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A compound contains $11.99 % N, 13.70 % O, 9.25 % B$ and $65.06 % F$ Its empirical formula is ( molar mass of B is $10.8 g {mol}^{- 1}$ )

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${NO}_{2} F_{2}$

${NOBF}_{4}$

${NOBF}_{2}$

$N_{2} {OF}_{2}$

answer is B.

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Detailed Solution

$N : O : B : F :: \frac{11.99}{14} : \frac{13.7}{16} : \frac{9.25}{10.8} : \frac{65.06}{19} :: 0.856 : 0.856 : 0.856 : 3.424 :: 1 : 1 : 1 : 4$

Empirical formula: ${NOBF}_{4}$

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