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 $\mathrm{C}: \mathrm{H}: \mathrm{O}=12: 1: 16$ ]
Solution
<p>$$\begin{array}{lllll}
\mathrm{C} & : & \mathrm{H} & : & \mathrm{O} \\
\frac{54.2}{12} & : & 9.2 & : & \frac{36.6}{16} \\
4.516 & : & 9.2 & : & 2.287 \\
\frac{4.516}{2.287} & : & \frac{9.2}{2.287} & : & \frac{2.287}{2.287} \\
1.97 & : & 4.02 & : & 1
\end{array}$$</p>
<p>$\mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \Rightarrow$ Empirical formula</p>
<p>E.F. mass $=24+4+16=44$</p>
<p>and molar mass $=132$</p>
<p>$$\begin{aligned}
\text { Hence molecular formula } & =\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)_3 \\
& =\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3
\end{aligned}$$</p>
<p>Correct Option (3)</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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