\(15 \mathrm{~cm}^3\) of a gaseous hydrocarbon required for the complete combustion of \(75 \mathrm{~cm} 3\) of oxygen and yielded \(45 \mathrm{~cm} 3\) of carbon (IV) oxide. Calculate the molecular formula of the hydrocarbon.
A. \(\mathrm{CH}_4 \)
B. \(\mathrm{C}_2 \mathrm{H}_6 \)
C. \(\mathrm{C}_3 \mathrm{H}_8\)
D. \(\mathrm{CaH}_{30}\)
Show Answer Show Explanation Correct Answer: D Explanation \begin{array}{l} \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(x+\frac{\mathrm{y}}{4}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O} \\ \downarrow \\ 15 \mathrm{~cm}^3{ }^1 \\ \frac{\mathrm{n}_{\mathrm{C}_x \mathrm{H}_y}}{\mathrm{~V}_{\mathrm{C}_x \mathrm{H}_y}}=\frac{\mathrm{n}_{\mathrm{O}_2}}{\mathrm{~V}_{\mathrm{O}_2}} \text { (Avogadro's law) } \\ \frac{1}{15}=\frac{x+\frac{\mathrm{y}}{4}}{75} \\ 75=15(x+\mathrm{y} / 4) \\ 5=x+y / 4 \ldots \ldots \ldots \ldots . \\ \text { Also, } \\ \frac{\mathrm{n}_{\mathrm{C}_x} \mathrm{H}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{C}_{\mathrm{x}}} \mathrm{H}_{\mathrm{y}}}=\frac{\mathrm{n}_{\mathrm{O}_2}}{\mathrm{~V}_{\mathrm{O}_2}} \\ \frac{1}{15}=\frac{x}{45} \\ 15 x=45 \\ x=45 / 15=3 \\ \text { Put } x=3 \text { in equation (i) } \\ 5=3+\mathrm{y} / 4 \\ \mathrm{y} / 4=2, \mathrm{y}=4 \times 2=8 \\ \text { thus, } \mathrm{C}_x \mathrm{H}_{\mathrm{y}}=\mathrm{C}_3 \mathrm{H}_8 . \end{array}