Sixteen millilitre of a hydrocarbon gas was exploded with an excess of `O_2`. On cooling ,t he volume of the resulting gaseous mixture was reduced by 48 ml. When `KOH` was added, there was a further decrease of 48 ml in bolume. Find the volucular formula of the compound.

Explanation

Since water vapour condenses to practically zero volume of water, the decrease in the volume on cooling i.e., contraction of volume is the volume of water vapour.
`underset(16" ml")underset(1" ml")(C_(x)H_(y))+underset((x+(y)/(4))ml)underset((x+(y)/(4))ml)((x+(y)/(4)))O_(2)tounderset(16x" ml")underset(x" ml")(xCO_(2))+underset(-)underset(-)((y)/(2)H_(2)O)`
Volume of `CO_2=16x=48` `x=3ml`
Volume of reactant `=`Volume of product `+` Contration
`16ml+V_(O_2)(excess)=16xx3ml+V_(O_2)(l eft)+48ml`
`V_(O_2)(excess)-V_(O_2)(l eft)=80ml`
volume of `O_2` used `=80ml`
`16(x+(y)/(4))=80ml`
Solving the equation by putting `x=3`, we get `y=8.`