Empirical formulas from combustion products identify possible alkanes Steps:

• Calculate carbon mass: (12/44) × CO₂ mass; hydrogen mass: (2/18) × H₂O mass.
• For X: C = 12 g, H = 3 g; empirical CH₃ (80% C), matches C₂H₆ (ethane, straight-chain alkane CnH₂n+₂).
• For Y: C = 12 g, H = 2 g; empirical CH₂ (85.7% C), exceeds alkane limit (<85.7% C for finite n).
• For Z: C = 6 g, H = 1 g; empirical CH₂ (same as Y, not alkane). Why D is correct:
• D identifies only X, whose CH₃ formula fits CnH₂n+₂ for n=2 (ethane), per alkane general formula. Why the others are wrong:
• A includes Y (CH₂, alkene-like, not alkane).
• B includes Y and Z (both CH₂, impossible for alkanes).
• C includes all (Y and Z mismatch CnH₂n+₂). Final answer: D