An organic compound contains $20.0 \% \mathrm{C}, 6.66 \% \mathrm{H}, 47.33 \% \mathrm{~N}$ and the rest was…
- $\mathrm{CH}_{4} \mathrm{~N}_{2} \mathrm{O}$
- $\mathrm{CH}_{2} \mathrm{NO}$
- $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{NO}$
- $\mathrm{CH}_{18} \mathrm{NO}$
Solution
\mathrm{C}: \mathrm{N}: \mathrm{O}: \mathrm{H}:: \frac{20}{12}: \frac{47.33}{14}: \frac{26.01}{16}: \frac{6.66}{1}:: 1.667: 3.381: 1.626: 6.66:: 1: 2: 1: 4
$$
Empirical formula $\mathrm{CN}_{2} \mathrm{OH}_{4}$
Molar empirical mass is $60 \mathrm{~g} \mathrm{~mol}^{-1}$ (same as the given molar mass). Hence, Molecular formula is $\mathrm{CN}_{2} \mathrm{OH}_{4}$.
Asked in: JEE-TOPICTESTS-CHEMISTRY
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