Question

Menthol, from the oil of mint, has a characteristic cool taste. the compound contains only c, h, and o. if 95.6 mg of menthol burns completely in o2, and gives 269 mg of co2 and 110 mg of h2o, what is the empirical formula of menthol?

Answer 1

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