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AFMC AFMC Solved Paper-2000

  • question_answer
    A compound has C = 40%, H = 13.33% and N =  46.67%. The empirical formula is:

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

    B) \[{{C}_{2}}{{H}_{5}}N\]

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

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

    Correct Answer: A

    Solution :

     Element Percent At. mass Relative number of moles Simple ratio
    C 40 12 \[\frac{40}{12}=3.3\] \[\frac{3.3}{3.3}=1\]
    H 13.3 1 \[\frac{13.3}{1}=13.3\] \[\frac{13.3}{3.3}\simeq 4.0\]
    N, 46.67 14 \[\frac{46.67}{14}=3.3\] \[\frac{3.3}{3.3}=1\]
    Empirical formula \[={{C}_{1}}{{H}_{4}}{{N}_{1}}\]


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