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(B) $1$. Calculate the molar mass of the hydrocarbon: At $STP$,$22400 \ mL$ corresponds to $1 \ mole$ of gas. Given $488 \ mL$ weighs $1.68 \ g$,so $2

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Alkene

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(B) $1$. Calculate the molar mass of the hydrocarbon: At $STP$,$22400 \ mL$ corresponds to $1 \ mole$ of gas. Given $488 \ mL$ weighs $1.68 \ g$,so $22400 \ mL$ weighs: $\frac{1.68 	imes 22400}{488} \approx 77.14 \ g/mol$. $2$. Determine the empirical formula: Carbon = $86\%$,Hydrogen = $100 - 86 = 14\%$. Moles of $C = \frac{86}{12} \approx 7.17$. Moles of $H = \frac{14}{1} = 14$. Ratio $C:H = 7.17:14 \approx 1:2$. Empirical formula is $CH_2$. $3$. Determine molecular formula: Empirical formula mass = $12 + 2 = 14 \ g/mol$. $n = \frac{	ext{Molar mass}}{	ext{Empirical mass}} = \frac{77.14}{14} \approx 5.5$. Since the hydrocarbon is a gas at $STP$ and follows $C_nH_{2n}$,it belongs to the alkene series.

$A$ hydrocarbon contains $86\%$ carbon. $488 \ mL$ of the hydrocarbon weighs $1.68 \ g$ at $STP$. Then the hydrocarbon is an:

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