Answer:
Empirical Formula = C₅H₇N₁
Solution:
Data Given:
Mass of Nicotine = 4.20 mg = 0.0042 g
Mass of CO₂ = 11.394 mg = 0.011394 g
Mass of H₂O = 3.266 mg = 0.003266 g
Step 1: Calculate %age of Elements as;
%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100
%C = (0.011394 ÷ 0.0042) × (12 ÷ 44) × 100
%C = (2.7128) × (12 ÷ 44) × 100
%C = 2.7128 × 0.2727 × 100
%C = 73.979 %
%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100
%H = (0.003266 ÷ 0.0042) × (2.02 ÷ 18.02) × 100
%H = (0.7776) × (2.02 ÷ 18.02) × 100
%H = 0.7776 × 0.1120 × 100
%H = 8.709 %
%N = 100% - (%C + %H)
%N = 100% - (73.979 % + 8.709%)
%N = 100% - 82.688%
%N = 17.312 %
Step 2: Calculate Moles of each Element;
Moles of C = %C ÷ At.Mass of C
Moles of C = 73.979 ÷ 12.01
Moles of C = 6.1597 mol
Moles of H = %H ÷ At.Mass of H
Moles of H = 8.709 ÷ 1.01
Moles of H = 8.6227 mol
Moles of N = %N ÷ At.Mass of O
Moles of N = 17.312 ÷ 14.01
Moles of N = 1.2356 mol
Step 3: Find out mole ratio and simplify it;
C H N
6.1597 8.6227 1.2356
6.1597/1.2356 8.6227/1.2356 1.2356/1.2356
4.985 6.978 1
≈ 5 ≈ 7 1
Result:
Empirical Formula = C₅H₇N₁
is the pressure of the gas,
is the volume of the gas,
is the number of moles of particles in this gas,
is the ideal gas constant, and
is the absolute temperature of the gas (in degrees Kelvins.)
;
;
.