Page 128 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 7] FORMULA CALCULATIONS 117

To get integer mole ratios, divide by the smallest, 0.956:
0.959 mol Mg 2.88 mol O
= 1.00 mol Mg = 3.00 mol O
0.959 0.959
0.959 mol S
= 1.00 mol S
0.959
The mole ratio is 1 mol Mg to 1 mol S to 3 mol O; the empirical formula is MgSO 3 .

7.54. Calculate the empirical formula for each of the following compounds: (a) 42.1% Na, 18.9% P, and
39.0% O. (b) 55.0% K and 45.0% O.
1 mol Na 1 mol O

Ans. (a) 42.1gNa = 1.83 mol Na 39.0gO = 2.44 mol O
23.0gNa 16.00gO

1 mol P
18.9gP = 0.610 mol P
31.0gP
Dividing by 0.610 yields 3.00 mol Na, 1.00 mol P, 4.00 mol O. The empirical formula is Na 3 PO 4 .

(b) 55.0gK 1 mol K = 1.41 mol K 45.0gO 1 mol O = 2.81 mol O
39.1gK 16.0gO
Dividing by 1.41 yields 1.00 mol K and 2.00 mol O. The empirical formula is KO 2 (potassium super-
oxide).

MOLECULAR FORMULAS
7.55. List five possible molecular formulas for a compound with empirical formula CH 2 .
Ans. C 2 H 4 ,C 3 H 6 ,C 4 H 8 ,C 5 H 10 ,C 6 H 12 (and any other formula with a C-to-H ratio of1:2).

7.56. Explain why we cannot calculate a molecular formula for a compound of phosphorus, potassium, and
oxygen.
Ans. The compound is ionic; it does not form molecules.

7.57. Which one of the following could possibly be defined as “the ratio of moles of each of the given elements
to moles of each of the others”?(a) percent composition by mass, (b) empirical formula, or (c) molecular
formula.
Ans. Choice (b). This is a useful definition of empirical formula. The molecular formula gives the ratio of moles
of each element to moles of the compound, plus the information given by the empirical formula. The percent
composition does not deal with moles, but is a ratio of masses.

7.58. A compound consists of 92.26% C and 7.74% H. Its molecular mass is 65.0 amu. (a) Calculate its
empirical formula. (b) Calculate its empirical formula mass. (c) Calculate the number of empirical
formula units in one molecule. (d) Calculate its molecular formula.
Ans. (a) The empirical formula is calculated to be CH, as presented in problem 7.51.
(b) The empirical formula mass is 13.0 amu, corresponding to 1 C and 1 H atom.
(c) There are
65.0 amu/molecule 5 empirical formula units
=
13.0 amu/empirical formula unit 1 molecule
(d) The molecular formula is (CH) 5 ,orC 5 H 5 .

7.59. The percent composition of a certain compound is 85.7% C and 14.3% H. Its molecular mass is 70.0
amu. (a) Determine its empirical formula. (b) Determine its molecular formula.

1 mol C 1 mol H
Ans. (a) 85.7gC = 7.14 mol C 14.3gH = 14.2 mol H
12.0gC 1.008gH
The ratio is1:2,andthe empirical formula is CH 2 .