Question

A compound containing only P  and F  was analyzed by converting 0.2324 grams of the compound into a gas.  The size of the gas cylinder was 378 cm3 .   The gas pressure was  97.3 mmHg  at 77.0 °C.  

All of the fluorine present in the gas was converted to CaF2 by mixing the gas with a CaCl2 solution.   The mass of CaF2 produced was 0.2631 grams.  Determine the molecular formula of the compound.

Answer 1

0.2631g CaF₂ x 37.9968g F = 0.1280 g F

37.9968g F

0.1280 g F x 100 = 55.08% F

0.2324g Compound

100 % - 55.08% F = 44.92% P

55.08 = 2.8991   2.8991 = 2 mol F

18.99 1.4503

44.92 = 1.4503 1.4503 = 1 mol P

30.97 1.4503

Empiric Formula = PF₂Molecular Weight of empiric formula = 68.95

P = 97.3 mmHg = 0.128 atm

V = 378 cm³ = 0.378 L

T =77ºC = 350.15ºK

R = 0.0821 L.atm/ºK.mol

P x V = n x R x T

0.128 atm x 0.378 L = n x 0.0821 L.atm/ºK.mol x 350.15ºK

0.0484 amt.L = n x 28.7473 L.atm/mol

n = 0.0484 atm.L / 28.7473 L.atm/mol

n = 0.0017 mol

Molecular weight = 0.2324 g / 0.0017 mol

Molecular weight = 136.7059 g/mol

136.7059 g/mol = 1.98 = 2

68.95 g/mol

(PF₂) x 2 = P₂F₄

Molecular Formula of the compound = P₂F₄