Combustion analysis finds empirical formula from C, H, O masses in products

Steps:

• Moles C = 4.00 g CO₂ / 44 g/mol = 0.091 mol
• Moles H = (1.80 g H₂O / 18 g/mol) × 2 = 0.20 mol
• Mass O = 3.00 g - (0.091 mol × 12 g/mol) - (0.20 mol × 1 g/mol) = 1.71 g
• Moles O = 1.71 g / 16 g/mol = 0.107 mol; divide by 0.091 gives C:H:O ≈ 1:2:1, so CH₂O

Why D is correct:

• Mole ratio ≈1:2:1 defines CH₂O as the simplest whole-number empirical formula from % composition.

Why the others are wrong:

• A. CH: H:C ratio 1:1, but data gives ~2:1
• B. CHO: H:C ratio 1:1, but data gives ~2:1
• C. CH₂: No oxygen, but 1.71 g O found in J

Final answer: D