An organic compound on analysis gave the following % composition C- 57.8%, H-3.6% and the rest is oxygen. The vapour density of compound was 83. Find empirical formula and molecular formula of the compound.
Answers
Composition of H = 3.6%
Total composition of organic compound = 100%
Composition of oxygen = 100 - (57.8 + 3.6)
= 38.6%
Element composition Atomic mass Ratio of atoms Simplest ratio
C 57.8% 12 57.8/12 = 4.82 4.82/2.41 = 2
H 3.6% 1 3.6/1 = 3.6 3.6/2.41 = 1
O 38.6% 16 38.6/16 = 2.41 2.41/2.41 = 1
Therefore the empirical formula of the compound = C2HO
Molecular weight = 2 * vapour density
= 2*83
= 166
Emperical formula weight = 24+1+16 = 41
n = Molecular weight/Emperical weight
n = 166/41
n = 4 (nearly)
Molecular formula = 4(C2HO) = C8H4O4
Hope this helps you.
Answer:
The above one answer is wrong, So don't rely on it
C8H6O4.
Explanation:
Element. C. H. O
% amount. 57.8 3.6. 100-57.8-3.6= 38.6
Relative number of atoms. (57.8/12=4.82) (3.6/1) = 3.6 (38.6/16) = 2.41
Ratio of number of atoms. (4.28/2.41=2) (3.6/4.21)=1.5 (2.41/2.41) = 1
Simplest ratio. 2x2=4. 1.5x2=3. 1x2=2
So, we can deduce the empirical formula of the organic compound = C4H3O2.
The empirical Formula mass = 4x12 + 3x1 + 2x16 = 83 u.
The molecular mass of compound can be calculated as:
Molecular mass = 2 x Vapour Density = 2 x 83 = 166 u.
n = Molecular mass / Empirical Formula Mass = 166/83 = 2.
Molecular Formula of the compound = n(Empirical Formula) = 2(C4H3O2) = C8H6O4.