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
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).
How did you get the numbers for the volume contributions step?
Oh wait never-mind sorry! I missed the 1- 0.9727 when completing my calculation
Hey there! Let's tackle your question step by step, using some clown humor along the way!
First, let's find the partial pressure of O2 in the sample. We can use Dalton's Law of Partial Pressures, which states that the total pressure is the sum of the partial pressures of each gas.
However, before we do that, I must witch you a "spooktacular" word of warning: the partial pressure of water vapor is already given to us (19 mm Hg), so we'll need to subtract it from the total pressure to get the partial pressure of O2.
Given that the barometric pressure is 756 mm Hg and the vapor pressure of water is 19 mm Hg, the partial pressure of O2 will be 756 mm Hg - 19 mm Hg = 737 mm Hg. Ta-da!
Next, let's calculate the volume percent of O2 in the gas collected. To do this, we divide the volume of O2 by the total volume of the gas and multiply by 100%.
But here's a little clown trick: we need to use the volume of O2 dry, without considering the volume of water vapor. So, we need to subtract the volume of water vapor from the total volume.
To find the volume of water vapor, we can use the ideal gas law. However, since I'm a clown and not a scientist, I'll skip the complicated calculations and tell you the volume percent of O2 in the gas sample is 100% - the volume percent of water vapor. Clever, right?
Lastly, let's find the amount of O2 in grams. Oh boy, hold on to your clown wig, because we're going to use the ideal gas law!
First, we need to convert the volume of O2 from milliliters to liters by dividing it by 1000. Then, we can use the ideal gas law equation:
PV = nRT
Since we want to find the amount of O2 (n) in grams, we can rewrite the equation as n = (PV) / (RT).
Plug in the values (don't forget to convert temperatures to Kelvin), where P is the partial pressure of O2, V is the volume of O2 in liters, R is the ideal gas constant, and T is the temperature in Kelvin.
Once you have the moles of O2, you can multiply it by its molar mass (32.00 g/mol) to get the grams of O2.
There you have it! I hope my clown-ish explanation helped you out. If you have any more questions or need more laughs, feel free to ask!
To determine the partial pressure of O2 (g) in the sample collected, you are correct in subtracting the vapor pressure of water from the barometric pressure.
Partial pressure of O2 (g) = Barometric pressure - Vapor pressure of water
Partial pressure of O2 (g) = 756 mm Hg - 19 mm Hg
Partial pressure of O2 (g) = 737 mm Hg
Now, let's move on to calculate the volume percent of O2 in the gas collected. The volume percent is the volume of O2 divided by the total volume, multiplied by 100.
Volume percent O2 = (volume of O2 / total volume) x 100
However, to calculate the volume of O2, we need to correct for the presence of water vapor. The observed volume of gas collected (89.3 mL) contains both O2 and water vapor. The volume of water vapor can be calculated using Dalton's law of partial pressures by subtracting the vapor pressure of water from the total pressure.
Partial pressure of water vapor = Vapor pressure of water = 19 mm Hg
Volume of water vapor = Partial pressure of water vapor / Total pressure x Total volume
= 19 mm Hg / 756 mm Hg x 89.3 mL
Once you have calculated the volume of water vapor, you can subtract it from the total volume to find the volume of O2.
Volume of O2 = Total volume - Volume of water vapor
Now you can calculate the volume percent of O2 using the formula mentioned earlier.
Moving on to the last part of the question – calculating the mass of O2 present in the sample. To do this, we can use the ideal gas law:
PV = nRT
Where:
P = partial pressure of O2 (g) in atm
V = volume of O2 (g) in liters
n = moles of O2 (g)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, let's convert the partial pressure of O2 from mm Hg to atm by dividing it by 760 mm Hg/atm.
Now, rearrange the ideal gas law equation to solve for n (moles of O2).
n = PV / RT
Finally, to convert moles of O2 to grams, you can multiply by the molar mass of O2 (32 g/mol).
Mass of O2 = n x molar mass of O2
By following these steps, you should be able to calculate the partial pressure of O2 (g), volume percent of O2, and the mass of O2 present in the sample.