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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_(2)H_(3)`

C

`CH_(2)`

D

`CH_(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
`%C=(12)/(44)xx(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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