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Punjab Medical Punjab - MET Solved Paper-2009

  • question_answer
    In an organic compound,\[C=68.5%\]and\[H=4.91%\]. Which empirical formula is correct for it?

    A) \[{{C}_{6}}{{H}_{10}}\]                  

    B) \[{{C}_{7}}{{H}_{6}}{{O}_{2}}\]

    C) \[{{C}_{5}}{{H}_{8}}O\]                 

    D) \[{{C}_{9}}{{H}_{3}}O\]

    Correct Answer: B

    Solution :

    Element At. wt. Per cent composition No. of moles Simple molar ratio
    \[C\] 12 68.5 \[\frac{68.5}{12}=5.708\] \[\frac{5.708}{1.67}=3.41\times 2=7\]
    \[H\] 1 4.91 \[\frac{4.91}{1}=4.91\] \[\frac{4.91}{1.67}=2.95\times 2=6\]
    \[O\] 16 \[(100-68.5+4.91)=26.59\] \[\frac{26.59}{16}=1.67\] \[\frac{1.67}{1.67}=1\times 2=2\]
    Hence, the empirical formula of the compound is\[{{C}_{7}}{{H}_{6}}{{O}_{2}}\].


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