Balancing Chemical Equations - Chemistry, Class 11 Video Lecture

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.
I have four carbons here.
How much hydrogen do I have?
I have two times six.
I have 12 hydrogens here.
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.