Chapter 8 Answers to Questions

Chapter 8
Answers to Questions
1.
(a) formula mass Na2S = (2 × 23.0 u) + (1 × 32.1 u) = 78.1 u
(b) formula mass CaCO3 = (1 × 40.1 u) + (1 × 12.0 u) + (3 × 16.0 u) = 100.1 u
(c) formula mass Al(NO3)3 = (1 × 27.0 u) + (3 × 14.0 u) + (9 × 16.0 u) = 213.0 u
2.
(a) molecular mass NO2 = (1 × 14.0 u) + (2 × 16.0 u) = 46.0 u
(b) molecular mass P4O6 = (4 × 31.0 u) + (6 × 16.0 u) = 220.0 u
(c) molecular mass IF7 = (1 × 126.9 u) + (7 × 19.0 u) = 259.9 u
3.
(a) molar mass SnI4 = (1 × 118.7 g) + (4 × 126.9 g) = 626.3 g∙mol−1
(b) molar mass PF5 = (1 × 31.0 g) + (5 × 19.0 g) = 126.0 g∙mol−1
(c) molar mass (NH4)2SO4 = (2 × 14.0 g) + (8 × 1.01 g) + (1 × 32.1 g) + (4 × 16.0 g)
= 132.1 g∙mol−1
4.
(a) molar mass SF6 = (1 × 32.1 g) + (6 × 19.0 g) = 146.1 g∙mol−1
(b) molar mass SnCl2∙2H2O = (1 × 118.7 g) + (2 × 35.5 g) + (4 × 1.01 g) + (2 × 16.0 g)
= 225.6 g∙mol−1
(c) molar mass Mg3(PO4)2 = (3 × 24.3 g) + (2 × 31.0 g) + (8 × 16.0 g) = 262.9 g∙mol−1
5.
(a)
First, calculate the molar mass of PbCl2 = (1 × 207.2 g) + (2 × 35.5 g) = 278.1 g∙mol−1
Strategy
mass PbCl2 → mol PbCl2
(b)
Relationship
1 mol ≡ 278.1 g PbCl2
First, calculate the molar mass of SiO2 = (1 × 28.1 g) + (2 × 16.0 g) = 60.1 g∙mol−1
Strategy
mass SiO2 → mol CaI2
6.
Relationship
1 mol ≡ 293.9 g SiO2
(a)
First, calculate the molar mass of MgSO4 = (1 × 24.3 g) + (1 × 32.1 g) + (4 × 16.0 g)
= 120.4 g∙mol−1
Strategy
Relationship
mass(kg) MgSO4 → mass(g) MgSO4 1 kg = 1000 g
mass MgSO4 → mol MgSO4
1 mol ≡ 120.4 g MgSO4
(b)
First, calculate the molar mass of Ag2S = (2 × 107.9 g) + (1 × 32.1 g) = 247.9 g∙mol−1
Strategy
Relationship
mass(mg) Ag2S → mass(g) Ag2S 1 kg = 1000 g
mass Ag2S → mol Ag2S
1 mol ≡ 247.9 g Ag2S
7.
(a)
First, calculate the molar mass of BaBr2 = (1 × 137.3 g) + (2 × 79.9 g) = 297.1 g∙mol−1
Strategy
Relationship
Mol of BaBr2 → mass BaBr2
1 mol ≡ 297.1 g BaBr2
(b)
First, calculate the molar mass of Cu2O = (2 × 63.5 g) + (1 × 16.0 g) = 143.0 g∙mol−1
Strategy
Mol of Cu2O → mass Cu2O
8.
Relationship
1 mol ≡ 132.1 g Cu2O
(a)
First, calculate the molar mass of PbCrO4 = (1 × 207.2 g) + (1 × 52.0 g) + (4 × 16.0 g)
= 323.2 g∙mol−1
Strategy
Relationship
Mol of PbCrO4 → mass PbCrO4 1 mol ≡ 323.2 g PbCrO4
(b)
First, calculate the molar mass of NCl3 = (1 × 14.0 g) + (3 × 35.5 g) = 120.4 g∙mol−1
Strategy
Mol of NCl3 → mass NCl3
Relationship
1 mol ≡ 132.1 g NCl3
9.
Strategy
Mol of N2O3 → # of molecules N2O3
# of molecules N2O3 → # of atoms O
Relationship
1 mol ≡ 6.02×1023 molecules
1 molecule N2O3 ≡ 3 atom O
10.
Strategy
Mol of Fe2O3 → # of formula units Fe2O3
# of formula units Fe2O3 → # of ions O2−
Relationship
1 mol ≡ 6.02×1023 formula units
1 formula unit Fe2O3 ≡ 3 ions O2−
11.
First, calculate the molar mass of SF4 = (1 × 32.1 g) + (4 × 19.0 g) = 108.1 g∙mol−1
Strategy
Mass of SF4 → mol SF4
Mol of SF4 → # of molecules SF4
# of molecules SF4 → # of atoms F
Relationship
1 mol ≡ 108.1 g SF4
1 mol ≡ 6.02×1023 molecules
1 molecule SF4 ≡ 4 atom F
12.
First, calculate the molar mass of PbF4 = (1 × 207.2 g) + (4 × 19.0 g) = 283.2 g∙mol−1
Strategy
Mass of PbF4 → mol PbF4
Mol of PbF4 → # of formula units PbF4
# of formula units PbF4 → # of ions F−
Relationship
1 mol ≡ 283.2 g PbF4
1 mol ≡ 6.02×1023 formula units
1 formula unit PbF4 ≡ 4 ions F−
13.
First, calculate the molar mass of PCl3 = (1 × 31.0 g) + (3 × 35.5) = 137.5 g∙mol−1
14.
First, calculate the molar mass of CaCO3 = (1 × 40.1 g) + (1 × 12.0 g) + (3 × 16.0 g)
= 100.1 g∙mol−1
15.
Assume 100.0 g of compound. This will contain 48.0 g of zinc and 52.0 g of chlorine.
Rounding to the nearest whole number of 1:2, gives the empirical formula of ZnCl2.
16.
Assume 100.0 g of compound. This will contain 19.0 g of tin and 81.0 g of iodine.
Rounding to the nearest whole number of 1:4, gives the empirical formula of SnI4.
17.
Percent O = 100.0% – 62.6% − 8.5% = 28.9%
Assume 100.0 g of compound. This will contain 62.6 g of lead; 8.5 g of nitrogen; and
28.9 g of oxygen.
Rounding to the nearest whole number of 1:2:6, gives the empirical formula of PbN2O6.
18.
Percent O = 100.0% – 36.5% − 25.4% = 38.1%
Assume 100.0 g of compound. This will contain 36.5 g of sodium; 25.4 g of sulfur; and
38.1 g of oxygen.
Rounding to the nearest whole number of 2:1:3, gives the empirical formula of Na2SO3.
19.
Assume 100.0 g of compound. This will contain 38.7 g of carbon; 9.8 g of hydrogen; and
51.5 g of oxygen.
Rounding to the nearest whole number of 1:3:1, gives the empirical formula of CH3O.
The molecular formula will be some multiple of this: (CH3O)n. To find n, the ratio of the
molar mass (62.1 g) and the empirical formula mass
(1 × 12.0 g + 3 × 1.01 g + 1 × 16.0 g = 31.0 g) is found. The value n is always an integer.
The molecular formula is (CH3O)2 or more correctly, C2H6O2
20.
Assume 100.0 g of compound. This will contain 48.6 g of carbon; 8.2 g of hydrogen; and
43.2 g of oxygen.
To obtain integers, each number must be multiplied by two: 3.00 C : 6.0 H : 2 O
Rounding to the nearest whole number of 3:6:2, gives the empirical formula of C3H6O2.
The molecular formula will be some multiple of this: (C3H6O2)n. To find n, the ratio of
the molar mass (74.1 g) and the empirical formula mass
(3 × 12.0 g + 6 × 1.01 g + 2 × 16.0 g = 74.1 g) is found. The value n is always an integer.
The molecular formula is (C3H6O2)1 or more correctly, C3H6O2
21.
First, the mass of oxygen can be found by subtraction = (4.626 g – 4.306 g) = 0.320 g
Then the empirical formula type solution is employed.
Empirical formula is Ag2O.
22.
First, the mass of oxygen can be found by subtraction = (7.59 g – 3.76 g) = 3.83 g
Then the empirical formula type solution is employed.
To obtain integers, each number must be multiplied by two: 2 Mn : 6.98 O
Rounding to the nearest whole number of 2:7, gives the empirical formula of Mn2O7.
23.
First, the mass of water is needed = (7.52 g – 5.54 g) = 1.98 g
Then the empirical formula type solution is employed.
Empirical formula is FePO4∙3H2O
24.
The empirical formula type solution is employed.
Empirical formula is CaCl2∙6H2O