Calculate the empirical formula of the compound having the following percentage composition: (a) C 6.7 %, H=74 %, N=25.9 %
Answers
Answer:
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Explanation:
Composition of C = 57.8%
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