A compound contains $11.99 \% \mathrm{~N}, 13.70 \% \mathrm{O}, 9.25 \% \mathrm{~B}$ and $65.06 \%…

A compound contains $11.99 \% \mathrm{~N}, 13.70 \% \mathrm{O}, 9.25 \% \mathrm{~B}$ and $65.06 \% \mathrm{~F}$. Its empirical formula is (molar mass of $\mathrm{B}$ is $10.8 \mathrm{~g} \mathrm{~mol}^{-1}$ )
  1. $\mathrm{NOBF}_{2}$
  2. $\mathrm{NOBF}_{4}$
  3. $\mathrm{N}_{2} \mathrm{OF}_{2}$
  4. $\mathrm{NO}_{2} \mathrm{~F}_{2}$

Solution

$$
\mathrm{N}: \mathrm{O}: \mathrm{B}: \mathrm{F}:: \frac{11.99}{14}: \frac{13.7}{16}: \frac{9.25}{10.8}: \frac{65.06}{19}:: 0.856: 0.856: 0.856: 3.424:: 1: 1: 1: 4
$$
Empirical formula: $\mathrm{NOBF}_{4}$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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