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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    An organic compound with \[C=40%\]and \[H=6.7%\]will have the empirical formula

    A) \[C{{H}_{2}}\]                                   

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

    C) \[{{C}_{3}}{{H}_{6}}{{O}_{3}}\]                                  

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

    Correct Answer: B

    Solution :

    \[C=40%,\] \[H=6.7%\] \[0%=100-(40\text{ }+\text{ }6.7)\] \[0%=100-46.7=53.3%\]
    Element % Atomic weight Relative no of otoms Simple ration of atom
    \[C\] \[40\] \[12\] \[\frac{40}{12}=3.33\] \[\frac{3.33}{3.33}=1\]
    \[H\] \[6.7\] \[1\] \[\frac{6.7}{1}=6.7\] \[\frac{6.7}{3.33}=2\]
    \[O\] \[53.3\] \[16\] \[\frac{53.3}{16}=3.33\] \[\frac{3.33}{3.33}=1\]
    So, the empirical formula is \[C{{H}_{2}}O\].


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