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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2015

  • question_answer
    \[0.30g\]of an organic compound containing C, H and O on combustion yields \[0.44g\,C{{O}_{2}}\]and\[0.18g\,{{H}_{2}}O\]. If one mole of compound weighs 60, then molecular formula of the compound is

    A)  \[{{C}_{3}}{{H}_{8}}O\]        

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

    C)  \[C{{H}_{2}}O\]          

    D)  \[{{C}_{4}}{{H}_{6}}O\]

    Correct Answer: B

    Solution :

    : Percentage of \[C=\frac{12}{44}\times \frac{0.44}{0.30}\times 100=40%\] Percentage of \[H=\frac{2}{18}\times \frac{0.18}{0.30}\times 100=6.6%\] Percentage of \[O=100-(40+6.6)=53.4%\]
    Element % Molar Ratio Simplest ratio
    C \[40\] \[\frac{40}{12}=3.3\] \[\frac{3.3}{3.3}=1\]
    H \[6.6\] \[\frac{6.6}{1}=6.6\] \[\frac{6.6}{3.3}=2\]
    O \[53.4\] \[\frac{53.4}{16}=3.3\] \[\frac{3.3}{3.3}=1\]
    Hence, empirical formula \[=C{{H}_{2}}O\] \[n=\frac{Molecular\,mass}{Empirical\,formula\,mass}=\frac{60}{30}=2\] \[\Rightarrow \] Molecular formula of the compound  \[={{(C{{H}_{2}}O)}_{2}}\] \[={{C}_{2}}{{H}_{4}}{{O}_{2}}\]


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