>> narrator: In this video,
we'll take a Closer Look
at Balancing Equations.
First, let's review
the basic set-up
for a chemical equation,
where the reactants are
on the left hand side
of the arrow
and the products are
on the right hand side
of the arrow.
The reactants are
the substances
that come together, collide,
and then after they
form new substances,
we call them products.
Here are the steps to balance
a chemical equation.
First, write out
the unbalanced equation.
Then, count the number
of atoms of each element
on each side of the equation.
To balance the equation,
to make the number of atoms
equal on both sides
of the equation,
you can add coefficients.
If you add a coefficient
in front of a molecule
or a compound, you must
multiply that coefficient
times the subscripts
in order to determine
the number of atoms present.
For example, placing
a coefficient of 2 in front
of the formula CaCl2
would mean that there would be
2 calciums and 4 chlorides.
Then, once you think
the equation is balanced,
check to see if all of the
coefficients can be reduced.
If they can,
reduce the coefficients
to the lowest possible ratio.
Finally, double check
that the number of atoms
is the same on both sides
of the equation,
that's when you know
it's balanced.
Here are some helpful
tips to use when you
balance equations.
Balance one element at a time.
Update the atom count
any time a coefficient
is changed.
If an element appears
more than once per side,
balance that element last.
And balance polyatomic ions
as single units.
For example, treat
the polyatomic ion sulfate
as 1 sulfate instead
of trying to balance it
as 1 sulfur and 4 oxygens.
Here's an example
of an equation that needs
to be balanced:
Aluminum and copper II chloride
react to form
copper and aluminum chloride.
The reason there are
two chlorides in the formula
for copper II chloride
is the copper ion
has a charge of 2+.
Since chloride's charge is 1-,
we will need two chlorides
to balance the charges
and make this a stable
ionic compound.
Similarly, the reason there
are 3 chlorides in the formula
for aluminum chloride
is the aluminum ion
has a charge of 3+,
so there will need to be
3 chlorides, each with
a charge of 1-,
in the formula to balance
the charges and make this
a stable ionic compound.
Something to keep in mind
is that although
this particular reaction
is a single replacement,
we could use these steps
to balance an equation for
any of the reaction types
we have previously discussed.
Going back to our problem.
When you look at the number
of atoms on both sides
of the equation, you see
that there are 3 chlorides
on the product side
and 2 chlorides
on the reactant side.
That's not balanced.
The lowest common multiple
of 3 and 2 is 6.
If you put a 3 in front
of copper II chloride
and a 2 in front
of aluminum chloride,
you have 6 chlorides
on both sides of the equation,
so the chlorides
would be balanced.
Now you can balance
the aluminum by putting
a 2 in front of the aluminum
on the reactant side,
and you can balance the copper
by putting a 3 in front of
the copper on the product side.
Remember to go back
and double-check to make sure
you have the same number
of each element present
on both sides of the equation.
There are 2 aluminums
on the reactant side
and 2 aluminums
on the product side.
There are 3 coppers
on the reactant side
and 3 coppers
on the product side.
And we have 6 chlorides
on the reactant side
and 6 chlorides
on the product side.
The equation has been balanced.
Here's another equation
that needs to be balanced.
In this reaction, there are
polyatomics that appear
on both sides of the equation,
so let's review
the helpful hint we discussed
earlier for dealing with
these types of equations.
You have 2 nitrates
on the product side
and 1 nitrate
on the reactant side.
Balance those nitrates
as complete polyatomic ions,
don't try to break them up
into individual nitrogen
and oxygen atoms.
With 2 nitrates
on the product side,
you need to put a 2
in front of silver nitrate
on the reactant side
in order to get 2 nitrates
on the reactant side.
And that will balance
the nitrates.
Now there are 2 silver
on the reactant side,
so you will need to put a 2
in front of silver chloride
on the product side.
And that will give you
2 silvers there as well.
That also happens
to balance the chlorides,
because there are 2 chlorides
present in the calcium chloride
on the reactant side,
and by placing a 2 in front
of the silver chloride,
that gives us 2 chlorides
on the products side as well,
which will balance
the chlorides in the reaction.
Let's double check the work.
There are 2 silvers
on the reactant side,
and 2 silvers
on the product side.
2 nitrates on
the reactant side,
and 2 nitrates
on the product side.
1 calcium on the reactant side,
and 1 calcium on
the product side.
2 chlorides on
the reactant side,
and 2 chlorides on
the product side.
The equation is now balanced.
Here's one for you to try.
Balance this equation,
which shows hydrogen peroxide
decomposing to form water
and oxygen gas.
When you're ready,
press pause, try to balance
the equation,
and then press play
and check your answer.
Hopefully your balanced
equation looks like this...
two moles of hydrogen peroxide
decomposing to form
two moles of water
and one mole of oxygen.
Here's how it was done...
The first thing you may
have noticed was that
there were 3 oxygens total
on the product side,
2 from the oxygen
and 1 from the water.
Since there were only
2 oxygens to begin with
on the reactant side
in the 1 hydrogen peroxide,
we needed more oxygen
on the reactant side.
So by placing a 2 in front
of the hydrogen peroxide,
and since you multiply
any coefficient by
the subscripts in the formula,
that gives you 4 oxygens
on the reactant side.
The 2 also distributes to
the 2 subscript for hydrogen,
so we have 4 hydrogens as well
on the reactant side.
For the hydrogen
on the product side,
put a 2 in front of water,
and that 2 gets multiplied
by the subscript 2
for the hydrogen in the water,
so that will give us
4 hydrogen.
The 2 oxygen from
the 2 water molecules
added to the 2 oxygen atoms
from the 1 oxygen molecule
add up to be 4 oxygen total.
So you now have 4 hydrogen
on the reactant side
and 4 hydrogen
on the product side.
You also have 4 oxygen
on the reactant side
and then 2 plus 2 is 4 oxygen
on the product side.
Remember, double-checking
to make sure that you have
the same number of each element
present on both sides
of the equation
is a very important step.
And the equation is balanced.
If you got it right,
good work!
That's a Closer Look
at balancing equations.
If you'd like more
Chemistry Matters,
go to our home page,
where the entire series
is available to you.