AP Chemistry Partial Pressure, Mol Fraction, & Vapor Pressure

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

In the last lesson, we introduced gas laws; we introduced the ideal gas law; and we talked a fair amount about pressure.0004

Today, we're going to continue on, and we're going to talk about mixtures of gases and something called partial pressure.0011

It's going to use exactly the ideal gas law--again, very, very simple mathematically--so let's just go ahead and jump in and see what we can do!0017

OK, it's very, very simple: let's just say that, if I have a mixture of gases--let's just say 3 gases--each one has a certain pressure.0028

Let's say, if I keep them in three separate containers--there is a pressure; there is a pressure; there is a pressure.0045

If I dump them all into one container, each one is contributing to the total pressure.0048

Well, it's exactly what you think: the total pressure of the system is equal to the partial pressure of 1, plus the partial pressure of 2, plus the partial pressure of the third.0053

Now, when we say partial, that means that it is part of a mixture--that is where the "partial" comes from.0065

It's not as if you are taking part of the pressure of the gas.0071

So, gas 1 contributes to the pressure; gas 2 contributes to the pressure; gas 3 contributes to the pressure; because now, you have a mixture of them; there is a total pressure that we can measure, but we can actually divide it up.0074

Each one actually contributes some kind of pressure to it.0089

That is all that is going on here; so we speak of these individual pressures of the individual gases as the partial pressure of that gas.0091

So, if I have a mixture of...well, for example, air: air is a mixture of nitrogen and oxygen gas; if I take a sample of it in a certain volume, the partial pressure of O₂ is going to be a certain something, and the partial pressure of N₂ is going to be something else.0100

That is all that is going on here--nothing strange.0115

Let's go ahead and just deal with some of the mathematics, because I think if you see the mathematics, where it comes from, it will make a little bit more sense as to what it is that is going on.0119

We, again, start off with our ideal gas law; so let me put that over here: PV=nRT: pressure times the volume equals the number of moles times the gas constant times the temperature in Kelvin (volume in liters, pressure in atmospheres).0129

For each gas in the container, for a given volume and a given temperature, we have the following.0144

The partial pressure of 1 is equal to the number of moles of 1, times RT, over V.0173

R is constant: given volume, given temperature--those are constant, so when we rearrange this equation, for each gas we can use that equation.0179

So, the number of moles of 1, times R, times T, divided by the volume, gives me the pressure of the one gas.0187

The partial pressure of 2 is equal to the number of moles of 2, times RT, over V: just a rearrangement of the ideal gas law.0197

And partial pressure of 3: exactly what you think: it's n₃RT over V.0205

Well, we have our basic relationship; if we just add the individual pressures, it is going to give us our total pressure.0211

So, total pressure is equal to P₁ + P₂ + P₃; now, I'll just substitute these values in for this.0217

I end up with n₁RT/V +n₂RT/V, + n₃RT/V.0228

Well, if you notice--RT/V--you can factor it out; so let me pull that out; RT/V times n₁ +n₂ + n₃.0241

So, the total pressure is equal to RT/V times the total number of moles.0255

The total number of moles comes from the moles of 1, plus the moles of 2, plus the moles of the other.0269

As you see, the number of moles and pressure are related.0274

Pressure is just a different expression--a different way of representing the number of moles of something.0279

That is what is nice about pressure; it is just the other side of the coin, if you will, of an amount.0285

So, we speak of the partial pressure; we speak of the total number of moles; the total number of moles is related to the partial pressure through this proportionality constant.0291

That is all that is going on.0303

That is what this right here is essentially saying: it's saying that the number of moles is related to the pressure through a proportionality constant.0305

Number of moles and pressure are just two sides of the same coin--amount; that is what it is representing--amount.0313

Moles is the standard unit by which we measure things in chemistry; well, when you are dealing with gases, pressure is more convenient--to speak about pressure and moles.0324

Here is our conversion; that is all that is going on here.0333

OK, so the total number of moles of a gas is what is important, not the identity; that is another thing that this is telling me.0337

It is the total number of moles that matter of the gas; the identity of the gas is completely irrelevant.0345

Now, let's introduce something called a mole fraction: for our purposes, we're probably not going to be using it all that much, as far as the problems that we do for gases, but mole fraction--I want to introduce it here.0353

It is something that is very important; it will show up again when we discuss solutions, not too long from now.0365

So, it's exactly what you would expect it to be when we talk about a fraction: a fraction is just a part over the whole.0370

Well, a mole fraction, which is symbolized by the Greek letter chi...the mole fraction of, let's say, a certain...something, is the number of moles of that something over the total number of moles.0378

It's just the part over the whole; that is it.0394

If I have 10 moles total--let's say I have 8 moles of nitrogen mixed with two moles of oxygen--well, the mole fraction of oxygen is 2/10, and the mole fraction is 8/10; it's just the number of moles of the part, over the total number of moles floating around.0399

That is it; that is all mole fraction is.0418

So, as it turns out, I'll go ahead and skip the math, but in terms of pressure...which--you can do this yourself if you want to...0420

Again, you have n₁; just expand this out with PV=nRT; just put the n₁=PV/RT, and you will see that the PV's over the RT's--they all cancel, and what you end up with is...0431

You can also find the mole fraction of a gas by measuring the pressure of that gas, individual gas, over the total pressure of the system; that is it.0448

You can work in moles, or you can work in pressure, because again, pressure and moles are just different ways of representing the same thing: an amount.0457

Notice, mole fraction doesn't have any units.0469

OK, we can also rearrange this; we can rearrange this to: the partial pressure of some individual gas is equal to the mole fraction of the gas, times the total pressure.0473

That is just expressing a fraction; that is all that is going on here--straight math--nothing you haven't seen before.0491

I did want to introduce it to you, because we will be talking about it more and more as we go on.0498

Now, I want to talk about something called vapor pressure; I'm going to use vapor pressure to segue into our first problem.0505

Let's talk--I'm going to draw a little bit of a thing here; if I were to take, let's say, a test tube, and I have a little trough of, let's say, water, and if I invert the test tube this way--and let's say I have some sample of a gas, and put a little fire under there, and I have a little tube that is going into the water and up, up...0516

So now, some gas is escaping; it is traveling through the tube; it is going underneath the water (and this tube is actually filled with water, to the top, just like we did with the mercury--except, in this case, we are dealing with just H₂O).0555

Well, the gas is going start bubbling up, right?--it's going to start bubbling, and it's going to make its way to the top.0570

When it makes its way to the top, the gas is actually going to start collecting at the top here.0578

As it does so, it's going to push the water level down; it's going to push--the gas is expanding; it's creating a volume up in the top container.0582

So, what happens is: the water level in there is going to drop as more gas collects.0592

Now, let's see: we have collected some gas, and let's say now the water level is there.0600

What is up here is the gas that we have collected.0605

When we collect a gas, using this method, over whatever liquid this happens to be (more often than not, it will be water; and for our purposes it will be water), something very interesting happens.0612

There is a certain amount of gas in here that is going to exert a pressure, because gas pressure--that is what is pushing the water level down, back into the trough of the water.0624

Well, something else interesting happens: any time I have some liquid in a closed container, some of the liquid particles at the surface escape and become gas particles--at any temperature (most standard temperatures).0635

So, in a closed container, if I have some water, it isn't just water-and-there-is-nothing-floating-around-on-top; as it turns out, water itself--some of the water molecules--will escape into the vapor phase, and so now, you have some water vapor on top of the liquid water.0654

That water vapor is a gas, and a gas exerts a pressure.0670

We call the gas that has escaped from the liquid phase of something, that is hovering above it in a container, the vapor pressure.0675

Let me give you a reasonable definition here.0685

We talk about the vapor pressure, and that is the pressure exerted by the gaseous molecules of a given liquid (which, in our case, is water) that have escaped the liquid phase.0690

That is it: if I have a container of a liquid, it isn't just an empty vacuum; as it turns out, some of the molecules actually escape, become gaseous, and--because they become gaseous--they actually exert a certain pressure on the walls of the container.0729

That pressure is what we call the vapor pressure; all liquids do this.0744

That is it; what this means for our purposes here is: we have collected a gas, but we have collected it over water.0748

Well, water, at different temperatures, has different vapor pressures, certainly; so here, we not only have gas particles, but we also have particles of water vapor.0757

This becomes a mixture of gases; and, because it is a mixture of gases, our Dalton's law of partial pressure applies.0769

The total pressure in here is going to be the pressure of the gas, plus the pressure of the water at that particular temperature.0776

There is a table of vapor pressure of water at different temperatures, and that is what we are going to use in our next example.0782

I just wanted to introduce this idea of vapor pressure; again, it will show up again and again, particularly when we talk about solutions.0789

We will discuss mole fraction in more detail, but I wanted you to see what was going on for the sake of this example.0800

Now, let's go ahead and do our Example #1.0806

This is a standard problem in chemistry, using gases, using stoichiometry, using...this is a typical problem that you will actually see on the AP exam in the free response section.0812

Solid KClO₃, which is potassium chlorate, was heated and decomposed to form solid potassium chloride and oxygen gas.0827

Well, the O₂ was collected over water with that thing that we just designed; so we heated up the potassium chlorate in the test tube; the oxygen gas was collected over water--pushed the water level down and collected in that little space, up above, at 25 degrees Celsius and a total pressure of 640 torr.0851

That means the total pressure in that space above is 640 torr; it consists of oxygen pressure and the water vapor pressure, which we're going to talk about next.0889

Now, the volume collected was 0.550 liters.0898

The vapor pressure of water at this temperature, at 25 degrees Celsius, is 23.8 torr; that means there is enough water vapor, floating around above the water, to create a pressure of 23.8 torricelli.0914

Our question is this: What is the partial pressure of O₂ collected, and the mass of KClO₃ decomposed?0937

Let's see what is going on here: if you refer back to the diagram, we burned the KClO₃; we produced oxygen gas.0970

Oxygen gas traveled through the tube, went up into the tube, and pushed the water level down; now, oxygen gas is collecting in that space above in the vertical tube.0978

It is also mixed with some water vapor, because again, any time you collect something over a liquid, some of that liquid turns into vapor.0987

So, it is saying that the total pressure is 640 torr, while the vapor pressure at this particular temperature that we ran this experiment is 23.8 torr.0994

We're asking for "What is the partial pressure of the oxygen?" and "How much KClO₃ was decomposed, based on the fact that I collected .550 liters?"1004

Wow, I collected .550 liters; how am I going to find out how much KClO₃ was actually decomposed?1016

Well, let's start with an equation; again, we are doing chemistry.1022

We do KClO₃; we heat it up; we end up with KCl + O₂.1027

I see a three--no, that is ozone; no, it's oxygen gas that we collected, not ozone.1038

When I see a three and I see a two, I immediately do this: just 3, 6, 3, 2, 6, and then just put a 2 here to balance out the Cl.1043

So, 2 moles of potassium chlorate produce 2 moles of potassium chlorite, solid, plus 3 moles of oxygen gas.1056

We know that the total pressure is equal to the partial pressure of the two gases: water vapor and oxygen: the partial pressure of O₂ plus the partial pressure of H₂O.1066

Well, we know the total pressure: that is 640 torr; we're looking for the partial pressure of O₂, and we are given the vapor pressure of water--it's 23.8 torr.1078

So, the first part of the problem is easy: the partial pressure of O₂ collected is just the 640 minus the 23.8, which gives us 616.2 torricelli.1094

That's it: I just use the standard: the total pressure is equal to the sum of the partial pressures.1110

I have 2 gases; I have the partial pressure of water vapor at 25 degrees Celsius; so, my partial pressure of O₂ is 616.2 torr.1115

Now, let's see if we can--the next thing we want to do is: we want to basically find the number of moles of oxygen.1125

From the moles of oxygen, we're going to use the reaction stoichiometry to find the number of moles of potassium chlorate.1133

From the moles of potassium chlorate, we're going to deduce the number of grams of potassium chlorate.1138

That is what we are doing here.1143

Now, let's go ahead and--again, we want to work in atmospheres, so let's convert this 616.2 torr; it's going to be times 1 atm=760 torricelli; so that gives us 0.811 atmospheres.1145

So, the partial pressure in atmospheres: the partial pressure of O₂...0.811 atmospheres is the partial pressure of O₂.1167

Now, we have PV=nRT; in this case, partial pressure of O₂, times V, equals the number of moles of O₂, times RT, because we're dealing only with the oxygen.1182

But, the equation applies.1202

We rearrange, because we want to find the number of moles, so the number of moles of O₂ is equal to the partial pressure of O₂, times the volume, divided by RT.1204

It is equal to 0.811; that is what we just found--that is the partial pressure--times 0.550 liters--that is the volume that it is actually encased in, the volume at the top of the tube; R is 0.08206, and we are at 298 Kelvin.1215

Well, when I do that, I end up with 0.0182 moles of oxygen; that is how many moles of oxygen I have.1241

Well, now, 0.0182 moles of O₂ (the mole relationship is 2 moles of potassium chlorate produce 3 moles of O₂).1250

Therefore, I know that this came from 0.0126 moles of KClO₃.1266

Now, I take 0.01216--I'm sorry, I forgot a 1 there--moles of KClO₃, times the molar mass of KClO₃, which happens to be 122.6 grams per mole, for a grand total of 1.49 grams KClO₃ decomposed.1276

This is no more than a standard stoichiometry problem, using mole ratios, except it's under gaseous conditions; therefore, I have to use partial pressures and ideal gas laws; that's all.1303

Moles--that is the nice thing about the ideal gas law: it contains moles, so we can use it in our stoichiometry.1317

Don't be fooled; this is a gas problem; it's not really a gas problem--this is a stoichiometry problem.1323

It's a standard chemistry problem--dealing with gases, therefore--for gases, we have one technique in our toolbox: PV=nRT.1329

Well, actually, we have two techniques; the other is total pressure is equal to the sum of the partial pressures of however many gases; that is it--those are the two things that we are using when we are dealing with gases.1338

Ideal gas law, Dalton's law of partial pressures: everything else is just classical stoichiometry.1351

OK, let's see what else we can do...let's move forward here, and let's deal with another example.1358

This example is going to be typical of the kind of example that you see in the AP exam: it's going to have a lot of wording; a lot is going on.1372

Don't that let that intimidate you.1380

A lot of times, half the problem will be just the description of what is happening, so that you have a sense of what is going on, so that you can deduce the chemistry, or deduce the math, from that.1383

Again, don't get intimidated.1394

Let's just start and see where we go.1395

The nitrogen content of an organic compound can be determined by something called the Dumas method.1398

The compound is passed over hot copper oxide, and reacts as follows (so I'm going to go ahead and write the reaction).1403

It is: you take the compound, you pass it over hot copper oxide, and you end up producing nitrogen gas; you end up producing CO₂ gas, and you end up producing water vapor ("water gas").1410

Let's see: total pressure, CO₂...OK.1430

Now, they say the product gases are passed through a solution of potassium hydroxide to remove the carbon dioxide.1438

You produce three gases, and then you pass it through a solution of potassium hydroxide; that binds the carbon dioxide and gets it out of the way.1444

The remaining gas is nitrogen, saturated with water vapor.1455

Nitrogen saturated with water vapor: that just means nitrogen and water vapor, in the same gas mixture; that is all that this means.1459

In a given experiment, 0.225 grams of the unknown compound produces 27.8 milliliters of nitrogen saturated with water vapor, at 25 degrees Celsius and 730 torr.1468

What is the mass percent of nitrogen in the compound?1483

The vapor pressure of water at this temperature is 23.8 torr.1486

What are they asking for?--they want to know what is the mass percent of nitrogen in this compound.1490

Well, we know mass percent...so let's just write out the equation so we don't get lost.1496

Mass percent equals the mass of nitrogen over the total mass, times 100; so that is all we want.1501

We have the total mass already: our compound is .225 grams--we already have one of the numbers.1516

All we need to do is find the mass of nitrogen.1522

Let's take a look and see how we are going to do this.1526

Again, mass of nitrogen--chances are, we are going to have to find the number of moles of nitrogen, and the number of moles is going to come from the ideal gas equation, so let's just work forward.1528

We are dealing with a mixture here: we have a mixture of water vapor and nitrogen, because the carbon dioxide has been removed by the potassium hydroxide solution.1539

Therefore, our total pressure of the system is equal to the partial pressure of the nitrogen gas, plus the partial pressure of the water vapor (I just wrote the nitrogen).1550

OK, well, let's see what we have.1564

In a given experiment, unknown compound, nitrogen saturated with water, 730 torr--that is our total pressure.1566

We have 730 torr; we need the partial pressure of nitrogen in order to find the number of moles of nitrogen, and they gave us the vapor pressure of water, which is 23.8 torr.1574

So again, we can find the partial pressure of nitrogen with a simple arithmetic problem: 730-23.8, and we end up with 702 (is that correct?--23, 24, 731)--yes, 702 torr.1588

Wait, is that correct?--730, minus...no, that is not correct; that is not 702 torr; let me see, let's do some quick arithmetic; arithmetic has never been my strong suit.1617

730, 23.8, 24, 25, 26, 27, 28, 29...it's going to be 706 (I think), point 2, torr--excellent.1631

Then, we're going to multiply that by...we're going to convert that to atmospheres, so 1 atmosphere over 760 torr...1649

OK, so I don't have a calculator at my disposal; my number was originally incorrect when I did this, but I'm going to use my original number.1664

I'm going to do it as if it were 702, but again, the division and the number...it's the process that is important, not the actual number here.1670

I'm going to use the original number that I got, which was 0.924 atmospheres.1680

That is the partial pressure of the nitrogen gas.1688

Well, PV=nRT; the partial pressure of the nitrogen gas, times the volume, equals the number of moles of nitrogen gas, times RT.1694

Rearrange: the number of moles of nitrogen gas equals the P of N₂, times the V, over RT.1707

The pressure is 0.924 atmospheres, times...and we collected 27.8 milliliters, so we want to convert that to liters: .0278 liters; the gas constant is .08206, and our temperature is 25 degrees Celsius; which is 298 Kelvin.1716

Again, liter, Kelvin, atmosphere.1744

We end up with a number of moles that is equal to 0.00105 moles of nitrogen gas.1748

Well, 0.00105 moles of nitrogen gas, times 28 grams per mole (it's 28 because this is N₂: nitrogen is 14, and 2 is 28)--we end up with 0.0294 grams of N₂.1759

OK, mass percent equals 0.0294 grams, divided by the total number of grams, which was 0.225, in our compound; when we multiply by 100%: 13% nitrogen by mass.1785

There we go.1810

We converted the nitrogen in a compound to a gas--to nitrogen gas, to carbon dioxide gas, to water gas.1816

We took care of the carbon dioxide gas; therefore, now we have a mixture of nitrogen and water vapor.1827

We know, at a certain temperature, that the vapor pressure of water is a certain number; we can look that up in a chart, or we are given it in the problem.1834

We can find the partial pressure of nitrogen.1841

From the partial pressure of nitrogen, we use our one basic technique, which is the ideal gas law, to find the number of moles of nitrogen.1843

From the number of moles of nitrogen, we find the number of grams of nitrogen.1851

Take the number of grams of nitrogen; divide by the number of grams of the total compound; and that gives us a mass percent.1855

So, again, the problem is routine, in the sense that it is nothing you haven't done before.1861

You are doing the same problems over and over again.1867

But now, we are doing it in the context of gases, so the technique that we use is Dalton's law of partial pressures and the ideal gas law.1869

Everything is going to be a variation of that: if you sort of look at it that way globally, I think a lot of this will start to make more sense.1878

And again, we start with the chemistry; start with an equation; balance the equation; see if you can understand what is going on--let the problem wash over you.1886

There is nothing strange that is happening here; it is basic math.1896

It just needs to be arranged in a certain way.1899

OK, so that is Dalton's law of partial pressures, and a little bit of mole fraction, and things like that.1902

With that, I will say "Thank you" for joining us here at Educator.com, and we'll see you next time--goodbye!1908