Answer:
Empirical Formula = C₃H₆O₁
Solution:
Data Given:
Mass of Ethyl Butyrate = 3.61 mg = 0.00361 g
Mass of CO₂ = 8.22 mg = 0.00822 g
Mass of H₂O = 3.35 mg = 0.00335 g
Step 1: Calculate %age of Elements as;
%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100
%C = (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100
%C = (2.277) × (12 ÷ 44) × 100
%C = 2.277 × 0.2727 × 100
%C = 62.09 %
%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100
%H = (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100
%H = (0.9279) × (2.02 ÷ 18.02) × 100
%H = 0.9279 × 0.1120 × 100
%H = 10.39 %
%O = 100% - (%C + %H)
%O = 100% - (62.09% + 10.39%)
%O = 100% - 72.48%
%O = 27.52 %
Step 2: Calculate Moles of each Element;
Moles of C = %C ÷ At.Mass of C
Moles of C = 62.09 ÷ 12.01
Moles of C = 5.169 mol
Moles of H = %H ÷ At.Mass of H
Moles of H = 10.39 ÷ 1.01
Moles of H = 10.287 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 27.52 ÷ 16.0
Moles of O = 1.720 mol
Step 3: Find out mole ratio and simplify it;
C H O
5.169 10.287 1.720
5.169/1.720 10.287/1.720 1.720/1.720
3.00 5.98 1
3 ≈ 6 1
Result:
Empirical Formula = C₃H₆O₁
