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

Answer

A

Explanation

Mole ratio of C & H in 1 mole of `C_(x)H_(y)O_(z)= (6//12)/(1//1)= 1:2= x: y` . As 2z mole of O (or z mole `O_(2)`) is used to combust one mole of `C_(x)H_(y)`
`overset("1 mole")(C_(x)H_(y)) + overset("z mole")(O_(2)) rarr CO_(2) + H_(2)O` , applying POAC for C, H and O, we get, `z=x + (y)/(4)`, if x=1 and y= 2` therefore z=1.5` i.e., `C_(2)H_(4)O_(3)` should be the empirical formula