Like the concept of the mole, balancing equations is one of
those ideas that you learn in first-year chemistry class.
It tends to give a lot of students a hard time, even
though it is a fairly straightforward concept.
I think what makes it difficult is that there's a
bit of an art to it.
So before we talk about balancing chemical equations,
what is a chemical equation?
Well here's some examples right here, and I have some
more in the rest of this video.
But it essentially just describes a chemical reaction.
You've got some aluminum.
You have some oxygen gas or a diatomic oxygen molecule.
And then you end up with aluminum oxide.
And you would say, ok, fine, that's an equation.
It looks nice.
I have my reactants, or the things that react.
These are the reactants.
And then I have the products of this reaction.
What's left there to do?
Well you have a problem here.
The way I've written it right now, I have one atom of
aluminum plus two atoms of oxygen, right?
They're bonded to each other, but there's two atoms of
oxygen here.
One molecule of diatomic oxygen, or one molecule of
oxygen, but they have two oxygen atoms here.
And when you add them together, I have two atoms of
aluminum and three atoms of oxygen.
So I have a different number of aluminums on both sides of
this equation.
On this side, I have one aluminum.
On this side, I have two aluminums. And then I have a
different number of oxygens.
On this side, I have two oxygens.
And on that side, I have three oxygens.
So balancing equations is all about fixing that problem, so
that I have the same number of aluminum on both sides of the
equation, and the same number of oxygens on both sides.
So let's try to do that.
I'll do it in orange.
So I said I have one aluminum here and I have
two aluminums there.
So maybe a simple thing is just to put a two out here.
So now I have two aluminums on this side and I have two
aluminums on this side.
The aluminums look happy.
Now let's look at the oxygen.
Here I have two oxygens on the
left-hand side of the equation.
And on the right-hand side of the equation
I have three oxygens.
What can I do here?
Well if I could kind of have half atoms, I could just
multiply this by one and a half.
1.5.
1.5 times 2 is 3.
So now I have three oxygens on both sides of this equation
and I have two aluminums on both sides of the equation.
Am I done?
Well, no, you can't have half an atom, or one and
a half of an atom.
That's not cool.
So what you do is you just multiply this so that you end
up with whole numbers.
So let's just multiply both sides of this
whole equation by two.
And we have four aluminums plus three oxygens yields-- we
multiplied everything by two-- two
molecules of aluminum oxide.
Now, you might have been tempted at some point in this
exercise to say, oh, well why don't I just tweak part of the
aluminum oxide?
Why don't I put a 2/3 in front of this oxygen.
You cannot do that.
The equation is as it is.
The molecule aluminum oxide is aluminum oxide.
You can't change the relative ratios of the aluminum and the
oxygen within the aluminum oxide molecule.
You can just change the number of aluminum oxide molecules as
a whole that you have, in this case, two.
So what did we do?
We looked at the aluminum.
We said, OK, we need two aluminums to have both sides
of that be two.
And then when we looked at oxygen, we said, well, if I
multiply this by one and a half, then that becomes three
oxygens here, because one and a half times two oxygens.
And three oxygens there.
And then all we said is, oh, I can't have a one half there,
so let me multiply both sides by two.
And I ended up with four aluminums plus three oxygen
molecules, or six oxygen atoms, yields two molecules of
aluminum oxide.
Let's see if we can do some more of these.
So here I have methane.
And that g in parentheses, I just wanted to
expose you to that.
That just means it's a gas.
So I have methane gas plus oxygen gas yields carbon
dioxide gas plus water.
That's an l there, so liquid water.
So what can we do here?
So the general thing is, do the complicated molecules
first, and then at the end you can worry about the single
atom molecules.
Because those are very easy to play with.
And the reason why you do that, whenever you change a
number here-- let's say we're trying to engineer how many
carbons we have on both sides of this equation-- if I set a
number here, I'm also changing the number of hydrogens.
Then I'll have to play with the hydrogens there.
And at the end, I'll have some number of oxygens.
And then I have to tweak this number right there.
So let's just start with the carbons.
It seems complicated, but when you go step by step and you
kind of play with things a little bit, it should proceed
fairly smoothly.
So here I have two carbons.
And here I have, on the right-hand
side, only one carbon.
So ideally I'd want two carbons on both
sides of this equation.
So let me put a two out here.
So there you go, my carbons are happy.
I now have two carbons and two carbons.
Now let me move to the hydrogens.
Remember, I wanted to do the oxygens last, because I can
just set this to whatever I want it to be without messing
up any of the other atoms. So on this side of the equation I
have four hydrogen atoms. How many hydrogen atoms do I have
on this side of the equation?
So I have four hydrogen atoms here.
How many do I have on this side?
Well, I have two hydrogen atoms right there.
So I want to have four on both sides, so let me put a two in
front of the water.
So now I have four hydrogen atoms. Cool.
Finally, oxygen.
On the left-hand side I have two oxygen atoms. And on the
right-hand side, what do I have?
I have two times this one oxygen, right here, so right
now that's two, right?
Two waters.
And in each water I have one oxygen atom, but I have two
water molecules, so I have two oxygens.
And then in the carbon dioxide I have two oxygens in each
carbon dioxide.
And I have two carbon dioxides, right?
When I put that magenta two out front.
So how many oxygens do I have?
Two times two.
I have four oxygens.
So on the left-hand side I have two oxygens.
On the right-hand side I have six oxygens, two in the two
molecules of water and four in the two
molecules of carbon dioxide.
So how do I make this two into six?
I want to have six oxygens on this side.
I want to them to have six oxygens on the
left-hand side as well.
Well, if I put a three out here, now I have six oxygens.
And our equation has been balanced.
I have two carbons on this side, two carbons on that
side, four hydrogens on this side, four hydrogens on this
side, six oxygens on this side, and then-- four plus
two-- six oxygens on that side.
Next equation to balance.
This actually becomes quite fun once you get
the knack of it.
So I have ethane plus oxygen gas yielding
carbon dioxide and water.
So this is a combustion process.
Let's look at the carbons first. I
have two carbons here.
I have one carbon there.
So let me put a two here.
So now I have two carbons.
Fair enough.
Remember, I'm going to worry about the oxygen last. This
one's actually not too different
than the last problem.
I have six hydrogens here.
I only have two hydrogens in the water.
So then we have three water molecules.
And now I've balanced out the hydrogens.
I have six hydrogens on both sides of the equation.
And now let's deal with the water.
Here on the right-hand side-- I'll do it in this orange
color-- I have two oxygens in each carbon dioxide molecule.
And I have two molecules, so I have four oxygens.
And I have one oxygen in each water molecule.
And I have three molecules, so I have three oxygens here.
Is that right?
Three oxygens on the right-hand side.
Yep, three water molecules.
And then I have four oxygens in the carbon dioxide.
Right.
So I have seven oxygens.
I have seven oxygens on this side and I only
have two on this side.
So how can I make this two into seven.
Well I could multiply it by three and a half.
Right?
Remember, I just want to have seven on both sides.
If I have three and a half of these diatomic oxygen
molecules-- three and a half times two is seven-- so now I
have seven oxygens on both sides of the equation.
Four plus three and seven here.
Two carbons.
And then six hydrogens.
And I'm almost done, except for the fact that you can't
have fractions of molecules.
So what you do is just multiply both sides of this
equation by two or three, or whatever you need to multiply
it to get rid of the fractions.
So if I multiply everything by two, I end up with two
molecules of ethane plus seven molecules of diatomic oxygen,
seven O2's, yielding two molecules of carbon dioxide--
oh, sorry, I'm multiplying everything by two, so four
molecules of carbon dioxide, plus six molecules of water.
And just to make sure that all still works, if you want to,
you can check.
How much carbon do we have?
I have four carbons here.
How much hydrogen do I have?
I have two times six.
I have 12 hydrogens here.
How much oxygen do I have?
Let me do a different color.
I have 14 oxygens here.
And here I have eight oxygens.
And here I have-- six times one-- six oxygens.
So six plus eight is 14.
So my equation has been balanced.
So that one, a lot of people might find that to be a hard
problem, because you have three and a half, and just
going straight to this might seem non-intuitive.
But if you just work from the more complicated molecules and
you go atom by atom, and if you end up with any fractions
you just multiply both sides by some number to get rid of
the fraction, of and then you're done.
All right.
Let's do another one.
So this one looks all hairy.
I have this iron oxide plus sulfuric acid yields all this
hairy stuff.
But the key here to realize is that this group right here,
this sulfate group, stays together.
Right?
You have this SO4 there and you have this SO4 there.
So to really simplify things for our head, you can kind of
treat that like an atom.
So let's make a substitution.
We'll substitute that for x.
So if we rewrite this as-- I'll do it in a vibrant
color-- iron oxide plus H2-- there's only one sulfate
group here-- x.
And then that yields two irons, this
molecule with two irons.
And it has three of these sulfate groups.
And then plus water.
All I did is replace the sulfate with an x.
And now we can treat that x like an atom, and we can just
balance the equation.
Let's see, on the left-hand side, how
many irons do we have?
We have two irons here.
We have two irons there.
So the irons look balanced, at first glance.
Let's deal with the oxygens.
So if we have three oxygens here-- let me do it in a
different color-- and we only have one oxygen here.
Remember, there were some oxygens in this x group, but
the sulfate group stayed together, so we can just treat
those separately.
We want to have three oxygens on the right-hand side, as
well, so let's stick a three here.
Oops.
So let me put that three there.
So now we have three oxygens, as well.
And then finally, let's look at the hydrogens.
Let's see, how many hydrogens do we have here.
We have six hydrogens now, on the right-hand side.
Six hydrogens here.
Three times two.
And we want to have six hydrogens here, so we have to
have three of these molecules.
And then finally, let's look at the sulfate group, that x.
We have three x's here.
And lucky for us, we have three x's over there.
So our equation has been balanced.
And if we want to write it back in terms of the sulfate
terms, we can just un-substitute the x, and we're
left with one molecule of iron oxide plus three molecules of
sulfuric acid, H2SO4 -- I just un-substituted the x with SO4
-- yields one of these molecules, SO4 3.
I just un-substituted the x plus three molecules of water.
Let's do one more.
And then I think we'll be all balanced out.
Carbon dioxide plus hydrogen gas yields methane plus water.
So let's deal with the carbon first. I have one carbon here,
one carbon there.
The carbons look happy.
Let's look at the oxygen.
I have two oxygens here.
I have one oxygen there.
So I want two oxygens here so let me stick a two there.
And then let's deal with the hydrogens last, because this
is the easiest one to play with because it doesn't affect
any of the other atoms. So if I have two here and I have
four, plus four here, right?
I have four hydrogens there.
And I have four hydrogens there.
So I need eight hydrogens on the left-hand side.
So I just put a four there.
We're all done balancing.
Anyway, hopefully you found that useful.