[narrator] Molar mass and Avogadro’s number are nothing more than conversion factors.
What we need to do is take one and set it up, for example nitrogen gas.
28.02 grams per mole and we want to write out the substance that we’re talking about
so we don’t get confused later.
This just means 1 mole of nitrogen gas molecules weigh 28.02 grams.
We can also say, for molar mass, 1 mole of nitrogen gas per 28.02 grams of N2.
We’ll see how to tell when to use which, coming up.
For Avogadro’s number we just take Avogadro’s number, 6.022 x 1023, and it is a something.
Whatever it happens to be; items, particles, what not, we’ll say particles because it
really could be anything per 1 mole of those things.
We could even simplify it and just say A, just call it whatever it is.
It could be an atom, it could be a rabbit, it could be anything.
The other way we could write it is 1 mole of A per Avogadro’s number.
Per number of A.
We can write it either way.
This is just the conversion factor.
Because it’s conversion factor we can use dimensional analysis
which just states that we recognize what units we’re given and what we want at the end.
What we’re given and what we want to pick the right conversion factor.
The way we do that is we say that we have our given unit, we have some number in our
given unit.
We always want our conversion factor to have the given unit plus some number on bottom.
We want a new unit on top.
We keep going as long as we need to get to our desired unit.
Forgive my handwriting.
The reason we do that is so we can cancel out top and bottom, note how this is all--we
could say given unit over 1 instead of just one big, long fraction here where we multiply
in the middle.
Some people like to do that.
Then they eventually get done so all these are just sequential fractions and we can divide
out like units and then divide out numbers that we get at the end.
We pick our conversion factor that we use based on what we’re given.
Let’s do a couple of examples.
What are we given?
How many nitrogen atoms are in 5.00 grams of nitrogen gas?
We are given 5.00 grams of N2.
We’ve got to know that nitrogen is a diatomic molecule.
Then we’ve got to know that we’ve got to get rid of grams of nitrogen gas, so we
put that on the bottom.
What do we have in our bag of tricks that gets us--that gives us grams of nitrogen gas?
Well, it’s the molar mass.
28.02 grams per 1 mole of N2.
The molar mass is our first conversion factor and then now that we’ve gotten rid of grams
of nitrogen gas we want to get rid of moles of nitrogen gas, so we say I want moles on
the bottom.
What gives us moles?
Well, what are we going for?
We want to end up with a number of nitrogen atoms.
We don’t want to end up with grams we want a number so we have to use Avogadro’s number.
But, it doesn’t ask us for molecules of nitrogen it asks us for nitrogen atoms and
so we have to do one final conversion factor that we’re not given.
Two nitrogens per one N2 makes sense.
It’s just a common sense so we can get rid of nitrogen, we can get rid of moles of N2
by cancelling what’s on the bottom so what we’ve got to do now is what’s leftover,
and I’ll come down here just because we have room.
When we multiply everything out and divide everything out we end up with a number, divide
our numbers to get the number, to three significant figures because we’re limited to three by
this number, the given number, 2.15 x 1023 N atoms.
The only unit we’re left with is nitrogen atoms
and the number is found by dividing out and multiplying by all the numbers in our fractions.
About the next example, how many grams of carbon are in 5.00 x 1024 molecules of methane?
If we have our given unit, methane, we know from our nomenclature, CH4, 5.00 x 1024 methane
molecules and we want to know how many grams of carbon?
Well, we want carbon so how many carbons per methane?
One C atom per CH4 molecule.
We can cancel out CH4 molecules and what do you know, it’s the same thing.
We have the same number, 5 x 1024.
That’s not very interesting.
But then to get to grams we have nothing that compares atoms to grams so we have to get
in terms of moles first--what our next unit should be.
It should put C atoms on the bottom.
It should have moles on top.
The only thing we can put, then, is 6.022 x 1023 C atoms.
We’d use Avogadro’s number.
Then we want to get rid of atoms and get grams.
Now we want to use the molar mass--excuse me, we’ve gotten rid of atoms and we want
to get rid of moles.
What’s on top must go on bottom in the next conversion factor.
We know 12.01 grams of carbon for 1 mole of carbon and when we multiply everything out
we get 99.7, to three significant figures, grams of carbon atoms.
Because, after the final cancellation moles cancels and we’re left with grams of carbon
and all the numbers divide out to 99.7.