JEE MAIN - Chemistry (2025 - 24th January Evening Shift - No. 6)
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 $\mathrm{C}: \mathrm{H}: \mathrm{O}=12: 1: 16$ ]
$\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2$
$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
$\mathrm{C}_4 \mathrm{H}_9 \mathrm{O}_3$
$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3$
Explanation
$$\begin{array}{lllll} \mathrm{C} & : & \mathrm{H} & : & \mathrm{O} \\ \frac{54.2}{12} & : & 9.2 & : & \frac{36.6}{16} \\ 4.516 & : & 9.2 & : & 2.287 \\ \frac{4.516}{2.287} & : & \frac{9.2}{2.287} & : & \frac{2.287}{2.287} \\ 1.97 & : & 4.02 & : & 1 \end{array}$$
$\mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \Rightarrow$ Empirical formula
E.F. mass $=24+4+16=44$
and molar mass $=132$
$$\begin{aligned} \text { Hence molecular formula } & =\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)_3 \\ & =\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3 \end{aligned}$$
Correct Option (3)
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