Empirical and Molecular Formulas from Percent Composition

Steps:

• Assume 100 g sample: 40 g C (40/12 = 3.33 mol), 6.7 g H (6.7/1 = 6.7 mol), 53.3 g O (53.3/16 = 3.33 mol).
• Divide moles by smallest (3.33): C = 1, H = 2, O = 1, so empirical formula CH₂O.
• Empirical mass = 30 g/mol; molecular mass ratio = 60/30 = 2.
• Multiply empirical by 2: molecular formula C₂H₄O₂.

Why B is correct:

• Matches empirical CH₂O (simplest ratio) and molecular C₂H₄O₂ (empirical × 2, where n = M / empirical mass).

Why the others are wrong:

• A: Molecular CH₂O has mass 30, not 60.
• C: Empirical CH₂O₂ incorrect; oxygen ratio is 1:1, not 1:2.
• D: Molecular C₃H₆O₃ has mass 90, not 60.

Final answer: B