States of Matter - Result Question 93
93. A hydrocarbon contains 10.5 g of carbon per gram of hydrogen. 1 L of the vapour of the hydrocarbon at $127^{\circ} \mathrm{C}$ and 1 atm pressure weighs 2.8 g . Find the molecular formula of the hydrocarbon.
(1980, 3M)
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Solution: 93.
$ \begin{aligned} &\text { First we determine empirical formula as }\\ &\begin{array}{llc} \hline & \mathrm{C} & \mathrm{H} \\ \hline \text { Weight } & 10.5 & 1 \\ \hline \text { Mole } & \frac{10.5}{12}=0.875 & 1 \\ \hline \text { Simple ratio } & 1 & 1 / 0.875=1.14 \\ \hline \text { Whole no. } & 7 & 8 \\ \hline \end{array} \end{aligned} $
$\Rightarrow$ Empirical formula $=\mathrm{C}_7 \mathrm{H}_8$
From gas equation : $p V=\left(\frac{w}{M}\right) R T$
$$ M=\frac{w R T}{p V}=\frac{2.8 \times 0.082 \times 400}{1 \times 1}=91.84 \approx 92 $$
$\because$ Molar mass $(M)$ is same as empirical formula weight.
Molecular formula $=$ Empirical formula $=\mathrm{C}_7 \mathrm{H}_8$
BETA
