10.68 g to mg

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