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In a compound C(x)H(y)O(z) , the mass % ...

In a compound `C_(x)H_(y)O_(z)` , the mass % of C and H is `6:1` and the amount of oxygen present is equal to the half of the oxygen required to react completely `C_(x)H_(y)`. Find the empirical formula of the compound.

Text Solution

Verified by Experts

The correct Answer is:
`C_(2)H_(4)O_(3)`
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Knowledge Check

  • A compound contains C=90% and H=10% Empirical formula of the compound is:

    A
    `C_(15)H_(30)`
    B
    `C_(15)H_(20)`
    C
    `C_3H_4`
    D
    `C_3H_(10)`

  • 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

    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)`

  • The ratio 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 contains half as much oxygen as required to burn one molecule of the 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

    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)`