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An organic compound contains \( \mathrm{C}=40 \%, \mathrm{H}=13.33 \% \) and \( \mathrm{N}=46.67 \% . \) Its emperical formula is
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Verified Answer
The correct answer is:
\( \mathrm{CH}_{4} \mathrm{~N} \)
For the empirical formula of the given compound
\begin{array}{|l|l|l|l|}
\hline Element & Percentage & Relative number of moles & Simplest ratio of moles \\
\hline C & \( 40 \) & \( \frac{40}{12}=3.33 \) & \( \frac{3.33}{3.33}=1 \) \\
\hline H & \( 13.33 \) & \( \frac{13.33}{1}=13.33 \) & \( \frac{13.33}{3.33}=4 \) \\
\hline N & \( 46.67 \) & \( \frac{46.67}{14}=3.33 \) & \( \frac{3.33}{3.33}=1 \) \\
\hline
\end{array}
Thus, the compound has the empirical formula \( \mathrm{CH}_{4} \mathrm{~N} \).
\begin{array}{|l|l|l|l|}
\hline Element & Percentage & Relative number of moles & Simplest ratio of moles \\
\hline C & \( 40 \) & \( \frac{40}{12}=3.33 \) & \( \frac{3.33}{3.33}=1 \) \\
\hline H & \( 13.33 \) & \( \frac{13.33}{1}=13.33 \) & \( \frac{13.33}{3.33}=4 \) \\
\hline N & \( 46.67 \) & \( \frac{46.67}{14}=3.33 \) & \( \frac{3.33}{3.33}=1 \) \\
\hline
\end{array}
Thus, the compound has the empirical formula \( \mathrm{CH}_{4} \mathrm{~N} \).
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