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:

Options

A. `C_(3)H_(6)O_(3)`
B. `C_(2)H_(4)O`
C. `C_(3)H_(4)O_(2)`
D. `C_(2)H_(4)O_(3)`

Answer

D

Explanation

`therefore x=1` and y=2
Combustion of `C_(x)H_(y)`
`C_(x)H_(y) + (x+y/4)O_(2) to xCO_(2) + (y/2)H_(2)O`
Oxygen atoms required for complete combustion of `C_(x)H_(y)`
`=2(x+y/4)`
`z=1/2[2(x+y/4)] =x + y/4`
`z=1 + 2/4 = 3/2`
Ratio of `x:y:z =1:2:3/2`
Formula : `C_(2)H_(4)O_(3)`