(a) Butyric acid contains only C, H, O. A 4.24 mg of sample of butyric acid is completely burnt. It gives 8.45 mg of CO2 and 3.46 mg of water. What is the mass % of each element in butyric acid. (b) The molecular mass of butyric acid was determined by experiment to be 88u. What is the molecular formula.

Calculate The mass % age of different elements

 

% of carbon

CO2=C

44=12

8.45Of Co2 contain C= 12x8.45/44= 2.30 mg

% of C= 2.30x100/4.24=54.3%

 

% Of Hydrogen

H2O=@H

18=2

3.46 mg of H2O contain H=0.384 mg

% of H= 0.384x100/4.24=9%

 

% of O= 100- (54.3+9)= 36.7%

 

Now calculate the empirical formula of the compound

 

C     54.3        12        54.3/12=4.52       4.52/2.29=1.97          2

 

H     9.0          1           9/1=9                 9/2.29=3.93              4

 

O     36.7        16          36.7/16=2.29       2.29/2.29=1             1

 

 

 

Empirical formula = C2H4O

 

 

Empirical formula mass = 2x12+4x1+16=24+4+16=44

 

Molecular mass =88

 

n= Molecular mass/ Empirical formula mass =88/44 =2

 

Molecular formula = nxEmpirical formula =2xC2H4O= C4H8O