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An organic compound on analysis gave the...

An organic compound on analysis gave the following percentage composition of C= 57.8, H= 3.6 and the rest is oxygen. Vapour density of the compound is 83. Find out the molecular formula of the compound.

Text Solution

Verified by Experts

We are given that
(i) Percentage composition of C= 57.8
Percentage composition of H= 3.6
(ii) Vapour density of the compound= 83
We are to find out the followings
(i) Percentage of `O_(2)`
Empirical formula and empirical formula mass.
(iii) Molecular formula and molecular formula mass.
Calculations
(i) Calculation of percentage of oxygen `=100- (% " of" C+ % "of " H) = 100- (57.8+ 3.6)= 38.6`
(ii) Calculation of empirical formula

`therefore` Empirical formula `=C_(4)H(+_(3)O_(2)`
Empirical formula mass `=4 xx 12 + 3 xx 1 + 2 xx 16= 83`
(iii) Calculation of molecular formula
Molecular formula mass `=2 xx` vapour density `=2 xx 83= 146`
Therefore, `n= (146)/(83)=2`
Hence, molecular formula `=2 xx (C_(4)H_(3)O_(2))= C_(8)H_(6)O_(4)`
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