Calculations using Avogadro's number (part 2) | Chemistry | Khan Academy

- [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.