(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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