8.4 ml of a gaseous hydrocarbon (A) was burnt with 50 ml of `O_(2)` in an eudiometer tube. The volume of the products after cooling to room temperature was 37.4 ml, when reacted with NaOH, the volume contracted to 3.8 ml. What is the molecular formula of A?

Explanation

Let `C_(x)H_(y (g))` be the hydrocarbon.
`C_(X)H_(Y(g))+ (x+(y)/(4))O_(2(g)) rarr xCo_(2(g))+(y)/(2)H_(2)O_(g)` [Remember this balanced combustion equation for `C_(x)H_(y)`]
From Gay Lussac’s Law of combining volume, we get : 1 vol. of `C_(x)H_(y) =(x+(y)/(4))` vol. of `O_(2)`
= x vol. of `CO_(2) = (y)/(2)` vol. of `H_(2)O`
Contraction in volume = `V_(R)- V_(P)` = (8.4 + 50) - (37.4) = 21 ml 
From equation, we have, the contraction `= 1 + (x + (y)/(4)) - (x + 0)` Contraction = 1 + `(y)/(4)implies` For 8.4 mL of `C_(x)H_(y)`, the contraction = 8.4 (1 + `(y)/(4)`)
`8.4 (1 + (y)/(4)) = 21 implies y = 6` 
After treating with NaOH, there is a contraction of (37.4 - 3.8) = 33.6 ml, which is equal to the volume of `CO_(2)` produced. Volume of `CO_(2)` produced by 8.4 ml of hydrocarbon = `8.4 x implies 8.4 x = 33.6 implies x = 4` So the molecular formula of hydrocarbon is `C_(4)H_(6)`.