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

So, X=1, Y=2
Equation for 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" = 2 (X+(Y)/(4))`
As mentioned,
`2(X+(Y)/(4))=2Z implies (1+(2)/(4))=Z implies Z=1.5`
`:.` molecule can be written as
`C_(X)H_(Y)O_(Z)`
`C_(1)H_(2)O_(3 // 2)`
`implies C_(2)H_(4)O_(3)`