On complete combustion,0.30 , g of an organic | vedclass

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(A) The mass of carbon in $CO_2$ is calculated as: $	ext{Mass of } C = \frac{12}{44} 	imes 	ext{mass of } CO_2$. Substituting the given values: $

Vedclass · Accepted Answer

$18$

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(A) The mass of carbon in $CO_2$ is calculated as: $	ext{Mass of } C = \frac{12}{44} 	imes 	ext{mass of } CO_2$. Substituting the given values: $	ext{Mass of } C = \frac{12}{44} 	imes 0.20 \, g = 0.05454 \, g$. The percentage of carbon in the organic compound is: $\% \, C = \frac{	ext{Mass of } C}{	ext{Mass of compound}} 	imes 100$. $\% \, C = \frac{0.05454}{0.30} 	imes 100 = 18.18 \, \%$. Rounding to the nearest integer,we get $18 \, \%$.

On complete combustion,$0.30 \, g$ of an organic compound gave $0.20 \, g$ of carbon dioxide and $0.10 \, g$ of water. The percentage of carbon in the given organic compound is $.....$ (Nearest Integer).

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