Solving Chemical Reactions Without Guessing

For those who have every been tormented by the evils of playing guessing games with balancing chemical equations, I have news for you. There is a way to solve them that involves real math, not adding magic/houdini numbers until both sides are balanced. The first and simplist way involves only simple algebra. For this reason I am surprised and somewhat shocked that even in College Chemistry, they teach the guessing method, which leads me to wonder if this is something that only math nerds know, if Chemists have some aversion to math, and how late, if ever, it is taught to Chemistry majors. Hopefully they do, but I wouldn't be surprised... :-/

Anyway, the first method works like this. Say you have an unbalanced reaction as follows:

H2 + O2 -> H2O

Begin by adding coefficients (variables representing the number of molecules of each part in the equation), which is what you are going to solve for:

A*H2 + B*O2 -> C*H2O

Now we can write a balanced equation representing each chemical contribution in the equation as follows:

H: 2A = 2C
O: 2B = C

Before proceeding, allow me to further explain how I arrived at the above equations. The equations are arrived at by isolating each element in the equation and the coefficients related to it. For Hydrogen, this is A and C on either side (A = C). To properly balance the equation, we have to multiply coefficients by the subscripts for the number of atoms in each molecule, to get the number of atoms on each side (which have to add up to the same thing for the equation to balance).

The next step is very simply, but conceptially probably the most advanced, having a connection to linear algebra. Essentially, the equation has an infinite number of answers (which makes sense, because once you balance the equation, you can multiply across the equation by any number, and still have a balanced equation). Furthermore, since we have 3 variables and 2 equations, we can not solve for any variable. To fix this problem, we simply assign an arbitrary value to one of the variables, and because there are an infinite amount of multiples which balance the equation, this "free" variable will lead us to a ratio of coefficients for which any multiple balances the equation. To keep things simple, it is usually easiest to simply assign A the value of 1 (A = 1). Which leads to the following conclusions:

2A = 2C : 2C = 2(1) : 2C = 2 : C = 2/2 = 1
2B = C : B = C/2 : B = (1)/2 = 1/2

Therefore:

A = 1
B = 1/2
C = 1

Which gives us the balanced equation:

A*H2 + B*O2 -> C*H2O
1 * H2 + 1/2*O2 -> 1*H2O

At this point, some of you are probably wondering what insanity leads me to think that a system that yields fractionally balanced equations (half a molecule you say?) is a great thing, because such answers are unacceptable. However, remember that this is simply a multiple of ratios for which their are an infinite number of alternative multiples, and that we can simply multiply across the reaction to scale to whatever we want. In this case, we want the smallest balanced equation with whole number coefficients. Therefore, the final step is to multiply the reaction by the smallest value which will yield all whole number coefficients. In this case, that number is 2:

2 * (1 + 1/2 -> 1)
2 + 1 -> 2
2 * H2 + 1 * O2 -> 2*H2O
2*H2 + O2 -> 2*H2O

Which is the final, balanced quation, having 4 atoms of H on the left and 4 on the right (2*2 = 2*2) and 2 atoms of O on each side (2 = 2). While this method might be a bit longer than just guessing for a reaction this simple, the guessing method can quickly become painful with reactions slightly more complex. That and methodical people like me cringe whenever we hear "just guess". :o

The second method is essentially the same thing in a different format, using matrices and linear algebra to speed the process up a bit. However, the theory is a little more detailed and draing matrices is a bit difficult with text. Somehow I strugged through Highschool Chem without this, but taking it over again at the University reminded me how evil it was. Also, from what I was learning in linear algebra, I was pretty sure their had to be a systematic way to solve these problems. I got pretty close to figuring it out by myself, but had to check the web (where I found the above way of doing it) and my linear algebra book (which had the matrix method). Of course, our instructor simply recommended a 3 step guessing plan (1 - Guess 2 - Add up 3 - Repeat).