On complete combustion of 0.492 g of an orga | vedclass

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(A) Mass of organic compound $= 0.492 \ g$. Moles of $CO_2 = \frac{0.7938 \ g}{44 \ g/mol} = 0.01804 \ mol$. Moles of $C = 0.01804 \ mol$,so mass of $

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$46$

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(A) Mass of organic compound $= 0.492 \ g$. Moles of $CO_2 = \frac{0.7938 \ g}{44 \ g/mol} = 0.01804 \ mol$. Moles of $C = 0.01804 \ mol$,so mass of $C = 0.01804 	imes 12 = 0.2165 \ g$. Moles of $H_2O = \frac{0.4428 \ g}{18 \ g/mol} = 0.0246 \ mol$. Moles of $H = 2 	imes 0.0246 = 0.0492 \ mol$,so mass of $H = 0.0492 	imes 1 = 0.0492 \ g$. Mass of $O = 0.492 - (0.2165 + 0.0492) = 0.492 - 0.2657 = 0.2263 \ g$. Percentage of $O = \frac{0.2263}{0.492} 	imes 100 \approx 46 \%$.

On complete combustion of $0.492 \ g$ of an organic compound containing $C, H$ and $O$,$0.7938 \ g$ of $CO_2$ and $0.4428 \ g$ of $H_2O$ were produced. The percentage composition of oxygen in the compound is $........$

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