎛ ⎞
= × × =
⎜ ⎟
⎜ ⎟
×
⎝ ⎠
4
? L 8.86 10 L
×

1.51
3
3
2.70 g 1 kg 1 cm
1000 g 0.01 m
1 cm
⎛ ⎞
= × × =
⎜ ⎟
⎝ ⎠
3 3
density 2.70 10 kg/m
×
1.52
3
0.625 g 1 L 1 mL
1 L 1000 mL 1 cm
= × × = 4 3
density 6.25 10 g/cm
−
×
1.53 Substance Qualitative Statement Quantitative Statement
(a) water colorless liquid freezes at 0°C
(b) carbon black solid (graphite) density = 2.26 g/cm
3
(c) iron rusts easily density = 7.86 g/cm
3
(d) hydrogen gas colorless gas melts at −255.3°C
(e) sucrose tastes sweet at 0°C, 179 g of sucrose dissolves in 100 g of H2O
(f) table salt tastes salty melts at 801°C
(g) mercury liquid at room temperature boils at 357°C
(h) gold a precious metal density = 19.3 g/cm
3
(i) air a mixture of gases contains 20% oxygen by volume
1.54 See Section 1.6 of your text for a discussion of these terms.
(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.
(e) Chemical property. The carbon dioxide is chemically converted into other molecules.
1.55 9
3
1 ton
(95.0 10 lb of sulfuric acid)
2.0 10 lb
× × =
×
7
4.75 10 tons of sulfuric acid
×
1.56 Volume of rectangular bar = length × width × height
52.7064 g
=
(8.53 cm)(2.4 cm)(1.0 cm)
= = 3
density 2.6 g/cm
m
V
1.57 mass = density × volume
(a) mass = (19.3 g/cm
3
) × [
4
3
π(10.0 cm)
3
] = 8.08 × 10
4
g
(b)
3
3 1 cm
(21.4 g/cm ) 0.040 mm
10 mm
⎛ ⎞
= × × =
⎜ ⎟
⎝ ⎠
6
mass 1.4 10 g
−
×
(c) mass = (0.798 g/mL)(50.0 mL) = 39.9 g

12
1.58 You are asked to solve for the inner diameter of the tube. If you can calculate the volume that the mercury
occupies, you can calculate the radius of the cylinder, Vcylinder = πr
2
h (r is the inner radius of the cylinder,
and h is the height of the cylinder). The cylinder diameter is 2r.
mass of Hg
volume of Hg filling cylinder
density of Hg
=
3
3
105.5 g
volume of Hg filling cylinder 7.757 cm
13.6 g/cm
= =
Next, solve for the radius of the cylinder.
Volume of cylinder = πr
2
h
volume
=
π ×
r
h
3
7.757 cm
0.4409 cm
12.7 cm
= =
π ×
r
The cylinder diameter equals 2r.
Cylinder diameter = 2r = 2(0.4409 cm) = 0.882 cm
1.59 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.
mass
volume
density
=
Mass of water = 87.39 g − 56.12 g = 31.27 g
3
mass 31.27 g
=
density 0.9976 g/cm
= = 3
Volume of the flask 31.35 cm
1.60
343 m 1 mi 3600 s
1 s 1609 m 1 h
× × = 767 mph
1.61 The volume of silver is equal to the volume of water it displaces.
Volume of silver = 260.5 mL − 242.0 mL = 18.5 mL = 18.5 cm
3
3
194.3 g
18.5 cm
= = 3
density 10.5 g/cm
1.62 In order to work this problem, you need to understand the physical principles involved in the experiment in
Problem 1.61. 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.

1.63
4
3 3
mass 1.20 10 g
volume 1.05 10 cm
×
= = =
×
3
density 11.4 g/cm
1.64
mass
Volume
density
=
3
3
1.20 10 g
0.53 g / cm
×
= = 3 3
Volume occupied by Li 2.3 10 cm
×
1.65 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.
known error in a measurement
Percent error 100%
value of the measurement
= ×
For the Fahrenheit thermometer,
0.056 C
100%
38.9 C
°
= × =
°
percent error 0.1%
For the Celsius thermometer,
0.1 C
100%
38.9 C
°
= × =
°
percent error 0.3%
Which thermometer is more accurate?
1.66 To work this problem, we need to convert from cubic feet to L. Some tables will have a conversion factor of
28.3 L = 1 ft
3
, but we can also calculate it using the dimensional analysis method described in Section 1.9 of
the text.
First, converting from cubic feet to liters:
3 3 3
7 3 9
3
12 in 2.54 cm 1 mL 1 10 L
(5.0 10 ft ) 1.42 10 L
1 ft 1 in 1 mL
1 cm
−
⎛ ⎞ ⎛ ⎞ ×
× × × × × = ×
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
The mass of vanillin (in g) is:
11
9 2
2.0 10 g vanillin
(1.42 10 L) 2.84 10 g vanillin
1 L
−
−
×
× × = ×
The cost is:
2 $112
(2.84 10 g vanillin)
50 g vanillin
−
× × = =
$0.064 6.4¢
1.67
9 F
? F = C + 32 F
5 C
°
⎛ ⎞
° ° × °
⎜ ⎟
°
⎝ ⎠
Let temperature = t
9
= + 32 F
5
°
t t

14
9
= 32 F
5
− °
t t
4
= 32 F
5
− °
t
t = −40°F = −40°C
1.68 There are 78.3 + 117.3 = 195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor.
195.6 C
100 S
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
D
D
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
°
⎛ ⎞
° ° − °
⎜ ⎟
°
⎝ ⎠
Solving for ? °S gives:
100 S
? S = (? C + 117.3 C)
195.6 C
°
⎛ ⎞
° ° ° ⎜ ⎟
°
⎝ ⎠
For 25°C we have:
100 S
(25 + 117.3) C
195.6 C
°
⎛ ⎞
= ° =
⎜ ⎟
°
⎝ ⎠
? S 73 S
° °
1.69 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.
240 mL of pure oxygen/min = (0.04)(volume of inhaled air/min)
240 mL of oxygen/min
Volume of inhaled air/min 6000 mL of inhaled air/min
0.04
= =
Since there are 12 breaths per min,
6000 mL of inhaled air 1 min
1 min 12 breaths
= × = 2
volume of air/breath 5 10 mL/breath
×
1.70 (a)
6000 mL of inhaled air 0.001 L 60 min 24 h
1 min 1 mL 1 h 1 day
× × × = 3
8.6 10 L of air/day
×
(b)
3 6
8.6 10 L of air 2.1 10 L CO
1 day 1 L of air
−
× ×
× = 0.018 L CO/day
1.71 The mass of the seawater is:
21 24 21
1 mL 1.03 g
(1.5 10 L) 1.55 10 g 1.55 10 kg seawater
0.001 L 1 mL
× × × = × = ×
Seawater is 3.1% NaCl by mass. The total mass of NaCl in kilograms is:
21 3.1% NaCl
(1.55 10 kg seawater)
100% seawater
= × × = 19
mass NaCl (kg) 4.8 10 kg NaCl
×

19 2.205 lb 1 ton
(4.8 10 kg)
1 kg 2000 lb
= × × × = 16
mass NaCl (tons) 5.3 10 tons NaCl
×
1.72 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 1.71.
mass
volume
density
=
1000 g
volume of 1 kg of seawater 970.9 mL 0.9709 L
1.03 g/mL
= = =
In other words, there are 1.3 g of Mg in every 0.9709 L of seawater.
Next, let’s convert tons of Mg to grams of Mg.
4 10
2000 lb 453.6 g
(8.0 10 tons Mg) 7.26 10 g Mg
1 ton 1 lb
× × × = ×
Volume of seawater needed to extract 8.0 × 10
4
ton Mg =
10 0.9709 L seawater
(7.26 10 g Mg)
1.3 g Mg
× × = 10
5.4 10 L of seawater
×
1.73 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.
mass
volume
density
=
3
860.2 g
Volume of crucible
21.45 g/cm
= = 3
40.10 cm
3
(860.2 820.2)g
Volume of water displaced
0.9986 g/cm
−
= = 3
40.1 cm
The volumes are the same (within experimental error), so the crucible is made of platinum.
1.74 Volume = surface area × depth
Recall that 1 L = 1 dm
3
. Let’s convert the surface area to units of dm
2
and the depth to units of dm.
2 2
8 2 16 2
1000 m 1 dm
surface area (1.8 10 km ) 1.8 10 dm
1 km 0.1 m
⎛ ⎞ ⎛ ⎞
= × × × = ×
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3 4
1 dm
depth (3.9 10 m) 3.9 10 dm
0.1 m
= × × = ×
Volume = surface area × depth = (1.8 × 10
16
dm
2
)(3.9 × 10
4
dm) = 7.0 × 10
20
dm
3
= 7.0 × 10
20
L

16
1.75 (a)
31.103 g Au
2.41 troy oz Au
1 troy oz Au
× = 75.0 g Au
(b) 1 troy oz = 31.103 g
1 lb 453.6 g
? g in 1 oz 1 oz
16 oz 1 lb
= × × = 28.35 g
A troy ounce is heavier than an ounce.
1.76 3
4
Volume of sphere
3
= πr
3
3 3
4 15 cm
Volume 1.77 10 cm
3 2
⎛ ⎞
= π = ×
⎜ ⎟
⎝ ⎠
3 3
3
22.57 g Os 1 kg
mass volume density (1.77 10 cm )
1000 g
1 cm
= × = × × × = 1
4.0 10 kg Os
×
1 2.205 lb
4.0 10 kg Os
1 kg
× × = 88 lb Os
1.77 (a)
|0.798 g/mL 0.802 g/mL|
100%
0.798 g/mL
−
× = 0.5%
(b)
|0.864 g 0.837 g|
100%
0.864 g
−
× = 3.1%
1.78 62 kg = 6.2 × 10
4
g
O: (6.2 × 10
4
g)(0.65) = 4.0 × 10
4
g O N: (6.2 × 10
4
g)(0.03) = 2 × 10
3
g N
C: (6.2 × 10
4
g)(0.18) = 1.1 × 10
4
g C Ca: (6.2 × 10
4
g)(0.016) = 9.9 × 10
2
g Ca
H: (6.2 × 10
4
g)(0.10) = 6.2 × 10
3
g H P: (6.2 × 10
4
g)(0.012) = 7.4 × 10
2
g P
1.79 3 minutes 43.13 seconds = 223.13 seconds
Time to run 1500 meters is:
1 mi 223.13 s
1500 m
1609 m 1 mi
× × = =
208.01 s 3 min 28.01 s
1.80 ? °C = (7.3 × 10
2
− 273) K = 4.6 × 10
2
°C
2 9 F
(4.6 10 C) 32 F
5 C
°
⎛ ⎞
= × ° × + ° =
⎜ ⎟
°
⎝ ⎠
2
? F 8.6 10 F
° × °
1.81 3 34.63% Cu 1000 g
(5.11 10 kg ore)
100% ore 1 kg
= × × × = 6
? g Cu 1.77 10 g Cu
×
1.82 4 2000 lb Au 16 oz Au $948
(8.0 10 tons Au) or
1 ton Au 1 lb Au 1 oz Au
× × × × = 12
$2.4 10 2.4 trillion dollars
×

1.83
12
21
4.0 10 g Au 1 mL
(1.5 10 L seawater)
1 mL seawater 0.001 L
−
×
= × × × = 12
? g Au 6.0 10 g Au
×
12 1 lb 16 oz $948
(6.0 10 g Au)
453.6 g 1 lb 1 oz
= × × × × = 14
value of gold $2.0 10
×
No one has become rich mining gold from the ocean, because the cost of recovering the gold would outweigh
the price of the gold.
1.84
22
1.1 10 Fe atoms
4.9 g Fe
1.0 g Fe
×
= × = 22
? Fe atoms 5.4 10 Fe atoms
×
1.85 21 19
0.50% crust
mass of Earth's crust (5.9 10 tons) 2.95 10 tons
100% Earth
= × × = ×
19 27.2% Si 2000 lb 1 kg
(2.95 10 tons crust)
100% crust 1 ton 2.205 lb
= × × × × = 21
mass of silicon in crust 7.3 10 kg Si
×
1.86 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
equal to the diameter of a Cu atom, which is (2)(1.3 × 10
−10
m). Let n be the number of times we can cut the
Cu wire in half. We can write:
10
1
0.1 m 2.6 10 m
2
−
⎛ ⎞
× = ×
⎜ ⎟
⎝ ⎠
n
9
1
2.6 10 m
2
−
⎛ ⎞
= ×
⎜ ⎟
⎝ ⎠
n
Taking the log of both sides of the equation:
9
1
log log(2.6 10 )
2
−
⎛ ⎞
= ×
⎜ ⎟
⎝ ⎠
n
n = 29 times
1.87 6 2
9.5 kg CO
5000 mi 1 gal gas
(40 10 cars)
1 car 20 mi 1 gal gas
× × × × = 10
2
9.5 10 kg CO
×
1.88 Volume = area × thickness.
From the density, we can calculate the volume of the Al foil.
3
3
mass 3.636 g
Volume 1.3472 cm
density 2.699 g / cm
= = =
Convert the unit of area from ft
2
to cm
2
.
2 2
2 2
12 in 2.54 cm
1.000 ft 929.03 cm
1 ft 1 in
⎛ ⎞ ⎛ ⎞
× × =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3
3
2
volume 1.3472 cm
1.450 10 cm
area 929.03 cm
−
= = = × = 2
thickness 1.450 10 mm
−
×

18
1.89 (a) homogeneous
(b) heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous.
1.90 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.
4 7
2 2
3.79 L 1 mL 1 g
(2.0 10 gallons H O) 7.58 10 g H 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
2 6
2
1 g chlorine
(7.58 10 g H O) 75.8 g chlorine
1 10 g H O
× × =
×
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.
100% soln 1 mL soln
75.8 g chlorine
6% chlorine 1 g soln
× × = 3
1.3 10 mL of chlorine solution
×
1.91 22
20
1 yr
(2.0 10 J)
1.8 10 J
× × =
×
2
1.1 10 yr
×
1.92 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.
2
2 5 2
1 cm
40 m 4.0 10 cm
0.01 m
⎛ ⎞
× = ×
⎜ ⎟
⎝ ⎠
0.10 mL = 0.10 cm
3
Volume = surface area × thickness
3
7
5 2
volume 0.10 cm
thickness 2.5 10 cm
surface area 4.0 10 cm
−
= = = ×
×
Converting to nm:
7
9
0.01 m 1 nm
(2.5 10 cm)
1 cm 1 10 m
−
−
× × × =
×
2.5 nm
1.93 The mass of water used by 50,000 people in 1 year is:
13
2
2
2
1.0 g H O
150 gal water 3.79 L 1000 mL 365 days
50,000 people 1.04 10 g H O/yr
1 person each day 1 gal 1 L 1 mL H 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:
13
2 6
2
1 g F 100% NaF 1 kg
(1.04 10 g H O)
45% F 1000 g
10 g H O
× × × × = 4
2.3 10 kg NaF
×

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%
569 L
× = 99%
1.94 (a)
3 3 3
3
$1.30 1 ft 1 in 1 cm 1 mL
12 in 2.54 cm 1 mL 0.001 L
15.0 ft
⎛ ⎞ ⎛ ⎞
× × × × =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3
$3.06 10 /L
−
×
(b)
3
3
0.304 ft gas $1.30
2.1 L water
1 L water 15.0 ft
× × = =
$0.055 5.5¢
1.95 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:
4
1609 m 1 cm
0.50 mi 8.05 10 cm
1 mi 0.01 m
× × = ×
3
12 in 2.54 cm
40 ft 1.22 10 cm
1 ft 1 in
× × = ×
volume of a cylinder = area × height = πr
2
× h
volume = π(8.05 × 10
4
cm)
2
× (1.22 × 10
3
cm) = 2.48 × 10
13
cm
3
Density of gases is usually expressed in g/L. Let’s convert the volume to liters.
13 3 10
3
1 mL 1 L
(2.48 10 cm ) 2.48 10 L
1000 mL
1 cm
× × × = ×
8
10
mass 1.0 10 g
volume 2.48 10 L
−
×
= = =
×
19
density 4.0 10 g/L
−
×
1.96 First, convert 10 μm to units of cm.
4
3
1 10 cm
10 m 1.0 10 cm
1 m
−
−
×
μ × = ×
μ
Now, substitute into the given equation to solve for time.
2 3 2
7 2
(1.0 10 cm)
2 2(5.7 10 cm /s)
−
−
×
= = =
×
0.88 s
x
D
t
It takes 0.88 seconds for a glucose molecule to diffuse 10 μm.
1.97 The mass of a human brain is about 1 kg (1000 g) and contains about 10
11
cells. The mass of a brain cell is:
8
11
1000 g
1 10 g/cell
1 10 cells
−
= ×
×

20
Assuming that each cell is completely filled with water (density = 1 g/mL), we can calculate the volume of
each cell. Then, assuming the cell to be cubic, we can solve for the length of one side of such a cell.
8 3
8 3
1 10 g 1 mL 1 cm
1 10 cm /cell
1 cell 1 g 1 mL
−
−
×
× × = ×
Vcube = a
3
a = (V)
1/3
= (1 × 10
−8
cm
3
)
1/3
= 0.002 cm
Next, the height of a single cell is a, 0.002 cm. If 10
11
cells are spread out in a thin layer a single cell thick,
the surface area can be calculated from the volume of 10
11
cells and the height of a single cell.
V = surface area × height
The volume of 10
11
brain cells is:
3
3
1 mL 1 cm
1000 g 1000 cm
1 g 1 mL
× × =
The surface area is:
2
3 2
5 2
1000 cm 1 10 m
5 10 cm
height 0.002 cm 1 cm
−
⎛ ⎞
×
= = = × × = ×
⎜ ⎟
⎜ ⎟
⎝ ⎠
1 2
Surface area 5 10 m
V
1.98 (a) A concentration of CO of 800 ppm in air would mean that there are 800 parts by volume of CO per
1 million parts by volume of air. Using a volume unit of liters, 800 ppm CO means that there are 800 L
of CO per 1 million liters of air. The volume in liters occupied by CO in the room is:
3
3 5
2 3
1 cm 1 L
17.6 m 8.80 m 2.64 m 409 m 4.09 10 L air
1 10 m 1000 cm
−
⎛ ⎞
× × = × × = ×
⎜ ⎟
⎜ ⎟
×
⎝ ⎠
2
5
6
8.00 10 L CO
4.09 10 L air
1 10 L air
×
× × =
×
327 L CO
(b) 1 mg = 1 × 10
−3
g and 1 L = 1000 cm
3
. We convert mg/m
3
to g/L:
3
3 2 3
3
0.050 mg 1 10 g 1 10 m 1000 cm
1 mg 1 cm 1 L
1 m
− −
⎛ ⎞
× ×
× × × =
⎜ ⎟
⎜ ⎟
⎝ ⎠
8
5.0 10 g / L
−
×
(c) 1 μg = 1 × 10
−3
mg and 1 mL = 1 × 10
−2
dL. We convert mg/dL to μg/mL:
2
3
120 mg 1 g 1 10 dL
1 dL 1 mL
1 10 mg
−
−
μ ×
× × =
×
3
1.20 10 μg / mL
×
1.99 This problem is similar in concept to a limiting reagent problem. We need sets of coins with 3 quarters,
1 nickel, and 2 dimes. First, we need to find the total number of each type of coin.
3 1 quarter
Number of quarters (33.871 10 g) 6000 quarters
5.645 g
= × × =

3 1 nickel
Number of nickels (10.432 10 g) 2100 nickels
4.967 g
= × × =
3 1 dime
Number of dimes (7.990 10 g) 3450 dimes
2.316 g
= × × =
Next, we need to find which coin limits the number of sets that can be assembled. For each set of coins, we
need 2 dimes for every 1 nickel.
2 dimes
2100 nickels 4200 dimes
1 nickel
× =
We do not have enough dimes.
For each set of coins, we need 2 dimes for every 3 quarters.
2 dimes
6000 quarters 4000 dimes
3 quarters
× =
Again, we do not have enough dimes, and therefore the number of dimes is our “limiting reagent”.
If we need 2 dimes per set, the number of sets that can be assembled is:
1 set
3450 dimes
2 dimes
× = 1725 sets
The mass of each set is:
5.645 g 4.967 g 2.316 g
3 quarters 1 nickel 2 dimes 26.534 g/set
1 quarter 1 nickel 1 dime
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
× + × + × =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Finally, the total mass of 1725 sets of coins is:
26.534 g
1725 sets
1 set
× = 4
4.577 10 g
×
1.100 We wish to calculate the density and radius of the ball bearing. For both calculations, we need the volume of
the ball bearing. The data from the first experiment can be used to calculate the density of the mineral oil. In
the second experiment, the density of the mineral oil can then be used to determine what part of the 40.00 mL
volume is due to the mineral oil and what part is due to the ball bearing. Once the volume of the ball bearing
is determined, we can calculate its density and radius.
From experiment one:
Mass of oil = 159.446 g − 124.966 g = 34.480 g
34.480 g
Density of oil 0.8620 g/mL
40.00 mL
= =
From the second experiment:
Mass of oil = 50.952 g − 18.713 g = 32.239 g
1 mL
Volume of oil 32.239 g 37.40 mL
0.8620 g
= × =

22
The volume of the ball bearing is obtained by difference.
Volume of ball bearing = 40.00 mL − 37.40 mL = 2.60 mL = 2.60 cm
3
Now that we have the volume of the ball bearing, we can calculate its density and radius.
3
18.713 g
Density of ball bearing
2.60 cm
= = 3
7.20 g/cm
Using the formula for the volume of a sphere, we can solve for the radius of the ball bearing.
3
4
3
= π
V r
3 3
4
2.60 cm
3
= πr
r
3
= 0.621 cm
3
r = 0.853 cm
1.101 It would be more difficult to prove that the unknown substance is an element. Most compounds would
decompose on heating, making them easy to identify. For example, see Figure 4.13(a) of the text. On
heating, the compound HgO decomposes to elemental mercury (Hg) and oxygen gas (O2).
1.102 We want to calculate the mass of the cylinder, which can be calculated from its volume and density. The
volume of a cylinder is πr
2
l. The density of the alloy can be calculated using the mass percentages of each
element and the given densities of each element.
The volume of the cylinder is:
V = πr
2
l
V = π(6.44 cm)
2
(44.37 cm)
V = 5781 cm
3
The density of the cylinder is:
density = (0.7942)(8.94 g/cm
3
) + (0.2058)(7.31 g/cm
3
) = 8.605 g/cm
3
Now, we can calculate the mass of the cylinder.
mass = density × volume
mass = (8.605 g/cm
3
)(5781 cm
3
) = 4.97 × 10
4
g
The assumption made in the calculation is that the alloy must be homogeneous in composition.
1.103 Gently heat the liquid to see if any solid remains after the liquid evaporates. Also, collect the vapor and then
compare the densities of the condensed liquid with the original liquid. The composition of a mixed liquid
would change with evaporation along with its density.
1.104 The density of the mixed solution should be based on the percentage of each liquid and its density. Because
the solid object is suspended in the mixed solution, it should have the same density as this solution. The
density of the mixed solution is:
(0.4137)(2.0514 g/mL) + (0.5863)(2.6678 g/mL) = 2.413 g/mL
As discussed, the density of the object should have the same density as the mixed solution (2.413 g/mL).

Yes, this procedure can be used in general to determine the densities of solids. This procedure is called the
flotation method. It is based on the assumptions that the liquids are totally miscible and that the volumes of
the liquids are additive.
1.105 When the carbon dioxide gas is released, the mass of the solution will decrease. If we know the starting mass
of the solution and the mass of solution after the reaction is complete (given in the problem), we can calculate
the mass of carbon dioxide produced. Then, using the density of carbon dioxide, we can calculate the volume
of carbon dioxide released.
1.140 g
Mass of hydrochloric acid 40.00 mL 45.60 g
1 mL
= × =
Mass of solution before reaction = 45.60 g + 1.328 g = 46.928 g
We can now calculate the mass of carbon dioxide by difference.
Mass of CO2 released = 46.928 g − 46.699 g = 0.229 g
Finally, we use the density of carbon dioxide to convert to liters of CO2 released.
2
1 L
Volume of CO released 0.229 g
1.81 g
= × = 0.127 L
1.106 As water freezes, it expands. First, calculate the mass of the water at 20°C. Then, determine the volume that
this mass of water would occupy at −5°C.
0.998 g
Mass of water 242 mL 241.5 g
1 mL
= × =
1 mL
Volume of ice at 5 C 241.5 g 264 mL
0.916 g
− ° = × =
The volume occupied by the ice is larger than the volume of the glass bottle. The glass bottle would crack!
ANSWERS TO REVIEW OF CONCEPTS
Section 1.3 (p. 9) (c)
Section 1.4 (p. 12) Elements: (b) and (d). Compounds: (a) and (c).
Section 1.5 (p. 14) (a)
Section 1.6 (p. 15) Chemical change: (b) and (c). Physical change: (d).
Section 1.7 (p. 22) (a)

CHAPTER 2
ATOMS, MOLECULES, AND IONS
Problem Categories
Conceptual: 2.31, 2.32, 2.33, 2.34, 2.61, 2.67, 2.68, 2.85, 2.88, 2.95.
Descriptive: 2.24, 2.25, 2.26, 2.49, 2.50, 2.62, 2.66, 2.73, 2.76, 2.77, 2.78, 2.79, 2.80, 2.82, 2.83, 2.84, 2.86, 2.90, 2.93.
Organic: 2.47, 2.48, 2.65, 2.97, 2.99, 2.100.
Difficulty Level
Easy: 2.7, 2.8, 2.13, 2.14, 2.15, 2.16, 2.23, 2.31, 2.32, 2.33, 2.43, 2.44, 2.45, 2.46, 2.47, 2.48, 2.83, 2.84, 2.91, 2.92.
Medium: 2.17, 2.18, 2.24. 2.26, 2.34, 2.35, 2.36, 2.49, 2.50, 2.57, 2.58, 2.59, 2.60, 2.61, 2.62, 2.63, 2.64, 2.65, 2.66,
2.67, 2.68, 2.69, 2.70, 2.73, 2.74, 2.75, 2.76, 2.77, 2.78, 2.80, 2.81, 2.82, 2.85, 2.86, 2.87, 2.88, 2.90, 2.93, 2.94, 2.95,
2.102, 2.104.
Difficult: 2.25, 2.71, 2.72, 2.79, 2.89, 2.96, 2.97, 2.98, 2.99, 2.100, 2.101, 2.103.
2.7 First, convert 1 cm to picometers.
10
12
0.01 m 1 pm
1 cm 1 10 pm
1 cm 1 10 m
−
× × = ×
×
10
2
1 He atom
(1 10 pm)
1 10 pm
= × × =
×
8
? He atoms 1 10 He atoms
×
2.8 Note that you are given information to set up the unit factor relating meters and miles.
4 4
nucleus
1 m 1 mi
10 10 2.0 cm
100 cm 1609 m
= = × × × =
atom 0.12 mi
r
r
2.13 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 Isotope 3
2He 4
2He 24
12Mg 25
12Mg 48
22Ti 79
35Br 195
78Pt
No. Protons 2 2 12 12 22 35 78
No. Neutrons 1 2 12 13 26 44 117
2.16 Isotope 7
15
N 16
33
S 29
63
Cu 38
84
Sr 56
130
Ba 74
186
W 80
202
Hg
No. Protons 7 16 29 38 56 74 80
No. Neutrons 8 17 34 46 74 112 122
No. Electrons 7 16 29 38 56 74 80

CHAPTER 2: ATOMS, MOLECULES, AND IONS 25
2.17 (a) 23
11Na (b) 64
28 Ni
2.18 The accepted way to denote the atomic number and mass number of an element X is as follows:
Z
AX
where,
A = mass number
Z = atomic number
(a) 74
186
W (b) 80
201
Hg
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.
(a) Li (0.53 g/cm
3
) K (0.86 g/cm
3
) H2O (0.98 g/cm
3
)
(b) Au (19.3 g/cm
3
) Pt (21.4 g/cm
3
) Hg (13.6 g/cm
3
)
(c) Os (22.6 g/cm
3
)
(d) Te (6.24 g/cm
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) This is a polyatomic molecule that is an elemental form of the substance. It is not a compound.
(b) This is a polyatomic molecule that is a compound.
(c) This is a diatomic molecule that is a compound.
2.32 (a) This is a diatomic molecule that is a compound.
(b) This is a polyatomic molecule that is a compound.
(c) This is a polyatomic molecule that is the elemental form of the substance. It is not a compound.
2.33 Elements: N2, S8, H2
Compounds: NH3, NO, CO, CO2, SO2
2.34 There are more than two correct answers for each part of the problem.
(a) H2 and F2 (b) HCl and CO (c) S8 and P4
(d) H2O and C12H22O11 (sucrose)
2.35 Ion Na
+
Ca
2+
Al
3+
Fe
2+
I
−
F
−
S
2−
O
2−
N
3−
No. protons 11 20 13 26 53 9 16 8 7
No. electrons 10 18 10 24 54 10 18 10 10

CHAPTER 2: ATOMS, MOLECULES, AND IONS
26
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
Ion K
+
Mg
2+
Fe
3+
Br
−
Mn
2+
C
4−
Cu
2+
No. protons 19 12 26 35 25 6 29
No. electrons 18 10 23 36 23 10 27
2.43 (a) Sodium ion has a +1 charge and oxide has a −2 charge. The correct formula is Na2O.
(b) The iron ion has a +2 charge and sulfide has a −2 charge. The correct formula is FeS.
(c) The correct formula is Co2(SO4)3
(d) Barium ion has a +2 charge and fluoride has a −1 charge. The correct formula is BaF2.
2.44 (a) The copper ion has a +1 charge and bromide has a −1 charge. The correct formula is CuBr.
(b) The manganese ion has a +3 charge and oxide has a −2 charge. The correct formula is Mn2O3.
(c) We have the Hg2
2+
ion and iodide (I
−
). The correct formula is Hg2I2.
(d) Magnesium ion has a +2 charge and phosphate has a −3 charge. The correct formula is Mg3(PO4)2.
2.45 (a) CN (b) CH (c) C9H20 (d) P2O5 (e) BH3
2.46 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 whole-
number subscripts?
Solution:
(a) Dividing both subscripts by 2, the simplest whole number ratio of the atoms in Al2Br6 is AlBr3.
(b) Dividing all subscripts by 2, the simplest whole number ratio of the atoms in Na2S2O4 is NaSO2.
(c) 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.
(d) 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.
2.47 The molecular formula of glycine is C2H5NO2.
2.48 The molecular formula of ethanol is C2H6O.
2.49 Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually
molecular.
Ionic: LiF, BaCl2, KCl
Molecular: SiCl4, B2H6, C2H4
2.50 Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually
molecular.
Ionic: NaBr, BaF2, CsCl.
Molecular: CH4, CCl4, ICl, NF3

2.57 (a) sodium chromate (h) phosphorus trifluoride
(b) potassium hydrogen phosphate (i) phosphorus pentafluoride
(c) hydrogen bromide (molecular compound) (j) tetraphosphorus hexoxide
(d) hydrobromic acid (k) cadmium iodide
(e) lithium carbonate (l) strontium sulfate
(f) potassium dichromate (m) aluminum hydroxide
(g) ammonium nitrite (n) sodium carbonate decahydrate
2.58 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
+
, Zn
2+
, Cd
2+
, and
Al
3+
.
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 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 chloride ion has a −1 charge, the Fe ion has a +2 charge.
The correct name is iron(II) chloride.
(d) potassium permanganate (e) cesium chlorate (f) hypoiodous acid
(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 (k) lithium nitride (l) sodium oxide
(m) This is an ionic compound in which the metal cation (Na
+
) has only one charge. The O2
2−
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.
(n) iron(III) chloride hexahydrate
2.59 (a) RbNO2 (b) K2S (c) NaHS (d) Mg3(PO4)2 (e) CaHPO4
(f) KH2PO4 (g) IF7 (h) (NH4)2SO4 (i) AgClO4 (j) BCl3
2.60 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

28
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), Ag
+
,
Zn
2+
, Cd
2+
, and Al
3+
.
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.
(b) 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.
(c) 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.
(d) Hydroiodic tells you that the anion of this binary acid is iodide, I
−
. The correct formula is HI(aq).
(e) Na is an alkali metal. It only forms a +1 cation. The polyatomic ion ammonium, NH4
+
, has a +1 charge
and the polyatomic ion phosphate, PO4
3−
, has a −3 charge. To balance the charge, you need 2 Na
+
cations. The correct formula is Na2(NH4)PO4.
(f) The Roman numeral II tells you that the Pb cation has a +2 charge. The polyatomic ion carbonate,
CO3
2−
, has a −2 charge. Since, the charges are numerically equal, no subscripts are necessary in the
formula. The correct formula is PbCO3.
(g) 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.
(h) 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.
(i) 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.
(j) The Roman numeral I tells you that the Hg cation has a +1 charge. However, this cation exists as
Hg2
2+
. Iodide has a −1 charge. You need two iodide ion to balance the +2 charge of Hg2
2+
. The
correct formula is Hg2I2.
(k) 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.
2.61 Uranium is radioactive. It loses mass because it constantly emits alpha (α) particles.
2.62 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.
2.63 The number of protons = 65 − 35 = 30. The element that contains 30 protons is zinc, Zn. There are two
fewer electrons than protons, so the charge of the cation is +2. The symbol for this cation is Zn
2+
.
2.64 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
−
.

2.65 (a) molecular, C3H8 (b) molecular, C2H2 (c) molecular, C2H6 (d) molecular, C6H6
empirical, C3H8 empirical, CH empirical, CH3 empirical, CH
2.66 NaCl is an ionic compound; it doesn’t form molecules.
2.67 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.
2.68 The species and their identification are as follows:
(a) SO2 molecule and compound (g) O3 element and molecule
(b) S8 element and molecule (h) CH4 molecule and compound
(c) Cs element (i) KBr compound
(d) N2O5 molecule and compound (j) S element
(e) O element (k) P4 element and molecule
(f) O2 element and molecule (l) LiF compound
2.69 (a) Species with the same number of protons and electrons will be neutral. A, F, G.
(b) Species with more electrons than protons will have a negative charge. B, E.
(c) Species with more protons than electrons will have a positive charge. C, D.
(d) A: 10
5B B: 14 3
7 N −
C: 39 +
19K D: 66 2+
30Zn E: 81
35Br−
F: 11
5B G: 19
9F
2.70 (a) Ne, 10 p, 10 n (b) Cu, 29 p, 34 n (c) Ag, 47 p, 60 n
(d) W, 74 p, 108 n (e) Po, 84 p, 119 n (f) Pu, 94 p, 140 n
2.71 When an anion is formed from an atom, you have the same number of protons attracting more electrons. The
electrostatic attraction is weaker, which allows the electrons on average to move farther from the nucleus. An
anion is larger than the atom from which it is derived. When a cation is formed from an atom, you have the
same number of protons attracting fewer electrons. The electrostatic attraction is stronger, meaning that on
average, the electrons are pulled closer to the nucleus. A cation is smaller than the atom from which it is
derived.
2.72 (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:
3 13 3 37 3
4 4
(3.04 10 cm) 1.177 10 cm
3 3
− −
= π = π × = ×
V r
The density of the nucleus can now be calculated.
23
37 3
3.82 10 g
1.177 10 cm
−
−
×
= = =
×
14 3
3.25 10 g/cm
m
V
×
d
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.

30
The mass of 11 electrons is:
28
26
9.1095 10 g
11 electrons 1.00205 10 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:
12
8
2
1 10 m 1 cm
186 pm 1.86 10 cm
1 pm 1 10 m
−
−
−
×
× × = ×
×
3 8 3 23 3
atom
4 4
(1.86 10 cm) 2.695 10 cm
3 3
− −
= π = π × = ×
V r
Velectrons = Vatom − Vnucleus = (2.695 × 10
−23
cm
3
) − (1.177 × 10
−37
cm
3
) = 2.695 × 10
−23
cm
3
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.
26
23 3
1.00205 10 g
2.695 10 cm
−
−
×
= = =
×
4 3
3.72 10 g/cm
m
V
−
×
d
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.
2.73 (a) This is an ionic compound. Prefixes are not used. The correct name is barium chloride.
(b) Iron has a +3 charge in this compound. The correct name is iron(III) oxide.
(c) NO2
−
is the nitrite ion. The correct name is cesium nitrite.
(d) 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.
2.74 (a) Ammonium is NH4
+
, not NH3
+
. The formula should be (NH4)2CO3.
(b) Calcium has a +2 charge and hydroxide has a −1 charge. The formula should be Ca(OH)2.
(c) Sulfide is S
2−
, not SO3
2−
. The correct formula is CdS.
(d) Dichromate is Cr2O7
2−
, not Cr2O4
2−
. The correct formula is ZnCr2O7.
2.75 Symbol 11
5B 54 2+
26Fe 31 3
15P − 196
79Au 222
86Rn
Protons 5 26 15 79 86
Neutrons 6 28 16 117 136
Electrons 5 24 18 79 86
Net Charge 0 +2 −3 0 0
2.76 (a) Ionic compounds are typically formed between metallic and nonmetallic elements.
(b) In general the transition metals, the actinides and lanthanides have variable charges.

2.77 (a) Li
+
, alkali metals always have a +1 charge in ionic compounds
(b) S
2−
(c) I
−
, halogens have a −1 charge in ionic compounds
(d) N
3−
(e) Al
3+
, aluminum always has a +3 charge in ionic compounds
(f) Cs
+
, alkali metals always have a +1 charge in ionic compounds
(g) Mg
2+
, alkaline earth metals always have a +2 charge in ionic compounds.
2.78 The symbol
23
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?
2.79 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.
2.80 Mercury (Hg) and bromine (Br2)
2.81 (a) Isotope 4
2He 20
10 Ne 40
18Ar 84
36Kr 132
54Xe
No. Protons 2 10 18 36 54
No. Neutrons 2 10 22 48 78
(b) neutron/proton ratio 1.00 1.00 1.22 1.33 1.44
The neutron/proton ratio increases with increasing atomic number.
2.82 H2, N2, O2, F2, Cl2, He, Ne, Ar, Kr, Xe, Rn
2.83 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.
2.84 They do not have a strong tendency to form compounds. Helium, neon, and argon are chemically inert.
2.85 Magnesium and strontium are also alkaline earth metals. You should expect the charge of the metal to be the
same (+2). MgO and SrO.
2.86 All isotopes of radium are radioactive. It is a radioactive decay product of uranium-238. Radium itself does
not occur naturally on Earth.
2.87 (a) Berkelium (Berkeley, CA); Europium (Europe); Francium (France); Scandium (Scandinavia);
Ytterbium (Ytterby, Sweden); Yttrium (Ytterby, Sweden).
(b) Einsteinium (Albert Einstein); Fermium (Enrico Fermi); Curium (Marie and Pierre Curie);
Mendelevium (Dmitri Mendeleev); Lawrencium (Ernest Lawrence).
(c) Arsenic, Cesium, Chlorine, Chromium, Iodine.

2.96 The change in energy is equal to the energy released. We call this ΔE. Similarly, Δm is the change in mass.
Because
2
=
E
m
c
, we have
3
11
2 8 2
1000 J
(1.715 10 kJ)
1 kJ
1.91 10 kg
(3.00 10 m/s)
−
× ×
Δ
Δ = = = × =
×
8
1.91 10 g
E
m
c
−
×
Note that we need to convert kJ to J so that we end up with units of kg for the mass.
2
2
1 kg m
1 J
s
⎛ ⎞
⋅
=
⎜ ⎟
⎜ ⎟
⎝ ⎠
We can add together the masses of hydrogen and oxygen to calculate the mass of water that should be
formed.
12.096 g + 96.000 = 108.096 g
The predicted change (loss) in mass is only 1.91 × 10
−8
g which is too small a quantity to measure. Therefore,
for all practical purposes, the law of conservation of mass is assumed to hold for ordinary chemical processes.
2.97 CH4, C2H6, and C3H8 each only have one structural formula.
C
H
H
H
H C
H
H
H
C
H
H
H C
H
H
H
C
H
H
C
H
H
H
C4H10 has two structural formulas.
C
H
H
H
C
H
H
C
H
C
H
H
H
H
C
H
H
H
C
H
C
H
H
H
C
H
H
H
C5H12 has three structural formulas.
C
H
H
H
C
H
H
C
H
C
H
C
H
H H
H
H
C
H
H C
H
C
H
C
C
H
H
H
H
H
H H H
C
H
H
H
C
C
C
H
H
H
C
H
H
H
H
H
H
2.98 (a) The volume of a sphere is
3
4
3
= π
V r
Volume is proportional to the number of nucleons. Therefore,
V ∝ A (mass number)

34
r
3
∝ A
r ∝ A
1/3
(b) Using the equation given in the problem, we can first solve for the radius of the lithium nucleus and
then solve for its volume.
r = r0A
1/3
r = (1.2 × 10
−15
m)(7)
1/3
r = 2.3 × 10
−15
m
3
4
3
= π
V r
15 3
4
(2.3 10 m)
3
−
= π × = 44 3
nucleus 5.1 10 m
−
×
V
(c) In part (b), the volume of the nucleus was calculated. Using the radius of a Li atom, the volume of a Li
atom can be calculated.
3 12 3 29 3
atom
4 4
(152 10 m) 1.47 10 m
3 3
− −
= π = π × = ×
V r
The fraction of the atom’s volume occupied by the nucleus is:
44 3
nucleus
29 3
atom
5.1 10 m
1.47 10 m
−
−
×
= =
×
15
3.5 10
V
V
−
×
Yes, this calculation shows that the volume of the nucleus is much, much smaller than the volume of the
atom, which supports Rutherford’s model of an atom.
2.99 Two different structural formulas for the molecular formula C2H6O are:
C
H
H
H
C O H
H
H
C
H
H
H
O C
H
H
H
In the second hypothesis of Dalton’s Atomic Theory, he states that in any compound, the ratio of the number
of atoms of any two of the elements present is either an integer or simple fraction. In the above two
compounds, the ratio of atoms is the same. This does not necessarily contradict Dalton’s hypothesis, but
Dalton was not aware of chemical bond formation and structural formulas.
2.100 (a) Ethane Acetylene
2.65 g C 4.56 g C
0.665 g H 0.383 g H
Let’s compare the ratio of the hydrogen masses in the two compounds. To do this, we need to start with
the same mass of carbon. If we were to start with 4.56 g of C in ethane, how much hydrogen would
combine with 4.56 g of carbon?
4.56 g C
0.665 g H 1.14 g H
2.65 g C
× =

We can calculate the ratio of H in the two compounds.
1.14 g
3
0.383 g
≈
This is consistent with the Law of Multiple Proportions which states that if two elements combine to
form more than one compound, the masses of one element that combine with a fixed mass of the other
element are in ratios of small whole numbers. In this case, the ratio of the masses of hydrogen in the
two compounds is 3:1.
(b) For a given amount of carbon, there is 3 times the amount of hydrogen in ethane compared to acetylene.
Reasonable formulas would be:
Ethane Acetylene
CH3 CH
C2H6 C2H2
2.101 (a) The following strategy can be used to convert from the volume of the Pt cube to the number of Pt
atoms.
cm
3
→ grams → atoms
3
3 22
21.45 g Pt 1 atom Pt
1.0 cm
1 cm 3.240 10 g Pt
−
× × =
×
22
6.6 10 Pt atoms
×
(b) Since 74 percent of the available space is taken up by Pt atoms, 6.6 × 10
22
atoms occupy the following
volume:
0.74 × 1.0 cm
3
= 0.74 cm
3
We are trying to calculate the radius of a single Pt atom, so we need the volume occupied by a single Pt
atom.
3
23 3
22
0.74 cm
volume Pt atom 1.12 10 cm /Pt atom
6.6 10 Pt atoms
−
= = ×
×
The volume of a sphere is 3
4
3
πr . Solving for the radius:
23 3 3
4
1.12 10 cm
3
−
= × = π
V r
r
3
= 2.67 × 10
−24
cm
3
r = 1.4 × 10
−8
cm
Converting to picometers:
8
12
0.01 m 1 pm
(1.4 10 cm)
1 cm 1 10 m
−
−
= × × × =
×
2
radius Pt atom 1.4 10 pm
×

36
2.102 The mass number is the sum of the number of protons and neutrons in the nucleus.
Mass number = number of protons + number of neutrons
Let the atomic number (number of protons) equal A. The number of neutrons will be 1.2A. Plug into the
above equation and solve for A.
55 = A + 1.2A
A = 25
The element with atomic number 25 is manganese, Mn.
2.103
2.104 The acids, from left to right, are chloric acid, nitrous acid, hydrocyanic acid, and sulfuric acid.
ANSWERS TO REVIEW OF CONCEPTS
Section 2.1 (p. 43) Yes, the ratio of atoms represented by B that combine with A in these two compounds is
(2/1):(5/2) or 4:5.
Section 2.3 (p. 50) (a) Hydrogen. The isotope is 1
1H .
(b) The electrostatic repulsion between the two positively charged protons would be too great
without the presence of neutrons.
Section 2.4 (p. 53) Chemical properties change more markedly across a period.
Section 2.6 (p. 59) (a) Mg(NO3)2 (b) Al2O3 (c) LiH (d) Na2S.
S N
B I

CHAPTER 3
MASS RELATIONSHIPS IN
CHEMICAL REACTIONS
Problem Categories
Biological: 3.29, 3.40, 3.72, 3.103, 3.109, 3.110, 3.113, 3.114, 3.117, 3.119.
Conceptual: 3.33, 3.34, 3.63, 3.64, 3.81, 3.82, 3.120, 3.123, 3.125, 3.148.
Descriptive: 3.70, 3.76, 3.78, 3.95, 3.96, 3.107, 3.121.
Environmental: 3.44, 3.69, 3.84, 3.109, 3.132, 3.138, 3.139, 3.141, 3.145.
Industrial: 3.28, 3.41, 3.42, 3.51, 3.67, 3.89, 3.91, 3.92, 3.94, 3.97, 3.108, 3.138, 3.139, 3.146, 3.147, 3.150.
Difficulty Level
Easy: 3.7, 3.8, 3.11, 3.14, 3.15, 3.16, 3.23, 3.24, 3.25, 3.51, 3.53, 3.65, 3.66, 3.67, 3.68, 3.72, 3.83, 3.100, 3.103,
3.118, 3.120, 3.125, 3.133, 3.134.
Medium: 3.5, 3.6, 3.12, 3.13, 3.17, 3.18, 3.19, 3.20, 3.21, 3.22, 3.26, 3.27, 3.28, 3.29, 3.30, 3.33, 3.39, 3.40, 3.41, 3.42,
3.43, 3.44, 3.45, 3.46, 3.47, 3.48, 3.49, 3.50, 3.52, 3.54, 3.59, 3.60, 3.63, 3.64, 3.69, 3.70, 3.71, 3.73, 3.74, 3.75, 3.76,
3.77, 3.78, 3.81, 3.82, 3.84, 3.85, 3.86, 3.89, 3.90, 3.91, 3.92, 3.93, 3.94, 3.101, 3.104, 3.105, 3.110, 3.111, 3.112,
3.114, 3.115, 3.116, 3.117, 3.119, 3.121, 3.124, 3.126, 3.127, 3.128, 3.129, 3.130, 3.131, 3.132, 3.140, 3.141, 3.142,
3.146, 3.147, 3.148, 3.152.
Difficult: 3.34, 3.95, 3.96, 3.97, 3.98, 3.99, 3.102, 3.106, 3.107, 3.108, 3.109, 3.113, 3.122, 3.123, 3.135, 3.136, 3.137,
3.138, 3.139, 3.143, 3.144, 3.145, 3.149, 3.150, 3.151, 3.153, 3.154, 3.155.
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.
It would seem that there are two unknowns in this problem, the fractional abundance of
6
Li and the fractional
abundance of
7
Li. However, these two quantities are not independent of each other; they are related by the
fact that they must sum to 1. Start by letting x be the fractional abundance of
6
Li. Since the sum of the two
abundance’s must be 1, we can write
Abundance
7
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
x = 0.075 corresponds to a natural abundance of
6
Li of 7.5 percent. The natural abundance of
7
Li is
(1 − x) = 0.925 or 92.5 percent.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
38
3.7
23
6.022 10 amu
The unit factor required is
1 g
⎛ ⎞
×
⎜ ⎟
⎜ ⎟
⎝ ⎠
23
1 g
13.2 amu
6.022 10 amu
= × =
×
23
? g 2.19 10 g
−
×
3.8
23
6.022 10 amu
The unit factor required is
1 g
⎛ ⎞
×
⎜ ⎟
⎜ ⎟
⎝ ⎠
23
6.022 10 amu
8.4 g =
1 g
×
= × 24
? amu 5.1 10 amu
×
3.11 In one year:
9 17
2 particles 3600 s 24 h 365 days
(6.5 10 people) 4.1 10 particles/yr
1 person each second 1 h 1 day 1 yr
× × × × × = ×
23
17
6.022 10 particles
4.1 10 particles/yr
×
= =
×
6
Total time 1.5 10 yr
×
3.12 The thickness of the book in miles would be:
23 16
0.0036 in 1 ft 1 mi
(6.022 10 pages) = 3.42 10 mi
1 page 12 in 5280 ft
× × × × ×
The distance, in miles, traveled by light in one year is:
8
12
365 day 24 h 3600 s 3.00 10 m 1 mi
1.00 yr 5.88 10 mi
1 yr 1 day 1 h 1 s 1609 m
×
× × × × × = ×
The thickness of the book in light-years is:
16
12
1 light-yr
(3.42 10 mi)
5.88 10 mi
× × =
×
3
5.8 10 light - yr
×
It will take light 5.8 × 10
3
years to travel from the first page to the last one!
3.13
23
6.022 10 S atoms
5.10 mol S
1 mol S
×
× = 24
3.07 10 S atoms
×
3.14 9
23
1 mol Co
(6.00 10 Co atoms) =
6.022 10 Co atoms
× ×
×
15
9.96 10 mol Co
−
×
3.15
1 mol Ca
77.4 g of Ca
40.08 g Ca
× = 1.93 mol 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.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS 39
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
197.0 g Au
= 15.3 mol Au =
1 mol Au
× 3
? g Au 3.01 10 g 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?
3.17 (a)
23
200.6 g Hg 1 mol Hg
1 mol Hg 6.022 10 Hg atoms
× =
×
22
3.331 10 g/Hg atom
−
×
(b)
23
20.18 g Ne 1 mol Ne
1 mol Ne 6.022 10 Ne atoms
× =
×
23
3.351 10 g/Ne atom
−
×
3.18 (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)
23
23
1 mol As 6.022 10 As atoms
and
1 mol As
6.022 10 As atoms
×
×
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
23
74.92 g As 1 mol As
1 mol As 6.022 10 As atoms
= × =
×
22
? g/As atom 1.244 10 g/As atom
−
×
(b) Follow same method as part (a).
23
58.69 g Ni 1 mol Ni
1 mol Ni 6.022 10 Ni atoms
= × =
×
23
? g/Ni atom 9.746 10 g/Ni atom
−
×
Check: Should the mass of a single atom of As or Ni be a very small mass?

40
3.19 12
23
1 mol Pb 207.2 g Pb
1.00 10 Pb atoms
1 mol Pb
6.022 10 Pb atoms
× × × =
×
10
3.44 10 g Pb
−
×
3.20 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
Avogadro's number is the key to the second conversion. We have
1 mol = 6.022 × 10
23
particles (atoms)
23
23
1 mol Cu 6.022 10 Cu atoms
and
1 mol Cu
6.022 10 Cu atoms
×
×
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
23
1 mol Cu 6.022 10 Cu atoms
3.14 g Cu
63.55 g Cu 1 mol Cu
×
= × × = 22
?atoms of Cu 2.98 10 Cu atoms
×
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 For hydrogen:
23
1 mol H 6.022 10 H atoms
1.10 g H
1.008 g H 1 mol H
×
× × = 23
6.57 10 H atoms
×
For chromium:
23
1 mol Cr 6.022 10 Cr atoms
14.7 g Cr
52.00 g Cr 1 mol Cr
×
× × = 23
1.70 10 Cr atoms
×
There are more hydrogen atoms than chromium atoms.
3.22 22
23
1 mol Pb 207.2 g Pb
2 Pb atoms = 6.881 10 g Pb
1 mol Pb
6.022 10 Pb atoms
−
× × ×
×
23 22
4.003 g He
(5.1 10 mol He) = 2.0 10 g He
1 mol He
− −
× × ×
2 atoms of lead have a greater mass than 5.1 × 10
−23
mol of helium.

3.23 Using the appropriate atomic masses,
(a) CH4 12.01 amu + 4(1.008 amu) = 16.04 amu
(b) NO2 14.01 amu + 2(16.00 amu) = 46.01 amu
(c) SO3 32.07 amu + 3(16.00 amu) = 80.07 amu
(d) C6H6 6(12.01 amu) + 6(1.008 amu) = 78.11 amu
(e) NaI 22.99 amu + 126.9 amu = 149.9 amu
(f) K2SO4 2(39.10 amu) + 32.07 amu + 4(16.00 amu) = 174.27 amu
(g) Ca3(PO4)2 3(40.08 amu) + 2(30.97 amu) + 8(16.00 amu) = 310.18 amu
3.24 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).
(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
3.25 To find the molar mass (g/mol), we simply divide the mass (in g) by the number of moles.
152 g
0.372 mol
= 409 g/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) = 30.068 g
The conversion factor needed is
2 6
2 6
1 mol C H
30.068 g C H
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:
23
6.022 10 ethane molecules
1 mol ethane
×

42
Let's complete the two conversions in one step.
grams of ethane → moles of ethane → number of ethane molecules
23
2 6 2 6
2 6
2 6 2 6
1 mol C H 6.022 10 C H molecules
0.334 g C H
30.068 g C H 1 mol C H
×
= × ×
2 6
? molecules of C H
= 6.69 × 10
21
C2H6 molecules
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
23
1 mol glucose 6.022 10 molecules glucose 6 C atoms
1.50 g glucose
180.2 g glucose 1 mol glucose 1 molecule glucose
×
× × ×
= 3.01 × 10
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
glucose as C atoms, so the number of O atoms = 3.01 × 10
22
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
C atoms, so the number of H atoms = 2(3.01 × 10
22
atoms) = 6.02 × 10
22
H atoms.
3.28 Strategy: We are asked to solve for the number of N, C, O, and H atoms in 1.68 × 10
4
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?
Solution: Let's first calculate the number of N atoms in 1.68 × 10
4
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.
23
4 1 mol urea 6.022 10 urea molecules 2 N atoms
= (1.68 10 g urea)
60.062 g urea 1 mol urea 1 molecule urea
×
× × × ×
?atoms of N
= 3.37 × 10
26
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.
Check: Does the answer seem reasonable? We have 1.68 × 10
4
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.
26 1 C atom
(3.37 10 N atoms)
2 N atoms
= × × = 26
? atoms of C 1.69 10 C atoms
×
26 1 O atom
(3.37 10 N atoms)
2 N atoms
= × × = 26
? atoms of O 1.69 10 O atoms
×
26 4 H atoms
(3.37 10 N atoms)
2 N atoms
= × × = 26
? atoms of H 6.74 10 H atoms
×
3.29 The molar mass of C19H38O is 282.5 g.
23
12 1 mol 6.022 10 molecules
1.0 10 g
282.49 g 1 mol
− ×
× × × = 9
2.1 10 molecules
×
Notice that even though 1.0 × 10
−12
g is an extremely small mass, it still is comprised of over a billion
pheromone molecules!
3.30
1.00 g
Mass of water = 2.56 mL = 2.56 g
1.00 mL
×
Molar mass of H2O = (16.00 g) + 2(1.008 g) = 18.016 g/mol
23
2 2
2
2 2
1 mol H O 6.022 10 molecules H O
= 2.56 g H O
18.016 g H O 1 mol H O
×
× ×
2
? H O molecules
= 8.56 × 10
22
molecules
3.33 Since there are only two isotopes of carbon, there are only two possibilities for CF4
+
.
(molecular mass 88 amu) and (molecular mass 89 amu)
12 19 13 19
6 9 4 6 9 4
C F C F
+ +
There would be two peaks in the mass spectrum.
3.34 Since there are two hydrogen isotopes, they can be paired in three ways:
1
H-
1
H,
1
H-
2
H, and
2
H-
2
H. There
will then be three choices for each sulfur isotope. We can make a table showing all the possibilities (masses
in amu):
32
S
33
S
34
S
36
S
1
H2 34 35 36 38
1
H
2
H 35 36 37 39
2
H2 36 37 38 40
There will be seven peaks of the following mass numbers: 34, 35, 36, 37, 38, 39, and 40.
Very accurate (and expensive!) mass spectrometers can detect the mass difference between two
1
H and one
2
H. How many peaks would be detected in such a “high resolution” mass spectrum?

44
3.39 Molar mass of SnO2 = (118.7 g) + 2(16.00 g) = 150.7 g
118.7 g/mol
100%
150.7 g/mol
= × =
%Sn 78.77%
(2)(16.00 g/mol)
100%
150.7 g/mol
= × =
%O 21.23%
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:
12.01 g/mol
%C 100%
119.4 g/mol
= × = 10.06%
1.008 g/mol
%H 100%
119.4 g/mol
= × = 0.8442%
3(35.45) g/mol
%Cl 100%
119.4 g/mol
= × = 89.07%
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)
(9)(12.01 g/mol)
100%
134.17 g/mol
= × =
%C 80.56%
(10)(1.008 g/mol)
100%
134.17 g/mol
= × =
%H 7.51%
16.00 g/mol
100%
134.17 g/mol
= × =
%O 11.93%
(b)
23
9 10 9 10
9 10
9 10 9 10
1 mol C H O 6.022 10 molecules C H O
0.469 g C H O
134.17 g C H O 1 mol C H O
×
× ×
= 2.11 × 10
21
molecules C9H10O
3.42 Compound Molar mass (g) N% by mass
(a) (NH2)2CO 60.06
2(14.01 g)
100% = 46.65%
60.06 g
×
(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.
3.43 Assume you have exactly 100 g of substance.
C
1 mol C
44.4 g C 3.697 mol C
12.01 g C
= × =
n
H
1 mol H
6.21 g H 6.161 mol H
1.008 g H
= × =
n
S
1 mol S
39.5 g S 1.232 mol S
32.07 g S
= × =
n
O
1 mol O
9.86 g O 0.6163 mol O
16.00 g O
= × =
n
Thus, we arrive at the formula C3.697H6.161S1.232O0.6163. Dividing by the smallest number of moles (0.6163
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 162.28 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.
1 mol C
= 19.8 g C = 1.649 mol C
12.01 g C
×
C
n
1 mol H
= 2.50 g H = 2.480 mol H
1.008 g H
×
H
n
1 mol N
= 11.6 g N = 0.8280 mol N
14.01g N
×
N
n
1 mol O
= 66.1 g O = 4.131 mol O
16.00 g O
×
O
n
Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript. The
formula is C1.649H2.480N0.8280O4.131. Dividing the subscripts by 0.8280 gives the empirical
formula, C2H3NO5.

46
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 120 g
1
= ≈
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.
1 mol C
(0.198) (120 g) 1.98 mol C
12.01 g C
= × × = ≈
C 2 mol C
n
1 mol H
(0.0250) (120 g) 2.98 mol H
1.008 g H
= × × = ≈
H 3 mol H
n
1 mol N
(0.116) (120 g) 0.994 mol N
14.01 g N
= × × = ≈
N 1 mol N
n
1 mol O
(0.661) (120 g) 4.96 mol O
16.00 g O
= × × = ≈
O 5 mol O
n
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.
3.45 2 3
2 3
2 3 2 3
1 mol Fe O 2 mol Fe
24.6 g Fe O
159.7 g Fe O 1 mol Fe O
× × = 0.308 mol Fe
3.46 Using unit factors we convert:
g of Hg → mol Hg → mol S → g S
1 mol Hg 1 mol S 32.07 g S
246 g Hg
200.6 g Hg 1 mol Hg 1 mol S
= × × × =
? g S 39.3 g S
3.47 The balanced equation is: 2Al(s) + 3I2(s) ⎯⎯
→ 2AlI3(s)
Using unit factors, we convert: g of Al → mol of Al → mol of I2 → g of I2
2 2
2
3 mol I 253.8 g I
1 mol Al
20.4 g Al
26.98 g Al 2 mol Al 1 mol I
× × × = 2
288 g I

3.48 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.
The percent by mass of fluorine in tin(II) fluoride, is calculated as follows:
2
2
mass of F in 1 mol SnF
mass % F 100%
molar mass of SnF
= ×
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)
1 mol H
2.1 g H 2.08 mol H
1.008 g H
× =
1 mol O
65.3 g O 4.081 mol O
16.00 g O
× =
1 mol S
32.6 g S 1.017 mol S
32.07 g S
× =
This gives the formula H2.08S1.017O4.081. Dividing by 1.017 gives the empirical formula, H2SO4.
(b)
1 mol Al
20.2 g Al 0.7487 mol Al
26.98 g Al
× =
1 mol Cl
79.8 g Cl 2.251 mol Cl
35.45 g Cl
× =
This gives the formula, Al0.7487Cl2.251. Dividing by 0.7487 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

48
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
C
1 mol C
40.1 g C 3.339 mol C
12.01 g C
= × =
n
H
1 mol H
6.6 g H 6.55 mol H
1.008 g H
= × =
n
O
1 mol O
53.3 g O 3.331 mol O
16.00 g O
= × =
n
Thus, we arrive at the formula C3.339H6.55O3.331, 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 (3.331).
3.339
1
3.331
≈
C :
6.55
2
3.331
≈
H :
3.331
1
3.331
=
O :
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:
C
1 mol C
18.4 g C 1.532 mol C
12.01 g C
= × =
n
N
1 mol N
21.5 g N 1.535 mol N
14.01 g N
= × =
n
K
1 mol K
60.1 g K 1.537 mol K
39.10 g K
= × =
n
Dividing by the smallest number of moles (1.532 mol) gives the empirical formula, KCN.
3.51 The molar mass of CaSiO3 is 116.17 g/mol.
40.08 g
%Ca
116.17 g
= = 34.50%
28.09 g
%Si
116.17 g
= = 24.18%
(3)(16.00 g)
%O
116.17 g
= = 41.32%
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 13.018 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.
molar mass
1 (integer values)
≥

In this case,
molar mass 78 g
6
= ≈
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.
C
1 mol C
35.51 g C 2.9567 mol C
12.01 g C
= × =
n
H
1 mol H
4.77 g H 4.732 mol H
1.008 g H
= × =
n
O
1 mol O
37.85 g O 2.3656 mol O
16.00 g O
= × =
n
N
1 mol N
8.29 g N 0.5917 mol N
14.01 g N
= × =
n
Na
1 mol Na
13.60 g Na 0.59156 mol Na
22.99 g Na
= × =
n
Thus, we arrive at the formula C2.9567H4.732O2.3656N0.5917Na0.59156, 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.
2.9567
= 4.9981 5
0.59156
≈
C :
4.732
= 7.999 8
0.59156
≈
H :
2.3656
= 3.9989 4
0.59156
≈
O :
0.5917
= 1.000
0.59156
N :
0.59156
= 1
0.59156
Na :
This gives us the empirical formula for MSG, C5H8O4NNa.
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)
≥

50
In this case,
molar mass 169 g
1
= ≈
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.
C
1 mol C
(0.3551) (169 g) 5.00 mol C
12.01 g C
= × × =
n
H
1 mol H
(0.0477) (169 g) 8.00 mol H
1.008 g H
= × × =
n
O
1 mol O
(0.3785) (169 g) 4.00 mol O
16.00 g O
= × × =
n
N
1 mol N
(0.0829) (169 g) 1.00 mol N
14.01 g N
= × × =
n
Na
1 mol Na
(0.1360) (169 g) 1.00 mol Na
22.99 g Na
= × × =
n
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 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
3.60 The balanced equations are as follows:
(a) 2N2O5 → 2N2O4 + O2 (h) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
(b) 2KNO3 → 2KNO2 + O2 (i) CO2 + 2KOH → K2CO3 + H2O
(c) NH4NO3 → N2O + 2H2O (j) CH4 + 2O2 → CO2 + 2H2O
(d) NH4NO2 → N2 + 2H2O (k) Be2C + 4H2O → 2Be(OH)2 + CH4
(e) 2NaHCO3 → Na2CO3 + H2O + CO2 (l) 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
(f) P4O10 + 6H2O → 4H3PO4 (m) S + 6HNO3 → H2SO4 + 6NO2 + 2H2O
(g) 2HCl + CaCO3 → CaCl2 + H2O + CO2 (n) 2NH3 + 3CuO → 3Cu + N2 + 3H2O

3.63 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.
2
2 mol CO
3.60 mol CO
2 mol CO
× = 2
3.60 mol CO
3.66 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 Q 1 mole SiCl4. From this relationship, we have two conversion factors.
2 4
4 2
2 mol Cl 1 mol SiCl
and
1 mol SiCl 2 mol Cl
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:
2
4
4
2 mol Cl
0.507 mol SiCl
1 mol SiCl
= × =
2 2
? mol Cl reacted 1.01mol Cl
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)

52
2
2 3
3
3 mol H
? mol H 6.0 mol NH
2 mol NH
= × = 2
9.0 mol H
2
2 3
3
1 mol N
? mol N 6.0 mol NH
2 mol NH
= × = 2
3.0 mol N
3.68 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)
2
2 4 10 2
4 10
8 mol CO
? mol CO 5.0 mol C H 20 mol CO
2 mol C H
= × = = 1
2
2.0 10 mol CO
×
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.
7 2
2
2 2
1 ton-mol SO 1 ton-mol S 32.07 ton S
(2.6 10 tons SO )
64.07 ton SO 1 ton-mol SO 1 ton-mol S
× × × × = 7
1.3 10 tons S
×
3.70 (a) 2NaHCO3 ⎯ →
⎯ Na2CO3 + H2O + CO2
(b) Molar mass NaHCO3 = 22.99 g + 1.008 g + 12.01 g + 3(16.00 g) = 84.008 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.
3 3
2
2
2 2 3
2 mol NaHCO 84.008 g NaHCO
1 mol CO
= 20.5 g CO
44.01 g CO 1 mol CO 1 mol NaHCO
× × ×
3
mass NaHCO
= 78.3 g NaHCO3
3.71 The balanced equation shows a mole ratio of 1 mole HCN : 1 mole KCN.
1 mol KCN 1 mol HCN 27.03 g HCN
0.140 g KCN
65.12 g KCN 1 mol KCN 1 mol HCN
× × × = 0.0581 g HCN
3.72 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 Q 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:

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