JEE Main & Advanced Chemistry Organic Chemistry: Some Basic Principles & Techniques Question Bank Chemical analysis of organic compounds

  • question_answer An organic compound contains C, H and O in the proportion of 6 : 1 : 8 by weight, respectively. Its vapour density is 30. Its molecular formula will be

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

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

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

    D) \[{{C}_{3}}HO\]

    Correct Answer: A

    Solution :

    Elements            No. of moles               Simple ratio
    C 6   6/12 = 0.5 = 1 1
    H 1   1/1 = 1 = 2 2
    O 8   8/16 = 0.5 = 1 1
    Thus, Empirical formula = \[\underset{\text{(enol form)}}{\mathop{{{C}_{6}}{{H}_{5}}-\underset{O}{\mathop{\underset{||}{\mathop{C}}\,}}\,-CH=\underset{OH}{\mathop{\underset{|\,\,\,\,}{\mathop{C\,\,\,}}\,}}\,-C{{H}_{3}}}}\,\] Empirical formula mass = 30 Mol. mass = 2 × V.D. = 2 × 30 = 60 \[n=\frac{60}{30}=2\] Mol. formula = \[{{(C{{H}_{2}}O)}_{2}}={{C}_{2}}{{H}_{4}}{{O}_{2}}\].


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