For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

### Related Articles:

### Partial Pressure, Mol Fraction, & Vapor Pressure

- The total pressure in a sample containing a mixture of gases is the sum of the individual pressures of each gas.
- Mol Fraction is like any other fraction: part over the whole. Mol Fraction of A is the mols of A divided by the total number of mols in the sample.
- Given any liquid, some molecules always escape into the gas phase. In a closed container, these gas-phase particles exert a pressure. This is called the Vapor Pressure of the liquid at a given temperature.
- Nitrogen content of an Organic compound can be determined by the Dumas Method. The compound is passed over hot Copper (2) Oxide and reacts.
- The product gases are passed through a solution of Potassium Hydroxide to remove the Carbon Dioxide. The remaining gas is Nitrogen saturated with water vapor. In a given experiment 0.225 g of the unknown compound produces 27.8 mL of Nitrogen saturated with water vapor at 25 degrees Celsius and 730 torr. What is the Mass % Nitrogen in the compound? (Vapor pressure of water at this temp is 23.8 torr.

### Partial Pressure, Mol Fraction, & Vapor Pressure

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Gases 0:27
- Gases
- Mole Fractions
- Vapor Pressure
- Example 1
- Example 2

### AP Chemistry Online Prep Course

### Transcription: 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 O _{2} is going to be a certain something, and the partial pressure of N_{2} 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 n _{3}RT 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 P _{1} + P_{2} + P_{3}; now, I'll just substitute these values in for this.*0217

*I end up with n _{1}RT/V +n_{2}RT/V, + n_{3}RT/V.*0228

*Well, if you notice--RT/V--you can factor it out; so let me pull that out; RT/V times n _{1} +n_{2} + n_{3}.*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 n _{1}; just expand this out with PV=nRT; just put the n_{1}=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 H _{2}O).*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 KClO _{3}, which is potassium chlorate, was heated and decomposed to form solid potassium chloride and oxygen gas.*0827

*Well, the O _{2} 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 O _{2} collected, and the mass of KClO_{3} decomposed?*0937

*Let's see what is going on here: if you refer back to the diagram, we burned the KClO _{3}; 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 KClO _{3} 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 KClO _{3} was actually decomposed?*1016

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

*We do KClO _{3}; we heat it up; we end up with KCl + O_{2}.*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 O _{2} plus the partial pressure of H_{2}O.*1066

*Well, we know the total pressure: that is 640 torr; we're looking for the partial pressure of O _{2}, 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 O _{2} 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 O _{2} 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 O _{2}...0.811 atmospheres is the partial pressure of O_{2}.*1167

*Now, we have PV=nRT; in this case, partial pressure of O _{2}, times V, equals the number of moles of O_{2}, 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 O _{2} is equal to the partial pressure of O_{2}, 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 O _{2} (the mole relationship is 2 moles of potassium chlorate produce 3 moles of O_{2}).*1250

*Therefore, I know that this came from 0.0126 moles of KClO _{3}.*1266

*Now, I take 0.01216--I'm sorry, I forgot a 1 there--moles of KClO _{3}, times the molar mass of KClO_{3}, which happens to be 122.6 grams per mole, for a grand total of 1.49 grams KClO_{3} 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 CO _{2} gas, and you end up producing water vapor ("water gas").*1410

*Let's see: total pressure, CO _{2}...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 N _{2}, 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 N _{2}: nitrogen is 14, and 2 is 28)--we end up with 0.0294 grams of N_{2}.*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

1 answer

Last reply by: Professor Hovasapian

Tue Aug 2, 2016 2:53 AM

Post by tae Sin on July 29, 2016

Hi, Mr. Hovasapian.

At the part about the vapor pressure, you were talking about vapor pressure with test tubes and the water. Should the gas react to the air when you insert the tube into the tube underwater? And does the air causes the gas to change the pressure of the test tube underwater?

I like the lesson. Thanks!

1 answer

Last reply by: Professor Hovasapian

Sat Nov 15, 2014 10:38 PM

Post by Datevig Daghlian on November 13, 2014

Dear Professor Hovasapian,

Thank you very much for your lecture! I had never seen the Ideal Gas law from that perspective. Thank you and God bless!

Thank You,

George Daghlian

3 answers

Last reply by: Professor Hovasapian

Tue Jan 28, 2014 3:04 AM

Post by Deborah Lee on January 27, 2014

Hi Professor Hovasapian,

I hope you are well.

We refer to Nitrogen as the gas (N_2) when calculating partial pressure, mol and mass. Is there a reason why you refer to Nitrogen as N when calculating Mass % ?

Thank you for your lectures, I find them extremely helpful!

1 answer

Last reply by: Professor Hovasapian

Wed Aug 14, 2013 2:44 PM

Post by Mark Andrews on August 14, 2013

Back when you introduced the Mol Fractions and said you'd skip over the maths, I was wondering if you would be able to do a video on that because while I was able to replicate your example with the numbers you provided, I kind of got lost with how you got to the 616.2.

I realise that 640-23.8 = 616.2

My thought was to approach the problem by converting 640 Torr as the P total

Then converting 23.8 Torr and subtracting that from the "total pressure" then using pV=nRT

When I followed that procedure I had P Total = 0.842

H2O Pressure worked out at 0.0313 which gave me the 0.8107 = 616.2

It has taken me a while to work out I ended up with the same answer but was that the mathematical process you didn't go into?

By the way it was after watching your FREE lectures that I decided to purchase a prescription. I think you're a very good teacher and far easier to understand than my current lecturer at Uni who has lost me in the subject about 4 weeks ago. Thank you for way you conduct your lessons.

1 answer

Last reply by: Professor Hovasapian

Wed Aug 7, 2013 1:26 AM

Post by Charles Zhou on August 6, 2013

For example 1, when I was dealing with the moles of oxygen, I use n=PV/RT. But I think the volume of 0.550L is the total volume, we need to subtract the Vapor volume by using the molar Fraction: n()2)/n(vapor)=P(O2)/P(vapor). So I got the n(O2) equals to 0.0181. It's really close to 0.0182. And the mass of KClO3 is 1.49. I don't know if I'm right. Thanks for your help and I really enjoy your class.