Complete combustion of 0.858 g of compound X | vedclass

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Question

(A) Mass of $C = \frac{12}{44} 	imes 2.63 = 0.717 \ g$. Mass of $H = \frac{2}{18} 	imes 1.28 = 0.142 \ g$. Total mass of $C + H = 0.717 + 0.142 = 0.

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(A) Mass of $C = \frac{12}{44} 	imes 2.63 = 0.717 \ g$. Mass of $H = \frac{2}{18} 	imes 1.28 = 0.142 \ g$. Total mass of $C + H = 0.717 + 0.142 = 0.859 \ g$,which is approximately equal to the mass of compound $X$ $(0.858 \ g)$. Thus,the compound contains only $C$ and $H$. Moles of $C = \frac{0.717}{12} = 0.05975$. Moles of $H = \frac{0.142}{1} = 0.142$. Ratio $C:H = 0.05975 : 0.142 \approx 1 : 2.37$. Converting to simple integer ratio: $1 : 2.37 \approx 3 : 7$. The empirical formula is $C_3H_7$. The empirical formula mass is $(3 	imes 12) + (7 	imes 1) = 43 \ g/mol$. The lowest possible molecular mass is $43 \ g/mol$.

Complete combustion of $0.858 \ g$ of compound $X$ gives $2.63 \ g$ of $CO_2$ and $1.28 \ g$ of $H_2O$. The lowest molecular mass $X$ can have is ............ $g$.

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