Amount of carbon in \(3.38 \mathrm{~g}\) of
\(\mathrm{CO}_2=\frac{12}{44} \times 3.38 \mathrm{~g}=0.9218 \mathrm{~g}\)
Amount of hydrogen in \(0.690 \mathrm{~g}\) of
\(\mathrm{H}_2 \mathrm{O}=\frac{2}{18} \times 0.690 \mathrm{~g}=0.0767 \mathrm{~g}\)
As compound contains only \(\mathrm{C}\) and \(\mathrm{H}\), therefore, total mass of the compound \(\quad=0.9218+0.0767\) \(=0.9985 \mathrm{~g}\)
\(\%\) of \(\mathrm{C}\) in the compound \(=\frac{0.9218}{0.9985} \times 100=92.32\)
\(\%\) of \(\mathrm{H}\) in the compound \(=\frac{0.0767}{0.9985} \times 100=7.68\)
\(\begin{array}{|c|c|c|c|c|c|}
\hline \text { Element } & \begin{array}{c}
\% \text { by } \\
\text { mass }
\end{array} & \begin{array}{c}
\text { Atomic } \\
\text { mass }
\end{array} & \begin{array}{c}
\text { Moles of } \\
\text { the element }
\end{array} & \begin{array}{c}
\text { Simplest } \\
\text { molar ratio }
\end{array} & \begin{array}{c}
\text { Simplest whole } \\
\text { number molar } \\
\text { ratio }
\end{array} \\
\hline \text { C } & 92.32 & 12 & \frac{92.32}{12}=7.69 & 1 & 1 \\
\hline \mathrm{H} & 7.68 & 1 & \frac{7.68}{1}=7.68 & 1 & 1 \\
\hline
\end{array}\)
\(\begin{aligned} \therefore & \text { Empirical Formula }=\mathrm{CH} \\ \therefore & 10.0 \mathrm{~L} \text { of the gas at STP weigh }=11.6 \mathrm{~g} \\ & 22.4 \mathrm{~L} \text { of the gas at STP will weigh } \\ &=\frac{11.6}{10.0} \times 22.4=25.984 \mathrm{~g} \text { or } 26 \text { approx } \\ & \text { Molar mass }=26 \mathrm{~g} \mathrm{~mol}{ }^{-1} \\ & \text { Empirical formula mass of } \mathrm{CH}=12+1=13 \\ \therefore \quad \mathrm{n}=& \frac{\text { Molecular mass }}{\text { E.F.mass }}=\frac{26}{13}=2 \\ \therefore \quad & \text { Molecular formula }=2 \times \mathrm{CH}=\mathrm{C}_2 \mathrm{H}_2 \end{aligned}\)