Question

A gaseous hydrocarbon gives upon combustion, \(0.72 \mathrm{~g}\) of water and \(3.08 \mathrm{~g}\) of \(\mathrm{CO}_{2}\). The empirical formula of the hydrocarbon is
(a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b)) \(\mathrm{C}_{3} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5}\) (d) \(\mathrm{C}_{7} \mathrm{H}_{3}\)

Answer 1

Correct option (d) \(\mathrm{C}_{7} \mathrm{H}_{8}\)
Explanation:
$$
\begin{aligned}
&18 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \text { contain } 2 \mathrm{~g} \mathrm{H} . \\
&\therefore 0.72 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \text { contain } 0.08 \mathrm{~g} \mathrm{H} \\
&\quad 44 \mathrm{~g} \mathrm{CO}_{2} \text { contain } 12 \mathrm{~g} \mathrm{C} . \\
&\therefore 3.08 \mathrm{~g} \mathrm{CO}_{2} \text { contain } 0.84 \mathrm{~g} \mathrm{C} . \\
&\therefore \quad \mathrm{C}: \mathrm{H}=\frac{0.84}{12}: \frac{0.08}{1}=0.07: 0.08=7: 8 \\
&\therefore \text { Empirical formula }=\mathrm{C}_{7} \mathrm{H}_{6}
\end{aligned}
$$

Answer 2

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