A gaseous compound of nitrogen and hydrogen c | vedclass

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(D) Given,percentage of $H = 12.5\%$.Therefore,percentage of $N = 100 - 12.5 = 87.5\%$.ElementPercentageAtomic ratioSimple ratio$H$$12.5\%$$\frac{12.5

Vedclass · Accepted Answer

$N_2H_4$

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(D) Given,percentage of $H = 12.5\%$.Therefore,percentage of $N = 100 - 12.5 = 87.5\%$.ElementPercentageAtomic ratioSimple ratio$H$$12.5\%$$\frac{12.5}{1} = 12.5$$\frac{12.5}{6.25} = 2$$N$$87.5\%$$\frac{87.5}{14} = 6.25$$\frac{6.25}{6.25} = 1$Empirical formula $= NH_2$.Empirical formula mass $= 14 + (2 	imes 1) = 16 \ g/mol$.Molecular mass $= 2 	imes 	ext{Vapour density} = 2 	imes 16 = 32 \ g/mol$.$n = \frac{	ext{Molecular mass}}{	ext{Empirical formula mass}} = \frac{32}{16} = 2$.Molecular formula $= n 	imes (	ext{Empirical formula}) = 2 	imes (NH_2) = N_2H_4$.

$A$ 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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