Answer:
Empirical Formula = C₅H₇N₁
Solution:
Data Given:
Mass of Nicotine = 4.20 mg = 0.0042 g
Mass of CO₂ = 11.394 mg = 0.011394 g
Mass of H₂O = 3.266 mg = 0.003266 g
Step 1: Calculate %age of Elements as;
%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100
%C = (0.011394 ÷ 0.0042) × (12 ÷ 44) × 100
%C = (2.7128) × (12 ÷ 44) × 100
%C = 2.7128 × 0.2727 × 100
%C = 73.979 %
%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100
%H = (0.003266 ÷ 0.0042) × (2.02 ÷ 18.02) × 100
%H = (0.7776) × (2.02 ÷ 18.02) × 100
%H = 0.7776 × 0.1120 × 100
%H = 8.709 %
%N = 100% - (%C + %H)
%N = 100% - (73.979 % + 8.709%)
%N = 100% - 82.688%
%N = 17.312 %
Step 2: Calculate Moles of each Element;
Moles of C = %C ÷ At.Mass of C
Moles of C = 73.979 ÷ 12.01
Moles of C = 6.1597 mol
Moles of H = %H ÷ At.Mass of H
Moles of H = 8.709 ÷ 1.01
Moles of H = 8.6227 mol
Moles of N = %N ÷ At.Mass of O
Moles of N = 17.312 ÷ 14.01
Moles of N = 1.2356 mol
Step 3: Find out mole ratio and simplify it;
C H N
6.1597 8.6227 1.2356
6.1597/1.2356 8.6227/1.2356 1.2356/1.2356
4.985 6.978 1
≈ 5 ≈ 7 1
Result:
Empirical Formula = C₅H₇N₁
