\( 0.30 \mathrm{~g} \) of an organic compound containing \( \mathrm{C}, \mathrm{H} \) and Oxygen on combustion yields \( 0.44 \mathrm{~g} \mathrm{CO}_{2} \)
and \( 0.18 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \). If one mol of compound weighs \( 60 \), then molecular formula of the compound
is

Percentage of \(C=\frac{12}{44} \times \frac{0.44}{0.30} \times 100=40 \%\)
Percentage of \(\mathrm{H}=\frac{2}{18} \times \frac{0.18}{0.30} \times 100=6.6 \%\)
Percentage of \(O=100-(40+6.6)=53.4 \%\)
\begin{array}{|l|l|l|l|}
\hline Element & \% & Molar ratio & Simplest ratio \\
\hline \mathrm{C} & 40 & \frac{40}{12}=3.3 & \frac{3.3}{3.3}=1 \\
\hline \mathrm{H} & 6.6 & \frac{6.6}{1}=6.6 & \frac{6.6}{3.3}=2 \\
\hline \mathrm{O} & 53.4 & \frac{53.4}{16}=3.3 & \frac{3.3}{3.3}=1 \\
\hline
\end{array}
hence, empirical formula \(=\mathrm{CH}_{2} \mathrm{O}\)
We know
\(n=\frac{\text { Molecular mass }}{\text { Empirial formula mass }}=\frac{60}{30}=2\) \(\Rightarrow\) Molecular formula of compound \(=\left(\mathrm{CH}_{2} \mathrm{O}_{2}\right)=\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) Hence, molar mass of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}=2 \times 12+1 \times 4+16 \times 2=60 \mathrm{gmol}^{-1}\)