An organic compound on analysis gave C = 48 , | vedclass

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Question

(A) Step $1$: Determine the empirical formula.ElementMass $(g)$MolesSimple Ratio$C$$48$$48/12 = 4$$1$$H$$8$$8/1 = 8$$2$$N$$56$$56/14 = 4$$1$Empirical

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$C_4H_8N_4$

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(A) Step $1$: Determine the empirical formula.ElementMass $(g)$MolesSimple Ratio$C$$48$$48/12 = 4$$1$$H$$8$$8/1 = 8$$2$$N$$56$$56/14 = 4$$1$Empirical formula = $CH_2N$.Empirical formula mass = $12 + 2(1) + 14 = 28 \,g/mol$.Step $2$: Determine the molar mass.At $NTP$, $22400 \,mL$ of gas corresponds to $1 \,mole$.Given $200 \,mL$ of the compound weighs $1 \,g$.Therefore, $22400 \,mL$ weighs $\frac{1}{200} 	imes 22400 = 112 \,g$.Molar mass = $112 \,g/mol$.Step $3$: Calculate the molecular formula.$n = \frac{	ext{Molar mass}}{	ext{Empirical formula mass}} = \frac{112}{28} = 4$.Molecular formula = $n 	imes (CH_2N) = 4 	imes (CH_2N) = C_4H_8N_4$.

An organic compound on analysis gave $C = 48 \,g$, $H = 8 \,g$ and $N = 56 \,g$. The volume of $1.0 \,g$ of the compound was found to be $200 \,mL$ at $NTP$. What is the molecular formula of the compound?

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