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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    A hydrocarbon has \[C\text{ }85.72%\]and remaining H. The hydrocarbon is

    A)  \[{{C}_{2}}{{H}_{4}}\]                                   

    B)  \[{{C}_{2}}{{H}_{6}}\]

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

    D)  \[C{{H}_{4}}\]

    Correct Answer: A

    Solution :

    \[C%=85.72%,\] (given)\[H%=100-85.72=14.28%\]
    Elements % Atomic Weight Relative  No of atoms Simples ratio
    \[C\] \[85.72\] \[12\] \[\frac{85.72}{12}=7.14\] \[\frac{7.14}{7.14}=1\]
    \[H\] \[14.28\] \[1\] \[\frac{14.2}{1}=14.2\] \[\frac{14.2}{7.14}=2\]
    Empirical formula \[=C{{H}_{2}}\] Empirical formula weight \[~=12+2\times 1=14\] \[n=\frac{molecular\text{ }weight\text{ }of\text{ }compound}{empirical\text{ }formula\text{ }weight}=2\]\[Molecular\text{ }formula={{(Empirical\text{ }formula)}_{n}}\]                                                 \[={{(C{{H}_{2}})}_{2}}={{C}_{2}}{{H}_{4}}\].


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