(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.

Asked by  | 20th Apr, 2012, 05:49: PM

Expert Answer:

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

Answered by  | 24th Apr, 2012, 06:28: PM

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