an organic compound contain 25%carbon,30%hydrogen,12%nitrogen, 16%sulphur, and remaining is chlorine calculate empirical and molecular formula of the organic compound. (given n=3).
Answers
Given: Carbon: 16.27=
12
16.27
=1.355
Hydrogen: 0.67=
1
0.67
=0.67
Chlorine: 72.2=
35.5
72.2
=2.03
Oxygen: 10.86=
16
10.86
=0.67
Mole ratio C:H:Cl:O=2:1:3:1
Emperical formula= C
2
HCl
3
O
E.F. weight= 24+1+106.5+16=147.5
Given, vapour density= 73.75
Molecular weight= 2×V.D
=2×73.75
=147.5 .
∴ Molecular Formula C
2
HCl
3
O
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Answer:
this is answer
Explanation:
Given: Carbon: 16.27=
12
16.27
=1.355
Hydrogen: 0.67=
1
0.67
=0.67
Chlorine: 72.2=
35.5
72.2
=2.03
Oxygen: 10.86=
16
10.86
=0.67
Mole ratio C:H:Cl:O=2:1:3:1
Emperical formula= C
2
HCl
3
O
E.F. weight= 24+1+106.5+16=147.5
Given, vapour density= 73.75
Molecular weight= 2×V.D
=2×73.75
=147.5 .
∴ Molecular Formula C
2
HCl
3
O