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

Answer

B

Explanation

Ratio of mass % of C and H in `C_(x)H_(y)O_(z)` is 6 : 1
Therefore,
Ratio of mole % of C and H in `C_(x)H_(y)O_(z)` will be 1 : 2.
Therefore x : y =1 : 2, Which is possible in options A, B and C.
Now oxygen required to burn `C_(x) H_(y) , " " C_(x)H_(y) + (x+(y)/(4))O_(2)rarr xCO_(2) + (y)/(2)H_(2)O`
Now z is half of oxygen atoms required to burn `C_(x)H_(y)`. `:. z = ((2x+(y)/(2)))/(2) = (x+(y)/(4))`
Now putting values of x and y from the given options:
OPtion (B), x = 2, y = 4 , `z = (2+(4)/(4)) = 3`
Option (C ), x = 3, y = 6 , `x = (3+(6)/(4)) = 4.5`
Therefore correct option is B `(C_(2)H_(4)O_(3))`