Question

Nicotine contains 74.9% carbon, 8.7% hydrogen and 17.3% nitrogen. it is known that this compound contains two nitrogen atoms per molecule. what are the empirical and molecular formulas of nicotine?

Answer 1

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