Answer:
Empirical Formula = C₅H₇N₁
Molecular Formula = C₁₀H₁₄N₂
Solution:
Step 1: Calculating Moles of Elements as;
For C;
Moles of C = %age of C ÷ At.Mass of C
Moles of C = 74.9 ÷ 12
Moles of C = 6.24
For H:
Moles of H = %age of H ÷ At.Mass of H
Moles of H = 8.7 ÷ 1
Moles of C = 8.7
For N;
Moles of N = %age of N ÷ At.Mass of N
Moles of N = 17.3 ÷ 14
Moles of N = 1.23
Step 2: Simplifying the mole ratio as;
C : H : N
6.24/1.23 : 8.7/1.23 : 1.23/1.23
5 : 7 : 1
So, The Empirical Formula is;
C₅H₇N₁
Step 3: Finding out Molecular Formula;
The Molecular mass of Nicotine is 162.23 g.mol⁻¹, So, the Molecular formula is calculated as,
n = Molecular Mass ÷ Empirical formula Mass
As,
Empirical formula Mass = C₅H₇N₁ = (12)₅+(1)₇+(14)₁ = 60+7+14 = 81
So,
n = 162 ÷ 81
n = 2
Now, Multiply Empirical Formula by 2 as;
(C₅H₇N₁) × 2 = C₁₀H₇N₂
