Question

An 89.3 mL sample of "wet" O2 (g) is collected over water at 21.3oC and a barometric pressure of 756 mm Hg(vapor pressure of water at 21.3oC = 19 mm Hg). What is the partial pressure of O2 (g) in the sample collected, in mm Hg? What is the volume percent O2 in the gas collected? How many grams of O2 are present in the sample?

So far I think I have figured out the partial pressure by doing barometric O2-partial of water. I am not sure how to do the other parts of this question

Answer 1

Given 89.3-ml gas mix of O₂(g) and H₂O(g) at 21.3ᵒC (=294.3K) and 756 mmHg ( = 0.9947 atm) TTL pressure, determine the following given also P(H₂O) = 19 mmHg at 756 mmHg/21.3ᵒC).

P(O₂) in mmHg, b. %V(O₂) contribution in mix & c. grams O₂(g) in mix.

P(O₂):
P(TTL) = P(O₂) + P(H₂O)
=> p(O₂) = 756 mmHg – 19 mmHg = 737 mmHg = (737 mmHg/760 mmHg/atm) = 0.9697 atm.

Volume Contribution of O₂(g) to total volume of mix (89.3-ml):
From PV = n(O₂)RT:
=> n(O₂) = PV/RT = (0.9697-atm)(0.0893-L)/(0.08206-L∙atm∙molˉ¹∙Kˉ¹)(294.3-K) = 0.0033 mole O₂(g)
=> n(H₂O) = PV/RT = [(19/760)-atm](0.0893-L)/(0.08206-L∙atm∙molˉ¹∙Kˉ¹)(294.3-K) = 0.0000924 mole H₂O(g)

Mole Fraction (X) = nᵢ/Σn:
=> X(O₂) = 0.0033/(0.0033 + 0.0000924) = 0.0033/0.0033924 = 0.9727
=> X(H₂O) = 0.0000924/0.0033924 = 0.0272; or, X(H₂O) = 1.000 – 0.9727 = 0.0272 (<=> ΣXᵢ = 1.000)

Volume Contributions => Xᵢ(TTL Volume):
=> V(O₂) = 0.9727(89.3-ml) = 86.9-ml O₂
=> V(H₂O) = 0.0282(89.3-ml) = 2.4-ml H₂O

Vol% Contribution of O₂(g)
Vol%(O₂) = [V(O₂)/V(O₂) + V(H₂O)]100% = [86.9-ml/(86.9-ml + 2.4-ml)] = (86.9-ml/89.3-ml)100% = 97.3 Vol% O₂

Grams of O₂(g):
=> Moles O₂ x formula wt O₂ = 0.0033 mole O₂(g) x 32 g/mole O₂(g) = 0.106 gram O₂(g).

Answer 2

How did you get the numbers for the volume contributions step?

Answer 3

Oh wait never-mind sorry! I missed the 1- 0.9727 when completing my calculation