Q33 / 180 · KCET 2015 · Chemistry

\( 0.30 \mathrm{~g} \) 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

  • A
    \( \mathrm{CH}_{2} \mathrm{O} \)
  • B
    \( \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O} \)
  • C
    \( \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O} \)
  • Correct answer
    D
    \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \)

Solution

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 $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}=2 \times 12+1 \times 4+16 \times 2=60 \mathrm{gmol}^{-1}$

Hence, the correct answer is option: 'D'.

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