Nicotine, a component of tobacco, is composed of c, h, and n. a 4.200-mg sample of nicotine was combusted, producing 11.394 mg o

Question

3 years ago

6

f co2 and 3.266 mg of h2o. what is the empirical formula for nicotine?

1 answer:

Rudiy27 · Answer 1 · 2022-06-26T23:40:23+03:00

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