Answer:
Empirical Formula = C₁₀H₂₀O
Solution:
Data Given:
Mass of Menthol = 9.045 × 10⁻² mg = 9.045 × 10⁻⁵ g
Mass of CO₂ = 0.2546 mg = 0.0002546 g
Mass of H₂O = 0.1043 mg = 0.0001043 g
Step 1: Calculate %age of Elements as;
%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100
%C = (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100
%C = (2.814) × (12 ÷ 44) × 100
%C = 2.814 × 0.2727 × 100
%C = 76.73 %
%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100
%H = (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100
%H = (1.153) × (2.02 ÷ 18.02) × 100
%H = 1.153 × 0.1120 × 100
%H = 12.91 %
%O = 100% - (%C + %H)
%O = 100% - (76.73% + 12.91%)
%O = 100% - 89.64%
%O = 10.36 %
Step 2: Calculate Moles of each Element;
Moles of C = %C ÷ At.Mass of C
Moles of C = 76.73 ÷ 12.01
Moles of C = 6.3888 mol
Moles of H = %H ÷ At.Mass of H
Moles of H = 12.91 ÷ 1.01
Moles of H = 12.7821 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 10.36 ÷ 16.0
Moles of O = 0.6475 mol
Step 3: Find out mole ratio and simplify it;
C H O
6.3888 12.7821 0.6475
6.3888/0.6475 12.7821/0.6475 0.6475/0.6475
9.86 19.74 1
≈ 10 ≈ 20 1
Result:
Empirical Formula = C₁₀H₂₀O₁
