Empirical formula from composition matches IR carbonyl peak Steps:

• Assume 100 g sample: 62.0 g C, 10.3 g H, 27.7 g O.
• Calculate moles: C = 62.0/12 = 5.17 mol, H = 10.3/1 = 10.3 mol, O = 27.7/16 = 1.73 mol.
• Divide by smallest (1.73): C ≈ 3, H ≈ 6, O = 1, giving C₃H₆O.
• IR spectrum shows strong absorption at ~1710 cm⁻¹, indicating C=O group in ketone.

Why A is correct:

• A is propanone (acetone), with molecular formula C₃H₆O and characteristic ketone C=O stretch matching the IR peak.

Why the others are wrong:

• B: Ethanol (C₂H₆O) has wrong composition (52.2% C, 13.0% H, 34.8% O) and O-H stretch, not C=O.
• C: Ethanal (C₂H₄O) has incorrect %C (53.3%) and aldehyde C-H stretches absent in IR.
• D: Propan-1-ol (C₃H₈O) has higher %H (13.3%) and broad O-H peak, not sharp C=O.

Final answer: A