Hi. It's Mr. Andersen and this is chemistry essentials video 28. It's on stoichiometry
which is really just measuring atoms. In other words in a reaction we're measuring the atoms
we have before the reaction and after. And because of the conservation of mass we have
to make sure those are the same. And so it's a really powerful tool. Chemistry teachers
love stoichometry. And students not so much. And so let's think of it as musical chairs.
As you're playing musical chairs you're walking around the circle. And so you're kind of bonded
to the floor. But when the music stops you have to quickly find a chair. And when you
do that it's like a chemical reaction. You're breaking your bonds with the floor, forming
new bonds with that chair. And if there's not enough chairs, then you're kind of out.
In other words chairs could be the limiting reactant in this reaction. And so sometimes
we don't have enough chairs. And sometimes we don't have enough people. And so reactions
are going to be the same way. And so basically stoichiometry allows us to measure chemical
reactions. But in order to do it we have to have a balanced chemical equation. And so
the coefficients in that equation are going to show us the number of moles. And so it's
the mole proportion. So just this one idea is so powerful. If you understand it, you
can answer so many different problems. But if you don't get it, then the following five
things, figuring out expected product, limiting reactants, percent yield, molar mass of
gases and titrations are going to be really confusing. But the one thing that ties them
all together is this idea that the coefficients in an equation represent the number of moles.
And so as you watch this video, look for the commonalities. What's going to be the one
thing that ties all five of these together? First thing, let's say we're trying to figure
out the expected products. And so in this reaction, in the burning of methane in a bunsen
burner, we're combining methane gas with oxygen to make carbon dioxide and water. And so let's
say you're given this problem. Given a certain amount of mass of methane, could you figure
out how much water is produced? Well the secret is going to look like this. We can go from
the number of grams to the number of moles. We can then go through a mole conversion and
then go back to grams again. And so you're going to see this over and over and over in
each of these problem sets. And so the first thing you have to do is make sure that that
equation is balanced. And so once it's balanced then we're ready to go. That should always
be your first step when you're doing stoichiometry. Make sure that equation is balanced. Then
we start with what we know. And I always write that over 1. Because it's easier to do these
factor-label methods. And so we've go 24.6 grams of methane. First thing you want to
do is convert that to moles. How do we do that? We're going to use the atomic mass of
methane. Remember it's going to be the carbon which is 12 and the hydrogen which is around
1. And so this is going to be the formula mass of methane. And so now we've converted
it to moles of methane. Next thing we want to do is we want to bridge the gap. We want
to go from the methane over to the water. And so we're going to use a mole conversion
to do that. So this 1 mole of methane is going to correspond to 2 moles of water. And so
this is the secret of stoichiometry. We can use the numbers in front, these coefficients.
And that's going to allow us to bridge the gap. So we've already converted it to moles
of water. Now we can just go back to grams again. So we're going to use that mole conversion,
in other words water is going to have an atomic mass of 18.02 grams. So now we've gone to
grams of water. And so we can quickly solve the problem. And so again, it's really easy
to predict what we're going to get in a reaction just using the equation and the periodic table.
Now what's a limiting reactant or a limiting reagent? That's going to be something in an
equation that we don't have enough of. And so right here what would be the limiting reactant
in this game of musical chairs? It would be the number of people. In other words once
they correspond we're going to have a lot extra of these chairs. And so let's look at
this reaction. Let's say that this is a real simplified model. So we've got 1 methane.
And then we've got 5 of these water molecules. This is before the reaction. And this is after
the reaction. What was the limiting reactant? Well the limiting reactant was methane. If
we would have had more methane we could have had more product. But since we only had 1
carbon we could only make 1 carbon dioxide out of it. And so limiting reactants are another
thing that we could measure using stoichiometry. So let's say you were presented with this
problem. In this equation you have 10 grams of methane, 10 gram of oxygen. Which of those
would be the limiting reactant? Well how do you solve this problem? What we're going to
have to do is kind of work it out using these two different alternatives. And so here it
is. It's familiar again. We're going to go from grams to moles to moles to grams. So
we're going to start with what we know. 10 grams of methane and 10 grams of oxygen. And
then we're simply going to compare it to one of the products on the right side. And so
what we can do is convert that to moles of methane and oxygen. So we've converted that.
And now what we're going to do is we're going to compare it to one of the products on the
right side. For this one I've just chosen to compare it to carbon dioxide. And so we're
going to bridge the gap between the moles of oxygen in methane to moles of carbon dioxide.
And so why do we have a 2 here? Because we have a coefficient of 2 on the left side.
We only have a 1 on this side because we have a coefficient of 1 on this side. So now we
can bridge the gap to carbon dioxide. We can then convert those back to grams again. And
so this is the expected products we would get using these two different reactants to
start with. And so in other words if we have 10 grams of methane, we could make 27.4 grams
of carbon dioxide, but we're not going to. Because the limiting reactant is the oxygen.
That's only going to provide us with 6.88 grams of carbon dioxide. And so oxygen would
be the limiting reactant in this reaction. Now, what we're looking at here is a predicted
yield. How much you think you could get. Now do you think in a reaction you're ever going
to get that amount? No. It's never going to go all the way to completion like that. We're
going to not have a lot of those atoms. Think of all those billions and billions of atoms
that are trying to combine together. And so this is simply a predicted yield. And so what
we can use is that predicted yield. And then the actual yield to figure out the percent
yield. In other words this is how much we could get, theoretically. This is how much we did
get. And so we can use this formula, actual yield divided by predicted yield and that
can give us the percent yield. And so we got 90 percent of what we could have got. And
we're never going to approach 100 percent. But we can get pretty close to it. Now let's
go to molar masses of gases. Let's say we start with 20 liters of methane. Okay. So
that would be 10 two liter bottles filled with methane gas. How many liters of water
could we get from that, that at STP? Now what does STP stand for? It's just standard temperature
and pressure. So we're moving into the gas laws here. And so you might think, here we go.
Grams to moles. Moles to grams. But what's our problem? We don't have grams at this point.
They're giving us liters to start with. And so you might think I'm stuck. But it's really
easy to go from liters to moles, especially at STP. And so how do we do that? We start
with 20 liters of methane in this case. We're now going to convert that to moles. And so
again using the gas law, we know that 1 mole of any gas is going to be 22.4 liters of that
gas. It doesn't matter what the gas is. And so now we can convert immediately to moles.
Once we go to moles of methane, we could convert that to moles of water. Again we have 1 on
this side and 2 on the side of the water because of the coefficient is 2. So now we've converted
it to moles of water. And now we could go back to liters again. Going to use that same
moles to liter conversion. And that's going to give us 40 liters of water that we could
get out of that reaction. And so again we're using the same thing. We're going to moles.
And once we get moles, we use our equation. And then we go to the other side. I hope that
makes sense so far. And now let's go to a titration. Titration is something I remember
doing in chemistry that I was totally confused with. But it's really simple. Let's say we
have this reaction. We're combining sodium hydroxide with oxalic acid. And so we're doing
a titration. So what you do is we're going to put the base up here. You're going to run
that base through. And then we're going to note what the volume is when it actually turns.
When we hit that equivalence point. And so why is that occurring? Well basically what's
happening is when we get to this equivalence point, we have run out of this oxalic acid.
In other words that's the limiting reactant at that point. And so if we just record the
amount of sodium hydroxide, we have to add, we could figure out, kind of backwards, how
much of this oxalic acid we're going to have. And so how do you do that? We start with the
known, milliliters. We can convert that to liters. Once we have the liters we can convert
that to moles. Because we're looking at a simple 1.0 molar solution. We could then go
from the moles to the moles of comparing the 2 together. So we've got the moles of the
oxalic acid. 1 here. And sodium hydroxide is 2. We can then go from there to grams.
And then we could figure out how much oxalic acid we had to start with. So once we get
to the moles, we can use these coefficients. And then we can compare anything in that equation
together. And so did you learn to use stoichiometry calculations to predict results of a reaction?
Remember that was that simple expected yield that we got. And then did you use it to relate
stoichiometry relationships including limiting reactants? Remember that's if we don't have
enough of it. And then going to completion. Are we going all the way to the end of that
reaction? And I hope that was helpful.