Determining molecular formula from mass percentages and molar mass range

Steps:

• Assume 100 g sample: 40 g C, 6.7 g H, 53.3 g O.
• Calculate moles: C = 40/12 = 3.33 mol, H = 6.7/1 = 6.7 mol, O = 53.3/16 = 3.33 mol.
• Simplify ratio by dividing by 3.33: C:H:O = 1:2:1, so empirical formula CH₂O.
• Empirical mass = 30 g/mol; n = molecular mass / empirical mass ≈ 60/30 = 2 (fits 55–65 range), so molecular formula C₂H₄O₂.

Why B is correct:

• C₂H₄O₂ has molar mass 60 (within 55–65) and exact percentage match: C = (24/60)×100 = 40%, H = (4/60)×100 ≈ 6.7%, O = (32/60)×100 ≈ 53.3%.

Why the others are wrong:

• A: Molar mass 30, below 55–65 range.
• C: Molar mass 78, above 65; C percentage ≈30.8%, not 40%.
• D: Molar mass 48, below 55–65 range; C percentage 25%, not 40%.

Final answer: B