Answer:
Empirical Formula = C₁₀H₂₀O₁
Solution:
Data Given:
Mass of Menthol = 95.6 mg = 0.0956 g
Mass of CO₂ = 269 mg = 0.269 g
Mass of H₂O = 110 mg = 0.110 g
Step 1: Calculate %age of Elements as;
%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100
%C = (0.269 ÷ 0.0956) × (12 ÷ 44) × 100
%C = (2.8138) × (12 ÷ 44) × 100
%C = 2.8138 × 0.2727 × 100
%C = 76.74 %
%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100
%H = (0.110 ÷ 0.0956) × (2.02 ÷ 18.02) × 100
%H = (1.1506) × (2.02 ÷ 18.02) × 100
%H = 1.1506 × 0.1120 × 100
%H = 12.89 %
%O = 100% - (%C + %H)
%O = 100% - (76.74% + 12.89%)
%O = 100% - 89.63%
%O = 10.37 %
Step 2: Calculate Moles of each Element;
Moles of C = %C ÷ At.Mass of C
Moles of C = 76.74 ÷ 12.01
Moles of C = 6.3896 mol
Moles of H = %H ÷ At.Mass of H
Moles of H = 12.89 ÷ 1.01
Moles of H = 12.7623 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 10.37 ÷ 16.0
Moles of O = 0.6481 mol
Step 3: Find out mole ratio and simplify it;
C H O
6.3896 12.7623 0.6481
6.3896/0.6481 12.7623/0.6481 0.6481/0.6481
9.85 19.69 1
≈ 10 ≈ 20 1
Result:
Empirical Formula = C₁₀H₂₀O₁
