On complete combustion of 0.492 text{ g} of a | vedclass

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(A) Step $1$: Mass of organic compound $= 0.492 	ext{ g}$.Step $2$: Moles of $CO_2 = \frac{0.7938 	ext{ g}}{44 	ext{ g/mol}} = 0.01804 	ext{ mol}$

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$46$

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(A) Step $1$: Mass of organic compound $= 0.492 	ext{ g}$.Step $2$: Moles of $CO_2 = \frac{0.7938 	ext{ g}}{44 	ext{ g/mol}} = 0.01804 	ext{ mol}$.Step $3$: Mass of $C = 0.01804 	ext{ mol} 	imes 12 	ext{ g/mol} = 0.21648 	ext{ g}$.Step $4$: Moles of $H_2O = \frac{0.4428 	ext{ g}}{18 	ext{ g/mol}} = 0.0246 	ext{ mol}$.Step $5$: Mass of $H = (2 	imes 0.0246 	ext{ mol}) 	imes 1 	ext{ g/mol} = 0.0492 	ext{ g}$.Step $6$: Mass of $O = 0.492 	ext{ g} - (0.21648 	ext{ g} + 0.0492 	ext{ g}) = 0.492 - 0.26568 = 0.22632 	ext{ g}$.Step $7$: Percentage of $O = \frac{0.22632}{0.492} 	imes 100 \approx 46 \%$.

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

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