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An organic compound containing $\mathrm{C}$ and $\mathrm{H}$ has $92.3 \%$ of carbon, Its empirical formula is
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Verified Answer
The correct answer is:
$\mathrm{CH}$
\begin{array}{c|l|l}
\hline Element & \begin{array}{l}\text { Percentage } \\
\text { atomic wt. }\end{array} & Simple ratio \\
\hline \mathrm{C} & \frac{92.3}{12}=7.69 & \frac{7.69}{7.69}=1 \\
\mathrm{H} & \frac{7.7}{1}=7.70 & \frac{7.70}{7.69} \approx 1 \\
\hline
\end{array}
$\therefore$ Empirical formula $=\mathrm{CH}$
\hline Element & \begin{array}{l}\text { Percentage } \\
\text { atomic wt. }\end{array} & Simple ratio \\
\hline \mathrm{C} & \frac{92.3}{12}=7.69 & \frac{7.69}{7.69}=1 \\
\mathrm{H} & \frac{7.7}{1}=7.70 & \frac{7.70}{7.69} \approx 1 \\
\hline
\end{array}
$\therefore$ Empirical formula $=\mathrm{CH}$
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