Page 126 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 7] FORMULA CALCULATIONS 115

(b) C 5 × 12.0 = 60.0 amu
H 10 × 1.0 = 10.0 amu
Total = 70.0 amu

60.0 amu C 10.0 amu H
%C = × 100% = 85.7% C % H = × 100% = 14.3% H
70.0 amu total 70.0 amu total
The percentages are the same as those in part (a). That result might have been expected. Since the ratio
of atoms of carbon to atoms of hydrogen is the same (1 : 2) in both compounds, the ratio of masses
also ought to be the same, and their percent by mass ought to be the same. From another viewpoint,
this result means that the two compounds cannot be distinguished from each other by their percent
compositions alone.

7.44. Calculate the percent composition of DDT (C 14 H 9 Cl 5 ).
Ans. C 14 × 12.01 = 168.1 amu
H 9 × 1.008 = 9.07 amu
Cl 5 × 35.45 = 177.2 amu
Total = 354.4 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%:
168.1 amu C

%C = × 100% = 47.43% C
354.4 amu total

9.07 amu H
%H = × 100% = 2.56% H
354.4 amu total

177.2 amu Cl
%Cl = × 100% = 50.00% Cl
354.4 amu total
The percentages add up to 99.99%. (The answer is correct within the accuracy of the number of significant
figures used.)
7.45. A forensic scientist analyzes a drug and finds that it contains 80.22% carbon and 9.62% hydrogen. Could
the drug be pure tetrahydrocannabinol (C 21 H 30 O 2 )?
Ans. 21 C 21 × 12.01 = 252.2 amu
30 H 30 × 1.008 = 30.24 amu
2O 2 × 16.00 = 32.00 amu
Formula mass = 314.4 amu
252.2 amu 30.24 amu
%C = × 100% = 80.22% C % H = × 100% = 9.618% H
314.4 amu 314.4 amu
Since the percentages are the same, the drug could be tetrahydrocannabinol. (It is not proved to be, however.
If the percent composition were different, it would be proved not to be pure tetrahydrocannabinol.)

7.46. A certain mixture of salt (NaCl) and sugar (C 12 H 22 O 11 ) contains 40.0% chlorine by mass. Calculate the
percentage of salt in the mixture.

Ans. In 100.0 g of sample (the size does not make any difference), there is 40.0 g of chlorine and therefore

58.5 g NaCl
40.0gCl = 65.9 g NaCl
35.5gCl
The percentage of NaCl (in the 100.0 g sample) is therefore 65.9%. When using percentages, be careful to
distinguish percentage of what in what!

EMPIRICAL FORMULAS
7.47. If each of the following mole ratios is obtained in an empirical formula problem, what should it be
multiplied by to get an integer ratio? (a)1.50:1,(b)1.25:1,(c)1.33:1,(d) 1.67 : 1, and (e)1.75:1.