An organic compound containing carbon,hydroge | vedclass

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Question

(A) $1$. Calculate the percentage of oxygen: $\%O = 100 - (52.20 + 13.04) = 34.76\%$.$2$. Calculate the molar ratio of elements:$C = 52.20 / 12 = 4.35

Vedclass · Accepted Answer

$C_2H_6O$

Vedclass · Answer

(A) $1$. Calculate the percentage of oxygen: $\%O = 100 - (52.20 + 13.04) = 34.76\%$.$2$. Calculate the molar ratio of elements:$C = 52.20 / 12 = 4.35$$H = 13.04 / 1 = 13.04$$O = 34.76 / 16 = 2.17$$3$. Divide by the smallest value $(2.17)$:$C = 4.35 / 2.17 \approx 2$$H = 13.04 / 2.17 \approx 6$$O = 2.17 / 2.17 = 1$$4$. The empirical formula is $C_2H_6O$. The empirical formula mass is $(2 	imes 12) + (6 	imes 1) + 16 = 46$.$5$. Molecular weight = $2 	imes 	ext{Vapour Density} = 2 	imes 23 = 46$.$6$. Since molecular weight equals empirical formula mass,the molecular formula is $C_2H_6O$.

An organic compound containing carbon,hydrogen,and oxygen contains $52.20\%$ carbon and $13.04\%$ hydrogen. The vapour density of the compound is $23$. Its molecular formula will be:

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