Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

  • question_answer
    \[64\text{ }g\]of an organic compound contains \[24\text{ }g\]of carbon, \[\text{8 }g\] of hydrogen and the rest oxygen. The empirical formula of the compound is

    A)  \[C{{H}_{2}}O\]        

    B)  \[{{C}_{2}}{{H}_{4}}O\]

    C)  \[C{{H}_{4}}O\]

    D)  \[{{C}_{2}}{{H}_{8}}{{O}_{2}}\]

    Correct Answer: C

    Solution :

    Weight of organic compound \[=64\text{ }g\] Weight of carbon \[=24\text{ }g\] Weight of liydrogen \[=8\text{ }g\] Weight of oxygen \[=64-[24+8]g\]                                                 \[=32g\] Percentage of carbon \[=\frac{24}{64}\times 100%\] \[\therefore \]   Percentage of carbon \[=37.5%\] Percentage of hydrogen \[=\frac{8}{64}\times 100%\] \[\therefore \] Percentage of hydrogen \[=12.5%\] Percentage of oxygen \[=\frac{32}{64}\times 100%\] \[\therefore \]  Percentage of oxygen \[=50%\]
    % of element \[C\] \[H\] \[O\]
    \[37.5%\] \[12.5%\] \[50%\]
    Mole \[\frac{37.5}{12}\] \[\frac{12.5}{1}\] \[\frac{50}{16}\]
    Relative number of atoms \[3.125\] \[12.5\] \[3.125\]
    Simplest atomic ratio   \[\frac{3.125}{3.125}\] \[\frac{12.5}{3.125}\] \[\frac{3.125}{3.125}\]
    \[=1\] \[=4\] \[=1\]
    Hence, the empirical formula of compound is\[C{{H}_{4}}O\].


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