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- In this episode of Real Chemistry
we're going to talk about Stoichiometry again,
and we're going to do some practice problems.
If you haven't yet watched my video
"Introduction to Stoichiometry,"
I recommend you start there.
In this video, we're just going to go through
a few more practice problems
so you can get a better hang of Stoichiometry.
Stoichiometry is basically just thinking about
our chemical reactions as recipes.
Just like this recipe below
is a recipe for cheese sandwiches,
we can have chemical recipes.
We can think about how much of different ingredients
we need to put in to get out a certain amount of product.
Let's go ahead and take a look
at our first Stoichiometry problem.
This problem says, "17.35 grams of aluminum
"react with excess oxygen, how many moles
"of aluminum oxide are formed?"
Here we have aluminum, the metal,
combining with oxygen to form aluminum oxide.
We know that we're starting out
with 17.35 grams of aluminum,
and we know we don't need to worry
about how much oxygen is around.
It says there's excess oxygen,
which basically just means
our oxygen's not going to run out,
and we want to know how many moles
of aluminum oxide are formed.
We're going to start with our quantity of aluminum,
17.35 grams of aluminum.
You'll notice that that is our very first step
for writing out our Stoichiometry problems.
We want to write out just our starting quantity.
We're always going to convert to moles.
If we don't have moles initially,
we're going to convert to moles,
because our chemical recipe, our reaction up there,
is in terms of moles, not in terms of grams.
We can't use grams.
The way we convert from grams to moles
is with the molar mass of our aluminum.
Here the molar mass of aluminum is 26.98 grams per mole.
Our 26.98 grams goes on the bottom,
and our one mole goes up top,
and that's just going to give us moles of aluminum.
Now we want to go to moles of aluminum oxide.
The way we do that is we write out
over on the right hand side just that desired quantity,
and that's going to help us write our conversion factor.
We know that we want to get to, at this point,
moles of aluminum oxide.
Moles of AL2O3.
Now we're going to write our conversion factor
using our chemical recipe.
We know that we want to get rid of our aluminum.
We want to have moles aluminum down here,
and we know that we want to get to aluminum oxide,
so we're going to put moles Al2O3 up here.
Then the numbers we fill in there
come from our chemical recipe.
We see that there's a two right here
in front of our aluminum oxide.
That's the number we put in for our conversion factor.
Every time we run this recipe,
we get two moles of aluminum oxide out.
To run the recipe,
we know we need to put four moles of aluminum in.
Since there's that four next to aluminum,
that's what we put in our conversion factor down here.
We plug all of that into our calculator,
and we'll get out 0.3215
moles of aluminum.
What we did there is we used our reaction
to figure out how much aluminum oxide we could get out
given 17.35 grams of aluminum going in.
Let's take a look at our next Stoichiometry problem.
Here we're reacting oxygen with excess hydrogen
to get out the grams of water formed.
Here we have another step to add.
In the past problems, we were going from grams to moles
and our final answer was going to be in moles.
In this case, we want our final answers in grams,
and that tacks one more step onto our Stoichiometry problem
where we have to convert from the moles of product we form
to the grams of product we form.
Once again we're going to start the same way.
We're going to start by writing out our starting quantity
which is 1.45 grams of oxygen.
We have 1.45 grams of O2.
That's step one.
Then we want to convert that to moles.
We wrote out our starting quantity.
Now we're going to convert to moles.
Oxygen has a molar mass of 32 grams per mole.
We're going to convert now from grams of oxygen to moles
by writing 32 grams on the bottom and one mole up top.
That's going to get us to moles of oxygen.
Our grams are going to cancel there,
and we're going to have moles of oxygen.
We know that eventually we want to get to way out here,
grams of H2O.
We write our starting quantity.
It's nice to write our desired quantity, I'm sorry,
because that just gives us an idea of where we're going,
and that helps us know what steps to take next.
We want to eventually get to grams of water.
Before we can do that, we need to get the moles of water,
and that's where we're going to use
our Stoichiometry conversion factor.
Here we're going from oxygen to water.
We're going to cancel out our oxygen,
and we're going to get our H2O.
It doesn't really matter in these conversion factors
if you write out mole explicitly or not.
These are just ratios that work in terms of moles,
and you can go ahead and drop the moles if you want
or write it out if that helps you keep track.
Next to our oxygen, we're going to write one,
because that is the number in front of oxygen
in our chemical reaction.
There's nothing actually written there.
There's just an implied one.
Then we look over at water, and we see a two.
That's going to get us to moles of water.
Since what we want is grams of water,
then we know that we need to convert to grams.
We wrote our conversion factor using the recipe,
and now we need to convert to grams of water.
When we're dealing with moles of water,
to convert to grams of water,
we know we need to use the molar mass of water.
Here it's really important to keep track
of what you're actually converting.
In the past, you might have just been able to keep up with
okay I'm converting moles to grams or grams to moles.
Now you need to actually think about,
"I'm converting from moles of water to grams of water,"
which is different than what we were doing
in the start up problem,
which was grams of oxygen to moles of oxygen.
What we're converting tells us what molar mass to use.
We're going to need to use here our molar mass of water,
because that's what we're left with at this point.
Our oxygen has been canceled out,
and since we're left with water,
we need to use the molar mass of water.
The molar mass of water is 18.02 grams.
We're going to put 18.02 grams up here
and one mole down there,
because currently you'll notice we have moles,
and if we write moles on the bottom
that's going to let that cancel out,
and we're going to be left with grams.
Now, I write that all into my calculator,
and I'm going to get 1.63 grams of water.
That is our final answer.
This Stoichiometry problem was just a touch
more complicated than the last one.
We went to moles first and then we had to convert
all the way to grams of our product.
These chemical recipes allow us to think about
how much mass of something we could put in
and how much mass of something we would get out.
All right?
Our last Stoichiometry problem
is going in a little bit of a different direction.
Here we're actually given how much product we want to form.
It says, "How many moles of oxygen," which are reactant,
"are needed to form 1.4 grams of aluminum oxide,"
which is a product.
We're going to go backwards.
This it totally fine.
It's just like if I walked into the kitchen,
and I saw ten grilled cheeses made,
and somebody said, "How much bread did they use?"
Well, they used 20 slices of bread.
In the same way,
we can say we want 1.4 grams of aluminum oxide,
so how much oxygen do we need to put in?
What we're still going to do
is we're still just going to write out
our starting quantity,
which is 1.4 grams of aluminum oxide.
Since our chemical recipes only work in terms of moles,
we're going to go ahead and convert to moles.
To convert our grams of aluminum oxide to moles,
we need to use the molar mass of aluminum oxide,
and that turns out to be 101.96.
We put 101.96 down here,
and that's grams,
and one mole up top.
I've already calculated the molar mass of aluminum oxide,
so that we can do this problem a little more quickly,
but sometimes you'll actually have to calculate
those molar masses as you go through.
Now we're left with moles of aluminum oxide.
Our grams, they've gone away.
Now we're going to want to use our chemical recipe
to write our conversion factor.
To know what conversion factor to write,
we're going to go ahead and write down our desired quantity,
and we know that we want to get moles of oxygen.
We want to get moles of oxygen,
and we want to get rid of moles of aluminum oxide,
which means aluminum oxide
is going to go down on the bottom.
Al2O3
and oxygen's going to go up top.
Then we just look at the recipe.
It doesn't matter whether we're going forwards or backwards,
the recipe tells us what conversion factors we need to use.
Notice that there's a two in front of our aluminum oxide,
and so we're going to put a two down here.
There's a three in front of our oxygen,
so we're going to put a three in front of our oxygen
and our conversion factor.
Because our chemical recipe just tells us the ratio of stuff
we put in and get out,
and we can use that to calculate
how much stuff we need to put in
or how much stuff we'll get out.
Now notice we're going to get rid of
our moles of aluminum oxide,
and we're going to be left with just oxygen.
Our Al2O3s, they go away,
and we're left with just moles and oxygen.
When I plug that into my calculator,
I'm going to get 0.021 moles of oxygen.
Notice now if the question had asked for grams of oxygen,
we could do that too.
All that we'd have to do now is just multiply
our .021 moles of oxygen by the molar mass of oxygen.
All sorts of different questions we can ask
involving Stoichiometry problems,
but hopefully now you have a good handle
on how to write the correct conversion factors down
to answer a wide range of Stoichiometry problems.
Thanks for watching this episode of Real Chemistry.
If you have any questions, please leave them below.
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