An organic compound contains 49.3% carbon 6.84 % hydrogen and its vapour density is 73. Molecular formula of the compound is

An organic compound contains 49.3% carbon 6.84 % hydrogen and its vapour density is 73. Molecular formula of the compound is

  1. A

    C3H5O2

  2. B

    C6H10O4

  3. C

    C3H10O2

  4. D

    C4H10O2

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    Solution:

    Element.No. of molesSimple ratio
    C     1249.3/12 = 4.14.1/2.7 = 1.3 x 2 = 2.6 = 3
    H     16.84/1 = 6.846.84/2.7 = 2.5 x 2 = 5
    O    16 43.86/16 = 2.72.7/2.7 = 1 x 2 = 2

    Empirical formula = C3H5O2   EF wt. = 12 × 3 + 1 × 5 + 16 × 2 = 73   Molecular weight = VD × 2 = 73 × 2 = 146   n =  M. wtE.F. wt = 14673 = 2   Molecular formula = EFn = C3H5O22 = C6H10O4.

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