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Question

A compound gave the following data of its composition by mass: C=57.82 %, O=38.58 % and the rest hydrogen. Its vapour density is 83. Empirical formula of compound is:

(C=12, O=16, H=1)

A
C6H6O4
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B
C2H4O2
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C
C4H3O2
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D
C2H3O4
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Solution

The correct option is C C4H3O2
Molecular formula = 2 x V.D. = 2 x 83 = 166
% of Hydrogen =100(57.82+38.58)=3.6
Element % Ratio Simplest Ratio Empirical formula
Carbon 57.82/12=4.82 4.82/2.41=2 C4H3O2
Hydrogen 3.6/1=3.6 3.6/2.41=1.49
Oxygen 38.58/16=2.41 2.41/2.41=1
The empirical formula weight =12×4+3×1+16×2=83
Formula ratio =166/83=2
The molecular formula of compound =C8H6O4

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