Empirical formula C2H4O limits possible structures to one functional group

Steps:

• Empirical C2H4O means molecular formulas C_{2k}H_{4k}O_k (k integer ≥1), e.g., k=1: C2H4O; k=2: C4H8O2.
• Alcohol (1): Saturated formula C_nH_{2n+2}O (e.g., C2H6O) doesn't match; C2H4O needs unsaturation, adding alkene group.
• Aldehyde (2): CH3CHO (acetaldehyde) is C2H4O (k=1), one functional group.
• Ester (3): CH3CO2C2H5 (ethyl acetate) is C4H8O2 (k=2), one functional group.
• Ketone (4): Needs ≥3 carbons, one O; k=1 (C2H4O) impossible; k=2 requires C4H8O (not O2).

Why C is correct:

• Aldehyde (CH3CHO, C2H4O) and ester (ethyl acetate, C4H8O2 empirical C2H4O) both fit with exactly one functional group per general formula.

Why the others are wrong:

• A: Alcohol in C2H4O adds alkene, violating one functional group.
• B: Ketone cannot match O_k ratio with one oxygen.
• D: Ketone impossible due to carbon minimum and oxygen mismatch.

Final answer: C