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$4 \mathrm{~g}$ of a hydrocarbon on complete combustion gave $12.571 \mathrm{~g}$ of $\mathrm{CO}_2$ and $5…
$4 \mathrm{~g}$ of a hydrocarbon on complete combustion gave $12.571 \mathrm{~g}$ of $\mathrm{CO}_2$ and $5.143 \mathrm{~g}$ of water. What is the empirical formula of the hydrocarbon?
$\mathrm{CH}$ $\mathrm{CH}_2$ $\mathrm{CH}_3$ $\mathrm{C}_2 \mathrm{H}_3$ Solution
Percentage of
$
\begin{aligned}
\mathrm{C} & =\frac{12}{14} \times \frac{\text { weight of } \mathrm{CO}_2 \times 100}{\text { weight of organic compound }} \\
& =\frac{12}{44} \times \frac{12.571}{4} \times 100=85.7 \%
\end{aligned}
$
Percentage of $\mathrm{H}$
$
\begin{aligned}
& =\frac{2}{18} \times \frac{\text { weight of water } \times 100}{\text { weight of organic compound }} \\
& =\frac{2}{18} \times \frac{5.143 \times 100}{4}=14.3 \%
\end{aligned}
$
\begin{array}{|c|c|c|c|c|}
\hline \begin{array}{c}\text { S. } \\
\text { No. }\end{array} & \% & \begin{array}{c}\text { Atomic } \\
\text { wt. }\end{array} & \begin{array}{c}\% \text { atomic } \\
\text { wt. }\end{array} & \begin{array}{c}\text { Simple } \\
\text { ratio }\end{array} \\
\hline 1 & C 85.7 & 12 & \frac{857}{12}=7.15 & \frac{715}{7.15}=1 \\
\hline 2 & H 14.3 & 1 & \frac{14.3}{1}=14.3 & \frac{14.3}{7.15}=2 \\
\hline
\end{array}
$\therefore$ Empirical formula $=\mathrm{CH}_2$
Asked in: AP EAMCET 2002
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