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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_(6)O_(3)`

B

`C_(2)H_(4)O`

C

`C_(2)H_(4)O`

D

`C_(3)H_(4)O_(2)`

Text Solution

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The correct Answer is:
A
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An organic compound contains C,H and O . If C (%):H^(%) = 6:1 , what is the simplest formula of the compound, given that one mole of the compound contains half as much oxygen as would be required to burn all the C and H atoms in it to CO_(2) and H_(2)O ?

Compound having molecular formula C_(3)H_(6)O_(2) is:-

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