Q.

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

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a

C3H5O2

b

C6H10O4

c

C3H10O2

d

C4H10O2

answer is B.

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Detailed Solution

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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