Question

Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol is combusted, producing 0.2546 mg of co2 and 0.1043 mg of h2o. what is the empirical formula for menthol? express your answer as a chemical formula.

Answer 1

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