WEBVTT
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>> Welcome to a screencast on an
introduction to chemical moles.
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If you were a cook or a baker making
food for lots and lots of people,
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you would certainly order something like eggs,
not by how many individual eggs you needed,
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but by how many dozens of eggs you needed.
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And if you were an office manager, you wouldn't
order how many sheets of paper you needed,
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you'd order paper by the
ream or possibly by the box.
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And things like a dozen, which means 12, of
course, or a ream, there's 500 pages in a ream
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of paper, those are useful quantities
for these sorts of situations.
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Now chemists, of course, deal with matter.
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The substance that makes up our world,
and even very small quantities of matter
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like this tiny droplet of water,
contain tremendously large numbers
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of the particles that make it up.
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In this case, water molecules.
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And because of this, chemists
use a unit somewhat analogous
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to a dozen, and it's called the mole.
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Now a dozen always means 12.
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You have 12 eggs.
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That's a dozen eggs.
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That's 12 eggs.
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You have a dozen horses.
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That's 12 horses.
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And a mole is also a number of things,
but it's a much, much bigger number.
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One mole of anything is 6.022 times 10 to
the 23rd of those things, whatever they are.
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This is kind of like the chemist dozen.
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Now, of course, you're typically not going to
count out individual atoms or molecules or ions.
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We typically weigh things out and the mole
has another meaning and the mass of a mole
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of any element or of any compound can be
determined from atomic masses, and those are,
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of course, found on the periodic table.
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Now one other thing to note.
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This really, really large number, 6.022 times
10 to the 23rd is also called Avogadro's number,
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and here it's symbolized by
n's of A. N stands for number.
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The subscript A stands for
Avogadro, and the definition
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of Avogadro's number is how many
atoms of the isotope carbon 12 are
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in exactly 12 grams of carbon 12?
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So, if we were able to pile carbon 12 atoms,
only the isotope carbon 12 on a scale and get it
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up to exactly 12 grams and then
count how many atoms we have,
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what we would end up with is a really, really,
really, large number, and it's exactly,
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or to a lot of significant figures,
6.022, 14129 times 10 to the 23rd,
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and we'll usually just use 6.02
or 6.022 times 10 to the 23rd
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for the Avogadro constant or Avogadro's number.
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Now this is usually easiest to
deal with by example when starting.
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So, let's say that we have one mole of copper.
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Just as a note, mole, mole our
abbreviation for that is mol.
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So, one mole of copper, how many atoms
is that and how many grams is it?
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Well, since it's a mole, it's 6.022
times 10 to the 23rd of those things.
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Just like if it was a dozen, it would be 12.
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So, one mole of copper is 6.022
times 10 to the 23rd copper atoms,
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and then the periodic table is useful because if
you look where copper is on the periodic table,
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and then look at its atomic mass,
that is the mass in grams of one mole
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or 6.022 times 10 to the 23rd copper atoms.
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So, notice that the 63.55 is the average
atomic mass of the naturally occurring isotopes
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of copper and in AMU, it's
the average mass of one atom.
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And in grams, it's the total
mass of a mole of atoms.
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Just for some perspective, a mole of
copper is about 22 of the old pennies,
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the pre-1983 pennies that were pure copper,
and if you weighed that, if you had one mole
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of copper you'd get about 63.55 or so of
grams, and if you were able to zoom in
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and count the atoms, you would end up with
this 6.022 times 10 to the 23rd atoms.
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This huge, huge, huge number.
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Second example.
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What if we don't have exactly one mole,
and we wanted either grams or atoms?
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Well, in this case we have
.0463 moles of krypton and we'd
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like to know how many grams we have.
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Then this is a conversion problem,
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but we're going to now be using the
periodic table to get our information.
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So, usual conversion problem method.
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We start with the known quantity.
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We look for a conversion factor
that will cancel out the unit
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that we're starting with and want to get rid of.
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So, moles will go on the bottom
of our conversion factor and then
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on the top we need to know what the mass is.
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So, we look on the periodic table.
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For krypton, the mass of
one mole is 83.80 grams.
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So, that goes on top.
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We do the math, cancel out the units, and .0463
times 83.80 gives us 3.88 grams of krypton.
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That's the mass that corresponds
to that number of moles.
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What if we wanted atoms?
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Well, in this case, we actually
don't need the periodic table,
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but we do the same type of process.
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We start with the known number of moles
of krypton, we look for conversion factor
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that converts from what we have,
moles of krypton, to what we want,
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which is krypton atoms, and if it's a mole it
contains 6.022 times 10 to the 23rd particles.
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And in this case, the particles
are krypton atoms.
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So, we do our math, .0463 times 6.022 times
10 to the 23rd, do our unit cancellation
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and we get 2.79 times 10
to the 22nd krypton atoms.
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So, notice, we have two ways of looking at
the amounts, number of moles of krypton,
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how many grams it is or how many atoms it is.
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Let's do one the other direction,
2.465 grams of copper.
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How many moles of copper is that?
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Well, we start with what we know.
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We look up the atomic mass
which we've used previously.
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This time we're cancelling grams.
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So, the mass goes on the bottom and the moles
goes on the top of our conversion factor.
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And 2.465 divided by 63.55 gives us .03879
moles of copper that that mass is equivalent to.
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Now what if instead we were trying to find
number of copper atoms starting with grams?
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Well, if this was the only problem we had
to convert from grams into number of atoms,
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we'd have to first convert from grams
to moles and then from moles to atoms.
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But we've already done the grams to moles
conversion so let's just continue from there.
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We have .083879 moles of copper.
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Our conversion factor is one mole of copper,
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contained 6.022 times 10
to the 23rd copper atoms.
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Cancel units, do the math, and we get
2.336 times 10 to the 22nd copper atoms.
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Now note this mole concept, this
mole idea is very useful in lots
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and lots of chemistry situations.
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It's going to help us determine the formula
compound and determining exact amounts
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of substances that react or are produced during
a chemical reaction and lots and lots more.
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So, you have to get to where you're pretty able
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to do certainly the basic
calculations and conversions readily.
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And that's it for an introduction
to chemical moles.