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A gaseous compound of nitrogen and hydrogen contains $12.5 \%$ (by mass) of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:
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Verified Answer
The correct answer is:
$\mathrm{N}_2 \mathrm{H}_4$
$\mathrm{N}_2 \mathrm{H}_4$
In an unknown compounds containing $\mathrm{N}$ and $\mathrm{H}$
given $\%$ of $\mathrm{H}=12.5 \%$
$$
\therefore \% \text { of } N=100-12.5=87.5 \%
$$
$$
\begin{array}{|c|c|c|c|}
\hline \text { Element } & \text { Percentage } & \text { Atomic ratio } & \text { Simple ratio } \\
\hline \mathrm{H} & 12.5 \% & \frac{12.5}{1}=12.5 & \frac{12.5}{6.25}=2 \\
\hline \mathrm{N} & 87.5 & \frac{87.5}{14}=6.25 & \frac{6.25}{6.25}=1 \\
\hline
\end{array}
$$
$2 \times$ vapour density $=$ Mol. wt $=16 \times 2=32$.
Molecular formula $=n \times$ empirical formula mass
$$
n=\frac{32}{16}=2
$$
$\therefore$ Molecular formula of the compound will be $=\left(\mathrm{NH}_2\right)_2$ $=\mathrm{N}_2 \mathrm{H}_4$
given $\%$ of $\mathrm{H}=12.5 \%$
$$
\therefore \% \text { of } N=100-12.5=87.5 \%
$$
$$
\begin{array}{|c|c|c|c|}
\hline \text { Element } & \text { Percentage } & \text { Atomic ratio } & \text { Simple ratio } \\
\hline \mathrm{H} & 12.5 \% & \frac{12.5}{1}=12.5 & \frac{12.5}{6.25}=2 \\
\hline \mathrm{N} & 87.5 & \frac{87.5}{14}=6.25 & \frac{6.25}{6.25}=1 \\
\hline
\end{array}
$$
$2 \times$ vapour density $=$ Mol. wt $=16 \times 2=32$.
Molecular formula $=n \times$ empirical formula mass
$$
n=\frac{32}{16}=2
$$
$\therefore$ Molecular formula of the compound will be $=\left(\mathrm{NH}_2\right)_2$ $=\mathrm{N}_2 \mathrm{H}_4$
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