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A 2.650 grams sample of gas occupies a volume of 428cm^3 at a pressure of 99.06 kPa and 24.3˚C. Analysis of this compound reveals a composition of 15.5% C, 23.0% Cl and61.5% F by mass. Determine the true, molecular, formula.

Sagot :

Eduard22sly

The true molecular formula of the compound is C₂ClF₅

Determination of the mole of the compound

  • Volume (V) = 428 cm³ = 428 / 1000 = 0.428 L
  • Pressure (P) = 99.06 KPa = 99.06 / 101.325 = 0.978 atm
  • Temperature (T) = 24.3˚C = 24.3 + 273 = 297.3 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Number of mole (n) =?

The number of mole can be obtained as by using the ideal gas equation as illustrated below:

n = PV / RT

n = (0.978 × 0.428) / (0.0821 × 297.3)

n = 0.017 mole

How to determine the molar mass

  • Mole = 0.017 mole
  • Mass = 2.650 g
  • Molar mass =?

Molar mass = mass / mole

Molar mass = 2.650 / 0.017

Molar mass = 155.88 g/mol

How to determine the empirical formula

  • Carbon (C) = 15.5%
  • Chlorine (Cl) = 23%
  • Fluorine (F) = 61.5%
  • Empirical formula =?

Divide by their molar mass

C = 15.5 / 12 = 1.292

Cl = 23 / 35.5 = 0.648

F = 61.5 / 19 = 3.237

Divide by the smallest

C = 1.292 / 0.648 = 2

Cl = 0.648 / 0.648 = 1

F = 3.237 / 0.648 = 5

Thus, the empirical formula of the compound is C₂ClF₅

How to determine the molecular formula

  • Molar mass of compound = 155.88 g/mol
  • Empirical formula = C₂ClF₅
  • Molecular formula =?

Molecular formula = empirical × n = molar mass

[C₂ClF₅]ₙ = 118.084

[(12×2) + 35.5 + (19×5)]ₙ = 155.88

154.5n = 155.88

Divide both side by 154.5

n = 155.88 / 154.5

n = 1

Molecular formula = [C₂ClF₅]ₙ

Molecular formula = [C₂ClF₅]1

Molecular formula = C₂ClF₅

Learn more about empirical and molecular formula:

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