Welcome to another Tutor Short provided by the Educational Support Services
Department
of the Lehigh Carbon Community College in Schnecksville, Pennsylvania which is just
outside of Allentown.
These videos review Key Learnings for the science courses provided here at
LCCC.
Please remember that the
Educational Support Services Department does provide
walk-in tutoring five days a week.
Let's look at balancing chemical equations when you have a combustion
reaction.
Combustion reactions all look the same.
You'll have
a complex Organic
molecule on the left
and there will always be a reaction with
Oxygen (it's diatomic molecule O2 -
Oxygen gas)
and the products of combustion are CO2
and H2O
(Carbon Dioxide and Water)
For our first example in balancing these types of equations,
let's take Methane (which is CH4).
If you start with the hints which are down here,
Hint 1
says to leave your diatomic molecules for last
so you leave your Oxygen for last (that might
help you in balancing the equation).
Hint number 3 says to start with the most complex molecule, which will be
the Organic molecule in this case (Methane).
So those two
Hints
are a good way to start with combustion equations.
Let's start here. We have 1
Carbon molecule here, then
look to the other side and we have 1 Carbon also so that looks balanced.
Then we have 4 Hydrogen in this molecule
so we go over to Water
and, in order to have 4 Hydrogen for Water,
we need to add a 2 in front of the Water molecule.
Now all that is left is Oxygen.
On this side, Carbon Dioxide
has 2
Oxygen atoms.
With Water,
each has 1 Oxygen atom but we have 2 Waters
so that's 2 Oxygen atoms.
Add those together,
You have 4 Oxygen atoms on the right.
To get 4 Oxygen atoms on the left,
we would put a 2 in front of the
diatomic Oxygen molecule here
and now you're balanced.
If we move on to a more complex
molecule......
Let's try to combust Butane.
Butane is
C4H10.
Let me
write that better...
So again,
using the same Hints, we're going to start with Butane.
We have
4 Carbon atoms on the left
so, to get 4 Carbons on the right,
we will put a 4 in front of the Carbon Dioxide.
Now we look at Hydrogen.
We have 10 Hydrogens on the left
so the Water molecule (in order to get 10
Hydrogen)
needs to
have a 5 in front of the Water molecule.
So Hydrogen is done.
But if you look at the Oxygen now,
with Carbon Dioxide molecules each one has 2 Oxygen
so we have
8 Oxygen with Carbon Dioxide
and then in the Water molecule,
we have 5 Water molecules that gives us 5 Oxygen molecules.
In total there will be 13.
We should recognize immediately that 13 is an odd number that is going to give us a little
trouble.
We come back left
to try to get to 13 Oxygen molecules. We have a diatomic molecule.
A simple way to look at this
is to
put 13 over 2 (a fraction).
In order to get 13
Oxygen molecules,
we have to multiplied O2 by 13 over 2.
Now this is
really saying that we need 6 and a half
diatomic Oxygen
molecules
in order to get 13 Oxygens.
Now we
should be aware that our
coefficients on all our molecules have to be whole numbers in a balanced equation.
So I'll erase
all this......
If we bring in Hint
number 4 from down here......
Hint number 4 - whenever we end up with partial
atoms or molecules, we need to
adjust the balance.
Look at the entire equation and use a common
factor of multiplication.
In this case,
if we look at the whole equation,
if we multiply it by 2,
we will cancel out
the division of 2 here
and will give us 13 Oxygen molecules.
You'll see that this will end up helping us balance this equation.
So now if we multiply the whole equation by 2,
in the new equation we have 2 C4H10
(2 Butane
molecules), you're going to have 13
Oxygen molecules.
2 times 4 or 8
Carbon Dioxide molecules and now 10
Water molecules.
I'll switch to another color here for you....
Maybe now we will be balanced.
So let's start again, for Carbon on this side we have 8 Carbons
(2 times the C4)
and on the right
we have 8 Carbons so the Carbons balance.
The Hydrogen has 2 molecules with each having 10 Hydrogen atoms
so we have 20 Hydrogens over here.
We go on the right
and 10 Water molecules times 2 atoms each
for 20 Hydrogens so that balances.
Now we look at the Oxygen,
we have 8
Carbon Dioxides each one has 2 Oxygens for 16 Oxygens in Carbon Dioxide.
For the Water,
ten times a single Oxygen in each
for 10 Oxygen and we have a total
of 26 Oxygens.
Then we go back to the left now.
13 molecules of O2 for 26 Oxygens.
Now we're balanced for Oxygen and
for the entire equation.
So in combustion equations you'll
commonly find that you'll need this multiplication factor from
Hint number 4 here in the Hints
in order to get a
balanced equation. Thank You.