A gaseous hydrocarbon gives upon combustion 0 | vedclass

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(D) The molar mass of $H_{2}O$ is $18 \ g/mol$,which contains $2 \ g$ of $H$.Therefore,$0.72 \ g$ of $H_{2}O$ contains $\frac{2}{18} 	imes 0.72 = 0.0

Vedclass · Accepted Answer

$C_{7}H_{8}$

Vedclass · Answer

(D) The molar mass of $H_{2}O$ is $18 \ g/mol$,which contains $2 \ g$ of $H$.Therefore,$0.72 \ g$ of $H_{2}O$ contains $\frac{2}{18} 	imes 0.72 = 0.08 \ g$ of $H$.The molar mass of $CO_{2}$ is $44 \ g/mol$,which contains $12 \ g$ of $C$.Therefore,$3.08 \ g$ of $CO_{2}$ contains $\frac{12}{44} 	imes 3.08 = 0.84 \ g$ of $C$.The molar ratio of $C:H$ is $\frac{0.84}{12} : \frac{0.08}{1} = 0.07 : 0.08 = 7 : 8$.Thus,the empirical formula is $C_{7}H_{8}$.

$A$ gaseous hydrocarbon gives upon combustion $0.72 \ g$ of water and $3.08 \ g$ of $CO_{2}$. The empirical formula of the hydrocarbon is:

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