The elemental composition of a compound is $54.2 \%, \mathrm{C}, 9.2 \% \mathrm{H}$ and $36.6 \% \mathrm{O}$. If the molar mass of the compound is $132 \mathrm{~g} \mathrm{~mol}^{-1}$, the molecular formula of the compound is : [Given : The relative atomic mass of $\mathrm{C}: \mathrm{H}: \mathrm{O}=$ 12 : 1 : 16]

(C) $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}$

(A) $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}_{3}$

(B) $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$

(D) $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}$

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Some Basic Concepts of Chemistry

The elemental composition of a compound is

54.2%, C, 9.2% H

and

36.6% O

. If the molar mass of the compound is

132 g mol^{- 1}

, the molecular formula of the compound is :
[Given : The relative atomic mass of

C : H : O =

12 : 1 : 16]

Answer: (C)

Converting mass percentages to mole ratios using the given atomic masses:

C : \frac{54.2}{12} = 4.52, H : \frac{9.2}{1} = 9.2, O : \frac{36.6}{16} = 2.29

Dividing through by the smallest value, 2.29:

C : 1.97 \approx 2, H : 4.02 \approx 4, O : 1

The empirical formula is

C_{2} H_{4} O

, with empirical formula mass

2 (12) + 4 (1) + 16 = 44 g/mol

.

Since the molar mass is

132 g/mol

n = 132/44 = 3

, so the molecular formula is

(C_{2} H_{4} O)_{3} = C_{6} H_{12} O_{3}

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option