Respuesta :
Answer:
Empirical Formula = C₃H₅O₃
Molecular Formula = C₆H₁₀O₆
Solution:
Data Given:
Mass of Sample = 10.68 mg = 0.01068 g
Mass of CO₂ = 15.83 mg = 0.01583 g
Mass of H₂O = 5.40 mg = 0.0054 g
Step 1: Calculate %age of Elements as;
%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100
%C = (0.01583 ÷ 0.01068) × (12 ÷ 44) × 100
%C = (1.482) × (12 ÷ 44) × 100
%C = 1.482 × 0.2727 × 100
%C = 40.42 %
%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100
%H = (0.0054 ÷ 0.01068) × (2.02 ÷ 18.02) × 100
%H = (0.505) × (2.02 ÷ 18.02) × 100
%H = 0.505 × 0.1120 × 100
%H = 5.656 %
%O = 100% - (%C + %H)
%O = 100% - (40.42 % + 5.656%)
%O = 100% - 46.076%
%O = 53.924 %
Step 2: Calculate Moles of each Element;
Moles of C = %C ÷ At.Mass of C
Moles of C = 40.42 ÷ 12.01
Moles of C = 3.365 mol
Moles of H = %H ÷ At.Mass of H
Moles of H = 5.656 ÷ 1.01
Moles of H = 5.60 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 53.924 ÷ 16.0
Moles of O = 3.370 mol
Step 3: Find out mole ratio and simplify it;
C H O
3.365 5.600 3.370
3.365/3.365 5.600/3.365 3.370/3.365
1 1.66 1.001
3 5 3
Hence, Empirical Formula = C₃H₅O₃
Step 4: 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 = 178.1 g.mol⁻¹
Empirical Formula Weight = 12 (C₃) + 1.01 (H₅) + 16 (O₃) = 89.05 g.mol⁻¹
So,
n = 178.1 g.mol⁻¹ ÷ 89.05 g.mol⁻¹
n = 2
Putting Empirical Formula and value of "n" in equation 1,
Molecular Formula = 2 × C₃H₅O₃
Molecular Formula = C₆H₁₀O₆
A. The empirical formula of the compound is C₃H₅O₃
B. The molecular formula of the compound is C₆H₁₀O₆
Data obtained from the question
- Mass of compound = 10.68 mg
- Mass of CO₂ = 15.83 mg
- Mass of H₂O = 5.4 mg
- Molar mass of compound = 178.1 g/mol
- Empirical formula =?
- Molecular formula =?
How to determine the mass of Carbon
- Mass of CO₂ = 15.83 mg
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 15.83
Mass of C = 4.317 mg
How to determine the mass of H
- Mass of H₂O = 5.4 mg
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 5.4
Mass of H = 0.6 mg
How to determine the mass of O
- Mass of compound = 10.68 mg
- Mass of C = 4.317 mg
- Mass of H = 0.6 mg
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 10.68 – (4.317 + 0.6)
Mass of O = 5.763 mg
A. How to determine the empirical formula
- C = 4.317 g
- H = 0. 6 mg
- O = 5.763 mg
- Empirical formula =?
Divide by their molar mass
C = 4.317 / 12 = 0.36
H = 0. 6 / 1 = 0.6
O = 5.763 / 16 = 0.36
Divide by the smallest
C = 0.36 / 0.36 = 1
H = 0.6 / 0.36 = 5/3
O = 0.36 / 0.36 = 1
Multiply by 3 to express in whole number
C = 1 × 3 = 3
H = 5/3 × 3 = 5
O = 1 × 3 = 3
Thus, the empirical formula of the compound is C₃H₅O₃
B. How to determine the molecular formula
- Empirical formula = C₃H₅O₃
- Molar mass of compound = 178.1 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[C₃H₅O₃]n = 178.1
[(12×3) + (1×5) + (16×3)]n = 178.1
89n = 178.1
n = 178.1 / 89
n = 2
Molecular formula = [C₃H₅O₃]n
Molecular formula = [C₃H₅O₃]₂
Molecular formula = C₆H₁₀O₆
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