Page 125 - Theory and Problems of BEGINNING CHEMISTRY

114 FORMULA CALCULATIONS [CHAP. 7

(prior problem), that atoms are too small to count. We can “count” them by combining them with a known
number of atoms of another element. For example, to count a number of carbon atoms, combine them with
a known number of oxygen atoms to form CO, in which the ratio of atoms of carbon to oxygen is1:1.

7.36. What mass of oxygen is combined with 4.13 × 10 24 atoms of sulfur in Na 2 SO 4 ?

1 mol S 4 mol O 16.0gO
24
Ans. 4.13 × 10 atoms S = 439gO
23
6.02 × 10 atoms S 1 mol S 1 mol O
7.37. A 106 g sample of an “unknown” element Q reacts with 32.0 g of O 2 . Assuming the atoms of Q react in
a1:1 ratio with oxygen molecules, calculate the atomic mass of Q.
Ans. Since the 106 g of Q reacts with 1.00 mol of O 2 in a 1 : 1 mole ratio, there must be 1.00 mol of Q in 106 g.
Therefore, 106 g is 1.00 mol, and the atomic mass of Q is 106 amu.

7.38. To form a certain compound, 29.57 g of oxygen reacts with 109.7 g of tin. What is the formula of the
compound?
Ans. The formula is the mole ratio:

1 mol O atoms
29.57 g O atoms = 1.848 mol O atoms
16.00gO

1 mol Sn atoms
109.7 g Sn atoms = 0.9242 mol Sn atoms
118.7gSn
The mole ratio is 1.848 mol O/0.9242 mol Sn = 2 mol O/1 mol Sn. The formula is SnO 2 .
7.39. If 29.57 g O 2 were used in Problem 7.38, how would the problem change?
Ans. The answer would be the same; 29.57 g of O 2 is 29.57 g of O atoms (bonded in pairs). See Problem 7.28.

7.40. How many P atoms are there in 1.13 mol P 4 ?
23

6.02 × 10 molecules P 4 4 atoms P 24
Ans. 1.13 mol P 4 = 2.72 × 10 P atoms
1 mol P 4 1 molecule P 4
7.41. How many hydrogen atoms are there in 1.36 mol NH 3 ?
23

6.02 × 10 molecules NH 3 3 H atoms 24
Ans. 1.36 mol NH 3 = 2.46 × 10 H atoms
1 mol NH 3 1 molecule NH 3
PERCENT COMPOSITION OF COMPOUNDS
7.42. A 10.0-g sample of water has a percent composition of 88.8% oxygen and 11.2% hydrogen. (a) What is
the percent composition of a 6.67-g sample of water? (b) Calculate the number of grams of oxygen in a
6.67-g sample of water.

Ans. (a) 88.8% O and 11.2% H. The percent composition does not depend on the sample size.
88.8gO

(b)6.67gH 2 O =5.92gO
100gH O
2
7.43. Calculate the percent composition of each of the following: (a)C 3 H 6 and (b)C 5 H 10 .
Ans. (a) C 3 × 12.0 = 36.0 amu
H 6 × 1.0 = 6.0 amu
Total = 42.0 amu
The percent carbon is found by dividing the mass of carbon in one molecule by the mass of the molecule
and multiplying the quotient by 100%:
36.0 amu C 6.0 amu H

%C = × 100% = 85.7%C % H = × 100% = 14.3% H
42.0 amu total 42.0 amu total
The two percentages add up to 100.0%.