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The ration of mass per cent of C and H o...

The ration of mass per cent of C and H of an organic compound `(C_(x)H_(y)O_(z)) "is"6:1`. If one molecule of the above compound `(C_(x)H_(Y)O_(z))` contains half as much oxygen as required to burn one molecule of compound `C_(x)H_(Y)` compleltely to `CO_(2) and H_(2)O`. The empirial formula of compound `C_(x)H_(y)O_(z)` is:

A

`C_3 H_4 O_2`

B

`C_2H_4O_3`

C

`C_3H_6O_3`

D

`C_2H_4O`

Text Solution

Verified by Experts

The correct Answer is:
B

So `C_X H_Y` have empirical formula : `CH_2`
For Burning a `CH_2` unit, oxygen required is `3/2` mol
`CH_2 + 3/2 O_2 to CO_2 + H_2O`
Empirical formula is `2 xx (CH_2O_(3//2)) rArr C_2H_4O_3`.
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