Question

The mass percent of a three component gas sample is 17.1% C4H8, 45.5% C2H2F4 and 37.4% S2F2. Calculate the partial pressure (atm) of C4H8 if the total pressure of the sample is 4712 torr.

Answer 1

Let there be 100g of the sample

Thus, mass of C₄H₈ = 17.1 g

mass of C₂H₂F₄ = 45.5 G

Mass of S₂F₂ = 37.4 G

Now, molar mass of C₄H₈ = 56 g/mole

Molar mass of C₂H₂F₄ = 102 g/mole

Molar mass of S₂F₂ = 102 g/mole

Ths, moles of C₄H₈ = mass/molar mass = 17.1/56 = 0.305

moles of C₂H₂F₄ = mass/molar mass = 45.5/102 = 0.446

moles of S₂F₂ = mass/molar mass = 37.4/102 = 0.367

Now, mole fraction of C₄H₈ = moles of C₄H₈/(total moles of the gas mixture) = 0.305/(0.305+0.446+0.367) = 0.273

mole fraction of C₂H₂F₄ = 0.446/1.118 = 0.399

mole fraction of S₂F₂ = 0.367/1.118 = 0.328

Thus, partial pressure of C₄H₈ = mole fraction*total pressure = 0.273*4712 = 1286.376 torr

partial pressure of C₂H₂F₄ =0.399*4712 = 1880.088 torr

partial pressure of S₂F₂ = 0.328*4712 = 1545.536 torr