Answer:
Empirical Formula = C₃H₈O₃
Molecular Formula = C₃H₈O₃
Solution:
Data Given:
Mass of Sample = 9.2 g
Mass of Carbon = 3.6 g
Mass of Hydrogen = 0.8 g
Mass of Oxygen = 9.2 - (3.6 + 0.8) = 4.8 g
Step 1: Calculate Moles of each Element;
Moles of C = Mass of C ÷ At.Mass of C
Moles of C = 3.6 ÷ 12.01
Moles of C = 0.2997 mol
Moles of H = Mass of H ÷ At.Mass of H
Moles of H = 0.8 ÷ 1.01
Moles of H = 0.7920 mol
Moles of O = Mass of O ÷ At.Mass of O
Moles of O = 4.8 ÷ 16.0
Moles of O = 0.3000 mol
Step 2: Find out mole ratio and simplify it;
C H O
0.2997 0.7920 0.3000
0.2997/0.2997 0.7920/0.2997 0.3000/0.2997
1 2.64 1.001
Multiply by 3,
3 7.92 ≈ 8 3
Hence, Empirical Formula = C₃H₈O₃
Step 3: Calculating Molecular Formula:
Molecular formula is calculated by using following formula,
Molecular Formula = n × Empirical Formula ---- (1)
Also, n is given as,
n = Molecular Weight / Empirical Formula Weight
Molecular Weight = 92 g.mol⁻¹
Empirical Formula Weight = 12 (C₃) + 1.01 (H₈) + 16 (O₃) = 92.08 g.mol⁻¹
So,
n = 92 g.mol⁻¹ ÷ 92 g.mol⁻¹
n = 1
Putting Empirical Formula and value of "n" in equation 1,
Molecular Formula = 1 × C₃H₈O₃
Molecular Formula = C₃H₈O₃