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Post by Professor Hovasapian on July 18, 2012

Link to the AP Practice Exam:

Take good Care


AP Practice Exam: Free Response Part II

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
  • Free Response 0:12
    • Free Response 3: Part A
    • Free Response 3: Part B
    • Free Response 3: Part C. i
    • Free Response 3: Part C. ii
    • Free Response 3: Part D
    • Free Response 4: Part A
    • Free Response 4: Part B
    • Free Response 4: Part C
    • Free Response 4: Part D
    • Free Response 4: Part E
    • Free Response 4: Part F
    • Free Response 4: Part G
    • Free Response 4: Part H
    • Free Response 5: Diagram
    • Free Response 5: Part A
    • Free Response 5: Part B
    • Free Response 5: Part C
    • Free Response 5: Part D
    • Free Response 5: Part E

Transcription: AP Practice Exam: Free Response Part II

Hello, and welcome back to, and welcome back to AP Chemistry.0000

We are going to continue on with our AP practice exam, the free response questions; this is going to be Part 2; so let's just dive right on in.0004

OK, so question #3 is a question about reaction rates; it is a question about kinetics; it gives you some kinetic data, and it asks you a bunch of questions about it.0014

We did spend a fair amount of time on kinetics, so it should be reasonably straightforward; let's see what we can do with it.0025

It says: A rate study of the reaction represented above was conducted at 25 degrees Celsius; the data that were obtained are shown in the table below.0033

Hopefully, you have a copy of "the table below"; I won't reproduce it here; but, of course, I am going to reproduce the reaction, because any time we do chemistry, we are always talking about some reaction.0042

2 nitrogen monoxide gas, plus bromine gas, goes to form 2 NOBr gas.0053

They give you some reaction data; they run three experiments, and they have the initial concentration of nitrogen monoxide, initial concentration of Br2, and then they have the initial rate of appearance of NOBr.0068

Notice how they give this to you: be very, very careful--read very carefully.0082

This actually says "the initial rate of appearance of the product"; most...I shouldn't say most; many experiments, especially the problems that we did, actually gave initial rates of disappearance of a given reactant or product.0088

So, we have to be very, very careful, because these numbers are not the same.0107

When we talk about a rate, we have to be very careful to make sure of what rate we are talking about.0114

The first part of this question will actually address that issue; so let's go ahead and get started.0120

A says: Calculate the initial rate of disappearance of Br2 in experiment 1.0126

Well, experiment 1...the rate of appearance of NOBr is 3.24x10-4; they want to know the rate of disappearance of Br2.0133

OK, so I'm going to write out a generic equation and describe the relationship that exists between the rates of disappearance of A, B, C, and D.0144

If you have an equation that is aA + bB (reaction) going to cC + dD, here is the relationship.0154

-1/a (the change in a, so -1/a) ΔA/Δt is equal to -1/b (ΔB/Δt), is equal to 1/c (negative because of disappearance) (ΔC/Δt), equals 1/d (ΔD/Δt).0172

What this equation does is: it gives us a relationship that exists between the appearance of product and the disappearance of reactant, based on the fact that these coefficients may not be 1.0213

So, based on this, what we write for experiment 1--it is this.0225

They want the rate of disappearance of bromine gas (I'm going to do this in black, actually); so they want the rate of disappearance of this, but what we are given is the rate of appearance of this in the experiment.0232

OK, so -1/1, so -ΔBr2/Δt (change in time)--this is the rate of disappearance; a rate is a change in concentration over the change in time.0245

This is the symbol for the rate.0261

It is equal to 1/2, times the rate at which this is appearing.0264

Well, the rate at which this is appearing is...well, let me just write it out.0276

ΔNOBr/Δt: again, we are just using these coefficients here to guide us, based on this relationship that exists between the reactants and the products.0281

But we know what this number is: this number is what the experiment gave us, so it is 1/2 of 3.24x10-4 moles per liter per second.0299

Units actually don't matter all that much here, unless they actually specify that the units matter; so you are not going to lose any points if you don't write the units.0322

When we do this 1/2 of this...well, 1/2 of that is equal to (let me move the negative sign over, and again, actually, the negative sign doesn't even matter here, either) -1.62x10-4; I'll go ahead and put the units in, just for good measure; there we go.0329

Make sure you take a look at what they are giving you the rate in: if it's appearance of products, and if they are asking you for the rate relative to another reactant or product, these coefficients--if they are not all 1--then this is the relationship that you use.0351

You just set one thing equal to the other thing--in this case, 1/1 ΔBr/Δt equals 1/2 ΔNOBr/ΔT; and you solve for this, which is the rate (the rate of disappearance of Br).0368

There you go; OK.0384

So now, B--Part B says: Determine the order of the reaction with respect to each reactant, Br2 and NO; in each case, explain your reasoning.0387

OK, so let's go ahead and take a look at the data in front of us (so again, hopefully you have the data in front of you).0397

We see that the initial concentrations of NO...we have, for experiment 1, .0160; for experiment 2, .0160 again; and experiment 3, .0320.0406

For the initial concentration of Br2, we have .0120, .0240, 0.0060; so we know that we are going to have some rate constant, times NO raised to some power and Br2 raised to another power.0418

Well, in this case, it looks like it's easier to do Br2, because experiments 1 and 2--the concentration of NO is held constant.0440

So, when I, if I hold those two constant, now let me take a look at the concentrations of Br2.0447

In the first experiment, it's .0120, and I have a certain rate; well, when I double the concentration from .0120 to .0240, I end up going to 6.38x10-4.0454

I double the rate: double the concentration, double the rate--that means it is an order 1.0466

1 for 1, 2 for 2; the relationship is linear; if I triple something, I triple the rate; if I quadruple something, I quadruple the rate; that is a linear relationship.0474

That means that m is equal to 1; so the order of bromine is 1.0483

So now, we have: the rate is equal to K, times NO, times Br2 to the 1; now, we have to worry about finding (oops, I used n for this one) n.0492

OK, so let's see what we can do: in this case, there are no two experiments where the initial concentration of bromine is constant, so we can't just look at the data and decide what happened with a simple mathematical procedure.0507

We have to write sort's simple math, but it looks a little more complicated than it actually is.0519

We are going to actually pick a couple of experiments, and literally plug the values in, and see if we can come up with some value for n.0525

Let's see--so let's set up a ratio here: so we are going to take rate 2, divided by rate 3; so we are going to take experiment 2 and experiment 3.0535

Actually, you know what, I'm going to do this on the next page, because I am getting some crazy lines here; and that is fine--we have some pages to spare.0547

Rate 2, divided by rate 3--well, that is going to equal 6.38x10-4, divided by 6.42x10-4; that equals...well, it equals K, times the concentration of NO to the n, times the concentration of Br2 to the 1 power, over K, the concentration of NO to the n power, Br2 to the 1 power; and now, we just plug in some values.0555

K times 0.0160 to the n power, times 0.0240 to the first power, over K times 0.0320 to the n power, times 0.0060 to the first power: when we do that, well, this number right here is equal to 1; that is equal to this number right here, which is...well, this over that is equal to 4; this over that is equal to 1/2 to the n power; when we solve this, we get 1/2 to the n power equals 1/4; therefore, n is equal to 2.0602

The order is--well, the order of the NO is actually equal to 2.0660

That is it--that is how you do it, nice and simple: you just set up the...take one rate over another rate, put in the numbers, and then this cancels with this; you end up getting something, and you just literally solve for the exponent; that is what you're doing.0671

We actually did this several times: it is called the method of initial rates.0690

OK, now let's do Part C--Part C says: For the reaction, write the rate law that is consistent with the data.0694

OK, so the rate law consistent with the data--well, we just did it: K, times NO to the second power and Br2 to the first power.0703

You don't necessarily need to put the 1 there--it's up to you--again, just sort of a habit that has stayed with me over the years.0713

That is it: that is part C, part 1.0720

Now, part 2 says: Calculate the value of the specific rate constant K, and specify units (so here they do want you to specify units; they want you to find this value of K).0725

Well, it is very simple: choose an experiment, either 1, 2, or 3; put the values in; and solve for K.0737

OK, so let's go ahead and do that: I think I'm probably just going to pick Experiment 1--yes, I think so; OK.0742

The rate is 3.24x10-4, equals K times 0.0160 squared, times 0.0120 to the first power; and what you get here is where you have to sort of be a little bit careful: I'm going to do the numbers first, and then I'm going to do the units after that...0755

K is equal to 105, and now let me do the units: the units of the rate were moles per liter per second; I'm going to divide that by molarity cubed.0784

I'm going to divide that by moles cubed, over liters cubed, and what you end up getting when you do this: this becomes a 2 and cancels that mole; this becomes a 2 and cancels that liter; these flip up top; you end up with liters squared over moles squared per second.0803

That is the unit: it is liters squared over moles squared per second.0829

That is the rate constant, including units.0837

My recommendation: do the mathematics first, and then do the units separately: this side of the equation is moles per liter per second--that is from the data; it actually gives it to you, right on the data.0840

Here, this is molarity squared, times molarity, is molarity cubed, which is moles cubed per liters cubed; put that down in the bottom, and then just start canceling things out until you are left with your final unit.0851

Rate constant units are always going to be kind of odd, simply by virtue of the science.0863

OK, so that is #C; now, let's go ahead and do D: The following mechanism was proposed for the reaction.0870

OK, so let's see what we have here: we have this mechanism, which says Br2 + NO goes to NOBr2; they are calling this the slow step, and here, we have NOBr...0880

Actually, you know what, I don't need to write this down; you have it in front of you!--that is not a problem.0908's going to take up extra space here.0914

OK, so we have: Br2 + NO goes to NOBr; they are saying that is the slow step; and then, we have NOBr2 + NO goes to 2 NOBr: that is the fast step.0916

They want to know: Is this mechanism consistent with the given experimental observations?--justify your answer.0926

OK, if you add those two equations...mechanism was one of the things that I actually did not cover when I discussed kinetics; it is actually very, very simple, so I'm going to go ahead and discuss it here, by just solving the problem.0932

All the problems are handled the same way.0948

When you are presented with a mechanism, there are two things that you have to check: you add the steps of the mechanism together, just like you would for Hess's Law; and if the final equation that you come up with actually equals the equation--the reaction--that is half of it.0951

However, what you have to do is: you take the slow step, and you actually use that equation to come up with a rate law.0968

If that rate law that you get from the slower step matches the rate law that you developed experimentally, it's a good mechanism--it's a viable mechanism.0976

If not, then the answer is no; so let's go ahead and do that now.0984

If we actually add these two equations (you know what, I'm going to have to write them anyway)...0989

Br2 + NO goes to NOBr2; and we have: NOBr2 + NO goes to 2 NOBr; they are saying that that is fast.0994

When I add these two together, NOBr2 and NOBr2 cancel; I get: Br2 + 2 NO goes to 2 NOBr--that part checks.1014

Now, I have to do the rate law: rate laws for mechanism steps--these things called elementary steps--are the steps of a mechanism; they are just called elementary steps.1025

These have coefficients that are actually based on molecularity; these--the coefficients, the n and the m on top of the differential rate law--they are these coefficients right here.1040

The rate...we take the slow one; the reason we take the slow elementary step: well, a reaction is only going to be as fast as its slowest step.1053

Just like, if you are on a freeway, and you have a 5-lane freeway, but only one lane is open--well, the freeway is only going to move as fast as cars that are going through that one lane; it is the slowest part of a mechanism--the slowest reaction--that controls the overall rate.1066

But you know this already: you know that you might have 10 fast steps, but if you have 1 slow step, the rate of the overall reaction (which might have 10 steps in it) is based on the slow step, because the slow step controls the flow of the reaction.1082

Since this is the slow step, we take it; the rate is equal to some rate constant, times...just take the coefficients of each reactant: NO to the first power; the other reactant is Br2 to the first power; that is it.1097

Is this the same as--is this equivalent to--what we got experimentally, KNO squared Br2 to the first power?--no!1114

2 doesn't match 1; so the answer is no, and the reason it isn't: because the derived rate law (in other words, this rate law that we got from the slowest step of the mechanism) does not match the experimental rate law (and it has to).1127

Again, your experimental rate law is a measure of the rate of the reaction; the rate of the reaction is only going to be as fast as its slowest step.1163

The slowest step...if we happen to propose a mechanism, and we say, "Oh, maybe it goes like this," we can just write the rate law straight from the equation itself, just like we do for equilibrium.1171

We can actually use the coefficients here; we don't have to determine the coefficients experimentally, the way we normally would.1182

We have to, for the reaction, if the reaction is made of multiple steps; they are saying that this reaction is made of 2 steps; so the answer would be no.1190

I hope that makes sense.1199

OK, let's move on to the next question: the next question is probably the most challenging part of the AP exam for most of the kids that take it, and the reason is: because it is the reaction section.1201

You are given eight reactions, and you have to choose five of them, and you have to literally write the reactants and the products; that is it.1215

What is nice about it is: you don't have to balance it; you don't have to write spectator ions; you don't have to write the states (like gas, liquid, aqueous, solid, things like that); you just have to write reactants and products.1222

There is actually a lot of leeway here: you can list other products, maybe, if it helps you to list them that way--if that is how you learned them; you are not going to be penalized for things that are extra.1233

You will only be penalized for things that are not there.1245

So, you actually have a lot of freedom here: you don't have to worry about balancing; you don't have to worry about spectator ions; molecular, total ionic, net just need to know what reactants and what products are going to end up.1248

Let's just jump right on in and see if we can make sense of this.1259

OK, so let's see: Write the formulas to show the reactants and products for any 5 of the laboratory situations described below (reactions).1265

Let me see: Assume that the solutions are aqueous, unless otherwise stated; represent substances in solution as ions, if the substances are extensively ionized.1277

Omit formulas for any ions or molecules that are unchanged by the reaction; you don't need to balance the reaction; OK.1291

The first one is--it says: Calcium oxide powder is added to distilled water.1298

So, calcium oxide powder (again, I am not going to write "solid"; I'm just going to write calcium oxide) plus water: when you have the oxide of an alkali or an alkaline-earth metal, and you add it to water, you are going to end up forming a basic solution.1304

You are going to end up forming a hydroxide; so, this one ends up being calcium hydroxide; it's that simple.1319

That is it; if you need to know what is actually happening here, what is happening here: calcium oxide is an ionic compound.1329

H2O (I'm going to write it as HOH)--OK, let's see what happens; you know what, I'm not going to actually discuss this one.1337

It's just: whenever you have an oxide of an alkaline-earth metal or an alkali metal, and it reacts with water, you are going to end up forming a hydroxide; you are going to form a basic solution.1359

If you want, you can write this as calcium + and OH-, but just go ahead and write it as calcium hydroxide.1371

OK, let's see: B says: Solid ammonium nitrate is heated to temperatures above 300 degrees Celsius.1378

So, ammonium nitrate is NH4NO3; let's see what we have.1390

We are going to end up heating it up (and again, we don't have to put any of these on here); we are going to heat it up to 300 degrees Celsius; chances are, when you are going to heat up some sort of a compound like this, this is one of those situations where it is just a reaction that you should just know.1404

Now, you don't have to know it; if you miss this, again, you have 5 reactions to choose from; you don't have to do all 8.1421

And if you miss a couple of them, it is not going to make a huge difference; yes, you are going to lose points, but it is not like it is going to have an effect on the overall grade that you get.1427

You can still miss a fair number of these and still walk away with a 5 on the AP exam, simply because it's one small part of the AP exam, not the whole AP exam.1437

Be very, very judicious in how you react to this and what you want to lose time over and what you don't want to lose time over.1448

Something like this--high temperatures,'s going to decompose; basically, what you are going to end up getting is nitrogen gas; you are going to end up with, probably, oxygen gas; and you are going to end up with water; that is it.1457

OK, C: Liquid bromine is shaken with a .5 Molar sodium iodide solution.1475

Liquid bromine--that is bromine (2)--is shaken with an iodide solution--iodide is -1; OK.1481

Here is where you are going to have to recognize something: what you have is 2 halogens, and what you have is a halogen that has a -1 charge, and you have a halogen (the bromine) on this--the oxidation state is 0.1492

This is going to be an oxidation-reduction reaction; this is one of those reactions where you are actually going to check the table of reduction potentials.1506

You are going to see which one has the highest reduction potential; that is the one that is going to be reduced--the other one is going to be oxidized.1514

So, when you look at the reduction potentials, you see the following: Br2 + 2 electrons goes to bromide: the reduction potential on that is 1.07 volts.1521

When you look at I2 + 2 electrons going to 2 iodide, the reduction potential for that is 0.53 volts.1542

Between these two, bromine has a higher reduction potential; in other words, in a battle for electrons, bromine will beat iodine.1555

What that means is that you have to reverse this reaction: when you reverse the reaction, that means whenever bromine comes across an iodide, bromine is going to steal the electrons from the iodide to become bromide, and the iodide is going to turn into iodine.1562

Br2 plus I- forms Br- plus I2 (again, we don't have to balance this; we just have to put the products).1583

You need to recognize that this is an oxidation-reduction reaction.1592

How would you do that?--well, part of it is just experience, doing a fair number of these reactions; another part of it is that you have two halogens--one of them is in oxidation state 0; this one--oxidation state -1; that should give you a hint that you need to check the reduction potentials on the chart to see which one will actually win.1596

In this case, bromine will steal the electrons that iodine has here and turn into bromide, and the iodine will end up binding to form I2.1615

That is it; in other words, the bromine will win the battle between the electrons; bromine is reduced; iodine is oxidized.1626

Now, this is a -1 charge on bromine; this is a 0 on the iodine.1637

OK, let's see, D: Solid lead carbonate (lead (2) carbonate, PbCO3) is added to hydrochloric acid; OK.1641

Hydrochloric acid and lead carbonate--something very interesting is going to happen here.1659

The acid is going to react, and it is going to actually bind with the carbonate.1664

This is going to be a 2-part reaction--I'm going to write it in two steps: you are going to end up with...they are going to switch partners here, and this is going to be lead sulfate, plus H2CO3 (hydrogen carbonate).1670

But we know that hydrogen carbonate is unstable, and it breaks down into H2O + CO2.1686

So, the final reaction we are going to write is: PbCO3 + H2SO4 goes to PbSO4 (lead sulfate) + CO2 + H2O.1693

That is it; any time you have some sort of an acid in contact with a carbonate or a bicarbonate, you are going to end up forming CO2, and you are going to end up forming H2O.1708

This is the reaction that you want to write.1719

OK, let's see--what else do we have?1724

C, D, E, all right: E says: A mixture of powdered iron (3) oxide (Fe2O3) and powdered aluminum is heated strongly.1733

Iron oxide and aluminum is heated strongly: this is going to end up giving you...this is the thermite reaction; so, you are going to end up with aluminum oxide...1748

Basically, they are just going to switch partners; aluminum oxide plus iron metal.1764

Again, it's just one of those reactions that you should sort of recognize.1771

OK, F: Methylamine gas is bubbled into distilled water.1776

Methylamine is CH3NH2 (and again, you don't have to do all of these; you have 5 to choose from; and if you miss a couple of those, you will still be absolutely fine): it's bubbled into distilled water.1783

We have a nitrogen-containing compound; this is going to be a hydrolysis reaction.1797

It is going to end up forming CH3NH3+, the methylamine; a lone pair on here; it is going to steal one of the OH-; that is it.1802

We have done enough of the acid-base chemistry, so we know this by now.1815

Carbon dioxide gas is passed over hot, solid sodium oxide: CO2 + sodium oxide (Na2O).1818

Again, another reaction: when you have an oxide plus...when you have an alkali metal oxide and you react it with CO2, you are going to end up with a carbonate.1832

You are going to end up with sodium carbonate; that is it.1844

H: .2 barium nitrate solution is added to an alkaline .2 Molar potassium chromate solution...barium nitrate.1850

Ba(NO3)2 + K2CrO4; alkaline just means it is in basic solution, so you have a bunch of hydroxide ion floating around.1859

Probably not a lot is going to happen here, as far as the hydroxide ion is concerned, but you are going to switch some partners, because these are both going to be soluble, because the potassium and because of the nitrate.1876

However, potassium nitrate is soluble; barium chromate is not; so what you are going to end up with is BaCrO4.1889

And again, you don't have to put the potassium nitrate, because it is soluble, and they are just going to be some spectator ions.1900

You would write: Ba(NO3)2 + K2CrO4 goes to BaCrO4.1906

OK, all right, let's do question #5: All right, question #5: I'm going to draw a little picture, and then we'll go ahead and start with the problem.1914

Actually, I'm going to read the problem first: A student performs an experiment to determine the molar mass of an unknown gas.1934

A small amount of the pure gas is released from a pressurized container and collected in a graduated tube over water at room temperature, as shown in the diagram above (and I'm going to go ahead and draw the diagram for you in just a second).1943

The collection tube containing the gas is allowed to stand for several minutes, and its depth is adjusted until the water levels inside and outside the tube are the same.1956

Assume that the gas is not appreciably soluble in water, that the gases collected in the graduated tube in the water are in thermal equilibrium, and a barometer, thermometer, and analytical balance (very important part) and a table of equilibrium vapor pressure of water at various temperatures are also available.1967

OK, so here is what is going on: we have this water, and we have this tube, and we have this container full of gas.1989

We are collecting, or pumping, some of this gas we are bubbling this up, and it is actually collecting on top, so it's pushing down this water level.1998

This is all water; so the gas is collecting on the top here, and it is being collected over water.2010

We allow it to settle for a little bit, and then we adjust the tube; we actually push it up or down until this water level is the same as this water level.2019

What we have at the end is: we have a water level; we have the tube; we have some gas in there; but this--the tube--has now been pushed down, so that the water levels are the same.2028

That is important.2042

We still have the container, and we still have the tube going into it, but now the water levels are the same; so now, we can start the problem.2045

OK, now: Part A says: Write the equations needed to calculate the molar mass of the gas.2054

OK, first of all, what is molar mass?2062

Molar mass is equal to the mass of a substance, divided by the number of moles (grams per mole).2068

Well, watch what happens: if I have PV=nRT, n is the number of moles; I could rearrange this equation to solve for the number of moles.2080

Moles equals the mass, divided by the molar mass; I can take this and put it in here, and I get PV=(mass/molar mass), times R, times T.2092

Now, I solve for the molar mass: I bring this up here, and I drop this down; I get: molar mass (this is a very typical procedure) equals mass, times R, times T, divided by P...2115

You know what, I'm not going to draw all of these time symbols; I'm just going to write them all out; this is ridiculous.2130

Mass, times, RT, divided by PV: so the only two equations that you need: you need molar mass equals mass over mole--or, if you want to write it this way, this equation is fine, too--either one is fine; and this equation--you need the ideal gas law, and you need the definition of molar mass.2138

Those two equations will allow you to rearrange the ideal gas law in order to solve for the molar mass of the gas--pretty cool.2159

OK, part B: part B says: List the measurements that must be made in order to calculate the molar mass of the gas.2168

So now, again, this is experimental procedure, so we are measuring something.2176

Well, take a look at the equation: the molar mass--I have a mass; I know what R is already--that is .08206 liter-atmosphere/mole-Kelvin; I need to measure the temperature; I need to measure the pressure--the atmospheric pressure; and I need to measure the volume of the gas; and I need to measure the mass.2179

There is your answer: I need to measure temperature (let me actually flip over here); I need to measure atmospheric pressure; I need to measure the volume; and I need to measure the mass of the gas.2205

Now, you are probably saying to yourself, "How do I measure the mass of the gas in terms of grams?"2235

Well, you remember, if you look back at that picture, we had a container, this pressurized container, like it said.2240

Well, if I weigh the container before I start the experiment, and if I pump a certain amount of gas out of the container, and then I weigh the container again, the difference in the mass of the container before and after I have actually removed gas from it--that is the mass of the gas--very, very simple.2245

OK, part C says: Explain the purpose of equalizing the water levels inside and outside the gas collection tube.2265

Well, what happens when I equalize the level inside and outside a tube?2276

When they are equal, that means the pressure outside is equal to the pressure inside; that is why--that is the answer.2282

The internal pressure (in the tube) and external pressure are the same; that way, when you measure the atmospheric pressure (which is the pressure outside pushing down, remember--this is our thing, and you have this tube, and the water level is here; the atmospheric pressure is here), well, that is the same as the pressure pushing down on that; that is why they are equal.2290

When we equalize them, we make the pressure on the outside equal to the pressure on the inside.2323

I need the pressure on the inside; by equalizing it, I can just measure the atmospheric pressure; that is the pressure on the inside.2327

That is the whole idea: you need the internal and external pressures to be the same--pretty nice.2334

Part D: The student determines the molar mass of the gas to be 64 grams per mole; write the expression (write the expression--you don't have to solve it) for calculating the percent error in the experimental value, assuming that the unknown gas is butane, which has a molar mass of 58 grams per mole.2342

Well, OK: so the expression for calculating the percent error: the percent error is the experimental value, minus the theoretical value, over the theoretical value.2360

She got 64 grams per mole; the actual mass is 58; divided by 58, times 100: that is the expression.2379

That is all you need to write: experimental error--what you get...the difference between what you get and what you should get--an absolute value.2390

Again, if you are short, your number is going to be negative; so that is why we take the absolute value.2399

In this case, her number was too high: 64--she got 64; well, the difference between 64 and what it should have been, 58--that difference, divided by what it should have been, gives you the experimental error.2406

How far do you deviate from what you should have gotten?2419

There you go.2423

And now, let's do part E: If the student fails to use information from the table of equilibrium vapor pressures of water in the calculation, the calculated value for the molar mass of the unknown gas will be smaller than the actual value; explain that.2427

OK, well, we said that the molar mass is equal to mass, times, mass, times RT, divided by PV.2443

OK, when you measure the external pressure, that is going to be the internal pressure.2461

However, this is the pressure of the gas alone; but remember, we are collecting this gas over water--water has a certain vapor pressure at a given temperature, which means that the pressure inside is actually made up of 2 pressures.2466

Part of that pressure comes from the gas itself; part of that pressure comes from the water vapor molecules.2481

I need to take the total pressure (which is equal to the atmospheric pressure, which is why the water levels are equalized); from that, I need to subtract the value of the vapor pressure of water, in order to get the pressure of the gas.2487

In this case, the pressure in the denominator will be too high if she doesn't, she is going to measure a certain pressure (which is the same as the atmospheric pressure).2502

However, that pressure is going to be too high; from this pressure that she measures, she has to subtract the vapor pressure of water.2514

Therefore, the pressure that she should get here should be lower than this; so, that is why she is going to end up with a pressure that is too high: she has forgotten to subtract the vapor pressure of water.2526

If this is too high, a big number in the denominator makes the molar mass smaller.2537

That is what that says: the student fails...calculate molar mass...that will be smaller than the actual value; and that is what happens--it is all based on the equation.2543

The equilibrium pressure--you have to subtract the vapor pressure of the water inside the tube.2554

We want just the pressure of the gas.2558

If this pressure is too high, then this number is going to be lower than it should be.2561

Pressure is higher than it should be; this is going to be lower than it should be, because there is an inverse relationship here; this is a denominator.2566

I hope that makes sense mathematically.2573

OK, well, thank you for joining us here at; we will see you next time for the final instalment of the free response questions.2576

Take good care; goodbye.2583