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

Calculation of empirical formula of `C_(X)H_(Y)`.
As `C:H=6:1`, we have

`thereforeX=1, Y=2`
Equation for combustion of `C_(X)H_(Y)` is