Our problem tells us that when 1.125 grams of a liquid
hydrocarbon, CxHy-- so we don't know how many C's and
H's there are, so it's a mystery liquid hydrocarbon.
It was burned in an apparatus like that shown in figure
4-7-- this is the apparatus, this is our mystery
hydrocarbon getting burned, or it's combusting with oxygen--
you need oxygen for the combustion to occur.
So when it burns in this apparatus, and it burns with
oxygen, it tells us -- and we have 1.125 of it to start
with-- it tells us that 3.447 grams of carbon dioxide and
1.647 grams of water were produced.
So let me write that reaction right there.
So let me just write this first sentence right here.
Write all of the information that they're giving us in that
first sentence.
They're saying that when we start off with that liquid
hydrocarbon that has x C's and y H's in it, and it's in its
liquid state, they say it's a liquid hydrocarbon.
And when we combust it using oxygen in its gaseous state,
we will produce carbon dioxide and water.
And you can't even really realistically balance this
equation because you don't know how many carbons and
hydrogens you have on the left-hand side.
But what they do tell us is how many grams of everything,
of at least this we start with, and how many grams of
each of these we end up with, and we're assuming that we
have an excess of oxygen.
That this hydrocarbon is the limiting reactant.
We just have an abundance of oxygen.
Just enough oxygen to turn all of this hydrocarbon, to
combust all of it into carbon dioxide and water.
And they tell us that we have 1.125 grams of the mystery
substance, and then it produces 3.447 grams of carbon
dioxide, and 1.647 grams of water.
Now to solve this-- well let me finish reading the problem.
It tells in a separate reaction or in a separate
experiment, the molar mass of the compound was found to be
86.2 grams per mole.
So I'll write that down here.
So CxHy molar mass is equal to-- let me write it this way
instead just to go with the convention that I've been
doing-- we have 86.2 grams of our mystery substance, CxHy
per mole of our mystery substance, CxHy.
And then they're asking us in this problem, they say
determine the empirical and molecular formulas for the
unknown hydrocarbon CxHy.
And just as a refresher, empirical means just the
simplest whole number ratio of the atoms in our molecule.
So the simplest ratio of our carbons to hydrogens, or
hydrogens to carbons, will give us the empirical formula.
The molecular formula is how many carbons, exactly how many
carbons and hydrogens do we have in each atom.
And I suspect we're going to be able to use this
information to come up, once we have the empirical
formulation, the empirical formula, we then can use this
molar mass right here to figure out the actual
molecular formula.
But let's figure out the empirical formula first.
And the way we can do it is we could say, well gee, how many
moles of carbon dioxide we get produced, how
many moles of hydrogen?
And then we could say, well in the final product, how many
moles of carbon do we have then?
And how many moles of hydrogen do we have there?
And then we can figure out the ratio.
And the ratio of carbons to hydrogens or hydrogens to
carbons in the product, that's going to have to be the same
as the ratio in the reactants.
And they all came from this reactant right there.
They didn't come from the water.
And then hopefully we'll be able to figure out the
simplest whole number ratio of hydrogens to carbons or
carbons to hydrogens is.
That sounds very complicated I think, but I think if we take
it step-by-step it won't be too bad.
So a good place to start, let's figure out how many
moles of each of these substances we have. I'll start
off with the carbon dioxide.
So if I start off with the carbon dioxide, what is the
molar mass of carbon dioxide?
So its atomic weight, carbon is 12 plus you have 2 oxygens,
so 2 times-- the atomic weight of oxygen is 16, you could
look that up if you like-- but that's equal to
44 atomic mass units.
That's its atomic weight, which is the weighted average
of all of the isotopes and all of that business.
But if you have a whole mole of carbon dioxide, 6.022 times
10 to the 23rd of these molecules, then you're going
to have 44 grams of the substance.
So we have 44 grams of CO2 per mole of CO2.
And if we want to figure out how many moles we have with
us, we know that we're starting off with
3.447 grams of CO2.
And we want to multiply this by something that has grams of
CO2 in the denominator.
So we want to multiply it by something that looks like
this-- and we want to turn it into moles of CO2.
So it should have moles of CO2 over grams of CO2.
Well that's the inverse of this right here.
This is grams per mole, we want moles per gram.
So instead of 44 grams per mole, we have 1 over 44 moles
per gram, or for every 44 grams we have 1 mole, for
every 1 mole we have 44 grams. I just inverted this.
So let's multiply.
The grams of CO2 cancel out with the grams of CO2, and you
are left with 3.447 divided by 44.
So let's figure out what that is.
So you have 3.447 divided by 44, which is equal to 0.07834.
So 7-8-3-4.
Let me remember that.
So that is 0.07834 moles of CO2-- that's
our only units left.
Now let's figure out how many moles of water we have. So I
could do the exact same thing over-- let's see,
where should I do that.
Let me write it over here.
So let's do the water in orange right here.
So we're starting off with 1.647 grams of H2O.
And what is water's molar mass?
Same argument, its atomic mass is-- each hydrogen is 1.
But you have 2 of them so it's 2 times 1 plus 16, which is
the atomic weight of oxygen.
So 2 times 1 plus 16 is 18.
That's also its molar mass.
You could say 18 grams of H2O per mole of H2O.
Now to convert this, the amount of grams we have to
moles, what we want to do is multiply it by something that
has moles in the numerator and grams in the denominator.
We want grams of H2O in the denominator and moles of H2O
in the numerator.
And that's the exact inverse of this.
So if we have 18 grams per mole, we have 1 over
18 moles per gram.
And what is this going to be equal to?
So this cancels out with that, so we're left with 1.647
divided by 18 moles of H2O.
Let me scroll over a little bit.
Get the calculator out.
Clear it out.
And so we have 1.647 divided by 18, which is 0.0915.
Let me do this in a new color.
We have a 0.0915 moles of H2O.
So so far we figured out in our products how many moles of
carbon dioxide we have, and how many moles of water.
Now, I said we want to figure out the ratio of hydrogens to
carbons or carbons to hydrogens to figure out the
empirical formula right here.
And to do that we can say hey look, all of the carbons and
the hydrogens in our products, they all had to come from our
mystery hydrocarbon.
So if we can figure out the ratio of hydrogens to carbons
in our products, we'll know the ratio of hydrogens to
carbons in our reactants.
Because they all had to come from this guy right here.
The oxygen, obviously, had no hydrogens and carbons in it.
So all of these hydrogens and carbons came
from our mystery substance.
So how many moles of carbon do we have?
How many moles of carbon do we have in the carbon dioxide, in
our products?
Well the only product with carbon is the carbon dioxide.
This has no carbon in it.
So if we have this many carbon dioxides, we also have the
same number of carbon atoms, moles of atomic carbon.
And you might say hey, that's just a number.
This is saying 0.07 times 6.022 times 10 to the 23rd
times that huge number.
A mole is just a number-- very important to remember.
So this means we have the same number of carbons.
And if we have that many waters, how many
hydrogens do we have?
Well for every water molecule we have 2 hydrogens.
So we're going to have twice as many hydrogen atoms.
So let's multiply that times 2.
So times 2 is equal to 0.183.
So we have 0.183.
I could even put a trailing 0, because we actually had-- well
I think I lost the significant digit here.
I could have put a zero right there.
But anyway, I won't worry too much about that.
0.183 moles of atomic hydrogen, because for every
water molecule we have 2 hydrogens.
So I just took this number and multiplied by 2.
And now let's figure out the ratio of hydrogens to carbons,
and that's pretty easy.
We just take a ratio of that to that.
So the ratio of hydrogens to carbons is 0.183 moles of
hydrogen in our product for every 0.07834 moles of the
carbon in our product.
And somehow we have to make this into some type of a whole
number ratio.
Let's see whether we can do anything here.
Let's just divide this right here and see what we get.
So if we just take-- well we have the 0.183 there, so let's
divide it by 0.07834, and you get 2.336.
So this is approximately equal to-- I'll do a new color-- is
approximately equal to 2.336 moles of hydrogen for every 1
mole of carbon.
This doesn't look like that neat of a ratio, but this is
pretty close to 1/3.
So maybe if we multiply the numerator and the denominator
by 3 what do we get?
Well 3 times 2 is 6 and 3 times 1/3 is another 1.
So this will be 7.
7 moles of hydrogen.
I'm approximating it.
I'm trying to get to some reasonable whole number.
7 moles of hydrogen for every 3 moles of carbon.
And this was a little bit of an art to it.
I said OK, 2.33:1 or 2.336:1, well gee, this looks like,
over here, this part of it looks like if we multiplied
that by 3 we're going to get pretty
close to a whole number.
So let me multiply the numerator and the
denominator by 3.
That's kind of the art there to recognize what do you have
to multiply this to turn it into a whole number.
And if you don't believe me, we could take that number and
multiply it by 3 and you get something very, very, very,
very close to 7.
So that's enough for me.
So you're multiplying both by 3, you get 7 moles of hydrogen
for every 3 moles of carbon.
And we've figured out our empirical formula.
We have 7 hydrogens for every 3 carbons.
Right?
If we have 7 moles, for every 3 moles that means we have 7
hydrogens for every 3 carbons.
And so the empirical formula of our mystery hydrocarbon is
for every 3 carbons we have 7 hydrogens.
Now the next thing we want to figure is the
exact molecular formula.
The exact number of-- this is just the minimum ratio.
But maybe we have 9 carbons for every 21 hydrogens.
We don't know.
So let's use the other information they gave us in
the problem where they told us that the molar mass is 86.2
grams per mole-- this information right here.
Let me copy that.
86.2-- that looks like an 80, but we'll
remember it says 86.2.
So let me paste it over here.
That's what they gave us in the problem.
So just for kicks, what would be the molar mass if this was
the molecular formula?
So as it's written right now-- well there's a couple of ways.
We could do it algebraically and all of that.
But the simplest way is to say as it's written right now what
would be the molecular mass of this?
Or the molar mass, I should say.
So it would be 3 times 12, plus 7, times 1.
3 times 12 is 36 plus 7 is equal to 43.
So if this was the molecular formula, its molecular mass
would be 43 grams per mole.
But they tell us in the problem that the molecular
mass is 86 grams per mole.
So they're saying that the mass is twice as much as this.
We must have twice as many atoms in the same ratio.
So the molecular formula for this-- we're in the home
stretch now-- the molecular formula must have twice as
many atoms in the exact same ratio.
So it must be C3H7 times 2, or we could write that as C6H14.
So this is still a ratio of 14:6, which is 7:3.
But using this information, we were able to figure out the
exact number of carbons and the exact number of hydrogens
in this molecule.
Anyway, hopefully you found that fun.