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- [Instructor] Let's
solve a few numerical on

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Avogadro number and moles.

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Here's the first one.

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How many glucose molecules
are in 2.37 moles of glucose?

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Let's quickly remind
ourselves what moles are.

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Moles are like dozen.

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Just like how one dozen equals 12,

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a mole represents an
Avogadro number of things.

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A dozen eggs equals 12 eggs.

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A dozen carrots means 12 carrots

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Similarly, a mole of eggs would
represent these many eggs.

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A mole of carrots would
represent these many carrots.

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A mole of glucose would represent

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these many glucose molecules, okay?

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So one mole of glucose equals

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these many molecules of glucose,

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or if you have two moles of glucose,

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it will be twice the amount.

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If you have three moles, it
will be thrice the amount,

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and so on and so forth.

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Now, what are we given?

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We are given, there are
2.37 moles of glucose,

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and we are asked how many
glucose molecules are there?

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Now, since one mole equals
these many molecules,

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2.37 moles of glucose
would just be 2.37 times

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these many molecules.

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Intuitively, that makes sense, right?

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But guess what?

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This is a slightly simple numerical,

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and that's why it's easier
to do it in our heads,

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and we can do it intuitively,

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but in general, there might
be more conversions required,

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more calculations needed.

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So a better way to do this
is to think in terms of

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conversions factors, okay?

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So we can say that, hey, we
are given moles of glucose,

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and we need to go, we
need to convert from moles

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to molecules.

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So how do we do that, okay?

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So here's how we do it.

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We start with what's given to us.

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We are given 2.37 moles of glucose.

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Now, from moles of glucose,

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how do we go to molecules of glucose?

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Well, for that we'll
build a conversion factor.

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And here's how I think about it.

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Since I want glucose molecules,

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my conversion factor should have

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molecules of glucose in the numerator.

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And since I want to get rid of moles,

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my moles should be in the denominator.

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So this is what my
conversion factor looks like,

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molecules per mole.

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But how many molecules per mole?

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Well, I know that there are 6.022

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times 10 to the 23 molecules per mole.

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So I can fill this in,

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and now have built my conversion factor.

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So look, the moles of glucose cancels out,

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and what I now end up with
is molecules of glucose.

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So I just need to plug
this in my calculator.

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So 2.37 times 6.022 times 10 to the 23.

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And the way you input that
is you use this EE, okay?

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Not this one, this one.

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10 to the 23, so you put 23.

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And that's it.

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So equals 1.427,

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which I can round it off to
three significant figures

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because there are three
significant figures here.

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So 1.43 times 10 to the 24.

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Don't forget this, okay?

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So 1.43 times 10 to the 24.

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And look, as we saw earlier,

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it makes intuitive sense as well.

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One mole has many molecules.

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So 2.37 moles will have 2.37
times these many molecules.

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But now let's see the true power

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of thinking it this way, okay?

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Let's say the question was
how many atoms of hydrogen

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are in 2.37 moles of glucose?

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How would we solve this?

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Well, again, let's think
about this intuitively.

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We already know that 2.37 moles of glucose

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has these many molecules,

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so have these many molecules of glucose.

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But now I'm asked how many
atoms of hydrogen are there

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in these many molecules of glucose?

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How do I do that?

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Well for that,

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I know that if I have
one molecule of glucose,

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then there are 12 hydrogen atoms.

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If I take two molecules of
glucose, I will have 12 plus 12.

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That is 12 times 2 hydrogen atoms.

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If I take three molecules of glucose,

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12 times 3 hydrogen atoms.

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So if I take these many
molecules of glucose,

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it'll be 12 times these
many atoms of hydrogen.

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Makes sense, right?

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But over here, we first calculate
the molecules of glucose,

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and then we are now calculating
the atoms of hydrogen.

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So multiple times we'll have
to crunch in the numbers

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in the calculator.

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But what if I had to do this from scratch?

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Not worry about the earlier calculations.

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Then again, if we think
in terms of conversions

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and building conversion factors,

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we can calculate very efficiently.

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So again, we start with
what's given to us.

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We are given 2.37 moles of glucose,

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and then we try to go from
moles to molecules of glucose.

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And I know how to do that.

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I need molecules in the numerator,
moles in the denominator.

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So Avogadro number.

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So I know there are these many molecules

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of glucose per mole.

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And so then this cancels out.

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Now comes the question,

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how do I go from molecules of
glucose to atoms of hydrogen?

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I need another conversion factor.

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I need atoms of hydrogen in the numerator

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because that's what I want to end up with,

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and I need molecules of
glucose in the denominator

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so that I can cancel it with this one.

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So my new conversion factor
would be atoms of hydrogen

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per molecules of glucose,

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and I know how much that is.

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12 atoms of hydrogen
per molecule of glucose.

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And so now I can cancel the
molecules of glucose out.

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And look, I'm done.

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I end up with atoms of hydrogen.

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And now in one shot I can crunch

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all of these numbers in my calculator.

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So 2.37 times 6.022

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times 10 to the 23.

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And the way you do that,
remember you use this one,

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to the 23 times 12 divided by 1.

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So that's it.

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And again, we now have
to round this off to

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how many significant figures?

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Well, there are three here.

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There are four here.

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So the minimum three significant figures.

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But wait a second, what about this one?

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Aren't there two significant figures here?

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No, because this is not a measurement.

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Every glucose molecule has
exactly 12 hydrogen atoms.

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So this is not a measurement,

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and therefore our significant figures

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are not limited by this number.

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It's limited by this number.

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So there are three significant figures.

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So we have to round it off
to three significant figures.

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So it would be 1.71 times
10 to their power 25.

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All right, let's kick
things up a little bit.

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Let's go to the next question.

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How many formula units
of calcium carbonate

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are in four grams of calcium carbonate?

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Alright, first things first,
what exactly are formula units?

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I used to get always confused with them.

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So let's clarify that.

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When we're dealing with non-metals,

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which are usually covalently bonded,

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we have individual molecules.

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So for example, if you
look inside a glucose,

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then we can say that, hey,
there is this glucose molecule,

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and there's that glucose molecules.

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You know, they're made of molecules, okay?

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You can identify individual molecules,

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and we can talk about them.

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But when it comes to ionic
compounds like calcium carbonate,

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or let's take a simpler one,
say sodium chloride, okay,

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because it's much simpler,

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then you don't find individual
molecules over there.

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If you were to look inside them,

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you'll probably see a crystal structure

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that looks like this.

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Okay, but if there are
no molecules over here,

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then what does this represent?

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Or over here, if you look at it,

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what does it mean to say that

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there are three oxygen atoms over here?

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What does it represent?

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Well, think about it this way, okay?

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Let's start with the sodium chloride.

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If I were to take a
chunk of sodium chloride,

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there are lots and lots of sodium

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and you know, sodium
cations and chlorine anions,

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but they will be in the
ratio one is to one.

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That's what it means to write it this way.

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Similarly, if I were to take
a chunk of calcium carbonate,

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its crystal structure is
much more complicated.

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That's why I didn't draw that over here.

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But if I were to take a
chunk of calcium carbonate,

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then I'll find that the ratio
of the amount of calcium

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to the amount of carbon
to the amount of oxygen

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that I'll find in that chunk
is one is to one is to three.

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There'll be thrice as many oxygen atoms

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as carbon or as calcium.

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That's what it means.

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So a formula unit is a representation

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of the relative ratios
of the different atoms

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or elements that make up your
ionic compound, all right?

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But more importantly,

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how does it change things
when it comes to the mole?

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The answer is it doesn't.

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See, over here when you're
dealing with the mole,

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we would say that a
mole of glucose would be

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Avogadro number of glucose molecules.

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Over here we would say a mole would be

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an Avogadro number of
formula units, and that's it.

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So it doesn't change anything, okay?

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All right, so we need to
find how many formula units

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of calcium carbonate are

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in four grams of calcium carbonate, okay?

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So this time we are given the
mass of the calcium carbonate.

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So look, this time we
need to convert from mass

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to formula units.

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How do we do that?

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Well, here's what I'm thinking.

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If we can convert from
mass to moles, we are done.

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If I can figure out how many
moles of calcium carbonate

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this represents, we are done.

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Because then just like what we did

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in our previous numerical, we
went from moles to molecules

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in the same way using the Avogadro number.

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In the same way over here we can go from

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moles to formula units.

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So the big question is how do I go from,

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you know, grams to moles?

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And we can figure that out
using the periodic table.

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See, the periodic table
will give me the molar mass

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of individual elements.

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So for example, here,
calcium has an atomic,

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an average atomic mass of 40.08 units.

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This means that one mole of calcium

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will have a mass of 40.08 grams.

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Remember that mole is a
conversion from U to grams, okay?

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So I know the molar mass of calcium.

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It's 40.08 grams per mole.

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Similarly carbon, 12.01 grams per mole.

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Oxygen, 16 grams per mole.

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So what will be the molar
mass of calcium carbonate?

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What will be the mass of one
mole of calcium carbonate?

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Well, one mole of calcium
carbonate will have

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one mole of calcium, one mole of carbon,

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and three moles of oxygen.

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So it'll be a great
idea to pause the video

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and see if you can use these
numbers to calculate first

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the molar mass of calcium carbonate.

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All right, here's how we can do it.

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The molar mass of calcium carbonate,

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meaning one mole of calcium carbonate,

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its mass will be the mass
of one mole of calcium

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plus mass of one mole of carbon,

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plus three times the mass
of one mole of oxygen

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because there are three over here.

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So you plug this into our calculator.

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00:10:32,190 --> 00:10:33,330
Let me just do this first

259
00:10:33,330 --> 00:10:34,470
because there's a
multiplication over here.

260
00:10:34,470 --> 00:10:39,470
So it's 3 times 16 plus 12.01 plus 40.08

261
00:10:43,170 --> 00:10:45,000
equals a 100.09.

262
00:10:45,000 --> 00:10:46,770
We'll not round it off right now

263
00:10:46,770 --> 00:10:48,840
because we are still in the
middle of the calculation.

264
00:10:48,840 --> 00:10:53,310
So its molar mass is a
100.09 grams per mole.

265
00:10:53,310 --> 00:10:54,870
So I can use that as a conversion factor

266
00:10:54,870 --> 00:10:56,820
to go from grams to moles.

267
00:10:56,820 --> 00:10:58,680
And then I know the Avogadro number.

268
00:10:58,680 --> 00:11:01,230
From there I can go from
moles to formula units.

269
00:11:01,230 --> 00:11:02,910
Again, it will be a great
idea to pause the video

270
00:11:02,910 --> 00:11:05,100
and see if you can try
this yourself first.

271
00:11:05,100 --> 00:11:06,030
All right, let's do it.

272
00:11:06,030 --> 00:11:07,470
So we'll start with what's given to us.

273
00:11:07,470 --> 00:11:10,470
We are given four grams
of calcium carbonate.

274
00:11:10,470 --> 00:11:12,360
Now, to convert this into moles,

275
00:11:12,360 --> 00:11:15,120
I need to have moles in the
numerator, my conversion factor,

276
00:11:15,120 --> 00:11:17,880
and grams in the denominator
to cancel this out.

277
00:11:17,880 --> 00:11:19,440
So I need moles per grams,

278
00:11:19,440 --> 00:11:22,890
moles of calcium carbonate
per gram of calcium carbonate.

279
00:11:22,890 --> 00:11:24,720
And I have that conversion
factor over here.

280
00:11:24,720 --> 00:11:26,010
I just have to do a reciprocal

281
00:11:26,010 --> 00:11:27,840
because I want the mole in the numerator.

282
00:11:27,840 --> 00:11:30,867
So I will write this as one
mole of calcium carbonate

283
00:11:30,867 --> 00:11:34,980
per 100.09 grams of calcium carbonate.

284
00:11:34,980 --> 00:11:36,210
You see why I wrote it that way,

285
00:11:36,210 --> 00:11:40,650
so that now I can cancel this.

286
00:11:40,650 --> 00:11:42,750
So now this gives me the moles.

287
00:11:42,750 --> 00:11:45,600
Now how do I go from
moles to formula units?

288
00:11:45,600 --> 00:11:47,610
Build a new conversion factor.

289
00:11:47,610 --> 00:11:49,890
I need a formula unit in the numerator,

290
00:11:49,890 --> 00:11:52,020
mole in the denominator.

291
00:11:52,020 --> 00:11:55,800
So how many formula
units do I have per mole?

292
00:11:55,800 --> 00:11:57,510
Well, the Avogadro number.

293
00:11:57,510 --> 00:11:59,460
I have Avogadro number of formula units

294
00:11:59,460 --> 00:12:03,000
of calcium carbonate per
mole of calcium carbonate.

295
00:12:03,000 --> 00:12:05,520
And boom, this cancels out.

296
00:12:05,520 --> 00:12:06,780
Let me cancel out slightly differently.

297
00:12:06,780 --> 00:12:08,040
Okay, this cancels out.

298
00:12:08,040 --> 00:12:08,940
And there you have it.

299
00:12:08,940 --> 00:12:11,850
I am now left with formula
unit of calcium carbonate.

300
00:12:11,850 --> 00:12:14,130
I can plug this into the calculator now.

301
00:12:14,130 --> 00:12:19,130
So it's 4 times 1 times
6.022 times 10 to the 23.

302
00:12:21,390 --> 00:12:23,430
So again, we'll use
the exponent over here,

303
00:12:23,430 --> 00:12:26,880
23 divided by 100.09.

304
00:12:26,880 --> 00:12:29,583
Divided by 100.09.

305
00:12:30,630 --> 00:12:33,030
This gives us 2.406.

306
00:12:33,030 --> 00:12:36,540
And again, we'll round it off
to three significant figures.

307
00:12:36,540 --> 00:12:38,040
That's the minimum one over here.

308
00:12:38,040 --> 00:12:42,570
So 2.41 times, don't forget this.

309
00:12:42,570 --> 00:12:45,540
2.41 times 10 to the power 22.

310
00:12:45,540 --> 00:12:46,373
There you have it.

311
00:12:46,373 --> 00:12:49,110
That many formula units
of calcium carbonate.

312
00:12:49,110 --> 00:12:51,090
Okay, here's the last question.

313
00:12:51,090 --> 00:12:52,680
How many oxygen atoms

314
00:12:52,680 --> 00:12:54,510
are in four grams of calcium carbonate?

315
00:12:54,510 --> 00:12:57,060
Again, pause the video and
see if you can do this.

316
00:12:57,060 --> 00:12:58,620
This is the last one.

317
00:12:58,620 --> 00:13:00,390
Okay, this is just like
the previous numerical.

318
00:13:00,390 --> 00:13:02,490
We already know four
grams of calcium carbonate

319
00:13:02,490 --> 00:13:03,960
has these many formula units,

320
00:13:03,960 --> 00:13:07,500
and we know a single formula
unit has three oxygen atoms.

321
00:13:07,500 --> 00:13:10,680
So these many formula units
will have three times this much.

322
00:13:10,680 --> 00:13:11,967
That's it.

323
00:13:11,967 --> 00:13:13,080
That's our answer, three times this much.

324
00:13:13,080 --> 00:13:15,810
But again, if this was asked
to us to do it from scratch,

325
00:13:15,810 --> 00:13:16,770
how would we do it?

326
00:13:16,770 --> 00:13:17,730
Well, we'll start from here.

327
00:13:17,730 --> 00:13:19,350
So we'll do all of this.

328
00:13:19,350 --> 00:13:22,080
So let me just copy and
paste that over here,

329
00:13:22,080 --> 00:13:24,540
and then we'll have one
more conversion factor

330
00:13:24,540 --> 00:13:26,130
that'll help me cancel our formula units

331
00:13:26,130 --> 00:13:27,750
and have oxygen atoms.

332
00:13:27,750 --> 00:13:31,320
So over here I would need
oxygen atoms in the numerator

333
00:13:31,320 --> 00:13:34,410
and formula units of calcium
carbonate in the denominator.

334
00:13:34,410 --> 00:13:36,300
How many oxygen atoms do I have

335
00:13:36,300 --> 00:13:38,220
per formula unit of calcium carbonate?

336
00:13:38,220 --> 00:13:39,053
Three.

337
00:13:39,053 --> 00:13:42,690
So my last conversion factor
would be three atoms of oxygen

338
00:13:42,690 --> 00:13:45,060
per formulate of calcium carbonate.

339
00:13:45,060 --> 00:13:46,473
This cancels out.

340
00:13:47,700 --> 00:13:48,533
And I'm done.

341
00:13:48,533 --> 00:13:51,480
And again, we can crunch
it in one last time.

342
00:13:51,480 --> 00:13:56,457
4 times 6.022 times 10 to the 23.

343
00:13:57,690 --> 00:14:02,690
So use the exponent, 23
times 3 divided by 100.09.

344
00:14:08,610 --> 00:14:09,600
That's it.

345
00:14:09,600 --> 00:14:11,790
So that gives us 7.219.

346
00:14:11,790 --> 00:14:15,900
Again, rounding it off to three
significant figures, 7.22.

347
00:14:15,900 --> 00:14:17,010
Don't forget this.

348
00:14:17,010 --> 00:14:20,463
Times 10 to the power 22 atoms of oxygen.

349
00:14:22,350 --> 00:14:24,180
And this is the same number as multiplying

350
00:14:24,180 --> 00:14:26,310
this number by three.

351
00:14:26,310 --> 00:14:28,110
But if we were asked from scratch

352
00:14:28,110 --> 00:14:29,430
and we didn't have this number,

353
00:14:29,430 --> 00:14:31,630
this is how we could
do it in just one shot.