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A compound (60 gm) on analysis gave C=24...

A compound (60 gm) on analysis gave `C=24gm` H=4gm` and `O=31gm`, its empirical formula is

A

`C_2H_2O`

B

`C_2H_4O_2`

C

`CH_2O`

D

`CH_2O_2`

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(C,,H,,O),((40)/(3),,(6.5)/(1),,(53.5)/(16)),(3.33,,6.5,,3.34),(1,,2,,1):}`
`EF=C_2H_4O_2`
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