A compound contains 69.5% oxygen and 30.5% ni | vedclass

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(C) To find the molecular formula,we first determine the empirical formula:Element$\%$At. wtAtomic ratioSimplest ratio$N$$30.5$$14$$\frac{30.5}{14} =

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$N_2O_4$

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(C) To find the molecular formula,we first determine the empirical formula:Element$\%$At. wtAtomic ratioSimplest ratio$N$$30.5$$14$$\frac{30.5}{14} = 2.18$$\frac{2.18}{2.18} = 1$$O$$69.5$$16$$\frac{69.5}{16} = 4.34$$\frac{4.34}{2.18} \approx 2$The empirical formula is $NO_2$.The empirical formula mass $= 14 + 2 	imes 16 = 46$.The value of $n = \frac{	ext{Molecular weight}}{	ext{Empirical formula mass}} = \frac{92}{46} = 2$.Therefore,the molecular formula $= n 	imes (NO_2) = 2 	imes NO_2 = N_2O_4$.

$A$ compound contains $69.5\%$ oxygen and $30.5\%$ nitrogen and its molecular weight is $92$. The molecular formula of the compound is

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