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NEET AIPMT SOLVED PAPER 1999

  • question_answer
                    An organic compound containing C, H and N gave the following analysis C = 40% H = 13.33% and N = 46.67%. Its empirical formula would be:                                                                                                                

    A)              CHN         

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

    C)              \[C{{H}_{4}}N\]

    D)              \[{{C}_{2}}{{H}_{7}}N\]

    Correct Answer: C

    Solution :

                              Table for empirical formula.       
    Element % At. wt. Rel. Number Ratio
    C 40.00 12 \[\frac{40}{12}=3.33\] \[\frac{3.33}{3.33}=1\]
    H 13.33 1 \[\frac{13.13}{1}=13.33\] \[\frac{13.33}{3.33}=4\]
    N 46.67 14 \[\frac{46.67}{16}=3.33\] \[\frac{3.33}{3.33}=1\]
                Hence, empirical formula \[=C{{H}_{4}}N\]


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