8.4 " mL of " gaseous hydrocarbon A was burnt with 50 " mL of " O_2 in a 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.

8.4 " mL of " gaseous hydrocarbon A was burnt with 50 " mL of " O_2 in

1.	8.4 " mL of " gaseous hydrocarbon A was burnt with 50 " mL of " O_2 in a 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.
Answer» Solution :Let the formula of hydrocarbon `(A)=C_(x)H_(y)` `thereforeundersetunderset(8.4)(1 vol) (C_(x)H_(y)(g))+undersetunderset(8.4(x+(y)/(4))(vol)(x+(y)/(4))vol)((x+(y)/(4))O_(2)(g))toundersetunderset(8.4 x vol)(x vol)(xCO_(2)(g))+(y)/(2)H_(2)(g)` Contraction in volume `=V_(R)-V_(P)` `=(8.4+50)-(37.4)` `=21mL` From equation the contraction is (volume of `H_(2)O` is not taken as it condenses) `therefore` Contraction`=(8.4[1+(cancel(x)+(y)/(4))-(cancel(x)+0)]` `implies8.4(1+(y)/(4))=21impliesy=6` After treating with `NaOH`, there is a contraction of `(37.4-3.8)=33.6mL` which is equal tot he VLUME of `CO_(2)` produced. Volume of `CO_(2)` produced by `8.4 mL` of hydrocarbon `=8.4x` `therefore8.4x` `therefore 8.4x=33.6` `ximplies4mL` Alternate METHOD: After `NaOH` tretment, the volume is reduced to `3.8 mL`, which CORRESPONDS to the volume of `O_(2)` usused. Volume of `O_(2)` (unused)`=`volume of `O_(2)` taken`-` volume of of `O_(2)` used `implies50-8.4(x+(y)/(4))=3.8` Sove to get, `y=6` Hence, hydrocarbon is `C_(4)H_(6)`.