In a compound,C, H, and N are present in the | vedclass

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(C) ElementRatioRelative MolesSimplest Ratio$C$$9$$9/12 = 0.75$$3$$H$$1$$1/1 = 1$$4$$N$$3.5$$3.5/14 = 0.25$$1$Empirical formula = $C_3H_4N$Empirical f

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$C_6H_8N_2$

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(C) ElementRatioRelative MolesSimplest Ratio$C$$9$$9/12 = 0.75$$3$$H$$1$$1/1 = 1$$4$$N$$3.5$$3.5/14 = 0.25$$1$Empirical formula = $C_3H_4N$Empirical formula mass = $(3 	imes 12) + (4 	imes 1) + (1 	imes 14) = 36 + 4 + 14 = 54 \ g/mol$$n = \frac{	ext{Molecular weight}}{	ext{Empirical formula mass}} = \frac{108}{54} = 2$Molecular formula = $n 	imes (C_3H_4N) = 2 	imes (C_3H_4N) = C_6H_8N_2$

In a compound,$C, H,$ and $N$ are present in the weight ratio of $9:1:3.5$. If its molecular weight is $108$,then its molecular formula will be:

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