An organic compound with a vapour density 56.5 has the following percentage composition carbon 53.1%,nitrogen 12.4% oxygen 28.3%,hydrogen 6.2% . the M.F of the compound is ?
Answers
Answer:
53.5
Explanation:
53.1
+12.4
+28.3
+6.2
100.0
100.0
-56.5
53.5
Given :
- C = 53.1 %
- N = 12.4 %
- H = 6.2 %
- O = 28.3 %
- Vapor density = 56.5
➢ Relative no of atoms
Divide the % composition of the elements with their atomic masses
➜ C = 53.1 / 12 = 4.42
➜ N = 12.4 / 14 = 0.88
➜ H = 6.2 / 1 = 6.2
➜ O = 28.3 / 16 = 1.76
➢ Simple ratio
Now divide all the values obtained above with the least possible value (i.e. 0.88 in this case )
➜ C = 4.42 / 0.88 = 5
➜ N = 0.88 / 0.88 = 1
➜ H = 6.2 / 0.88 = 7
➜ O = 1.76 / 0.88 = 2
Empirical formula = C5 N H7 O2
➜ Empirical mass = 5 x C +N + 7 x H +2 x O
➜ Empirical mass = 60 + 14 + 7 + 32
➜ Empirical mass = 113 u
Molecular mass = 2 x Vapor density
Molecular mass = 2 x 56.5
Molecular mass = 113 u
We know that ,
➜ Molecular mass = n x empirical mass
➜ n = 113 / 113
➜ n = 1
Molecular formula = C5 N H7 O2