### Video Transcript

In this video, we will learn how
to balance chemical equations. And we will discover some shortcuts
that we can use to balance multiatom groups.

Let’s start by looking at what we
mean by the word “balanced” when we’re talking about reaction equations. Here we have two reaction
equations. Both of these equations have N2 and
H2 on the reactant side. And they have NH3, ammonia, on the
right-hand side. The difference between these two
reaction equations is that the one on the left is not balanced, while the one on the
right is balanced.

A balanced equation follows the law
of conservation of mass. The law of conservation of mass
says that, in an isolated system, mass is neither created nor destroyed by a
chemical reaction. What this means is that the mass of
our reactants must exactly equal the mass of our products. But when writing reaction
equations, we don’t really talk about grams, which is what we would usually
associate with mass. So how does this apply?

When writing a balanced reaction
equation, we need to have the exact same number of atoms of each element on the
reactant side as we do on the product side. When we look back at our unbalanced
reaction equation, we can see that we have two nitrogen atoms on the left, but only
one nitrogen atom on the right. Likewise, we have two hydrogen
atoms on the left, but three hydrogen atoms on the right. So this is why our first equation
here is unbalanced.

In the balanced example, we can see
that we’ve added coefficients in front of one of the reactants and the product. This means that we now have two
nitrogen atoms on both sides and six hydrogen atoms on both sides. So this is why our second equation
is balanced. So now that we understand what a
balanced equation is and why it’s important, let’s have a look at how we go about
balancing an equation.

The process of balancing a reaction
equation can sometimes be a case of trial and error. Luckily, with practice, this will
become a faster and more efficient process. But we’ll go through some key steps
to get us started.

We’ll use the reaction of sodium
metal with water as an example to go through the process. It can be useful to begin with the
key reactant. This could be the most complex
reactant or the one we’re most interested in. In this example, let’s begin with
sodium. We can set up a little table, a bit
like this, and count the number of sodium atoms in each of our reactants and then
again for each of our products. In this case, we have one sodium
atom on our reactants and one sodium atom on our products. So the sodium atoms are balanced
for now.

Let’s move on to the next
element. If we look at hydrogen, we have two
hydrogen atoms in our water. So that’s a total of two hydrogen
atoms in the reactants. But we need to be careful when
looking at our products since two of our products contain hydrogen atoms. There is one hydrogen atom in our
sodium hydroxide and two hydrogen atoms in our hydrogen gas. This gives us a total of three on
the product side versus only two on the reactant side. So this is clearly not balanced
yet.

Now, let’s go through and do our
oxygen atoms. We can see that we have one oxygen
atom in our reactants and one on our products. So the oxygen atoms are balanced
for now. The next step to create a balanced
equation is to add coefficients in front of some of the reactants and all products
in order to change the number of the hydrogen atoms. Be mindful though that if you add a
coefficient in front of water, for example, you’re going to change the number of
hydrogen atoms but also change the number of oxygen atoms. So after adjusting any
coefficients, you should always go back and recheck the number of all of your
atoms.

The only unbalanced element in this
equation at the moment is hydrogen. We have one extra hydrogen atom in
our products than we do in our reactants. In an ideal world, we would place
coefficients so that we only alter the number of hydrogen atoms, though of course
this may not always be possible.

Our last product is the only
molecule in our reaction which only contains hydrogen. But can we add a coefficient to
this H2 to make our reaction balanced? Technically, we could add a
coefficient of one-half. This would reduce the number of
hydrogen atoms on the product side and, therefore, leave us with two hydrogens, both
sides, making it balanced. At the same time, it doesn’t affect
the numbers of any of the other elements.

While this is technically a
balanced equation, it’s always nicer if we can have whole-number coefficients
wherever possible. A really easy way to do this is to
simply multiply the entire reaction by two. This results in a coefficient of
two being placed in front of sodium, water, and sodium hydroxide. And the half coefficient in front
of our hydrogen has disappeared. When we recheck the number of each
element on both sides of our equation, we can see that we are still balanced.

Now that we know the basics of how
to balance an equation, let’s see if there are any tricks or tips for making this
process easier. It turns out that there is a little
shortcut for reaction equations which involve polyatomic ions. Remember that polyatomic ions are
made up of more than one atom. Examples of polyatomic ions are
nitrate, NO3−; sulfate, SO42−; and ammonium, NH4+.

We can modify our method for
balancing an equation when it involves polyatomic ions. We are able to treat polyatomic
ions as a single unit and balance them as such. But note that this only works if
the polyatomic ion is present both in the products and the reactants. To understand this more clearly,
let’s have a go at an example. Let’s have a look at this example
reaction.

Aqueous solutions of Pb(NO3)2 and
NaCl are mixed, producing aqueous NaNO3 and a PbCl2 precipitate. Write a balanced equation with
state symbols.

Let’s begin writing our balanced
equation by simply putting the reactants and the products down. We have lead nitrate and sodium
chloride as our reactants. And we have sodium nitrate and lead
chloride as our products. We can already add our state
symbols as well. We are told that both of our
reactants are aqueous, as well as our sodium nitrate, but our lead chloride is a
precipitate, making it a solid.

Now we just need to balance our
equation. So let’s set up our table for
counting the atoms. Our most complex reactant is, of
course, our lead nitrate. So let’s begin there. We have one lead atom on the left
and one lead atom on the right. So for now, our lead atoms are
balanced.

Next, our shortcut can come in
handy. When we look closely, we can see
that we have the nitrate ion on both sides of our equation. So rather than counting this
nitrate ion as nitrogen atoms and oxygen atoms, we can simply count it as a single
unit. When we do this, we see that we
have two nitrate ions in our reactant and only one in our product. So this is not yet balanced.

Next, let’s finish counting the
rest of our elements. We can see that our sodium atoms
are balanced, but our chlorine atoms are not. So here we have two elements which
need to become balanced. Let’s begin by balancing our
nitrate ions. By adding two as a coefficient in
front of our sodium nitrate, we have balanced our nitrate ions. However, don’t forget this also
changes the number of sodium atoms.

Now we need to rebalance our sodium
atoms. We have two on the products and
only one on the reactants. So let’s add a coefficient of two
in front of our sodium chloride reactant. Remember that this also alters the
number of chlorine atoms. So now our sodium atoms are
balanced, let’s move on to our chlorines.

When we come to recheck our
chlorines, we can see that actually we’ve already balanced them. And a quick recheck of all of our
numbers shows that this is now our final balanced equation with state symbols.

We now have a process by which we
can try to balance equations and a handy little shortcut for when there are
polyatomic ions involved. So let’s use this on some
questions.

Gallium(III) bromide can be
produced by the reaction of gallium with bromine. Write a balanced molecular equation
for this reaction including state symbols.

This question is asking us to write
a chemical equation from the information about a reaction given. More importantly, it’s asking for a
balanced molecular equation. When an equation is balanced, it
obeys the law of conservation of mass. This means in practice that we need
to have the same number of atoms of each element on both the reactant side and the
product side.

Let’s begin writing our equation by
extracting the reactants and the products from the information given. We are told that this reaction
produces gallium(III) bromide. So this must be our product. Now we just need the reactants. We are told in the question that
these are gallium and bromine. The chemical symbol for gallium is
Ga. It can be found on the periodic
table just below aluminum. The symbol for bromine is Br. However, don’t forget that bromine
exists as a diatomic. So this is really Br2.

Now we just need to work out the
formula for gallium(III) bromide. Remember that bromine forms a
negative anion. This Br− anion is called
bromide. We are told that we have
gallium(III) in our gallium bromide. So this means that we have
gallium(III) plus as a cation. The formula for gallium bromide
can’t be GaBr. This is because the charges don’t
balance. We need two more negative charges
from the bromide to balance out the three plus charge on the gallium. We can do this by adding two extra
bromide anions. So our formula for gallium(III)
bromide is GaBr3.

Next, we need to balance our
equation. To balance our equation, let’s
start with the main reactant. In this case, let’s go with
gallium. We have one gallium on the
reactants and one on the products. So for now, the gallium atoms are
balanced. Next, we come to bromine. We can see that we have two bromine
atoms on the left but three on the right. So this is not yet balanced.

We need to alter our coefficients
to balance our bromines. Ideally, what we want is one more
bromine atom on our reactant side. So we could place a coefficient of
three over two in front of our Br2. This would give us three bromine
atoms on both sides of our reaction, making it balanced.

However, it’s not always nice to
have fractional coefficients. So let’s get rid of our fraction by
multiplying the whole reaction by two. This simply doubles the number of
all of the atoms on both sides of our equation. So it’s still balanced. But we can always recheck just to
be sure.

Finally, we need to add our state
symbols. State symbols tell us whether our
reactants or products are a solid, liquid, gas, or an aqueous solution. Bromine is, of course, a liquid at
room temperature. Gallium is technically a solid at
room temperature. But for this reaction, it’s going
to need to be heated. Interestingly, the melting point of
gallium is only slightly above room temperature. So it’s a metal that can melt in
your hand. So we’ll label gallium as a
liquid. Gallium bromide is a white powder,
so we can label that as a solid. So here is our balanced molecular
equation.

Let’s summarize our key points. Balanced reaction equations obey
the law of conservation of mass. This means that a balanced reaction
will have the same number of atoms of each element on the reactant side as it does
on the product side. The process of balancing equations
is sometimes a trial-and-error approach. But with practice, you’ll get
better at it. The general method is to start with
your main reactant, adjust the coefficients, and recheck, repeating this process
until your equation is balanced.

We’ve also seen that there is a
shortcut you can use when dealing with polyatomic ions. We can balance polyatomic ions like
nitrate and sulfate as a single unit, making life much simpler. Remember though that this only
works if the polyatomic ion is present on both sides of our reaction equation.