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BCECE Engineering BCECE Engineering Solved Paper-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 \[\frac{49.3}{12}=4.10\] \[\frac{4.10}{2.74}=1.5\times 2=3\]
    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 \[=2\times \text{vapour}\,\text{density}\] \[=2\times 73=146\] \[n=\frac{\text{molecular}\,\text{weight}}{\text{empirical}\,\text{formula}\,\text{weight}}\] \[=\frac{146}{73}=2\] Molecular formula = empirical formula x 2 \[=2({{C}_{3}}{{H}_{5}}{{O}_{2}})={{C}_{6}}{{H}_{10}}{{O}_{4}}\]


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