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EAMCET Medical EAMCET Medical Solved Paper-2005

  • question_answer
    4 g of hydrocarbon \[({{C}_{x}}{{H}_{y}})\] on complete combustion gave 12 g of \[\text{C}{{\text{O}}_{\text{2}}}\text{.}\]What is the empirical formula of the hydrocarbon? (C=12,H=1)

    A)  \[C{{H}_{3}}\]                                  

    B)  \[{{C}_{4}}{{H}_{9}}\]

    C)  \[CH\]                                 

    D)  \[{{C}_{3}}{{H}_{g}}\]

    Correct Answer: D

    Solution :

    % of carbon in compound \[=\frac{12}{44}\times \frac{\text{weight of C}{{\text{O}}_{\text{2}}}}{\text{weight of comp}\text{.}}\times 100\] \[=\frac{12}{44}\times \frac{12}{4}\times 100\] \[=81.81%\] % of \[H=100-81.81=18.19%\]
    Element \[\frac{\text{percentage}}{\text{at}\text{. wt}\text{.}}\] Simple ratio
    C \[\frac{81.81}{12}=6.81\] \[\frac{6.81}{6.81}=1.0\times 3=3\]
    H \[\frac{18.19}{1}=18.19\] \[\frac{18.19}{6.81}=2.67\times 3=8\]
    Hence, empirical formula of the compound\[={{C}_{3}}{{H}_{8}}\]


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