A compound is 40.0% C, 6.7% H, 53.3% O by mass. Determine its empirical formula.

• A. CH2O
• B. C2H4O2
• C. CHO
• D. C3H6O

Answer: (A)

The question tests turning mass percentages into an empirical formula by converting to moles and reducing to the smallest whole-number ratio. Assume a 100 g sample: carbon 40.0 g, hydrogen 6.7 g, oxygen 53.3 g. Convert to moles using approximate atomic masses: C ≈ 40.0/12.01 ≈ 3.33 mol, H ≈ 6.7/1.008 ≈ 6.64 mol, O ≈ 53.3/16.00 ≈ 3.33 mol. The smallest number of moles is about 3.33, so divide all by this: C ≈ 1.00, H ≈ 2.00, O ≈ 1.00. This gives the simplest whole-number ratio C:H:O = 1:2:1, i.e., CH2O.

The empirical formula is the simplest ratio of elements in a compound; larger formulas like C2H4O2 or C3H6O share the same empirical formula CH2O, but the question asks for the empirical one, which is CH2O. A formula like CHO would imply a 1:1:1 ratio and doesn’t match the data.