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