Question

# $$500\ mL$$ of a gaseous hydrocarbon when burnt in excess of $$\displaystyle O_{2}$$ gave $$2.5\ L$$ of $$\displaystyle CO_{2}$$ and $$3.0\ L$$ of water vapour under same conditions. The molecular formula of the hydrocarbon is :

A
C4H8
B
C4H10
C
C5H6
D
C5H12

Solution

## The correct option is C $$\displaystyle C_{5}H_{12}$$The balanced chemical equation for the combustion reaction is$$\underset {\underset {\displaystyle 1 \: L}{\displaystyle 0.5 \: L}}{C_xH_y} +O_2 \rightarrow x \underset {\underset {\displaystyle 5 \: L}{\displaystyle 2.5 \: L}}{CO_2} + \frac {y}{2} \underset {\underset {\displaystyle 6 \: L}{\displaystyle 3 \: L}}{H_2O}$$The volume is directly proportional to the number of moles.Thus, 1 mole of hydrocarbon on combustion will give 5 moles of $$CO_2$$ and 6 moles of $$H_2O$$.Thus, the molecular formula of the hydrocarbon is $$\displaystyle C_{5}H_{12}$$.Chemistry

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