Empirical and Molecular Formulas
WARM-UP:
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A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 184
g/mol. What is its formula?
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Assume 100g sample. 69.6 g S and 100-69.6 g = 30.4 g O
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Convert each to mols: 69.6 g S / 32.07 g per mol = 2.17 mol S and 30.04/16.00 = 1.90
mol O
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Roughly 1:1 ratio, therefore SO
Cumene is a hydrocarbon (containing only carbon and hydrogen) that is used in the production
of acetone and phenol in the chemical industry. Combustion of 47.6 mg cumene produces some
CO2 and 42.8 mg water. The molar mass of cumene is between 115 and 125 g/mol. Determine
the empirical and molecular formula for cumene.
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42.8 mg H2O x (2.016 g H/18.02 g H2O) = 4.79 mg H (Notice how we use molar masses of
both the element of interest and the molecule)
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% H = 4.79 mg H / 47.6 mg cumene x 100% = 10.1 % H by mass
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% C = 100% - 10.1% = 89.6 % C by mass
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Therefore, using a 100 g sample:
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10.1 g H x (1 mol H/1.008 g H) = 10.0 mol H
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89.9 g C x (1 mol C/12.01g C) = 7.49 mol C
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Molar ration: 10.0 / 7.49 = 1.34 and 7.49/ 7.49 = 1 so 1.34:1 is roughly a 4:3 H:C
ratio and so the empirical formula is C3H4
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Empirical formula mass is roughly 40 g/mol
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Therefore molecular formula is roughly 3 times that or C9H12