Empirical formula from molecular formula of ethanoic acid

Steps:

• Recall molecular formula of ethanoic acid (CH₃COOH) is C₂H₄O₂.
• Count atoms: 2 carbon, 4 hydrogen, 2 oxygen.
• Divide by smallest number (2): C=1, H=2, O=1.
• Write simplest ratio as CH₂O.

Why A is correct:

• Empirical formula is the lowest whole-number ratio of atoms in a compound; for C₂H₄O₂, it simplifies to CH₂O by dividing subscripts by 2.

Why the others are wrong:

• B (CH₃O): Gives C:H:O ratio of 1:3:1, not matching ethanoic acid's 1:2:1.
• C (CH₂O): Identical to A, but designated correct as A.
• D (CH₃O₂): Gives C:H:O ratio of 1:3:2, exceeding ethanoic acid's hydrogen and oxygen proportions.

Final answer: A