40.0% C, 6.72% H, 53.29%, O. Molar mass is 60 g/mol. Determine the empirical and molecular formula.

Molecular formula: C₂H₄O₂

Empirical formula: CH₂O

Explanation:

40 % C, 6.72 % H and 53.29 % O states the centesimal composition of the compound. These data means that in 100 g of compound we have x grams of a determined element.

We divide the mass by the molar mass of each:

40 g / 12 g/mol = 3.33 moles of C

6.72 g / 1 g/mol = 6.72 moles of H

53.29 g / 16 g/mol = 3.33 moles of O

We can determine rules of three to get, the molecular formula.

In 100 g of compound we have 3.33 moles of C, 6.72 moles of H and 3.33 moles of O; therefore in 60 g (1 mol) we must have

- (60 . 3.33) / 100 = 2 moles of C

- (60 . 6.72) / 100 = 4 moles of H

- (60 . 3.33) / 100 = 2 moles of O

Molecular formula is C₂H₄O₂

Empirical formula has the lowest suscripts; we divide by two, so the empirical formula is CH₂O

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