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(D) For gases,the ratio of volumes is equal to the ratio of moles.Let the hydrocarbon be $C_xH_y$.The combustion reaction is: $C_xH_y + (x + \frac{y}{

Vedclass · Accepted Answer

$C_5H_{12}$

Vedclass · Answer

(D) For gases,the ratio of volumes is equal to the ratio of moles.Let the hydrocarbon be $C_xH_y$.The combustion reaction is: $C_xH_y + (x + \frac{y}{4}) O_2 \longrightarrow x CO_2 + \frac{y}{2} H_2O(g)$.Given volumes: $0.5 \, L$ of $C_xH_y$ produces $2.5 \, L$ of $CO_2$ and $3.0 \, L$ of $H_2O$.Dividing by $0.5 \, L$: $1 \, mol$ of $C_xH_y$ produces $5 \, mol$ of $CO_2$ and $6 \, mol$ of $H_2O$.Comparing coefficients: $x = 5$ and $\frac{y}{2} = 6 \implies y = 12$.Therefore,the molecular formula is $C_5H_{12}$.

When $500 \, mL$ of a gaseous hydrocarbon is burned with sufficient $O_2$,it produces $2.5 \, L$ of $CO_2$ and $3.0 \, L$ of water vapor. What is the molecular formula of the hydrocarbon?

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