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We're going to wrap up our work now on atomic masses and molecular formula from

concept development study 2, by talking about one more example of chemical

stoichiometry. You might recall that what we have been

trying to do is count the number of particles which are participating in a

chemical reaction. That's important to be able to do because

we know that atoms and molecules react with each other in simple energy ratios.

So we have to be able to count the particles to know how many of them are

going to react with each other. Let's consider now, rather than the way we

have done things, which is to simply take a mass.

What do we do when we have a reaction taking place that's in a solution?

In that case, we're going to have to deal with something that we call the

concentration, or chemists will call the molarity.

And in this, we're going to need to distinguish what we mean by a solution.

So, what I have here, actually, is a solution of sodium chloride dissolved in

150 milliliters of water in this beaker. And the question that I might ask is, how

many moles of sodium chloride are there in this Erlenmeyer flask?

The answer is, we don't have a means to determine that yet.

I could, for example, just weigh the water, or the solution rather, which is in

this Erlenmeyer flask and try to determine the number of moles in a way that we have

but that won't do me any good. Because overwhelmingly what I am weighing

is the water, which is the solvent in this case, and not the sodium chloride, the

solute, which is dissolved in the water. In many instances I need to know how many

moles of solute are there in a particular solution.

Let's give an example here. Here's a chemical algebra, a chemical

stoichiometry problem that we need to work out.

We're going to try to dissolve some solid copper in some nitric acid.

So here's a nitric acid solution, copper dissolves in nitric acid to form copper

nitrate, so nitrogen dioxide. And some water as the products here.

Let's imagine that we're doing the same kind of chemical stoichiometry that we've

done before, where we weigh out the mass of one of the reactants.

In this particular case, we're going to take 10 grams of copper metal.

And the question is, how much nitric acid do I need to react with that 10 grams of

metal? And how would I find that number of moles

of nitric acid? If in fact I can't simply weight it.

So how do we know how many moles of nitric acid there are in a solution of nitric

acid? That fundamentally is the question we're

going to try to address and will address by the end of this particular lecture.

To do that, we're going to consider this idea of concentration that I discussed a

moment ago, by defining the concentration as being now the number of moles of solute

in a liter of solution. The formula that we're going to focus on

here is this one. The concentration is given by this bracket

so if S is the solute, might be sodium chloride for example dissolved in the

water, then. Putting S in brackets is the concentration

of the solute in the solution. And it's pretty clear when we look at this

formula that in this particular formula here, the number of moles of solute is a

number of moles, and the volume is going to be measured in liters.

So the units here are going to be moles per liter.

We'll describe this in units of moles of solute per liter of solution.

And sometimes that unit is simply given the capital M or molar.

We might refer to a 1 molar solution or a 0.5 molar solution.

For example. So let's imagine then that what we want to

be able to do is count the number of moles of solute using the concentration of the

solution. Here's a particular problem.

Let's say that I have a 0.1 molar solution, again here's that unit capital M

that means over here moles per liter. And let's say that I've measured out then

200 milliliters of the solution. What is n?

In other words, how many moles of solute are there?

If we look at the formula on the screen there the concentration of the solute is

equal to the number of moles of the solute which is what we're trying to find out.

Divided by the volume of the solute. I can easily rearrange that equation to

say that the number of moles is equal to the concentration multiplied by the

volume. And we've been given both of those

numbers. Let's look back at the problem here.

Here's the volume, there's the concentration.

So the number of moles is going to be equal to the concentration which is 0.10

molar or moles per liter. Multiplied by the volume.

And here we have to be just slightly careful because the problem has given us

the volume in terms of milliliters, whereas the units on molarity are moles

per liter. So we need to rewrite the volume as 0.200

liters. If we multiply those two together, we get

0.020 moles because moles per liter times liter gives us just units of moles.

So we've answered the question now. We've actually been able to determine how

many moles there are in a 200 milliliter sample of a solution, which has

concentration 0.1 molar. We could do this slightly differently as

well. What if we turned it around?

What if we needed to measure out 0.5 moles of the solute, what volume would be

required? That might be useful.

That's the same as trying to say what mass of material do I need.

To have a certain number of moles for the chemical reaction to take place.

How are we going to do this? Well, again, our formula back over here

says that the concentration is equal to the number of moles divided by the volume.

What have I been given in the problem here?

What I've been given is a, a number of moles needed.

And a concentration. So let's see.

I need the concentration to be equal to 0.10 moles.

I need the number of moles to be 0.50 moles.

The unknown here is the volume. I can rewrite this equation fairly

straightforwardly, just simply say that the volume, then, is equal to the number

of moles, 0.50 moles, divided by the concentration, 0.10 molar.

And that simply then turns out to be 5.0 liters.

So that tells us that if we need a certain amount of moles, we can find that number

of moles by measuring a particular volume, or if we have a particular volume of a

known concentration solution, we know the number of moles which are in that

solution. Let's apply this now in a particular

chemical contub, context. Here's a chemical reaction.

We're going to react two solutions now, each of which has a dissolved solute in

it. The first solution's going to have sodium

chloride in it. The second solution's going to have silver

nitrate in it. When sodium chloride and silver nitrate

are both in a solution together, they react.

And the product of the reaction is a solution of sodium nitrate and solid

silver chloride. Notice what we've produced here is a solid

which is actually no longer in the solution.

The sodium nitrate remains behind in the solution.

Let's imagine now that what we're going to do is mix together two solutions.

We are going to mix together a solution of silver nitrate and a mixture of sodium

chloride, but they don't have the same concentration of what we have here, don't

have the same concentration, and we're not taking the same volumes either.

How are we going to work this particular problem?

Well, it turns out to be a limited reactant problem, of the type we have

solved before. What we need to do is to calculate how

many moles, of the sodium, of the silver nitrate do I have.

And then I should know how many moles of the sodium chloride do I have so that I

can know whether or not I have equal numbers of moles or if one of the moles is

in greater proportion than the other. Because we can go back and notice that

they react in a one to one ratio, so I need to know which one of these I have

more moles of. Well, according to the formula we had a

little while ago the number, the number of moles, remember we had the concentration

of a solute is equal to the number of moles divided by the volume.

So in each one of these cases I can say that the number of moles of silver nitrate

is equal to the concentration of the silver nitrate, which was given, times the

volume of the silver nitrate. And let's see, what are those numbers?

The concentration of the silver nitrate, back from before, recall, is actually 0.15

molar. And the volume of the silver nitrate is

given here as well, is 0.5 liters. And if we multiply those together we get

0.075 moles of silver nitrate, which are in the solution.

How about the sodium chloride? Well again, we need the concentration of

the sodium chloride, which was given, we can go back over and look at that.

It is 0.20 moles, and the volume of the sodium chloride, which was given as 0.25

liters. Plug those numbers in.

The concentration of the sodium chloride is 0.20 molar; the volume of the sodium

chloride is 0.25 liters. And if we multiply those two numbers

together, we get 0.050 moles of sodium chloride.

So, do we have enough sodium chloride to reaction with the sil, silver nitrate?

Yes, we do. Uh,er, actually, no, sorry, we don't.

We need 0.075 moles of silver nitrate to react, I'm sorry, of sodium chloride to

react with 0.075 moles of the silver nitrate.

We don't have it. Therefore, we can conclude that sodium

chloride is the limiting reactant in this particular reaction.

Because we'll run out of it first. So if I now ask the question let's say how

much silver chloride am I going to produce?

The answer than turns out to be, let's see.

From back in the original formula here there's clearly a one to one ratio between

the number of moles of sodium chloride which react and the number of moles of

silver chloride which are produced. So, if in fact I have 0.50 moles of sodium

chloride. The number of moles of silver chloride,

which will be produced will be equal to 0.050 moles of solid which will be

produced. So if we ask the question now, how much is

produced, we've got the answer. What remains in the solution behind?

Well, we precipitated out quite a bit of silver chloride.

But notice, there was a lot more silver in that solution than there was chlorine.

Buy 0.25, by 0.025 moles. So left in the solution is some sodium

ions, some nitrate ions. But also some silver ions.

There are silver ions left behind in the solution.

And we can actually calculate the numbers of silver ions left over in solution.

But also, importantly, we could calculate the concentrations of the silver ions

which remain in the solution. So, to do that, we have to take into

account a new phenomena as well. Something we're going to call dilution.

Dilution's a pretty familiar kind of thing in everyday life.

We take some sort of a solution and we add some solvent to it.

This might be, for example, taking a soap solution and watering it down.

Taking an alcohol solution and watering it down.

What we're doing, is we're adding solvent. But we are not changing the number of

moles of the solute, because we're only adding solvent.

The cons, consequence of that, is that we've actually changed the concentration

of the solution because the concentration of the solution is the number of moles of

the solute, divided by the volume of the solution.

I don't change the number of moles in the solute and I do change the volume of the

solution, then I have changed the concentration.

But note the number of moles of the solute is fixed so doing a dilution, when I in

fact dilute something before or after dilution.

The number of moles of the solute is fixed.

But recall the concentration, let's call it the concentration before a dilution is

equal to the number of moles of the solute divided by the volume before dilution.

And likewise the concentration after the dilution is the number of moles of the

solute divided by the volume after the dilution.

If I set the number of moles equal to each other then I can actually write down a

formula and here's the rather simple formula.

That the concentration before dilution. Multiplied by the volume before dilution,

equals the concentration after the dilution, multiplied by the volume after

the dilution. So for example, if I start off with a

solution which is let's say 0.1 molar, and I have 1 liter of that solution.

And I dilute it to double it's original volume and I want to know what the final

concentration it's equal to after I double the volume.

Well all I gotta do is divide both sides of this equation by 2 and the final

concentration is half of the original concentration.

Which makes sense. And it comes out to, to be 0.05 molar

after a dilution by a factor of 2. So if we go back and think about the

previous problem here. And we know that we have left behind from

our previous calculations, we'll go back and recover those.

We know that we still have the number of moles of silver, which are in the

solution, is still equal to now, comparing the numbers up above I had 0.075 moles of

silver nitrate and 0.05 moles of silver, sodium chloride.

And all of the sodium chloride all of the chlorine has precipitated out with the

chloride. There still remains 0.025 moles of silver

back in the solution. The concentration of the silver ions in

the solution is then equal to the number of moles of the silver, divided by the

volume of the solution. But notice that the volume of the solution

didn't remain fixed. Because we mixed together a half a liter

of a solution. Go back here and take a look.

Here's a half a liter with a quarter of a liter.

So now we've got 0.75 liters of solution. So we could write this now as being equal

to 0.025 molar divided by 0.75 liters. And if we actually divide those two things

out, we discover that the concentration of the solution in the end, checking my notes

here, is actually equally to 0.033 molar, moles per liter, of silver ions in

solution. In other words not only have we

precipitated out some of the silver, with the chlorine.

We've also diluted the solution, which remains, because there's not nearly the

concentration that we had started off with.

Now let's finally go back and look at the problem we've described at the very

beginning here. At the outset, one of the things we were

interested in is. How much, nitric acid is required to

dissolve 10 grams of copper? Well, to work that problem out, what we

would need to know is what is the number of moles of copper that we started off

with. Well let's see, because it's a solid, I

can find its number of moles by taking its mass divided by its molar mass.

Its mass is equal to 10 grams. Its molar mass, I can look that up in the

periodic table, is 63.55 grams per mole, the number of moles turns out to be 0.167

moles. From the stoichiometry, the balance to

chemical equation here. I know I need 4 moles of nitric acid for

every mole of copper that I started off with.

So, the number of moles of nitric acid which are needed.