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InterviewSolution
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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 compound `C_(X)H_(Y)O_(Z)` isA. `C_(2)H_(4)O`B. `C_(3)H_(4)O_(2)`C. `C_(2)H_(4)O_(3)`D. `C_(3)H_(6)O_(3)` |
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Answer» Correct Answer - C `{:("Element",,C:H),("Mass ratio",,6:1),("Mole ratio",,6//12:1 rArr = 1:2):}` So `C_(X) H_(Y)` have empirical formula : `CH_(2)` For Burning a `CH_(2)` unit, oxygen required `1 (3)/(2)` mol `CH_(2)+(3)/(2)O_(2) rarr CO_(2) + H_(2)O` Empirical formula is `2 xx (CH_(2)O_(3//2)) rArr C_(2)H_(4)O_(3)`. |
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