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An organic compound having $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$ has $13.13 \% \mathrm{H}, 52.14 \% \mathrm{C}$ and $34.73 \%$ O. Its molar mass is $46.068 \mathrm{~g}$. What are its empirical and molecular formulae?
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Verified Answer
The correct answer is:
$\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}, \mathrm{C}_2 \mathrm{H}_6 \mathrm{O}$
$$
\begin{aligned}
& \therefore \text { Empinical formula }=\mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \\
& \begin{aligned}
\text { Empirical formula mass } & =2(12)+6(1)+1(16) \\
& =46
\end{aligned}
\end{aligned}
$$
Molecular mass $=46$
$$
\begin{aligned}
& \because n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{46}{46}=1 \\
& \begin{aligned}
\therefore \text { Molecular formula } & =(\text { empirical formal)n } \\
& =\left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}\right)_1 \\
& =\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}
\end{aligned}
\end{aligned}
$$
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