16 mL of a hydrocarbon gas was exploded with excess of oxygen. On cooling, the volume of the resulting gaseous mixture was reduced by 48 mL. When KOH was added, there was a further decrease of 48 mL in the volume. Find the molecular formula of the compound.

Explanation

Since water vapour condenses to practically zero volume of water, the decrease in volume on cooling is the volume of water vapour. `CO_(2)` is absorbed by KOH and so volume of `CO_(2)` is equal to 48 mL.
Thus,
`{:(,C_(x)H_(y)"(say)",+,O_(2),rarr,CO_(2),+,H_(2)O),(,"16 mL",,,,"48 mL",,"48 mL"),("or","16 moles",,,,"48 moles",,"48 moles"):}`
Since C and H are completely converted to `CO_(2)` and `H_(2)O` respectively, applying POAC for C and H atoms we get respectively,
`x xx` moles of `C_(x)H_(y)=1xx` moles of `CO_(2)`
`16x=48. therefore x=3`
`yxx` moles of `C_(x)H_(y)=2xx` moles of `H_(2)O`
`16y=2xx48. therefore y=6`.
Hence, the formula of the hydrocarbon is `C_(3)H_(6)`.