CHAPTER
The Mole Concept
8
Section 8.1 Avogadro’s Number
2.
4.
(a)
(c)
Element
C
Se
Average Mass
12.01 amu
78.96 amu
Element
(b) S
(d) I
Average Mass
32.07 amu
126.90 amu
(a)
(c)
Element
C
Se
Mass
12.01 g
78.96 g
Element
(b) S
(d) I
Mass
32.07 g
126.90 g
Section 8.2 Mole Calculations I
6.
(a)
(b)
1 mol Si = 6.02 × 102 3 atoms
1 mol SiH4 = 6.02 × 102 3 molecules
8.
(a)
(b)
6.02 × 102 3 atoms S = 1 mol S
6.02 × 102 3 molecules SO2 = 1 mol SO2
10.
(a)
0.250 mol Ca
×
6.02 × 102 3 atoms Ca
1 mol Ca
(b)
0.500 mol F2
×
6.02 × 102 3 molecules F2
1 mol F2
(c)
0.750 mol CaF2 ×
= 1.51 × 102 3 atoms Ca
= 3.01 × 102 3 molecules F2
6.02 × 102 3 formula units CaF2
1 mol CaF2
= 4.52 × 1023 formula units CaF2
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The Mole Concept
45
12.
1 mol Fe
6.02 × 102 3 atoms Fe
(a)
2.50 × 102 2 atoms Fe ×
(b)
5.00 × 102 3 molecules CO2 ×
= 0.0415 mol Fe
1 mol CO2
6.02 × 102 3 molecules CO2
= 0.831 mol CO2
(c)
7.50 × 102 4 formula units FeCO3 ×
1 mol FeCO3
23 6.02 × 10 formula units FeCO
3
= 12.5 mol FeCO3
Section 8.3 Molar Mass
14.
16.
Element
(a) As
(b) Ge
(c) O3
(d) P4
(a)
(b)
(c)
Compound
BrI
Br3 O8
C 3 H5 (OH)3
(d) C 3 H5 O3 (NO2 )3
Molar Mass
74.92 g/mol
72.61 g/mol
3(16.00 g O) = 48.00 g/mol
4(30.97 g P) = 123.88 g/mol
Molar Mass
79.90 g Br + 126.90 g I
= 206.80 g/mol
3(79.90 g Br) + 8(16.00 g O) = 367.70 g/mol
3(12.01 g) + 5(1.01 g) + 3(16.00 g + 1.01 g)
= 92.11 g/mol
3(12.01 g) + 5(1.01 g) + 3(16.00 g) +
3(14.01 g + 16.00 g + 16.00 g) = 227.11 g/mol
Section 8.4 Mole Calculations II
18.
(a)
MM of Kr = 83.80 g/mol
1 mol Kr
1.21 × 102 4 atoms Kr × 6.02 × 102 3 atoms Kr
(b)
×
83.80 g Kr
1 mol Kr = 168 g Kr
MM of N2 O = 2(14.01 g N) + 16.00 g O = 44.02 g/mol
1 mol N2 O
44.02 g N2 O
6.33 × 102 2 molecules N2 O × 6.02 × 102 3 molecules N O × 1 mol N O
2
2
= 4.63 g N2 O
46
Chapter 8
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(c)
MM of Mg(ClO4 )2
= 24.31 g Mg + 2(35.45 g Cl) + 8(16.00 g O)
= 223.21 g/mol
1 mol Mg(ClO4 )2
4.17 × 102 1 formula units Mg(ClO4 )2 × 6.02 × 102 3 formula units Mg(ClO )
4 2
×
20.
(a)
223.21 g Mg(ClO4 )2
1 mol Mg(ClO4 )2 = 1.55 g Mg(ClO4 )2
MM of Pt = 195.08 g/mol
1 mol Pt
6.02 × 102 3 atoms Pt
7.57 g Pt × 195.08 g Pt ×
= 2.34 × 102 2 atoms Pt
1 mol Pt
(b)
MM of C2 H6 = 2(12.01 g C) + 6(1.01 g H) = 30.08 g/mol
1 mol C2 H6
6.02 × 102 3 molecules C2 H6
3.88 g C2 H6 × 30.08 g C H ×
1 mol C2 H6
2 6
= 7.77 × 102 2 molecules C2 H6
(c)
MM of AlCl3 = 26.98 g Al + 3(35.45 g Cl) = 133.33 g/mol
1 mol AlCl3
6.02 × 102 3 formula units AlCl3
0.152 g AlCl3 × 133.33 g AlCl ×
1 mol AlCl3
3
= 6.86 × 102 0 formula units AlCl3
22.
(a)
MM of CH4 = 12.01 g C + 4(1.01 g H) = 16.05 g/mol
16.05 g
1 mol CH4
(b)
1 mol CH4
6.02 × 102 3 molecules
= 2.67 × 10–23 g/molecule
MM of NH3 = 14.01 g N + 3(1.01 g H) = 17.04 g/mol
17.04 g
1 mol NH3
(c)
×
×
1 mol NH3
6.02 × 102 3 molecules
= 2.83 × 10–23 g/molecule
MM of SO3 = 32.07 g S + 3(16.00 g O) = 80.07 g/mol
80.07 g
1 mol SO3
×
1 mol SO3
6.02 × 102 3 molecules
= 1.33 × 10–22 g/molecule
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The Mole Concept
47
(d) MM of NO2 = 14.01 g N + 2(16.00 g O) = 46.01 g/mol
46.01 g
1 mol NO2
×
1 mol NO2
6.02 × 102 3 molecules
= 7.64 × 10–23 g/molecule
Section 8.5 Molar Volume
24.
STP conditions: 273 K and 1 atm
26.
(a)
MM of Xe = 131.29 g/mol
Density of Xe (at STP):
(b)
MM of F2 = 2(19.00 g F) = 38.00 g/mol
Density of F2 (at STP):
(c)
131.29 g 1 mol
1 mol × 22.4 L = 5.86 g/L
38.00 g
1 mol
×
1 mol
22.4 L
= 1.70 g/L
MM of C3 H8 = 3(12.01 g C) + 8(1.01 g H) = 44.11 g/mol
Density of C3 H8 (at STP):
44.11 g
1 mol
×
1 mol
22.4 L
= 1.97 g/L
(d) MM of C4 H1 0 = 4(12.01 g C) + 10(1.01 g H) = 58.14 g/mol
28.
Density of C4 H1 0 (at STP):
58.14 g
1 mol
×
1 mol
22.4 L = 2.60 g/L
(a)
MM of ozone:
2.14 g
22.4 L
×
L
mol = 47.9 g/mol
(b)
MM of silane:
1.43 g
22.4 L
×
L
mol = 32.0 g/mol
(c)
MM of nitric oxide:
1.34 g
22.4 L
×
L
mol = 30.0 g/mol
(d) MM of Freon–22:
30.
Gas
48
3.86 g
22.4 L
×
L
mol = 86.5 g/mol
Molecules
Mass
Volume at STP
nitrogen, N2
6.02 × 102 3
28.02 g
22.4 L
hydrogen, H2
6.02 × 102 3
2.02 g
22.4 L
ammonia, NH3
6.02 × 102 3
17.04 g
22.4 L
Chapter 8
2014 © Pearson Education, Inc.
Section 8.6 Mole Calculations III
32.
(a)
1 mol CH4 22.4 L CH4
2.22 × 102 2 molecules CH4 × 6.02 × 102 3 molecules CH × 1 mol CH
4
4
= 0.826 L CH4
(b)
1 mol C2 H6
22.4 L C2 H6
4.18 × 102 4 molecules C2 H6 × 6.02 × 102 3 molecules C H × 1 mol C H
2 6
2 6
= 156 L C2 H6
34.
(a)
MM of C3 H8 = 3(12.01 g C) + 8(1.01 g H) = 44.11 g/mol
1 mol C3 H8
5.42 × 102 2 molecules C3 H8 × 6.02 × 102 3 molecules C H ×
3 8
44.11 g C3 H8
1 mol C3 H8
= 3.97 g C3 H8
(b)
MM of C4 H1 0 = 4(12.01 g C) + 10(1.01 g H) = 58.14 g/mol
1 mol C4 H1 0
1.82 × 102 3 molecules C4 H1 0 × 6.02 × 102 3 molecules C H
×
4 10
58.14 g C4 H1 0
1 mol C4 H1 0
= 17.6 g C4 H1 0
36.
(a)
MM of CO = 12.01 g C + 16.00 g O = 28.01 g/mol
0.150 g CO
×
1 mol CO
28.01 g CO
×
6.02 × 102 3 molecules H2
1 mol CO
= 3.22 × 102 1 molecules CO
(b)
MM of NO = 14.01 g N + 16.00 g O = 30.01 g/mol
2.75 g NO
1 mol NO
× 30.01 g NO
×
6.02 × 102 3 molecules NO
1 mol NO
= 5.52 × 102 2 molecules NO
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The Mole Concept
49
38.
Gas
Molecules
Mass
Volume at STP
ozone, O3
2.50 × 102 2
1.99 g
0.930 L
carbon dioxide, CO2
2.50 × 102 2
1.83 g
0.931 L
carbon monoxide, CO
2.50 × 102 2
1.16 g
0.930 L
Section 8.7 Percent Composition
40.
The law of definite composition states that a compound always contains the
same elements in the same proportion by mass. Thus, the percent oxygen by
mass in a liter of water is 89%.
42.
MM of C1 4H1 0O4
= 14(12.01 g C) + 10(1.01 g H) + 4(16.00 g O)
= 168.14 g C + 10.10 g H + 64.00 g O
= 242.24 g/mol
168.14 g C
242.24 g C7 H6 O3 × 100% = 69.41% C
10.10 g H
242.24 g C7 H6 O3 × 100% = 4.17% H
64.00 g O
242.24 g C7 H6 O3 × 100% = 26.42% O
44.
MM NaC5 H8 NO4
= 22.99 g Na + 5(12.01 g C) + 8(1.01 g H) + 14.01 g N + 4(16.00 g O)
= 22.99 g Na + 60.05 g C + 8.08 g H + 14.01 g N + 64.00 g O
= 169.13 g/mol
22.99 g Na
169.13 g NaC5 H8 NO4 × 100% = 13.59% Na
60.05 g C
169.13 g NaC5 H8 NO4 × 100% = 35.51% C
8.08 g H
169.13 g NaC5 H8 NO4 × 100% = 4.78% H
14.01 g N
169.13 g NaC5 H8 NO4 × 100% = 8.284% N
64.00g O
169.13 g NaC5 H8 NO4 × 100% = 37.84% O
50
Chapter 8
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46.
MM of C5 5H7 0MgN4 O6
= 55(12.01 g C) + 70(1.01 g H) + 24.31 g Mg + 4(14.01 g N) + 6(16.00 g O)
= 660.55 g C + 70.7 g H + 24.31 g Mg + 56.04 g N + 96.00 g O
= 907.6 g/mol
660.55 g C
907.6 g C5 5H7 0MgN4 O6 × 100% = 72.78% C
70.7 g H
907.6 g C5 5H7 0MgN4 O6 × 100% = 7.79% H
24.31 g Mg
907.6 g C5 5H7 0MgN4 O6 × 100% = 2.678% Mg
56.04 g N
907.6 g C5 5H7 0MgN4 O6 × 100% = 6.175% N
96.00 g O
907.6 g C5 5H7 0MgN4 O6 × 100% = 10.58% O
Section 8.8 Empirical Formula
48.
1 mol Ni
0.500 g Ni × 58.69 g Ni = 0.00852 mol Ni
0.704 g NixOy – 0.500 g Ni = 0.204 g O
1 mol O
0.204 g O × 16.00 g O = 0.0128 mol O
Ni 0.00852 O 0.0128
0.00852 0.00852
50.
1.435 g Hg ×
= Ni1.00O1.50
The empirical formula is Ni2 O3 .
1 mol Hg
200.59 g Hg = 0.007154 mol Hg
1.550 g HgxOy – 1.435 g Hg = 0.115 g O
0.115 g O ×
1 mol O
16.00 g O = 0.00719 mol O
Hg 0.007154 O 0.00719
0.007154
0.007154
= Hg1.000O1.01
The empirical formula is HgO.
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The Mole Concept
51
52.
1 mol Ti
47.88 g Ti = 0.0149 mol Ti
×
0.715 g Ti
2.836 g TixCly – 0.715 g Ti = 2.121 g Cl
×
2.121 g Cl
1 mol Cl
35.45 g Cl
Ti 0.0149 Cl 0.05983
54.
= Ti1.00Cl4.02
0.0149
0.0149
68.0 g V
×
1 mol V
50.94 g V
32.0 g O
×
1 mol O
16.00 g O = 2.00 mol O
V 1.33
1.33
56.
O 2.00
1.33
89.7 g Bi ×
1 mol Bi
208.98 g Bi = 0.429 mol Bi
10.3 g O ×
1 mol O
16.00 g O = 0.644 mol O
0.429
0.429
= Bi1.00O1.50
30.7 g C ×
1 mol C
12.01 g C
7.74 g H ×
1 mol H
1.01 g H
20.5 g O ×
1 mol O
16.01 g O = 1.28 mol O
×
1 mol S
32.07 g S = 1.28 mol S
41.0 g S
52
Chapter 8
1.28
1.28
The empirical formula is V2 O3 .
The empirical formula is Bi2 O3 .
= 2.56 mol C
= 7.66 mol H
C 2.56 H 7.66 O 1.28 S 1.28
1.28
The empirical formula is TiCl4 .
= 1.33 mol V
= V1.00O1.50
Bi 0.429 O 0.644
58.
= 0.05983 mol Cl
1.28
= C2.00H5.98O1.00S1.00
The empirical formula is C2 H6 OS.
2014 © Pearson Education, Inc.
Section 8.9 Molecular Formula
60.
MM of C1 0H1 2NO = 10(12.01 g C) + 12(1.01 g H) + 14.01 g N + 16.00 g O
= 120.10 g C + 12.12 g H + 14.01 g N + 16.00 g O
= 162.23 g/mol
Quinine:
(C1 0H1 2NO)n
325 g/mol
=
C 1 0H1 2NO
162.23 g/mol
n≈2
The molecular formula of quinine is (C1 0H1 2NO)2 or C2 0H2 4N2 O2 .
62.
MM of C3 H5 O2
= 3(12.01 g C) + 5(1.01 g H) + 2(16.00 g O)
= 36.03 g C + 5.05 g H + 32.00 g O
= 73.08 g/mol
(C3 H5 O2 )n
147 g/mol
=
C 3 H5 O2
73.08 g/mol
Adipic acid:
n≈2
The molecular formula of adipic acid is (C3 H5 O2 )2 or C6 H1 0O4 .
64.
Empirical Formula
54.5 g C
×
1 mol C
12.01 g C
1 mol H
9.15 g H × 1.01 g H
36.3 g O
×
2.27
= 9.06 mol H
1 mol O
16.00 g O = 2.27 mol O
C 4.54 H 9.06 O 2.27
2.27
= 4.54 mol C
2.27
= C2.00H3.99O1.00
The empirical formula is C2 H4 O.
Molecular Formula
MM of C2 H4 O = 2(12.01 g C) + 4(1.01 g H) + 16.00 g O = 44.06 g/mol
Dioxane:
(C2 H4 O)n
88 g/mol
C 2 H4 O = 44.06 g/mol
n≈2
The molecular formula of dioxane is (C2 H4 O)2 or C4 H8 O2 .
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The Mole Concept
53
66.
Empirical Formula
1 mol Hg
× 200.59 g Hg
85.0 g Hg
15.0 g Cl
1 mol Cl
× 35.45 g Cl
Hg 0.424 Cl 0.423
0.423
0.423
= 0.424 mol Hg
= 0.423 mol Cl
= Hg1.00Cl1.00
The empirical formula is HgCl.
Molecular Formula
MM of HgCl = 200.59 g Hg + 35.45 g Cl = 236.04 g/mol HgCl
Mercurous chloride:
(HgCl)n
470 g/mol
=
HgCl
236.04 g/mol
n≈2
The molecular formula of mercurous chloride is (HgCl)2 or Hg2 Cl2 .
68.
Empirical Formula
40.0 g C ×
1 mol C
12.01 g C = 3.33 mol C
6.72 g H ×
1 mol H
1.01 g H = 6.65 mol H
53.3 g O ×
1 mol O
16.00 g O = 3.33 mol O
C 3.33 H 6.65 O 3.33
3.33
3.33
3.33
= C1.00H2.00O1.00
The empirical formula is CH2 O.
Molecular Formula
MM of CH2 O = 12.01 g C + 2(1.01 g H) + 16.00 g O = 30.03 g/mol
Allose:
(CH2 O)n
180 g/mol
=
CH2 O
30.03 g/mol n ≈ 6
The molecular formula of allose is (CH2 O)6 or C6 H1 2O6 .
54
Chapter 8
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General Exercises
1 mol Cu 63.55 g Cu
70.
0.0001 g Cu ×
72.
1 mol furry moles ×
×
6.02 × 102 3 atoms Cu
= 9 × 101 7 atoms Cu
1 mol Cu
6.02 × 102 3 furry moles
17 cm
1 m
× 1 furry mole × 100 cm ×
1 mol furry moles
1 km
1000 m
= 1.0 × 102 0 km
Note: 9.5 × 101 2 km (1 light year) < < < 1.0 × 102 0 km (1 mol of furry moles).
Thus, a mole of furry moles is about 10,000,000 times longer than the
distance light travels in a year.
74.
0.500 g Ge ×
1 mol Ge
72.61 g Ge
= 0.00689 mol Ge
1.456 g GexCly – 0.500 g Ge = 0.956 g Cl
0.956 g Cl ×
1 mol Cl
35.45 g Cl
Ge 0.00689 Cl 0.0270
0.00689
76.
0.00689
= 0.0270 mol Cl
= Ge1.00Cl3.92
The empirical formula is GeCl4 .
MM of C2 H5 OH = 2(12.01 g C) + 6(1.01 g H) + 16.00 g O = 46.08 g/mol
1 mol C2 H5 OH
46.08 g C2 H5 OH
1 molecule C2 H5 OH × 6.02 × 102 3 molecules C H OH × 1 mol C H OH ×
2 5
2 5
1 mL C2 H5 OH
0.789 g C2 H5 OH
= 9.70 × 10–23 mL C2 H5 OH
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The Mole Concept
55
78.
MM of C6 H1 2O6 = 6(12.01 g C) + 12(1.01 g H) + 6(16.00 g O) = 180.18 g/mol
1 mol C6 H1 2O6
6 mol C
6.02 × 102 3 atoms C
1.00 g C6 H1 2O6 × 180.18 g C H O × 1 mol C H O ×
1 mol C
6 12 6
6 12 6
= 2.00 × 102 2 atoms C
80.
1 mol Fe2 O3
159.70 g Fe2 O3
1 mol Fe
10.0 g Fe × 55.85 g Fe ×
× 1 mol Fe O
= 14.3 g Fe2 O3
2 mol Fe
2 3
Challenge Exercises
82.
1 atom Al
(0.255 nm)3 ×
⎛⎜ 1 × 109 nm⎞⎟ 3
1 m
⎝
⎠
×
⎛⎜ 1 m ⎞⎟ 3 × 1 cm3 ×
2.70 g Al
⎝ 100 cm⎠
=
6.03 × 102 3 atoms Al
mol Al
Online Exercises
84.
56
1 Faraday = 1 mol of electrons = 6.02 × 102 3 electrons
Chapter 8
2014 © Pearson Education, Inc.
26.98 g Al
1 mol Al