The elemental composition of a compound is $54.2 \% \mathrm{C}$, $9.2 \% \mathrm{H}$ and $36.6 \% \m

Question

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 $\mathrm{C}: \mathrm{H}: \mathrm{O}= 12: 1: 16$ ]\

Professor PC Thomas Classes · Answer

(a) :  Empirical formula $=\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}$\ Molecular formula $=\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}ight)_{n}$\ Empirical formula mass $=24+4+16$\ = 44\ Given molar mass of the compound $= 132 \mathrm{~g} \mathrm{~mol}^{-1}$\ $	herefore \quad n=\frac{132}{44}=3$\ ∴ Molecular formula of the compound $=\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}$\