Stoichiometry

Step 2: Balancing chemical equations

Now that you know the mole the second thing you need to do stoichiometry is balanced chemical equations. This is just the basic way. There are more complex equations that can't be balanced this way.

First lets take a simple and common reaction. Combustion. The definition of combustion is a fuel when burned with oxygen produces only H2O and CO2.
Here we have the combustion reaction of sugar (which happens in your cells and powers your body).
C6H12O6 + O2 --> H2O + CO2

To balance this all you need to do is get the same number of atoms on both sides. First off just pick a molecule(any of them work but the biggest is usually the best) and assign it a number (again any number works but to keep it easy use one).

So far we have:
C6H12O6 + O2 --> H2O + CO2
       (1)

So on the left side we have 6 carbon atoms 12 hydrogen atoms and an unknown amounts of oxygen atoms.
To balance we need the same number on each side. Since we know the left sides number of carbon and hydrogen atoms we know the right sides number. We have to balance the molecules to have the same number of atoms.

6 H2O- Since we have 12 hydrogen atoms and each molecule has 2 atoms we 6 molecules of water.                                      

6 CO2- Since we have 6 carbon atoms and each molecule has one atom in it we need 6
molecules of CO2

So now the equation looks like that and we are almost done:
C6H12O6 + O2 --> H2O + CO2
        (1)           x           (6)       (6)

Now that we know the number of molecules on the right we know the number of oxygen atoms in the equation. We have six H2O molecules each with one oxygen atom, so we have 6 oxygen atoms in those molecules. We also have six CO2 molecules each with two oxygen atoms in it, so we have 12 oxygen atoms in those molecules. So we have a total of 18 oxygen atoms on the left side.

6 H2O= 6 oxygen atoms          
6 CO2= 12 oxygen atoms
6+12=18 atoms

If we know the number on the right side we know the number on the left side. Since we have one molecule of sugar that needs six atoms of oxygen; we subtract six from 18.
18-6=12

We have an unknown amount of oxygen molecules each with 2 oxygen atoms each and we have 12 oxygen atoms left. To finish just divide the number of oxygen atoms left by the number you need in each molecule.
12/2=6 oxygen molecules

That is the last number we need and the equation is balanced
C6H12O6 + O2 --> H2O + CO2
      (1)            (6)         (6)       (6)

That one was really easy but sometimes you will get an equation that doesn't work out so well. Sometimes one of the number wont come out as a whole number and that just wont work, because you can't have half of a molecule. Don't despair though all your hard work wasn't in vain just multiple by whatever makes that number a whole number

Example:
C3H7NO2 + O2 --> CO2 + H2O + NO2
     (1)
3  carbon, 7 hydrogen, 1 nitrogen dioxide, x oxygen

C3H7NO2 + O2 --> CO2 + H2O + NO2
      (1)              x          (3)       (3.5)     (1)
3 CO2- 6 oxygen atoms   3.5 H2O- 3.5 oxygen atoms
6+3.5=9.5 oxygen atoms
9.5/2=4.75 oxygen molecules

C3H7NO2 + O2 --> CO2 + H2O + NO2
      (1)          (4.75)      (3)     (3.5)       (1)

As you can see not all the numbers came out as whole numbers, but it can easily be fixed by multiplying it by the least common multiple. In this case its 4.
  C3H7NO2 + O2 --> CO2 + H2O + NO2
      (1)            (4.75)      (3)      (3.5)      (1)   * 4
=
C3H7NO2 + O2 --> CO2 + H2O + NO2
      (4)           (19)       (12)     (14)       (4)

Always remember to check your work. Once you get this down its on to the actual stoichiometry. Oh and one last thing once you have balanced the equation write it like this:
4 C3H7NO2 + 19 O2 --> 12 CO2 + 14 H2O + 4 NO2