Empirical formula from mass percentages Steps:

• Calculate oxygen mass percent: 100% - 48.6% - 8.1% = 43.3%.
• Convert to moles assuming 100 g sample: C = 48.6/12 = 4.05 mol, H = 8.1/1 = 8.1 mol, O = 43.3/16 = 2.706 mol.
• Divide each by smallest moles (2.706): C = 1.5, H = 3.0, O = 1.0.
• Multiply by 2 for whole numbers: C₃H₆O₂.

Why none match:

• Calculated formula C₃H₆O₂ (48.6% C, 8.1% H, 43.3% O) fits data exactly per mole ratio definition.

Why the others are wrong:

• A and C (CH₂O): Yields 40% C, 6.7% H; C percentage too low.
• B (C₂H₃O₂): Yields 40.7% C, 5.1% H; both C and H percentages mismatch.
• D (C₂H₅O₂): Yields 39.3% C, 8.2% H; C percentage too low.

Final answer: C₃H₆O₂