Chapter 12
Answers to Questions
1.
(a) nitrogen dioxide
(b) hydrogen sulfide (c) hydrogen chloride
2.
(a) hydrogen cyanide
3.
X = sulfur, Y = oxygen, Z = sulfur dioxide
4.
X = carbon monoxide, Y = oxygen, Z = carbon dioxide
(b) dinitrogen oxide
(c) ammonia
5.
Strategy
Press/mol/temp of gas → vol gas
V=
Relationship
V = nRT/P
nRT
1.00 mol (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(743 K)
=
= 0.66 L
P
(9.3 × 103 kPa)
6.
Strategy
Press/mol/temp of gas → vol gas
V=
Relationship
V = nRT/P
nRT
1.00 mol (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(263 K)
=
= 3.6 × 103 L
P
(0.60 kPa)
7.
Strategy
Press/vol/temp of O2 → mol O2
Mol O2 → mass O2
n=
Relationship
n = PV/RT
1 mol O2 ≡ 32.0 g
PV
100 kPa (25.0 L)
=
= 1.01 mol
RT (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(298 K)
32.0 g
= 32.3 g O2
1 mol
Mass O2 = 1.01 mol O2 ×
8.
Strategy
Mass CO2 → mol CO2
Mol/press/temp CO2 → vol CO2
Mol CO2 = 30.0 g CO2 ×
V=
Relationship
1 mol CO2 ≡ 44.0 g
V = nRT/P
1 mol
= 0.682 mol CO2
44.0 g
nRT
0.682 mol (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(298 K)
=
= 16.9 L
P
(100 kPa)
9.
Strategy
Press/vol/temp of C2H2 → mol C2H2
Mol C2H2 → mass C2H2
n=
Relationship
n = PV/RT
1 mol C2H2 ≡ 26.0 g
PV
1.72 × 103 kPa (87.0 L)
=
= 59.8 mol
RT (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(301 K)
26.0 g
= 1.56 × 103 g C2 H2 = 1.56 kg C2 H2
1 mol
Mass C2 H2 = 59.8 mol C2 H2 ×
10.
Strategy
Press/vol/temp of He → mol He
Mol He → mass He
n=
Relationship
n = PV/RT
1 mol He ≡ 4.00 g
PV
102 kPa (5.74 × 106 L)
=
= 2.43 × 105 mol
RT (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(290 K)
Mass He = 2.43 × 105 mol He ×
= 0.972 tonne He
4.00 g
= 9.72 × 105 g He
1 mol
11.
Strategy
Press/vol/temp of X2H6 → mol X2H6
Mol/mass X2H6 → molar mass X2H6
n=
Relationship
n = PV/RT
mm X2H6 = m/n
PV
196 kPa (1.26 L)
=
= 8.35 × 10−2 mol
RT (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(356 K)
molar mass =
5.20 g
= 62.3 g ∙ mol−1
8.35 × 10−2 mol
2  X = [62.3  6(1.01)] gmol1 = 56.2 gmol1
X = 28.1 gmol1
According to the Periodic Table, this molar mass corresponds to the element silicon
12.
Strategy
Press/vol/temp of XH3 → mol XH3
Mol/mass XH3 → molar mass XH3
n=
Relationship
n = PV/RT
mm XH3 = m/n
PV
115 kPa (0.775 L)
=
= 3.44 × 10−2 mol
RT (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(312 K)
molar mass =
2.68 g
= 77.9 g ∙ mol−1
3.44 × 10−2 mol
X = [77.9  3(1.01)] gmol1 = 74.9 gmol1
According to the Periodic Table, this molar mass corresponds to the element arsenic
13.
Strategy
Press/mol/temp of gas → vol gas
Vol/mass of gas → density of gas
Relationship
V = nRT/P
d = m/V
nRT
1.00 mol (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(298 K)
Molar V =
=
= 24.8 L
P
(100 kPa)
Molar mass, H2 = 2.02 g
Density =
2.02 g
= 8.15 × 10−2 g ∙ L−1
24.8 L
14.
Strategy
Press/mol/temp of gas → vol gas
Vol/mass of gas → density of gas
Relationship
V = nRT/P
d = m/V
nRT
1.00 mol (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(298 K)
=
= 24.8 L
P
(100 kPa)
Molar mass, UF6 = 352.0 g
Molar V =
Density =
15.
352.0 g
= 14.2 g ∙ L−1
24.8 L
Part 1: Find Empirical Formula
Assume 100.0 g of compound. This will contain 92.3 g of carbon and 7.7 g of hydrogen.
Mol of C = 92.3 g C ×
1 mol
= 7.69 mol C
12.0 g
Mol of H = 7.7 g H ×
1 mol
= 7.7 mol H
1.01 g
Ratio =
7.69 mol C 7.7 mol H
∶
= 1 C ∶ 1.0 H
7.69 mol
7.69 mol
Rounding to the nearest whole number of 1:1, gives the empirical formula of CH.
Part 2: Find molar mass of Compound
Strategy
Relationship
Press/vol/temp → mol
Mol/mass → molar mass
n=
n = PV/RT
mm = m/n
PV
101 kPa (0.226 L)
=
= 7.36 × 10−3 mol
RT (8.31 kPa ∙ L ∙ mol−1 ∙ K −1 )(373 K)
molar mass =
0.573 g
= 77.8 g ∙ mol−1
7.36 × 10−3 mol
Part 3: Find Molar Mass and Molecular Formula
The molecular formula will be some multiple of the empirical formula: (CH)n. To find n,
the ratio of the molar mass (77.8 g) and the empirical formula mass
(1 × 12.0 g + 1 × 1.01 g = 13.0 g) is found. The value n is always an integer.
Mass ratio, 𝑛 =
77.8 g
=6
13.0 g
The molecular formula is (CH)6 or more correctly, C6H6
16.
Part 1: Find Empirical Formula
Assume 100.0 g of compound. This will contain 64.56 g of carbon, 10.86 g of hydrogen,
and 24.58 g of oxygen.
Mol of C = 64.56 g C ×
1 mol
= 5.376 mol C
12.01 g
Mol of H = 10.86 g H ×
1 mol
= 10.77 mol H
1.008 g
Mol of O = 24.58 g H ×
1 mol
= 1.536 mol O
16.00 g
Ratio =
5.376 mol C 10.86 mol H 1.536 mol O
∶
∶
= 3.500 C ∶ 7.070 H ∶ 1 O
1.536 mol
1.536 mol
1.536 mol
Rounding off, it is: 3.5 C : 7 H : 1 O
Multiplying through by 2 gives: 7 C : 14 H : 2 O
giving the empirical formula of C7H14O2.
Part 2: Find molar mass of Compound
Strategy
Press/vol/temp → mol
Mol/mass → molar mass
n=
Relationship
n = PV/RT
mm = m/n
PV
156.5 kPa (1.880 L)
=
= 7.897 × 10−2 mol
−1
−1
RT (8.314 kPa ∙ L ∙ mol ∙ K )(448.1 K)
molar mass =
10.29 g
= 130.3 g ∙ mol−1
7.897 × 10−2 mol
Part 3: Find Molar Mass and Molecular Formula
The molecular formula will be some multiple of the empirical formula: (C7H14O2)n. To
find n, the ratio of the molar mass (130.3 g) and the empirical formula mass
(7 × 12.01 g + 14 × 1.008 g + 2  16.00 = 130.2 g) is found. The value n is always an
integer.
Mass ratio, 𝑛 =
130.3 g
=1
130.2 g
The molecular formula is (C7H14O2)1 or more correctly, C7H14O2