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

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(C) $1$. Calculate the relative number of moles of each element by dividing the weight ratio by their respective atomic weights: $C = 9/12 = 0.75$,$H

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$C_6H_8N_2$

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(C) $1$. Calculate the relative number of moles of each element by dividing the weight ratio by their respective atomic weights: $C = 9/12 = 0.75$,$H = 1/1 = 1$,$N = 3.5/14 = 0.25$. $2$. Determine the simplest molar ratio by dividing by the smallest value $(0.25)$: $C = 0.75/0.25 = 3$,$H = 1/0.25 = 4$,$N = 0.25/0.25 = 1$. $3$. The empirical formula is $C_3H_4N$. $4$. Calculate the empirical formula weight: $(3 	imes 12) + (4 	imes 1) + (1 	imes 14) = 36 + 4 + 14 = 54$. $5$. Determine the value of $n = 	ext{Molecular weight} / 	ext{Empirical formula weight} = 108 / 54 = 2$. $6$. The molecular formula is $n 	imes (C_3H_4N) = 2 	imes (C_3H_4N) = C_6H_8N_2$.

In a compound,$C$,$H$,and $N$ atoms are present in $9 : 1 : 3.5$ by weight. The molecular weight of the compound is $108$. What is the molecular formula of the compound?

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