Page 118 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 7] FORMULA CALCULATIONS 107

Ans. The percent composition of each compound may be determined:
FeO Fe 2 O 3
55.85gFe 2 × 55.85gFe
%Fe = × 100.0% = 77.73% Fe % Fe = × 100.0% = 69.94% Fe
71.85 g FeO 159.70gFe O 3
2
16.00gO 3 × 16.00gO
%O = × 100.0% = 22.27% O % O = × 100.0% = 30.06% O
71.85 g FeO
159.70gFe O 3
2
The compound is Fe 2 O 3 .

7.6. EMPIRICAL FORMULAS
The formula of a compound gives the relative number of atoms of the different elements present. It also gives
the relative number of moles of the different elements present. As was shown in Sec. 7.5, the percent by mass
of each element in a compound may be computed from its formula. Conversely, if the formula is not known, it
may be deduced from the experimentally determined composition. This procedure is possible because once the
relative masses of the elements are found, the relative numbers of moles of each may be determined. Formulas
derived in this manner are called empirical formulas or simplest formulas. In solving a problem in which percent
composition is given, any size sample may be considered, since the percentage of each element does not depend
on the size of the sample. The most convenient size to consider is 100 g, for with that size sample, the percentage
of each element is equal to the same number of grams.

EXAMPLE 7.12. What is the empirical formula of a compound which contains 52.96% carbon and 47.04% oxygen by
mass?
Ans. In each 100.0 g of the substance, there will be 52.96 g of carbon and 47.04 g of oxygen. The number of moles of
each element in this sample is given by

1 mol C
52.96gC = 4.410 mol C
12.01gC

1 mol O
47.04gO = 2.940 mol O
16.00gO
Therefore, the ratio of moles of C to moles of O
4.410 mol C 1.500 mol C 3 mol C
= =
2.940 mol O 1.000 mol O 2 mol O
is 3 to 2. The ratio of 3 mol of carbon to 2 mol of oxygen corresponds to the formula C 3 O 2 , a compound called
carbon suboxide.

If more than two elements are present, divide all the numbers of moles by the smallest to attempt to get an
integral ratio. Even after this step, it might be necessary to multiply every result by a small integer to get integral
ratios, corresponding to the empirical formula.

EXAMPLE 7.13. Determine the empirical formula of a compound containing 29.09% Na, 40.55% S, and 30.36% O.
1 mol Na 1.265 mol Na
Ans. 29.09gNa = 1.265 mol Na = 1.000 mol Na
22.99gNa 1.265

1 mol S 1.265 mol S
40.55gS = 1.265 mol S = 1.000 mol S
32.06gS 1.265

1 mol O 1.898 mol O
30.36gO = 1.898 mol O = 1.500 mol O
16.00gO 1.265
Multiplying each value by 2 yields 2 mol Na, 2 mol S, and 3 mol O, corresponding to Na 2 S 2 O 3 .