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 Relationship

mass PbCl2 → mol PbCl2 1 mol ≡ 278.1 g PbCl2

(b)

First, calculate the molar mass of SiO2 = (1 × 28.1 g) + (2 × 16.0 g) = 60.1 g∙mol−1

Strategy Relationship

mass SiO2 → mol CaI2 1 mol ≡ 293.9 g SiO2

6. (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 Relationship

Mol of Cu2O → mass Cu2O 1 mol ≡ 132.1 g Cu2O

8. (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 Relationship

Mol of NCl3 → mass NCl3 1 mol ≡ 132.1 g NCl3

9.

Strategy Relationship

Mol of N2O3 → # of molecules N2O3 1 mol ≡ 6.02×1023

molecules

# of molecules N2O3 → # of atoms O 1 molecule N2O3 ≡ 3 atom O

10.

Strategy Relationship

Mol of Fe2O3 → # of formula units Fe2O3 1 mol ≡ 6.02×1023

formula units

# of formula units Fe2O3 → # of ions O2−

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 Relationship

Mass of SF4 → mol SF4 1 mol ≡ 108.1 g SF4

Mol of SF4 → # of molecules SF4 1 mol ≡ 6.02×1023

molecules

# of molecules SF4 → # of atoms F 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 Relationship

Mass of PbF4 → mol PbF4 1 mol ≡ 283.2 g PbF4

Mol of PbF4 → # of formula units PbF4 1 mol ≡ 6.02×1023

formula units

# of formula units PbF4 → # of ions F− 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