A compound contains 40.0% C, 6.7% H and 53.3% O. If the molecular mass is 180, its molecular formula is [C = 12, H = 1, O = 16]

Correct answer: C₆H₁₂O₆ (Option D)

Step 1: Empirical formula from percentages (assume 100 g)

• C: 40.0 g → \(40.0/12 = 3.33\) mol
• H: 6.7 g → \(6.7/1 = 6.7\) mol
• O: 53.3 g → \(53.3/16 = 3.33\) mol

Divide by the smallest (~3.33): C : H : O ≈ 1 : 2 : 1 → empirical formula = CH₂O (M_r = 12 + 2 + 16 = 30).

Step 2: Molecular formula using the molar mass

\(\dfrac{180}{30} = 6\) → multiply subscripts in CH₂O by 6 → C₆H₁₂O₆.

Why others are incorrect

• CH₂O — empirical unit only (M_r = 30), not 180.
• C₃H₆O₃ — 3× empirical (M_r = 90), still too small.
• C₆H₆O₃ — wrong H:C ratio (doesn’t match the given percentages).
• C₆H₁₂O₆ — matches both composition and molar mass.