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connermichaela
10.07.2019 •
Chemistry
Acompound contains only carbon, hydrogen, and oxygen. combustion of 10.68 mg of the compound yields 15.83 mg and 5.40 mg . the molar mass of the compound is 178.1 g/mol. what are the empirical and molecular formulas of the compound? (enter the elements in the order: c, h, o.)
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Ответ:
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
Ответ:
Each component in a mixture contributes a fraction to the total number of moles in the mixture. This fraction is called the _mole_ fraction of the component. In a mixture of gases, this fraction is proportional to the _mole_ pressure for each gas present.