We have already seen that a chemical change is a change in which the atoms in the molecules are rearranged in some way. For example if methane is burned the result is the formation of carbon dioxide and water and the methane molecule is no more. Chemical equations afford us a way of keeping track of what we started with and with what we end. We can write such an equation in general as

R_{1} + R_{2} P_{1} + P_{2}

where the R's are the reactants and the P's are the products. This equation will tell us nothing about the time required to go from reactants to products, how we get from reactants to products, or how complete the reaction is (may still have a significant amount of reactants left after a reasonable time). So the chemical reaction just gives us the "big picture". Many details are lacking and some of the details will be discussed when we study Chemical Kinetics. The reaction may also give off (or absorb) heat and we can indicate that as:

R_{1} + R_{2} P_{1} + P_{2} + heat

or

R_{1} + R_{2} P_{1} + P_{2} - heat

It is extremely important to understand that the total mass of all reactants entering into a chemical reaction must equal the total mass of all the products formed in the reaction. We can't create mass by a chemical reaction! We make sure that this
mass balance occurs by *balancing* the chemical equation which means that we must have the same number of each type of atom on each side of the equation -- thus ensuring mass balance.

So let's look at some examples.

H_{2}(g) + O_{2}(g) H_{2}O(l)

I have also introduced another terminology with the (g) and (l -- this is a small "L") which tell us the *physical state* of the substance (solid, liquid, or gas). Does this equation satisfy our mass balancing requirement? We have two hydrogen
atoms on the left and two on the right, but we have two oxygen atoms on the left (remember that oxygen and hydrogen exist as a diatomic) and only one oxygen atom on the right. So we need to *balance* the equation. This one is easy for we can put a
two in front of the water on the right and a two in front of the hydrogen on the left and the equation is balanced. The numbers that we put in front of the atoms or compounds are called the *stoichiometric coefficients*. **You can't change the
subscripts for if you do you are changing what the substance is and you don't have that kind of power!**

2H_{2}(g) + O_{2}(g) 2H_{2}O(l)

We can check the masses and we see on the left (2*(2.02g h_{2}) + 32.00 g O_{2}) = 36.04 g reactants and on the right (2*(18.02 g H_{2}O)) = 36.04 g products. So the equation is mass balanced.

So let's balance some more equations and perhaps learn some other concepts.

Al(s) + Br_{2}(g) Al_{2}Br_{6}(s)

This is not balanced so think about how to balance it before looking at how I do it.

This is balanced as follows:

2Al(s) + 3Br_{2} Al_{2}Br_{6}

Check it out yourself.

Here is one a little harder:

C_{4}H_{10}(g) + O_{2}(g) CO_{2}(g) + H_{2}O(g)

This one is obviously not balanced so play with it a little to see what you can do toward balancing it.

The easiest way to balance it is to not think that you have to use integers all the time. I first balance the atoms which only appear once on each side (C and H in this case). Attacking C first is done by putting a 4 in front of CO_{2}. Then I
balance the hydrogens by putting a 5 in front of H_{2}O so that now I have balanced the C's and H's but now I have 13 oxygen atoms on the right and 2 on the left.

C_{4}H_{10}(g) + O_{2}(g) 4CO_{2}(g) + 5H_{2}O(g)

Now I put 13/2 in front of the O_{2} on the left and the equation is balanced!

C_{4}H_{10}(g) + 13/2O_{2}(g) 4CO_{2}(g) + 5H_{2}O(g)

Even though this is a perfectly legitimate way to end this problem, we just don't like to see fractions sitting in the equation. So we multiply both sides by 2 and then we have the neat result:

2C_{4}H_{10}(g) + 13O_{2}(g) 8CO_{2}(g) + 10H_{2}O(g)

We did that so well, let's try another one:

B_{4}H_{10}(s) + O_{2}(g) B_{2}O_{3}(s) + H_{2}O(g)

I will start here with the elements which only appear once on each side (B and H). We can balance the B by putting a two before the B_{2}O_{3}(s). Now I see that if I put a 5 in front of the H_{2}O the H's will be balanced which
leaves me with my ol' nemesis oxygen. I see that I have 11 oxygens on the right and 2 on the left. Want to guess what I am going to do? I will put 11/2 in front of the O_{2} to balance the equation:

B_{4}H_{10}(s) + 11/2O_{2}(g) 2B_{2}O_{3}(s) + 5H_{2}O(g)

And I will leave this in a neat form by multiplying both sides by 2 to give:

2B_{4}H_{10}(s) + 11O_{2}(g) 4B_{2}O_{3}(s) + 10H_{2}O(g)

Now you need to **practice, practice, practice**. You have homework problems to work and the site at the top of the page has a lot of equations on which to practice.

I also recommend practicing HERE for balancing equations.

Now take a practice quiz to help you understand if you understand the basic concepts. |

You must use your real name when it asks for a name. |

The test will only submit when you have answers all of the questions correctly. |

If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests. |

*Web Author: Dr. Leon L. Combs*

*Copyright ©1999 by Dr. Leon L. Combs - ALL RIGHTS RESERVED*