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 $\mathrm{C}: \mathrm{H}: \mathrm{O}=12: 1: 16$ ]
- $\mathrm{C}_4 \mathrm{H}_9 \mathrm{O}_3$
- $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
- $\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2$
- $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3$
Solution

Molecular mass of compound $=132 \mathrm{~g} \mathrm{~mol}^{-1}$ Molecular formula of compound is $\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)_n$
$\mathrm{n}=\frac{\text { Molecular mass }}{\mathrm{EF} \text { mass }}=\frac{132}{44}=3$
$\therefore$ Molecular formula of compound is $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3$
Asked in: JEE Main 2025 (24 Jan Shift 2)
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