Mass percentage of C, H and O in a compound a | vedclass

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(D) To find the molecular formula,we first determine the empirical formula:ElementPercentageAtomic MassMolesSimplest Ratio$C$$60\%$$12$$60 / 12 = 5$$5

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$C_3H_8O$

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(D) To find the molecular formula,we first determine the empirical formula:ElementPercentageAtomic MassMolesSimplest Ratio$C$$60\%$$12$$60 / 12 = 5$$5 / 1.667 \approx 3$$H$$13.33\%$$1$$13.33 / 1 = 13.33$$13.33 / 1.667 \approx 8$$O$$26.67\%$$16$$26.67 / 16 = 1.667$$1.667 / 1.667 = 1$Empirical formula = $C_3H_8O$Empirical formula mass = $(3 	imes 12) + (8 	imes 1) + (1 	imes 16) = 36 + 8 + 16 = 60$Given Molecular weight = $60$$n = 	ext{Molecular weight} / 	ext{Empirical formula mass} = 60 / 60 = 1$Molecular formula = $n 	imes (	ext{Empirical formula}) = 1 	imes (C_3H_8O) = C_3H_8O$.

Mass percentage of $C, H$ and $O$ in a compound are $60\%$,$13.33\%$ and $26.67\%$ respectively. The molecular weight of the compound is $60$. The molecular formula of the compound is

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