On complete combustion,1.0 g of an organic c | vedclass

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(A) Moles of $C = n_{CO_2} = \frac{1.46}{44} \approx 0.03318 \ mol$. Mass of $C = 0.03318 	imes 12 = 0.398 \ g$. Moles of $H = 2 	imes n_{H_2O} = 2

Vedclass · Accepted Answer

$30$

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(A) Moles of $C = n_{CO_2} = \frac{1.46}{44} \approx 0.03318 \ mol$. Mass of $C = 0.03318 	imes 12 = 0.398 \ g$. Moles of $H = 2 	imes n_{H_2O} = 2 	imes \frac{0.567}{18} = 0.063 \ mol$. Mass of $H = 0.063 	imes 1 = 0.063 \ g$. Mass of $O = 1.0 - (0.398 + 0.063) = 0.539 \ g$. Moles of $O = \frac{0.539}{16} \approx 0.0337 \ mol$. Ratio of $C:H:O = 0.03318 : 0.063 : 0.0337 \approx 1:2:1$. Empirical formula $= CH_2O$. Empirical formula mass $= 12 + (2 	imes 1) + 16 = 30 \ g \ mol^{-1}$.

On complete combustion,$1.0 \ g$ of an organic compound $(X)$ gave $1.46 \ g$ of $CO_2$ and $0.567 \ g$ of $H_2O$. The empirical formula mass of compound $(X)$ is $.......... \ g \ mol^{-1}$. (Given molar mass in $g \ mol^{-1}: C: 12, H: 1, O: 16$)

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