2. Questions Bank
  3. JEE Main & Advanced
  4. Chemistry
  5. Organic Chemistry: Some Basic Principles & Techniques / कार्बनिक रसायन : कुछ मूलभूत सिद्धांत एवं तकन

JEE Main & Advanced Chemistry Organic Chemistry: Some Basic Principles & Techniques / कार्बनिक रसायन : कुछ मूलभूत सिद्धांत एवं तकन Question Bank Chemical analysis of organic compounds

  • question_answer
    An organic compound on analysis gave C = 48 gm, H =  8 gm and N = 56 gm. Volume of 1.0 g of the compound was found to be 200 ml at NTP. Molecular formula of the compound is [MP PET 1986]

    A) \[{{C}_{4}}{{H}_{8}}{{N}_{4}}\]

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

    C) \[{{C}_{12}}{{H}_{24}}{{N}_{12}}\]

    D) \[{{C}_{16}}{{H}_{32}}{{N}_{16}}\]

    Correct Answer: A

    Solution :

    Element      %         No. of Moles        Simple Ratio
    C 48 48/12 = 4 1
    H 8 8/1 = 8 2
    N 56 56/14 = 4 1
    Empirical formula = \[C{{H}_{2}}N\] Empirical formula mass = 28 Now, 200 ml of compound = 1 gm 22400 ml of compound \[\frac{1}{200}\times 22400=112\] \[n=\frac{12}{2}=6\] Therefore, Molecular formula \[={{(C{{H}_{2}}N)}_{4}}={{C}_{4}}{{H}_{8}}{{N}_{4}}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner