Answer to Question #325311 in General Chemistry for Becky

Question

A 45.60g sample of a gaseous flurocarbon contains 7.9g of carbon and 37.66g of fluorine and occupies 7.40L at STP(P=1.00atm and T=273.15k). Determine the approximate molar mass of the flurocarbon and give it's molecular formula

Answer 1

Solution:

The Ideal Gas equation (PV = nRT) can be used to calculate the moles of a gaseous flurocarbon:

n = PV / RT

n = (1.00 atm × 7.40 L) / (0.082 L atm K⁻¹ mol⁻¹ × 273.15 K) = 0.33 mol

Moles of a gaseous flurocarbon = 0.33 mol

Moles = Mass / Molar mass

Molar mass of a gaseous flurocarbon = Mass / Moles = (45.60 g) / (0.33 mol) = 138.2 g/mol

The approximate molar mass of the flurocarbon is 138.2 g/mol

The molar mass of carbon (C) is 12.0107 g/mol

The molar mass of fluorine (F) is 18.9984 g/mol

Calculate the moles of each element in the flurocarbon:

Moles of C = (7.9 g C) × (1 mol C / 12.0107 g C) = 0.6577 mol C

Moles of F = (37.66 g F) × (1 mol F / 18.9984 mol F) = 1.9823 mol F

Divide both moles by the smallest of the results:

C: 0.6577 mol / 0.6577 mol = 1

F: 1.9823 mol / 0.6577 mol = 3

Therefore, the empirical formula of the flurocarbon is CF₃

Empirical formula mass = Ar(C) + 3 × (F) = 12.0107 + 3×18.9984 = 69.0 (g mol⁻¹)

The molar mass of the flurocarbon is 138.2 g mol⁻¹

 

Molar mass / Empirical formula mass = n formula units/molecule

(138.2 g mol⁻¹) / (69.0 g mol⁻¹) = 2 formula units/molecule

Finally, derive the molecular formula for the flurocarbon from the empirical formula by multiplying each subscript by two: (CF₃)₂ = C₂F₆

The molecular formula of the flurocarbon is C₂F₆

Answers:

The approximate molar mass of the flurocarbon is 138.2 g/mol

The molecular formula of the flurocarbon is C₂F₆