0.53 g of an organic compound (x) when heate | vedclass

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(D) $1$. Calculate percentage of $C$: Moles of $CO_2 = \frac{1.32 \ g}{44 \ g \ mol^{-1}} = 0.03 \ mol$. Mass of $C = 0.03 \ mol 	imes 12 \ g \ mol^{

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$4$

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(D) $1$. Calculate percentage of $C$: Moles of $CO_2 = \frac{1.32 \ g}{44 \ g \ mol^{-1}} = 0.03 \ mol$. Mass of $C = 0.03 \ mol 	imes 12 \ g \ mol^{-1} = 0.36 \ g$. $\% C = \frac{0.36 \ g}{1.0 \ g} 	imes 100 = 36\%$. $2$. Calculate percentage of $Br$: Moles of $AgBr = \frac{0.75 \ g}{188 \ g \ mol^{-1}} \approx 0.00399 \ mol$. Mass of $Br = 0.00399 \ mol 	imes 80 \ g \ mol^{-1} \approx 0.3192 \ g$. $\% Br = \frac{0.3192 \ g}{0.53 \ g} 	imes 100 \approx 60.2\%$. $3$. Calculate percentage of $H$: $\% H = 100 - (\% C + \% Br) = 100 - (36 + 60.2) = 3.8\%$. Rounding to the nearest integer, we get $4\%$.

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