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4g of hydrocarbon on complete combustion...

4g of hydrocarbon on complete combustion gave 12.571g of `CO_(2)` and 5.143g of water. What is the empirical formula of the hydrocarbon?

A

CH

B

`C_2H_3`

C

`CH_2`

D

`CH_3`

Text Solution

Verified by Experts

The correct Answer is:
C
`%C=12/44xx(12.571)/(4.0) xx100=85.7`
`% H = 2/18 xx (5.143)/(4.0) xx100 = 14.3`
The mole ratio of C to H is `(85.7)/12 : (14.3)/(1) = 7.14 : 14.3= 1: 2 =CH_2`
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