- [Instructor] Let's solve a few numerical on
Avogadro number and moles.
Here's the first one.
How many glucose molecules are in 2.37 moles of glucose?
Let's quickly remind ourselves what moles are.
Moles are like dozen.
Just like how one dozen equals 12,
a mole represents an Avogadro number of things.
A dozen eggs equals 12 eggs.
A dozen carrots means 12 carrots
Similarly, a mole of eggs would represent these many eggs.
A mole of carrots would represent these many carrots.
A mole of glucose would represent
these many glucose molecules, okay?
So one mole of glucose equals
these many molecules of glucose,
or if you have two moles of glucose,
it will be twice the amount.
If you have three moles, it will be thrice the amount,
and so on and so forth.
Now, what are we given?
We are given, there are 2.37 moles of glucose,
and we are asked how many glucose molecules are there?
Now, since one mole equals these many molecules,
2.37 moles of glucose would just be 2.37 times
these many molecules.
Intuitively, that makes sense, right?
But guess what?
This is a slightly simple numerical,
and that's why it's easier to do it in our heads,
and we can do it intuitively,
but in general, there might be more conversions required,
more calculations needed.
So a better way to do this is to think in terms of
conversions factors, okay?
So we can say that, hey, we are given moles of glucose,
and we need to go, we need to convert from moles
to molecules.
So how do we do that, okay?
So here's how we do it.
We start with what's given to us.
We are given 2.37 moles of glucose.
Now, from moles of glucose,
how do we go to molecules of glucose?
Well, for that we'll build a conversion factor.
And here's how I think about it.
Since I want glucose molecules,
my conversion factor should have
molecules of glucose in the numerator.
And since I want to get rid of moles,
my moles should be in the denominator.
So this is what my conversion factor looks like,
molecules per mole.
But how many molecules per mole?
Well, I know that there are 6.022
times 10 to the 23 molecules per mole.
So I can fill this in,
and now have built my conversion factor.
So look, the moles of glucose cancels out,
and what I now end up with is molecules of glucose.
So I just need to plug this in my calculator.
So 2.37 times 6.022 times 10 to the 23.
And the way you input that is you use this EE, okay?
Not this one, this one.
10 to the 23, so you put 23.
And that's it.
So equals 1.427,
which I can round it off to three significant figures
because there are three significant figures here.
So 1.43 times 10 to the 24.
Don't forget this, okay?
So 1.43 times 10 to the 24.
And look, as we saw earlier,
it makes intuitive sense as well.
One mole has many molecules.
So 2.37 moles will have 2.37 times these many molecules.
But now let's see the true power
of thinking it this way, okay?
Let's say the question was how many atoms of hydrogen
are in 2.37 moles of glucose?
How would we solve this?
Well, again, let's think about this intuitively.
We already know that 2.37 moles of glucose
has these many molecules,
so have these many molecules of glucose.
But now I'm asked how many atoms of hydrogen are there
in these many molecules of glucose?
How do I do that?
Well for that,
I know that if I have one molecule of glucose,
then there are 12 hydrogen atoms.
If I take two molecules of glucose, I will have 12 plus 12.
That is 12 times 2 hydrogen atoms.
If I take three molecules of glucose,
12 times 3 hydrogen atoms.
So if I take these many molecules of glucose,
it'll be 12 times these many atoms of hydrogen.
Makes sense, right?
But over here, we first calculate the molecules of glucose,
and then we are now calculating the atoms of hydrogen.
So multiple times we'll have to crunch in the numbers
in the calculator.
But what if I had to do this from scratch?
Not worry about the earlier calculations.
Then again, if we think in terms of conversions
and building conversion factors,
we can calculate very efficiently.
So again, we start with what's given to us.
We are given 2.37 moles of glucose,
and then we try to go from moles to molecules of glucose.
And I know how to do that.
I need molecules in the numerator, moles in the denominator.
So Avogadro number.
So I know there are these many molecules
of glucose per mole.
And so then this cancels out.
Now comes the question,
how do I go from molecules of glucose to atoms of hydrogen?
I need another conversion factor.
I need atoms of hydrogen in the numerator
because that's what I want to end up with,
and I need molecules of glucose in the denominator
so that I can cancel it with this one.
So my new conversion factor would be atoms of hydrogen
per molecules of glucose,
and I know how much that is.
12 atoms of hydrogen per molecule of glucose.
And so now I can cancel the molecules of glucose out.
And look, I'm done.
I end up with atoms of hydrogen.
And now in one shot I can crunch
all of these numbers in my calculator.
So 2.37 times 6.022
times 10 to the 23.
And the way you do that, remember you use this one,
to the 23 times 12 divided by 1.
So that's it.
And again, we now have to round this off to
how many significant figures?
Well, there are three here.
There are four here.
So the minimum three significant figures.
But wait a second, what about this one?
Aren't there two significant figures here?
No, because this is not a measurement.
Every glucose molecule has exactly 12 hydrogen atoms.
So this is not a measurement,
and therefore our significant figures
are not limited by this number.
It's limited by this number.
So there are three significant figures.
So we have to round it off to three significant figures.
So it would be 1.71 times 10 to their power 25.
All right, let's kick things up a little bit.
Let's go to the next question.
How many formula units of calcium carbonate
are in four grams of calcium carbonate?
Alright, first things first, what exactly are formula units?
I used to get always confused with them.
So let's clarify that.
When we're dealing with non-metals,
which are usually covalently bonded,
we have individual molecules.
So for example, if you look inside a glucose,
then we can say that, hey, there is this glucose molecule,
and there's that glucose molecules.
You know, they're made of molecules, okay?
You can identify individual molecules,
and we can talk about them.
But when it comes to ionic compounds like calcium carbonate,
or let's take a simpler one, say sodium chloride, okay,
because it's much simpler,
then you don't find individual molecules over there.
If you were to look inside them,
you'll probably see a crystal structure
that looks like this.
Okay, but if there are no molecules over here,
then what does this represent?
Or over here, if you look at it,
what does it mean to say that
there are three oxygen atoms over here?
What does it represent?
Well, think about it this way, okay?
Let's start with the sodium chloride.
If I were to take a chunk of sodium chloride,
there are lots and lots of sodium
and you know, sodium cations and chlorine anions,
but they will be in the ratio one is to one.
That's what it means to write it this way.
Similarly, if I were to take a chunk of calcium carbonate,
its crystal structure is much more complicated.
That's why I didn't draw that over here.
But if I were to take a chunk of calcium carbonate,
then I'll find that the ratio of the amount of calcium
to the amount of carbon to the amount of oxygen
that I'll find in that chunk is one is to one is to three.
There'll be thrice as many oxygen atoms
as carbon or as calcium.
That's what it means.
So a formula unit is a representation
of the relative ratios of the different atoms
or elements that make up your ionic compound, all right?
But more importantly,
how does it change things when it comes to the mole?
The answer is it doesn't.
See, over here when you're dealing with the mole,
we would say that a mole of glucose would be
Avogadro number of glucose molecules.
Over here we would say a mole would be
an Avogadro number of formula units, and that's it.
So it doesn't change anything, okay?
All right, so we need to find how many formula units
of calcium carbonate are
in four grams of calcium carbonate, okay?
So this time we are given the mass of the calcium carbonate.
So look, this time we need to convert from mass
to formula units.
How do we do that?
Well, here's what I'm thinking.
If we can convert from mass to moles, we are done.
If I can figure out how many moles of calcium carbonate
this represents, we are done.
Because then just like what we did
in our previous numerical, we went from moles to molecules
in the same way using the Avogadro number.
In the same way over here we can go from
moles to formula units.
So the big question is how do I go from,
you know, grams to moles?
And we can figure that out using the periodic table.
See, the periodic table will give me the molar mass
of individual elements.
So for example, here, calcium has an atomic,
an average atomic mass of 40.08 units.
This means that one mole of calcium
will have a mass of 40.08 grams.
Remember that mole is a conversion from U to grams, okay?
So I know the molar mass of calcium.
It's 40.08 grams per mole.
Similarly carbon, 12.01 grams per mole.
Oxygen, 16 grams per mole.
So what will be the molar mass of calcium carbonate?
What will be the mass of one mole of calcium carbonate?
Well, one mole of calcium carbonate will have
one mole of calcium, one mole of carbon,
and three moles of oxygen.
So it'll be a great idea to pause the video
and see if you can use these numbers to calculate first
the molar mass of calcium carbonate.
All right, here's how we can do it.
The molar mass of calcium carbonate,
meaning one mole of calcium carbonate,
its mass will be the mass of one mole of calcium
plus mass of one mole of carbon,
plus three times the mass of one mole of oxygen
because there are three over here.
So you plug this into our calculator.
Let me just do this first
because there's a multiplication over here.
So it's 3 times 16 plus 12.01 plus 40.08
equals a 100.09.
We'll not round it off right now
because we are still in the middle of the calculation.
So its molar mass is a 100.09 grams per mole.
So I can use that as a conversion factor
to go from grams to moles.
And then I know the Avogadro number.
From there I can go from moles to formula units.
Again, it will be a great idea to pause the video
and see if you can try this yourself first.
All right, let's do it.
So we'll start with what's given to us.
We are given four grams of calcium carbonate.
Now, to convert this into moles,
I need to have moles in the numerator, my conversion factor,
and grams in the denominator to cancel this out.
So I need moles per grams,
moles of calcium carbonate per gram of calcium carbonate.
And I have that conversion factor over here.
I just have to do a reciprocal
because I want the mole in the numerator.
So I will write this as one mole of calcium carbonate
per 100.09 grams of calcium carbonate.
You see why I wrote it that way,
so that now I can cancel this.
So now this gives me the moles.
Now how do I go from moles to formula units?
Build a new conversion factor.
I need a formula unit in the numerator,
mole in the denominator.
So how many formula units do I have per mole?
Well, the Avogadro number.
I have Avogadro number of formula units
of calcium carbonate per mole of calcium carbonate.
And boom, this cancels out.
Let me cancel out slightly differently.
Okay, this cancels out.
And there you have it.
I am now left with formula unit of calcium carbonate.
I can plug this into the calculator now.
So it's 4 times 1 times 6.022 times 10 to the 23.
So again, we'll use the exponent over here,
23 divided by 100.09.
Divided by 100.09.
This gives us 2.406.
And again, we'll round it off to three significant figures.
That's the minimum one over here.
So 2.41 times, don't forget this.
2.41 times 10 to the power 22.
There you have it.
That many formula units of calcium carbonate.
Okay, here's the last question.
How many oxygen atoms
are in four grams of calcium carbonate?
Again, pause the video and see if you can do this.
This is the last one.
Okay, this is just like the previous numerical.
We already know four grams of calcium carbonate
has these many formula units,
and we know a single formula unit has three oxygen atoms.
So these many formula units will have three times this much.
That's it.
That's our answer, three times this much.
But again, if this was asked to us to do it from scratch,
how would we do it?
Well, we'll start from here.
So we'll do all of this.
So let me just copy and paste that over here,
and then we'll have one more conversion factor
that'll help me cancel our formula units
and have oxygen atoms.
So over here I would need oxygen atoms in the numerator
and formula units of calcium carbonate in the denominator.
How many oxygen atoms do I have
per formula unit of calcium carbonate?
Three.
So my last conversion factor would be three atoms of oxygen
per formulate of calcium carbonate.
This cancels out.
And I'm done.
And again, we can crunch it in one last time.
4 times 6.022 times 10 to the 23.
So use the exponent, 23 times 3 divided by 100.09.
That's it.
So that gives us 7.219.
Again, rounding it off to three significant figures, 7.22.
Don't forget this.
Times 10 to the power 22 atoms of oxygen.
And this is the same number as multiplying
this number by three.
But if we were asked from scratch
and we didn't have this number,
this is how we could do it in just one shot.