# Gas collected over water & Fick's Law

• skaai
In summary: I was also able to verify that 769.23 \frac{l·atm}{mol} is a reasonable value for Henry's law constant ( I saw a few sources that gave it as 777). In summary, the conversation discusses a simple problem of collecting oxygen gas over water and calculating the amount of oxygen collected. However, there is debate about whether or not to consider the oxygen dissolved in the water, which leads to different calculations and results. The use of Henry's Law is suggested to accurately account for the dissolved oxygen, and it is found that the dissolved amount is around 16% of the collected amount. There is also an error in the math provided by Purdue University, which leads to a wrong answer.
skaai
ok, something's been bugging me

consider this simple problem: 193 mL or O2 was collected over water with pressure = 762 mmHg at 23°C. How many grams of oxygen were collected?

The strategy is simple enough: PT = Po2 + Pwater
Poxygen = 741 mmHg →n=$\frac{pV}{RT}$=7.69x10-4 mol = 2.46 x 10-2 g (see example in full)

what about the O2 dissolved in the water?
• granted, the question gives you an easy out by NOT asking for the total O2, but just the O2 collected, but assume I want to know the total O2 evolved... which in reality I need to know if I want to calculate the stoichiometry of a decomposition (such as KClO2 that produces O2.

the textbooks tell me to ignore it with gasses that do not dissolve appreciably in water, but then many such books (like the one in the link) ask you to calculate the O2 evolved... suggesting O2 does not dissolve appreciably in water...

but,

Fick's Law and Henry's Law both tell me a significant amount of the gas WILL dissolve in the water, and if I'm not counting it, then I'm going to be off by that amount, right? I mean, we rely on Fick's law to calculate oxygen diffusion across the alveoli of the lungs!

so why don't the Dalton's "gas over water" examples include the amount of oxygen in the water? (another example of this). I almost thought I had this when I assumed the PO2(gas) = PO2(liquid) but this would suggest I add a third term to the above equation, right/wrong?

won't ignoring the dissolved gas in the water produce an underestimate?
(remember the book accepts O2 as being minimally soluble but Fick and Henry both say it is)...

I'd be happy to do the math, but I don't know where to start to prove this to myself, as the above equation seems incomplete, or I'm missing something.

thanks so much for considering my dilemma!

1. There is another source of error that you ignore so far - water vapor pressure.

2. Why don't you assume some water volume and try to calculate how much oxygen can be dissolved, to be able to compare amount produced with amount "lost" to dissolving?

3. Please remember water was most likely already saturated with the oxygen before you started collecting it. How does it change the situation?

1 person
Borek said:
1. There is another source of error that you ignore so far - water vapor pressure.

2. Why don't you assume some water volume and try to calculate how much oxygen can be dissolved, to be able to compare amount produced with amount "lost" to dissolving?

3. Please remember water was most likely already saturated with the oxygen before you started collecting it. How does it change the situation?
As follow up to Borek's item #2, why don't you just look up the Henry's law constant for oxygen dissolution in water, and then calculate how much oxygen could possibly dissolve in the amount of water that you have if no dissolved oxygen were initially present, and if oxygen then came to equilibrium with the partial pressure of the oxygen in the gas phase. This will remove all your guesswork and speculation.

Chet

1 person
Henry's Law gets us closer, or messes things up worse?

Borek, Chestermiller

but I think the Law I forgot to apply was Henry's Law (Pgas=kH·[gas])
For O2 this value is 769.23 $\frac{l·atm}{mol}$.

if we assume PO2=762 mmHg (0.975 atm) the problem becomes:
[O2] = $\frac{0.975 atm}{769.23 \frac{l·atm}{mol}}$ = 1.2675 x 10-3 $\frac{mol}{l}$

if now we assume (it's not in the problem, but I add it in for accuracy) we have 1 liter of water:
1.2675 x 10-3 mol O2 · $\frac{32g}{1 mol}$O2 = 4.056 x 10-2 g O2

we can now add this to our mass of oxygen that displaced the water (2.46 x 10-2 g O2)

but NOTICE:the dissolved oxygen is APPRECIABLY DISSOLVED IN WATER, as a matter of fact (unless I'm wrong somewhere here, and I could be), with 1 liter, we have 1.6 times as much oxygen in the water as we do in the tube!

so use less water... ok, we can use 1/3rd of a liter but that still leaves 50% of the oxygen in the water ... an appreciable amount.

so as I see it:
• The "collection of gas over water" and Dalton's law (a staple of HS chemistry books and video examples) is missing a significant component- while HS chem is simplified, this seems way off
• If the water already has oxygen dissolved in it, this might reduce the total oxygen "generated" by an experiment (by mixing it with oxygen already present), but I doubt by much... we could always let the measuring cylinder sit for a while to see how much oxygen collects on its own.

either way, it seems either the books are all doing it wrong, or I am...
I'm pretty sure I am, but I can't see where I'm wrong

Your calculations are OK, although you still have not accounted for the fact water was already saturated with the atmospheric oxygen.

However, Purdue chemists are wrong, mass of the collected oxygen is not 2.4×10-2 g, but 2.4×10-1 g. I will try to contact them about it.

Simple sanity check - we are not far from STP, so molar volume is not far from 24 L. 0.193 L/24 L times 32 g/mol is 0.around 26 g. We are surely slightly wrong, but not an order of magnitude, so 0.0246 is out of the question.

1 person
Yes, there is an error in Purdue's math, and maybe the book is right?

first of all, thank you so much Borek,

I recalculated and realized there WAS an error... I can't believe I missed it the first time around!

it seems the dissolution of oxygen in water turns out to be 16% of the amount of the oxygen in the graduated cylinder... still a significant amount if you were experimenting, but not as huge as it had been before.

$\frac{(0.975)(0.193)(32)}{(0.08206)(298)}$= 0.2462 g O2 in gas​

0.24624 + 0.04056 = .2868 g 16.4% greater​

You're right that I have not accounted for oxygen already in the water prior to the experiment, and this should reduce that value further... If I am reasoning correctly, I can ignore ALL oxygen dissolved in water if the pressure of the surroundings matched the pressure of the system and tested everything at equilibrium...

if so, this would validate the books = ignore oxygen dissolved in water as long as PO2_system = PO2surroundings

I don't mind taking a 3 mile trip to validate the trip across the street... if that is what I just did.

skaai said:
first of all, thank you so much Borek,

I recalculated and realized there WAS an error... I can't believe I missed it the first time around!

it seems the dissolution of oxygen in water turns out to be 16% of the amount of the oxygen in the graduated cylinder... still a significant amount if you were experimenting, but not as huge as it had been before.

$\frac{(0.975)(0.193)(32)}{(0.08206)(298)}$= 0.2462 g O2 in gas​

0.24624 + 0.04056 = .2868 g 16.4% greater​

You're right that I have not accounted for oxygen already in the water prior to the experiment, and this should reduce that value further... If I am reasoning correctly, I can ignore ALL oxygen dissolved in water if the pressure of the surroundings matched the pressure of the system and tested everything at equilibrium...

if so, this would validate the books = ignore oxygen dissolved in water as long as PO2_system = PO2surroundings

I don't mind taking a 3 mile trip to validate the trip across the street... if that is what I just did.
Hey guys, what about the nitrogen that's also already dissolved in the water and gets stripped out by the pure oxygen bubbling through (adding to the gas flow)?

skaai said:
You're right that I have not accounted for oxygen already in the water prior to the experiment, and this should reduce that value further... If I am reasoning correctly, I can ignore ALL oxygen dissolved in water if the pressure of the surroundings matched the pressure of the system and tested everything at equilibrium...

This is more complicated, as the system is not at equilibrium by definition - you have atmosphere on the outside, and pure oxygen inside, so definitely some of the collected oxygen will dissolve.

if so, this would validate the books = ignore oxygen dissolved in water as long as PO2_system = PO2surroundings

Which as I explained above is not true (basically partiall pressure of the oxygen is .21 atm outside and 1 atm inside).

However, there is another factor to account for - how fast does the solution saturate. If the dissolution is slow enough, amount of gas dissolved can be negligible. Honestly, I have no idea how fast it is and how it should be treated.

Chestermiller said:
Hey guys, what about the nitrogen that's also already dissolved in the water and gets stripped out by the pure oxygen bubbling through (adding to the gas flow)?

Hard to deny it is another possible source of errors.

Interestingly, there is another, parallel thread that discusses very similar problem: https://www.physicsforums.com/showthread.php?t=732638

## 1. What is gas collected over water?

Gas collected over water refers to the process of collecting a gas sample by displacing the water in a container with the gas. This method is commonly used in chemistry experiments to measure the volume of a gas produced or released.

## 2. Why is gas collected over water?

Gas collected over water is used because it allows for the measurement of the volume of a gas at atmospheric pressure. The pressure of the gas is equal to the atmospheric pressure plus the pressure of the water vapor produced, making it easier to calculate the volume of the gas.

## 3. What is Fick's Law?

Fick's Law is a mathematical equation that describes the rate of diffusion of a gas. It states that the rate of diffusion is directly proportional to the surface area and concentration gradient of the gas, and inversely proportional to the distance it must travel.

## 4. How is Fick's Law used in gas collection over water?

In gas collection over water, Fick's Law is used to calculate the rate of diffusion of the gas being collected. This information can then be used to determine the volume of the gas produced, as well as to study the properties of the gas.

## 5. What are some factors that can affect the accuracy of gas collection over water?

The accuracy of gas collection over water can be affected by factors such as temperature, atmospheric pressure, and the solubility of the gas in water. These variables can alter the volume of the gas collected and should be taken into consideration when conducting experiments.

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