Our second balancing chemical equations video shows you a different method - an algebraic
method. Some people love algebra - it has a power to reduce all sorts of hairy problems
to systems of equations with variables - knowns and unknowns, and if that approach appeals
to you, you should try this method of balancing chemical equations. Let's not forget why we're
doing this in the first place. Every time we write a balanced chemical equation,
we acknowledge the Law of Conservation of Mass. We can thank Antoine Lavoisier for this
fundamental principle. He said: "We may lay it down as an incontestable axiom that, in
all the operations of art and nature, nothing is created; an equal quantity of matter exists
both before and after the experiment." (Except he said it in French.)
The point is, matter is neither created nor destroyed during a chemical reaction. You
can't write some number of oxygen atoms on one side of a chemical equation and poof,
they disappear. That same number of oxygen atoms have to appear on the right-hand side
of the equation as well. You can't have, say, extra hydrogen atoms appearing on the right-hand
side of your equation. That violates the fundamental laws of the universe, and you'll go to Physics
jail. So don't even try it. We're going to do 5 examples of balancing
chemical equations using the algebraic method. We're going to balance the same chemical equations
as in the other video about the inspection method, so you can see both methods, and make
up your mind which method works better for you.
Here's our first example. Hydrogen plus oxygen yields water. Remember hydrogen and oxygen
are both diatomic molecules. H2 + O2 → H2O.
This is an unbalanced chemical equation. To balance it, we'll have to place coefficients
in front of the hydrogen, oxygen, or water molecules. Remember, we can't change any subscripts.
Now, we don't know what those coefficients are, so let's write them as variables, x,
y, and z. xH2 + yO2 → zH2O
We can use the chemical reaction to establish some algebraic relationships. For instance,
how many hydrogen atoms are there? We know the number of hydrogen atoms on the left MUST
equal the number of hydrogen atoms on the right. So
2x = 2z Do you see where those 2s came from? That's
right, the subscripts. H2 on the left, H2O on the right.
Let's do the same for oxygen. The number of oxygen atoms on the left equals the number
of oxygen atoms on the right, so 2y = z
Again, I got that 2 in front of the y from the subscript 2 in O2.
Those are the relationships we know about these variables.
What if we start with 1 molecule of oxygen gas? That means we set y = 1.
2(1) =z 2=z
2x = 2(2) 2x = 4
so x=2 Let's rewrite our chemical reaction, using
these coefficients: 2H2 + 1O2 → 2H2O the coefficient of 1 is
understood, so we rewrite as 2H2 + O2 → 2H2O. It's a good idea to double-check
your work and make sure there are the same number of atoms on both sides:
On the left we have 4H and 2O, and on the right we have 4H and 2O. It's balanced.
I started with one molecule of oxygen gas. What if we start with 1 molecule of hydrogen
gas instead? That means x is 1. 2(1) = 2z
2 = 2z z=1
Plug that into the second equation. 2y=1
y=1/2 1H2 + ½ O2 → 1H2O
Chemists don't like to see fractions as coefficients. That's because it doesn't make a lot of sense
to talk about half a molecule or half an atom in a reaction. To get rid of this fraction,
multiply EVERYTHING by 2. 2 times the quantity [H2 + ½ O2 → 1H2O]
2H2 + 1O2 → 2H2O remember the coefficient of 1 is understood, so we write this as
2H2 + O2 → 2H2O
Here's our second example. Nitrogen gas and hydrogen gas react to form ammonia. We write
that as N2 + H2 → NH3. That's unbalanced.
Let's put in some unknown coefficients. xN2 + yH2 → zNH3.
Write down the relationships we know. Let's look at nitrogen - the number of nitrogen
atoms on the left must equal the number of nitrogen atoms on the right.
2x = z Next, look at hydrogen.
2y = 3z Let's set x=1. That means we're starting with
1 molecule of nitrogen gas. 2(1) =z
2=z Substitute that into the second equation.
2y = 3z 2y = 3(2)
2y = 6 y=3
plug those coefficients into our equation xN2 + yH2 → zNH3, and we get
1N2 + 3H2 → 2NH3 Remember the coefficient of 1 is understood,
so we rewrite this as N2 + 3H2 → 2NH3
Double check your work by counting up the atoms on both sides.
On the left, we have 2N and 6H. On the right, we have 2N and 6H. That's balanced.
Here's our third example: ammonia reacting with oxygen.
NH3 + O2 → NO + H2O Now we'll have 4 coefficients. aNH3 + bO2 → cNO + dH2O
Write out the algebraic relationships we can see from the equation.
The number of nitrogen atoms must be equal on both sides, so
a = c The number of hydrogen atoms on both sides
are equal, so 3a = 2d
Finally, the number of oxygen atoms on both sides must be the same, so
2b = c + d Be careful with that part - see how oxygen
is in two places on the right hand side.
Let's say we start with 1 molecule of ammonia, so a = 1
If a=1, c=1. Plugging a=1 into the second equation gives
us 3(1) =2d
3=2d d=3/2
The final equation is then 2b = 1 + 3/2 = 2/2 + 3/2 = 5/2
Divide both sides by 2 b = 5/4
Put in our coefficients into aNH3 + bO2 → cNO + dH2O
That gives us 1NH3 + 5/4 O2 → 1NO + 3/2 H2O. Remember we don't want fractional coefficients,
so to get rid of those fractions, let's multiply the ENTIRE equation by 4.
4 times the quantity [1NH3 + 5/4 O2 → 1NO + 3/2 H2O] = 4NH3 + 5O2 → 4NO + 6H2O.
Double-check your work by counting the number of atoms on both sides.
4N 12H 10O on the left....4N 12H (4+6 is 10O on the right) this equation is balanced.
Our next example is hydrogen peroxide decomposing into water and oxygen gas.
H2O2 → H2O + O2 Assign variables to the coefficients:
xH2O2 → yH2O + zO2 Look at hydrogen on both sides. That gives
us 2x = 2y
The oxygen relationship is 2x = y + 2z
Let's set x =1 2(1) = 2y
2=2y y=1
and in the second equation, 2(1) = 1 + 2z
we can solve for z 2=1 + 2z
1=2z z=1/2
Put in these values for the coefficients: 1H2O2 → 1H2O + ½ O2
Multiply everything by 2 to get rid of that fraction
2 times the quantity [1H2O2 → 1H2O + ½ O2]
gives us 2H2O2 → 2H2O + O2 check the # of atoms on both sides:
4H and 4O on the left, 4H and 4O on the right. That's balanced.
Our last example is a combustion reaction - the combustion of glucose.
C6H12O6 + O2 → CO2 + H2O We have 4 variables, so let's use a,b,c, and
d. aC6H12O6 + bO2 → cCO2 + dH2O
The number of carbon atoms must balance, so 6a = c
The number of hydrogen atoms must balance, so
12a=2d The number of oxygen atoms must balance, so
6a + 2b = 2c + d. That part was a little complicated. Just be methodical.
Now, pick one of these to set to 1. Let's say we start with 1 molecule of glucose, C6H12O6.
so a=1. In the first equation, 6(1) =c
6=c In the second equation, we then have 12(1)
= 2d 12=2d
d=6
Plug what we know into the third equation. 6(1) + 2b = 2(6) + 6, solve for b.
6 + 2b = 12 + 6 2b = 12
b=6 Plug in these coefficients into our first
equation: C6H12O6 + 6O2 → 6CO2 + 6H2O Check your work
by checking the number of atoms on each side: 6C 12H 18O on the left 6C 12H and(12 + 6 = 18
O) on the right... BALANCED.