In an organic compound,the weight ratio of C, | vedclass

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(C) The ratio of atoms is given by $C:H:N = \frac{9}{12} : \frac{1}{1} : \frac{3.5}{14}$.Simplifying the ratio: $C:H:N = 0.75 : 1 : 0.25$.Dividing by

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$C_6H_8N_2$

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(C) The ratio of atoms is given by $C:H:N = \frac{9}{12} : \frac{1}{1} : \frac{3.5}{14}$.Simplifying the ratio: $C:H:N = 0.75 : 1 : 0.25$.Dividing by the smallest value $(0.25)$: $C:H:N = \frac{0.75}{0.25} : \frac{1}{0.25} : \frac{0.25}{0.25} = 3 : 4 : 1$.Thus,the empirical formula is $C_3H_4N$.The empirical formula weight is $(3 	imes 12) + (4 	imes 1) + (1 	imes 14) = 36 + 4 + 14 = 54$.Calculate $n = \frac{	ext{Molecular weight}}{	ext{Empirical formula weight}} = \frac{108}{54} = 2$.Molecular formula = $n 	imes (	ext{Empirical formula}) = 2 	imes (C_3H_4N) = C_6H_8N_2$.

In an organic compound,the weight ratio of $C$,$H$,and $N$ atoms is $9 : 1 : 3.5$. If the molecular weight of the compound is $108$,what is its molecular formula?

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