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BCECE Medical BCECE Medical Solved Papers-2010

  • question_answer
    An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of the compound is

    A)  \[{{C}_{3}}{{H}_{5}}{{O}_{2}}\]

    B)  \[{{C}_{4}}{{H}_{10}}{{O}_{2}}\]

    C)  \[{{C}_{6}}{{H}_{10}}{{O}_{4}}\]

    D)  \[{{C}_{3}}{{H}_{10}}{{O}_{2}}\]

    Correct Answer: C

    Solution :

    Element % Relative number of atoms Simplest ration
    C 49.3 \[\frac{49.3}{12}=4.10\] \[\frac{49.10}{2.74}=1.5\times 2=3\]
    H 6.84 \[\frac{6.84}{1}=6.84\] \[\frac{6.84}{2.74}=2.5\times 2=5\]
    O 43.86 \[\frac{43.86}{16}=2.74\] \[\frac{2.74}{2.74}=1\times 2=2\]
                    \[\therefore \] The empirical formula is \[{{C}_{3}}{{H}_{5}}{{O}_{2}}\]. Empirical formula weight                 \[=12\times 3+1\times 5+16\times 2\]                 = 73 Molecular weight of the compound                 \[=2\times \] vapour density                 \[=2\times 73=146\]                 \[n=\frac{molecular\text{ }weight}{empirical\text{ }formula\text{ }weight}\]                 \[=\frac{146}{73}=2\] Molecular formula = empirical formula \[\times 2\] \[=2\,({{C}_{3}}{{H}_{5}}{{O}_{2}})={{C}_{6}}{{H}_{10}}{{O}_{4}}\]


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