Balancing Equations

A very important aspect of this lesson is for you to learn how to balance equations. It may come very easily and quickly to you, or it may take an awful lot of hard work. You may have had previous experience with this and only need review. In any case you need to make sure you get sufficient practice. I will show you how to balance some equations and work through several examples and then have you take the time to get a lot of practice. Each succeeding example adds some degree of difficulty or complexity.

What we need to do is come up with the numbers that have to go in front of the formulas for each of these chemicals in order to represent the same total number of atoms on each side of the arrow. Those numbers are called coefficients. We'll look at six examples, and then part of your lab work is to do more balancing equations practice.

Example 1

Balance the equation below:

K + Cl₂ → KCl

In the first example potassium plus chlorine becomes potassium chloride. First of all let's take a quick inventory of what elements are involved and what is already shown. There is one potassium on the left and one potassium on the right. No problem there now. It may change, but for the moment we don't have to do anything with the potassium. There are two chlorine atoms shown on the left and one chlorine shown on the right.

potassium: 1 chlorine: 2 K + Cl₂ → KCl potassium: 1 chlorine: 1

We will have to balance that. If we start with two chlorine atoms, we are going to need finish with two chlorine atoms ... they can't just disappear.

To accomplish that we cannot just change the formula for potassium chloride! We can't put a two in after the Cl and make it KCl₂. That would balance the equation, but KCl₂ is not the formula of the compound that is formed when these two elements combine. KCl is the formula, and you cannot change the formula. Remember, the formula shows the ratio in which the atoms actually combine. Nature determines the ratio. We cannot change it just to make it convenient to balance the equation. KCl is the formula, and we must leave it as KCl! The atoms combine in a one to one ratio in this compound.

What we have to do instead is put a two in front of the KCl.

potassium: 1 chlorine: 2 K + Cl₂ → 2 KCl potassium: 2 chlorine: 2

That gives us two chlorines on each side, but it also gives us two potassium on the right side. If we end up with two potassium, we need to start with two potassium. So, we go back and put a two in front of the K on the left side.

potassium: 2 chlorine: 2 2 K + Cl₂ → 2 KCl potassium: 2 chlorine: 2

This gives us a properly balanced equation.

Here is an important thing to note. Notice that there is not room to put a two between the K and the Cl in the formula of the potassium chloride, and that is because KCl is the formula of a compound. You don't insert anything into the formula. KCl is a formula unit. You can, however, put a number in front of the formula to show more than one of those units.

• Do change coefficients in front of formulas.

• Don't change the formulas or insert numbers in between symbols in the formulas.

• Remember that no coefficient means "1" not "0" of that chemical.

Example 2

Balance the equation below:

Al + O₂ → Al₂O₃

Let’s start with an inventory. We have one aluminum on the left and two aluminums on the right. (We could just put a two in front of the aluminum on the left to balance that, but let's finish the inventory first.) Also, there are two oxygens on the left and three on the right. Another way of saying that is oxygens come in pairs on the left and triplets on the right. Any time this is the case, balancing will be a bit more complicated, for the same reason that it is somewhat tricky to add the fractions 1/2 and 1/3.

aluminum: 1 oxygen: 2 Al + O₂ → Al₂O₃ aluminum: 2 oxygen: 3

As you may remember from learning to add fractions, the key tool to use is the least common multiple. The least common multiple of 2 and 3 is 6, so to add 1/2 + 1/3, you convert the fractions so you are adding 3/6 + 2/6, for a total of 5/6. In this case, we will also use this least common multiple idea. To make the pairs on the left line up with the triplets on the right, we will need six oxygen atoms on each side. So we put a 3 on the O₂ and a 2 on the Al₂O₃.

aluminum: 1 oxygen: 6 Al + 3 O₂ → 2 Al₂O₃ aluminum: 4 oxygen: 6

Having done that we end up with four aluminums in the 2 Al₂O₃'s (two formula groups times two aluminum atoms per Al₂O₃ formula group).If we finish with four aluminums on the right, we must start with four aluminums on the left, so we put a four in front of the Al on the left side.

aluminum: 4 oxygen: 6 4 Al + 3 O₂ → 2 Al₂O₃ aluminum: 4 oxygen: 6

Note that in this particular equation I skipped over the aluminum and did not try to balance it first. I started with the oxygen. You can start anywhere you want; and as you keep going back and forth, you will eventually get the equation balanced. However, if you start with the combination that is going to give you the most problems and get that sorted out first, then the easier ones fall into place and you don't have to change things so many times.

Example 3

Balance the equation below:

Ag₂SO₄ + CaCl₂ → AgCl + CaSO₄

First, a quick inventory. Silver: we have two silver on the left and one on the right. Sulfur: one sulfur on the left and one on the right. Oxygen: four oxygen on the left and four on the right. Calcium: one calcium on the left and one on the right. Chlorine: two chlorine on the left and one on the right. So the only things that are really out of whack at the moment are the silver and the chlorine.

We start with Ag₂ on the left so we need to have two silver on the right. If you put a two in front of the AgCl, that gives us two silver on each side. That also gives us two chlorine on the right, which is what is already shown on the left. We didn't have to change the coefficients for the sulfur or oxygen or calcium, just the silver and the chlorine. Let’s double check to make sure everything is balanced. We have Ag₂ on the left and 2 Ag on the right; that's the same number. We have an SO₄ on the left and an SO₄ on the right; that's the same. We have one Ca on the left and one Ca on the right. Finally, we have Cl₂ on the left and 2 AgCl gives us two chlorine on the right.

Ag₂SO₄ + CaCl₂ → 2 AgCl + CaSO₄

Even though the equation is quite complex, it was fairly simple to balance: the coefficient of 2 on AgCl was all that was needed.

Example 4

Balance the equation below:

Mg₃N₂ + H₂O → MgO + NH₃

First let’s take a quick inventory. We have three magnesium on the left and only one on the right, so that is going to take some work. Two nitrogen on the left and one on the right; that is going to take some work. Two hydrogens on the left and three on the right; so that is going to take some work. One oxygen on the left and one on the right. Only the oxygen matches, and it may not stay that way when we start changing everything else.

There are two places where you could start balancing this equation. Notice that you have two hydrogens on the left and three on the right. We could find the common multiple and take three groups of two and two groups of three. This would work out quickly.

Instead, I will start with magnesium nitride, Mg₃N₂, because it is the most complicated formula. With Mg₃, we have three Mg on the left, so we need three on the right; so put a three in front of the MgO.

Mg₃N₂ + H₂O → 3 MgO + NH₃

The three MgO also gives us three oxygen on the right. If we end up with three oxygen on the right, we have to start with three on the left; so put a three in front of the H₂O.

Mg₃N₂ + 3 H₂O → 3 MgO + NH₃

That gives us three oxygens but it also gives us six hydrogen. If we start with six hydrogen on the left, we have to end up with six on the right. We have three in a group in the ammonia on the right, so we have to put a two in front of the NH₃. Two groups of three makes six hydrogen. It also gives us two nitrogen. Looking back on the left, the formula is Mg₃N₂, so we have two nitrogen on the left and two on the right.

Mg₃N₂ + 3 H₂O → 3 MgO + 2 NH₃

Taking our inventory again, we can see that we have successfully balanced the equation. More complicated equations like this one often require a couple times going back and forth across the arrow before they are fully balanced.

Example 5

Balance the equation below:

C₃H₈ + O₂ → CO₂ + H₂O

This equation introduces two new features. Look where those elements are going. Notice that the oxygen from the O₂ on the left ends in two different compounds on the right. When something like that happens, wait until the very last to figure out what the coefficient of the O₂ should be. The amount of oxygen needed on the left will depend on the amount of carbon dioxide and water together, so we won't be able to determine the amount of oxygen until we have figured out the others. Also notice that the carbon and the hydrogen from the C₃H₈ go to two different places. The carbon goes to the CO₂ and the hydrogen goes to the H₂O. Equations of this kind are easiest to balance by focusing on the C₃H₈ and noting that the three carbons go to the CO₂ and the eight hydrogens go to the H₂O. So the place to start is with the C₃H₈. OK, let's do it. Three carbons from C₃H₈ will give 3 CO₂'s. Three carbons on the left have to give three carbons on the right, so we need three CO₂.

C₃H₈ + O₂ → 3 CO₂ + H₂O

Also, eight hydrogens from C₃H₈ will have to give eight hydrogens on the right. To get eight hydrogens, we will need four waters, each with two hydrogens. That gives eight hydrogens.

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

Now, we have three CO₂ and four H₂O. How many oxygens is that? Let's see. There are six oxygens from the CO₂ and four oxygens each from the 4 H₂O's.That makes a total of ten oxygen atoms on the right, so we need to start with ten on the left. They come in pairs, so we need five of them.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

The equation is properly balanced.

Example 6

Balance the equation below:

C₄H₁₀ + O₂ → CO₂ + H₂O

This sixth equation introduces another dilemma. We will see what it is in a moment. The way of approaching this equation is very much the same as the previous one. The carbon and hydrogen on the left split up and go into the other compounds, so let's start with that. C₄ will have to become 4 CO₂ to give four carbons on each side. If we start with ten hydrogens in C₄H₁₀, we have to end with ten hydrogens. They come in pairs in H₂O, so it is going to be 5 H₂O.

C₄H₁₀ + O₂ → 4 CO₂ + 5 H₂O

That takes care of the carbon and the hydrogen. What about the oxygen? We have 4 CO₂'s, so that is eight oxygen atoms and 5 H₂O's is five more oxygen atoms. That is a total of 13 oxygen atoms, so we need 13 oxygen atoms on the left. This is where the problem comes in. Thirteen is an odd number. The formula O₂ shows that oxygen comes in pairs. You cannot have an odd number arranged in pairs. It just doesn't work out. There are a couple ways of getting around this dilemma, and I think this is the simplest.

We need 13 oxygen atoms, which would be the same as having 6.5 O₂'s. We can go ahead and write that down for the moment.

C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

However, we don't like to have fractional numbers in a balanced equation. Just think about what O₂ means. It means that oxygen comes as two atoms attached to one another. If that is the way they come, then you can't have half of two of them. You have to have two of them at a time. So somehow we must get rid of that fraction. What we have now is one C₄H₁₀ plus 6.5 O₂ becomes 4 CO₂ plus 5 H₂O. We have the right ratio of numbers, we just don't have the right numbers. To preserve our ratios while eliminating the fraction, we can double all of our coefficents (including the implied 1 in front of the C₄H₁₀).

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

The lab workbook gives you ample opportunities, both graded and ungraded, to practice balancing equations. I suggest you do so now, and I'd suggest tackling a given problem with multiple approaches if it is a complex one. Because there is no single right way to balance an equation, you can often get a lot of practice out of just one example.