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Lecture Comments (50)

1 answer

Last reply by: Professor Hovasapian
Thu May 7, 2015 4:12 AM

Post by R Abdullah on May 7, 2015

Hello Professor Hovasapian

Before I ask my question, I'd like to thank you for providing us with these amazing lessons. They have helped me so much in my academic career and I believe I've even developed a passion for mathematics from watching your lessons on Linear Algebra and Multivariable Calculus.

I took the AP Chemistry exam on May 5th and on the free-response section, there was a simple question asking for the percent yield of a reaction. I solved it with the actual yield over theoretical yield being in grams. However, most of my peers solved the percent yield through mols of actual yield over mols of percent yield. I'm not sure which is correct and which one AP considers correct.

Once again, thank you for all that you have done.

Rasheed Abdullah

1 answer

Last reply by: Professor Hovasapian
Thu Jan 8, 2015 1:48 AM

Post by Lyngage Tan on January 5, 2015

hi professor in example 2 i believe the molar mass of silicon is 28.09 g/mol. and going through the solution i ended up with this ratios Na= 1.06568731/0.530437878 = 2 Si= 0.530437878/0.530437878= 1 and F 3.189473684/0.530437878 = 6. so the empirical formula is Na2SiF6.  

1 answer

Last reply by: Professor Hovasapian
Fri Nov 7, 2014 9:15 PM

Post by Noorhan A. on November 6, 2014

Hello Prof. Hovasapian
Thank you so much for your detailed videos which explain some difficult concepts very thoroughly. Your videos helped me go from a C to an A+ on my Chemistry tests! Currently, I am taking Honors Chemistry, which is a weighted course and is very rigorous but does not cover some concepts from AP Chemistry. My school does not offer AP Chemistry but I would still like to take the exam in May. Do you have any suggestions for how I can self-study? Also, do sig figs matter in the AP Chem test? My teacher takes points off our tests if we do not use the correct number of sig figs.

0 answers

Post by Milan Ray on September 29, 2014

haha, pen really was acting up!

1 answer

Last reply by: Professor Hovasapian
Sun Sep 28, 2014 11:23 PM

Post by Quazi Niloy on September 28, 2014

what happens wlhen the empirical formula does not equal to 100%? Is this just a matter of error or is there a way to go about doing a problem like it?

1 answer

Last reply by: Professor Hovasapian
Tue Aug 5, 2014 6:04 PM

Post by Jason Kim on August 5, 2014

So I am assuming that at 14:38 you rounded 2.3*3 which is 6.9 to just 7 professor?

1 answer

Last reply by: Professor Hovasapian
Tue Jun 10, 2014 8:55 PM

Post by Alice Rochette on June 10, 2014

for example 2 you said 1 mol of Si = 32.07g, how did you find that? because in my periodic table it says that the atomic weight for silicon is 28.08g... anyway thank you!!

1 answer

Last reply by: Professor Hovasapian
Fri Mar 14, 2014 7:50 PM

Post by sadia sarwar on February 26, 2014

so how did you get the last answer for example 4?? it was great tutorial though, thanks!!

1 answer

Last reply by: Professor Hovasapian
Thu Dec 26, 2013 3:14 PM

Post by Nada Al Bedwawi on December 25, 2013

I didn't quite understand how in example 3 while converting from mol of CO^2 to mole of C.I thought that could only be done if by molar ratio(as in a chemical reaction)

1 answer

Last reply by: Professor Hovasapian
Tue Aug 27, 2013 9:15 PM

Post by Marian Iskandar on August 27, 2013

Excellent tutorial! Regardless of the tiny errors made, the solutions were laid out so clearly...left no room for ambiguity. Thank you again, Professor! :)

1 answer

Last reply by: Professor Hovasapian
Thu Aug 22, 2013 5:04 PM

Post by Venugopal Ghanta on August 22, 2013

you look and sound a lot like albert einstein

1 answer

Last reply by: Professor Hovasapian
Tue Jul 2, 2013 6:00 PM

Post by Sarawut Chaiyadech on July 2, 2013

You are Very good teacher thank you so much Professor

1 answer

Last reply by: Professor Hovasapian
Fri Jun 21, 2013 6:15 PM

Post by Angela Patrick on June 21, 2013

Is the answer to example four in significant figures?
I got the same answer but thought that there were only two significant figures and got the answer 10.
If the answer 10.17 is in significant figures can you explain why there are four significant figures?

1 answer

Last reply by: Professor Hovasapian
Sat Dec 1, 2012 5:47 PM

Post by Andrew Stewart on December 1, 2012

I do not understand how to obtain the final answer of 2.67 x 10^23.?

1 answer

Last reply by: Professor Hovasapian
Sat Oct 27, 2012 4:25 PM

Post by Max Mayo on October 26, 2012

Thanks for the help. Has anyone ever told you that you look a lot like Albert Einstein?

2 answers

Last reply by: sadia sarwar
Wed Feb 19, 2014 2:50 AM

Post by Revanth Guptha on September 8, 2012

so what is the correct answer for Example 2?

thanks for the great tutorial!! :D :)

1 answer

Last reply by: Professor Hovasapian
Mon Sep 3, 2012 6:26 PM

Post by Anuradha Kumar on September 3, 2012

Didn't he use the atomic mass of sulfur instead of silicon? Great tutorial nonetheless! :)

3 answers

Last reply by: Professor Hovasapian
Sun Aug 19, 2012 8:59 PM

Post by Arshin jain on August 17, 2012

It's not necessary to take general chemistry before AP chemistry, as long as you understand everything, right?

1 answer

Last reply by: Professor Hovasapian
Sun Aug 12, 2012 3:47 PM

Post by Suresh Sundarraj on August 12, 2012

that was a great lecture, thanks a lot!

4 answers

Last reply by: Professor Hovasapian
Sun Sep 1, 2013 10:47 PM

Post by noha nasser on August 5, 2012

and i think that there is something wrong in the final answer of example 4 too... tried to calculate it many times and i still couldn't get it :)

0 answers

Post by mateusz marciniak on May 5, 2012

i agree that was the atomic mass of sulfur not silicon, great tutorial nonetheless

1 answer

Last reply by: Professor Hovasapian
Mon Jul 16, 2012 12:42 AM

Post by chenyu liu on April 25, 2012

i think you made a mistake in14:04 by using sulfur's atomic number for silicon.

0 answers

Post by Trace Shapiro on March 18, 2012

Visually defining (mapping) the solution path through the stoichiometric problems is absolutely genius! I can't believe I have never been taught this before! Thank you so much!!!

Related Articles:


  • All Stoichiometric calculations pass through mols, the basic unit of Chemistry
  • There is always a path connecting data you are given and data requested. Express this path pictorially, then convert to mathematics.


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
  • Stoichiometry 0:25
    • Introduction to Stoichiometry
    • Example 1
    • Example 2
    • Example 3
    • Example 4
    • Example 5: Questions
    • Example 5: Part A - Limiting Reactant
    • Example 5: Part B
    • Example 5: Part C

Transcription: Stoichiometry

Hello, and welcome back to Educator.com0000

Today, we are going to continue our quick review of the basics of chemistry before we jump into some solid AP work.0002

We're going to go over stoichiometry today.0010

I'm going to talk about several varieties of stoichiometry problems.0013

Stoichiometry, essentially, is about amounts; that is what it is.0018

In chemistry, the basic unit of amount is the mole.0026

This is something that I know you all have heard before, so really quickly: a mole is 6.02x1023 particles.0031

"Particles," because in chemistry, we can talk about atoms; we can talk about molecules; we can talk about ions, electrons, protons...whatever.0042

I tend to use just the generic term "particles"; and the problem itself will specify what particle we're talking about.0049

In one mole (so this is a conversion factor) of something, we have 6.02x1023 particles.0057

Now, the periodic table is arranged based on the mole; a molar mass, that you see on the periodic table--that is one mole of oxygen atoms, one mole of magnesium atoms, one mole of iron atoms--those weigh those particular numbers.0065

So, the mole is the standard unit that we use in chemistry--again, because we can't count individual atoms (they're too tiny), so we have to choose a unit, kind of like a dozen or a century--things like that.0079

Stoichiometry basically concerns conversion factors: setting up a bunch of conversion factors based on what you know versus what you want.0093

There is a series of steps to that.0103

So, I'm going to introduce this notion of how to deal with stoichiometry problems.0106

Because the mole is always what we're talking about, the mole is the central hub; you start from the mole and you go to the mole, or you pass through the mole getting to where you want.0111

One way or the other, you are going to have to use a mole.0120

The reason is: the equations in chemistry (like, for example, 2H2+O2 → 2H2O, the reaction of hydrogen and oxygen gas to produce water)--well, these numbers here--the coefficients--the 2, 1, and the 2--they represent moles.0122

What this says is, "Two moles of hydrogen gas need to react with one mole of oxygen gas in order to produce two moles of H2O."0143

So, all equations--balanced equations--represent the standard that we use to work from. 0150

It's all in moles.0158

Now, in any stoichiometry problem, any species that you're dealing with always has (this is the best way to think about it) moles (you know what, I'm going to actually draw this up here) of X.0160

Every species that is mentioned in a stoichiometry problem--something like that; from the mole, we can go to grams of X, or we can go to the number of particles of X.0179

Again, X might be a molecule; it might be an atom; it might be an ion; it depends.0198

Later on, we might need to go to liters of X.0202

But, they all have to pass through the mole.0209

Again, every species under discussion has one of these little things associated with it.0213

If a stoichiometry problem has two species that are mentioned, and there is some relationship among the two species (which we will get to in a minute), that species is also going to have something like this.0218

Let's call it mol y, and again, you can go to grams of y; number of particles of y; or maybe liters of y.0231

Now, here is what is interesting: the reason that I say that each species--you should write something like this: when you do something like this, this will give you a pathway to actually find out how to write the problem, stoichiometrically; how to solve it, mathematically.0250

This gives you a solution path.0264

Any time you are moving from one species to another--like, for example, in a minute we're going to sort of choose a couple of reactants and a couple of products and stoichiometric problems--movement this way or that way is based on the mole ratio, and these mole ratios are precisely the coefficients in the balanced reaction.0267

Now, this will make more sense in a minute, when we actually do a problem, and I'll show you the way to actually do this.0286

Hopefully, you are already good at this; but if not, this will be a good review and a little bit of practice.0293

Having said that, let's just jump into a problem, and I think it will make sense.0299

OK, so example number 1: How many Cl atoms are there in 18.50 grams of phosphorus pentachloride?0304

OK, let's see: so now, let's identify the species: one of the species is chlorine; one of the species is PCl5; and so, I'm going to write that.0333

I'm going to write mol PCl5, and I'm going to write mol Cl; those are the two species.0344

Now, what is it that I'm interested in?--I'm actually interested in the number of atoms, so I need the number of Cl atoms.0354

OK, what am I given?--I'm given grams of PCl5.0365

I'm given grams of PCl5; I'm supposed to find the number of chlorine atoms.0375

Here is how you do it: you go...this is the path that you follow, and each arrow represents a conversion factor.0380

I need to go from grams of PCl5 to moles of PCl5.0390

From moles of PCl5, I need to go to moles of Cl; this is going to be the mole ratio.0394

From moles of Cl, I can go to the number of chlorine atoms.0400

Here is the setup (actually, let me draw this a little bit higher, so that I can make some room): grams PCl5...0405

This is why we say each species has this little setup here: when you set it up like this, you can actually go ahead and draw the pathway; so here, I'm going from grams of y, to moles of y, to moles of x, to number of particles of x.0418

That's it; I'm following a pathway, and I can go anywhere I want, but I'm always passing through moles; there is a mole ratio.0434

OK, you always write what you are given: 18.50 grams of PCl5.0441

This is one conversion factor, two conversion factors, three conversion factors: I know I'm going to have three conversion factors--I don't even need to fill in the numbers yet.0451

I can just automatically do this.0461

Now, the thing that you are coming from is going to be the unit on the bottom; the thing that you are going to is going to be the unit on the top.0463

So, I'm going to have grams of PCl5, moles of PCl5; well, how many grams in a mole of PCl5?0471

Well, the molar mass is 208.22 in one mole of PCl5.0481

Now, my second conversion: what is the relationship between moles of PCl5 and moles of Cl?0487

One mole of PCl5 releases 5 moles of Cl; that is just based on the molecular formula--it is PCl5.0493

All right, PCl5: one of these produces five of these.0501

So, again, the thing that you are going to is on top--mole of Cl; the thing you are coming from is down at the bottom--mole of PCl5.0507

Well, the relationship is: one mole of this produces five moles of that; that is the mole ratio; that is this step.0524

Now, I'm looking for number of Cl atoms in one mole of Cl.0532

Well, one mole of Cl--one mole of anything contains 6.02x1023 (let me actually write "atoms").0544

So again, here, the particle we're talking about is atoms.0558

When I do this multiplication, I end up with (OK, I can write it down here) 2.67x1023 atoms of Cl; that is my final answer.0561

Again, it's just based on this relationship; every species has a relationship; from mole, you can go to grams, to particles, to liters.0578

Then, among the species, there is a mole ratio--a relationship.0588

In this one, it was 1 mole of this produces 5 moles of that.0592

That is how you proceed through this network.0595

You have something that you are given; you have another thing that you want; you find a path through there.0598

If you're given this and you want that, you find a path through there, and it always passes through moles, because the mole is the basic unit of chemistry.0603

OK, we'll get to some more stoichiometry in just a minute.0614

The next part of the review: I'm going to cover an empirical formula problem.0618

Oftentimes, you are given a certain amount of data, and you need to find the empirical formula of a compound.0623

The empirical formula is the formula that gives the lowest number ratio of those things.0630

So, for example, if I had something like C6H12O6, which is glucose; and let's say I had another sugar, C12H24O12; well, these are two different sugars.0636

But notice, the base formula--I can actually divide it by 6 here; I end up with CH2O, and I end up with CH2O, right?0656

That is the lowest number ratio of atoms to each other.0666

In other words, all sugars have a ratio of one carbon to two hydrogens to one oxygen.0670

That is the empirical formula; it's a general umbrella formula that covers all of the molecules of a given compound.0676

Each individual compound has its own molecular formula.0684

These are molecular formulas; they are specific to the compound.0687

But, an empirical formula covers both; that's it.0690

So, let's go ahead and do an empirical formula problem.0695

Let's see...let's write this as example 2: all right, I have a compound that is 24.5% sodium, and 14.9% silicon, and 60.6% fluorine.0698

I want to know what the empirical formula is.0722

OK, here is how you handle it: because you are given percentages...percentage is based on 100; so, since they give us the percentages, let's just presume that we have 100 grams of something.0724

So, we can convert these percentages: 24.5% of 100 grams is 24.5 grams.0735

What I'm going to do is: I'm going to convert these to moles.0741

So, sodium: I take 24.5 grams, and I convert to moles: well, one mole of that is about 23 grams, so I get 1.07 moles of sodium.0744

Now, I do the same for silicon: silicon--I have 14.9%, which is 14.9 grams if we take a 100-gram sample; we're just doing this so we can make our math easy.0760

Well, one mole of silicon is 32.07 grams, so I get 0.46 moles of silicon.0771

Then, when we do fluorine, we have 60.6 grams of fluorine; one mole of fluorine is 19 grams, and I end up with 3.2 moles of that.0782

Well, now that I have it in moles, I divide by the lowest number of moles; so I divide all of them by 0.46, 0.46, 0.46, and when I do that, I end up with 2.3 here; I end up with 1 here, of course, and here I end up with 7.0796

I need whole-number ratios; so 2.3--in order to make this a whole number, I have to multiply by 3, which means I have to multiply everything else by 3.0822

This becomes 21; this becomes 3; and this becomes 7.0832

So, this is my empirical formula: I have: Na7Si3F21.0837

That is my empirical formula.0851

It could be Na14Si6F42, or any other multiple, but the basic relationship, the ratio, of sodium to silicon to fluorine is 7:3:21.0853

This procedure always allows you to find the empirical formula.0866

You change the percentage to gram, gram to mole for all of them; you divide by the lowest number of moles to standardize, so that you have at least one of them with a 1; and then you multiply by an appropriate constant to make sure that all of them give you whole numbers, because we can't leave it as 2.3; you can't have a fraction of an atom.0872

That is a standard empirical formula problem.0891

OK, so now we're going to take the next step, and I'm going to show you a molecular formula problem.0894

Once we find the empirical formula, if we happen to have the molar mass of a particular compound, we can find the actual molecular formula.0901

Let's do that one.0908

Let's go ahead and make this example 3: Menthol is composed of carbon, hydrogen, and oxygen.0913

1.005 grams is combusted to produce 2.829 grams of CO2 and 1.159 grams of H2O.0932

If menthol is 156 grams per mole, what is its molecular formula?0969

Here, we're going to end up having to find the mass of the individual components--the carbon, the hydrogen, the oxygen.0994

We're going to use the empirical formula, and then, from there, we're going to use the fact that they give us the molar mass.1003

We're going to divide that molar mass, the 156, by the mass of the empirical formula to see how many empirical units actually go into the whole formula.1010

That is how we get the molecular formula.1018

Let's just jump right on in.1021

Notice, this didn't give you the percentages straight on; this is a slightly different problem--slightly more complicated.1023

Let's see: 2.89 grams of CO2; let's go ahead and...let me write this equation down: menthol, when you combust it, it's with O2, and it's going to produce CO2 + H2O.1029

All right, now: 2 many grams of carbon...OK, so now we have 2.829 grams of CO2.1054

One mole of CO2 weighs 44 grams; now we're going to we know that all of the carbon ends up in CO2; all of the hydrogen in menthol ends up in H2O; it's the oxygen that is split between the two.1077

We need to find the number of grams of carbon, the number of grams of hydrogen; they give us the total number of grams of menthol that were combusted, so the balance of that is going to be the oxygen.1092

Then I can start my problem.1104

2.829 grams of CO2) times...again, this is a mole ratio: 1 mole of CO2 produces one mole of carbon, right?1107

One mole of CO2 produces one mole of carbon, times 12 grams of carbon per mole, and I end up with 0.7715 grams of carbon.1119

Now, I have 1.159 grams of water times...1 mole of water is 18 grams: 1 mole of H2O produces 2 moles of hydrogen, right?1135

Erase that: 2 moles of hydrogen--1 mole of this produces 2 moles of hydrogen.1159

Hydrogen is 1 gram per mole, so I get 0.1288 grams of (oops, this isn't carbon; this is) hydrogen.1165

Now, if I want to find the mass of oxygen, I take the 1.005 grams of menthol, and I subtract the amount of carbon, 0.7715, plus 0.1288, and I end up with a mass of 0.1047 grams of oxygen.1182

Now I have my masses of each, based on a combustion analysis.1211

I hope you understand how it is that I came up with that.1216

Now, I can start my empirical formula problem.1218

I have 0.7715 grams of carbon, times 1 mole over 12 grams; that gives me 0.06429--often, with these empirical formula problems, you want to carry out the decimals as far as you can, to get a good number.1222

Then, if I do hydrogen, I have 0.1288 grams, times 1 mole, which is 1 gram, and I get 0.1288--this is in moles.1242

Oxygen: I have 0.1047 grams; 1 mole of that is 16 grams of oxygen (again, we'll use the oxygen atom), and this is going to equal 0.0065438 moles.1258

Now that we have moles, we need to divide by the smallest, and it looks like the smallest here is this one.1284

When we divide by that, when we divide by 0.0065438, and we divide by 0.0065438, we end up with 10 here; we end up with 20 here; and we end up with 1 here.1289

So, the empirical formula for menthol is C10H20O1.1312

Now, (OK, so here is where we have to...) what we do is: we take the molar mass of the compound, the 156 (let me actually write this out), and we divide it by the molar mass of the empirical formula.1324

When we do that, again, the empirical formula is the lowest number ratio, so in this particular case, we had C10H20O1; that means the carbon to hydrogen to oxygen ration is 10:20:1.1363

Well, it might be C20H40O2; we don't know.1380

So, we divide the mass of the empirical formula into the molar mass of the compound, which we do know.1385

This ends up being...they said it was 156 grams per mole, and the molar mass of the empirical happens to also be 156 grams per mole; therefore, the ratio is 1.1391

Therefore, menthol is C10H20O1.1405

If this number were 3, it would be C30H60O3; that's it.1411

This is the actual...not only empirical formula, but it happens to be the molecular formula for menthol.1416

This process will always work for molecular formula.1422

You run through it; you find the empirical formula; and then you divide the molar mass of the compound by the molar mass of the empirical formula, and that will give you the ratio--the number by which to multiply all of these little subscripts in order to get your final molecular formula.1425

Let's do a stoichiometric calculation based on an equation.1443

The fermentation of glucose produces ethyl alcohol and carbon dioxide according to the following equation: C6H12O6, and fermentation (so no oxygen involved), goes to 2C2H5OH + 2 CO2.1450

The question is, "How many grams of glucose are required to produce 5.2 grams of ethanol, C2H5OH?1474

So, how many grams of glucose do I need in order to produce 5.2 grams of C2H5OH?1502

OK, so let's take a look: the two species that we are talking about are glucose and C2H5OH; we need to pass through moles.1508

They want...we are given grams of C2H5OH that we want; we want to find the number of grams of glucose.1519

Here is the process: based on this equation (I'll write it like this), I need grams of glucose.1526

Grams of glucose comes from moles of glucose.1538

Moles of glucose I can get from moles of ethanol.1543

Moles of ethanol comes from grams of ethanol.1551

That is the process: I am given grams of this--I am given 5.2 grams of this; I have to go to the moles--from grams to moles, mole to mole (I use mole ratio--that one), and then from mole back to gram.1559

This is my path: I go here to here to here to here; that's it--I have 1, 2, 3 arrows; that is three conversion factors.1574

Let's go ahead and write this down; we always start with what we're given.1586

5.2 grams of EtOH (EtOH is just a shorthand for ethanol); that is 1, that is 2, that is 3; grams of EtOH here; I'm going to moles.1590

Moles of EtOH--notice, I'm not going to do the numbers until afterward--I have moles of EtOH here, because, again, they have to cancel, and it's also the thing that I'm coming from.1605

I'm going to moles of glucose, and this is moles of glucose; grams of glucose...1617

5.2 grams of ethanol; ethanol happens to be 46 grams per mole; the mole ratio--how many moles of ethanol to moles of glucose?1629

Moles of ethanol to moles of glucose: 1; that is why you need a balanced equation.1638

Glucose is 180 grams per mole; therefore, my final answer is 10.17 grams of glucose.1645

10.17 grams of glucose will produce 5.2 grams of ethanol--standard stoichiometry.1655

The two species were glucose and CH3OH; there is my 1 mole of glucose; mole of ethanol; these are the relationships: I need grams of glucose coming from grams of ethanol; this is my path.1664

If you can't just do it automatically, draw it out; see the solution path, and the path will tell you what the mathematics looks like.1681

Let's finish off this lesson with a limiting reactant problem.1692

Limiting reactant problems are going to come up very, very often, so basically, it's just...a balanced equation is such that it says, "All of this will react with all of that; nothing is left over."1697

But in real life, it doesn't work like that; often, when we run a reaction, one of the reactants is going to be left over; one of them is going to run out.1709

The thing that runs out--when it runs out, the reaction stops.1717

It's a limiting reactant--it limits the outcome of the reaction.1720

It controls how much product you actually get, because when it runs out, again, the reaction stops.1725

Let's take a look at this equation here: 2 aluminum hydroxide, plus 3 moles of sulfuric acid, produce 1 mole of aluminum sulfate, plus 6 waters.1732

I'm given 0.350 moles of aluminum hydroxide, and I'm given 0.450 moles of sulfuric acid.1756

My questions are: a) which is the limiting reactant? part b) how many grams of aluminum sulfate (Al2(SO4)3) can be recovered, theoretically?1769

That is what we are calculating; we are calculating a theoretical yield, as if everything went right, theoretically.1805

And part c) is: How many moles of excess reactant are left over?1811

Let's do part a: what is the limiting reactant?1831

We need to find out which one of these runs out first.1834

You can choose either one to start with: basically, what you're going to do is what it looks like: I'm going to go ahead and choose the aluminum hydroxide.1837

0.350 moles of Al(OH)3; the mole ratio between this and this is 2:3.1847

Moles of Al(OH)3 (this is the problem with chemistry--there is just so much writing as far as these symbols are concerned--the names!) to moles of H2SO4; I need to do a mole ratio calculation.1863

It is 2 moles of this to 3 moles of that; this actually tells me that I require 0.525 moles of H2SO4.1885

In other words, if I have .350 moles of aluminum hydroxide, this calculation tells me that I actually need .525 moles of H2SO4 to completely react with it.1897

Well, this is required; the question is: I require .525--do I have .525? I do not (let me do this in red).1908

I don't have .525--I only have .50; therefore, H2SO4 is limiting.1921

Because it's limiting, it controls the rest of the reaction; this is the number that I use to find out how much of this I'm going to get; not this.1932

When this runs out, the reaction stops; there is going to be excess of this.1940

Therefore, this is what controls how much I have.1944

Let's go ahead and do part b, which is, "How many grams of aluminum sulfate can be recovered?"1947

All right, well, since this is my limiting reactant, that is the one that I take: 0.450 moles of H2SO4 times...well, what is the mole ratio here?--it's 3:1; 3 moles of H2SO4...this is not going to work...1954

3 moles of H2SO4...oh, these crazy lines; OK, let me try this one more time.1987

Let's start from the top: 1 mole of aluminum sulfate comes from 3 moles of H2SO4, and then we have 342.7 grams of aluminum sulfate per mole.2002

There we go; this cancels that; that cancels this; my final answer is going to be in grams, and it's going to be =51.32 grams of Al2(SO4)3 (it's tedious writing all of those symbols, isn't it?).2035

There you go; no, I'm not going to have...I do not want those lines to show up; therefore, I'm going to write this one more time.2061

=51.32 grams of Al2(SO4)3; there we go!2073

Based on .450 moles of H2SO4, the most aluminum sulfate I can recover is 51.32 grams.2088

The reason is: because that is what controls the reaction.2096

OK, now: this final problems says, "How many moles of excess reactant will be left over?"2101

This is our limiting, so this is our excess: we need to find out how much is used, and then subtract it from the amount that we started with, and that will give us the amount left over.2107

The amount used is, again, going to be based on the limiting reactant; so, part C: 0.450 moles of H2SO4...2117

Well, I have that 3 moles of H2SO4 requires 2 moles of aluminum hydroxide.2132

I end up with 0.30 moles of aluminum hydroxide used.2145

0.350 moles-0.30 moles used (let me actually write these out: moles of Al(OH)3 minus 0.03 mol used up) leaves me with 0.05 moles of aluminum hydroxide to end up with.2156

You see what we did: once you know the limiting reactant, that controls the rest of the reaction.2191

.450 moles of H2SO4 uses up .3 moles of aluminum hydroxide.2196

I started with .350; I subtract, and I'm left with .05 moles of aluminum hydroxide; that is my excess reactant.2201

OK, so in this lesson, we have done some basic stoichiometry coverage.2212

We have talked about empirical formula; we have talked about molecular formula; and we finished off with a limiting reactant problem.2215

That, and a previous lesson with naming, should provide sort of a good, general foundation for the rest.2222

Starting with the next lesson, we're going to jump straight into some good, solid AP chemistry.2230

Until then, thank you for joining us here at; we'll see you next time; goodbye.2234