1.

The ratio of mass percent 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) completely to CO_(2) and H_(2)O, the empirical formula of the compound C_(X)H_(Y)O_(Z) is

Answer»

`C_(3)H_(6)O_(3)`
`C_(2)H_(4)O`
`C_(3)H_(4)O_(2)`
`C_(2)H_(4)O_(3)`

Solution :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


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