Lesson 3: 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.
The equations to be balanced in these examples are also shown in example 16 in your workbook. If you already have experience balancing equations, you might want to look at those first and try balancing them. Then look at the worked-out examples of those (if any) that give you trouble.
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 (the part that counts as your "lab report" this week) is to do more balancing equations practice on the lab computers.
First Example | Second Example | Third Example | Fourth Example
Fifth Example | Sixth Example | Practice
First Example
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.
We will have to balance that. If we start with two chlorine atoms,
we are going to finish with two chlorine atoms. |
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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. |
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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 is put a two in front of the KCl. |
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That gives us two chlorines on each side, but it also gives us two potassiums on the right side. If we end up with two potassiums, we need to start with two potassiums. |
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So, we go back and put a two in front of the K on the left side. That balances
the equation (2 K + Cl2 |
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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.
Second Example
The next example is Al (aluminum) plus O2 (oxygen) becomes Al2O3 (aluminum oxide). This one is a little bit more involved even though you still have just two elements combining to form a compound.
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. So, if you have something that starts out in pairs and ends up in triplets, it is not going to be as easy to balance as what we had before. It is a lot like trying to add fractions. If you can remember what you had to struggle through when you learned to add 1/2 to 1/3, you used something called the lowest common denominator, the least common denominator, or perhaps it was called the least common multiple or something like that.
The least common multiple of 2 and 3 is 6. If we started
with 6 oxygens on the left, we could end up with 6 oxygens on the right and still have
pairs and triplets. |
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To do this we could put a three in front of the O2 that gives 6 oxygens, and put a two in front of the Al2O3 and that gives 6 oxygens on the right. That balances the oxygen. |
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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). |
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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. |
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Thus, the
balanced equation for that reaction is 4 Al + 3 O2 |
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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.
It might be useful to start over with that equation and try balancing the aluminum first and carry it through until you get it balanced. See if it makes a difference to you, whether you start with aluminum or oxygen.
Third Example
In this third example, you are not expected to know the names of these things; but we have silver sulfate (Ag2SO4) reacting with calcium chloride (CaCl2) to become silver chloride (AgCl) and calcium sulfate (CaSO4).
Later in this course you will become more familiar with polyatomic ions (or clusters of atoms) like this SO4. Notice that we have an SO4 on the left and an SO4 on the right. We don't have to add up the sulfurs separately and the oxygens separately. We could do that but it is usually easier to treat them as groups, as long as you can find identical groups on both sides of the equation.
First, a quick inventory. Silver: we have two silvers on the left and one on the right. Sulfur: one sulfur on the left and one on the right. Oxygen: four oxygens on the left and four on the right. Calcium: one calcium on the left and one on the right. Chlorine: two chlorines 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. |
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| We start with Ag2 on the left so we need to have two silvers on the right. If you put a two in front of the AgCl, that gives us two silvers on each side. That also gives us two chlorines 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 chlorines on the right. | |||||
Everything is the same. The balanced
equation is Ag2SO4 + CaCl2 |
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Fourth Example
The next example is more involved. It almost always seems that when you have a 3 or any other odd number for a subscript it is going to give you some trouble.
Here we have magnesium nitride reacting with water to form magnesium oxide and ammonia.
First let’s take a quick inventory. We have three magnesiums on the left and only one on the right, so that is going to take some work. Two nitrogens 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. |
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Only the oxygen matches, and it may not stay that way when we start changing
everything else. |
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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. I will start with magnesium nitride, Mg3N2, because it is the most complicated formula. |
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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. |
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The three MgO also gives us three oxygens on the right. If we end up with three oxygens on the right, we have to start with three on the left; so put a three in front of the H2O. |
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That gives us three oxygens but it also gives us six
hydrogens. If we start with six hydrogens on the left, we have to end up with six on the
right. |
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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 hydrogens. It also gives us two nitrogens. Looking back on the left, the formula is Mg3N2, so we have two nitrogen on the left and two on the right. |
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The balanced equation then is Mg3N2 + 3 H2O |
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To double check the answer we take another inventory. There are three magnesiums on the left and three on the right. There are two nitrogens on the left and two on the right. There are six hydrogens on each side. Finally oxygen, there are three oxygens from three waters on the left and three MgO's give three oxygens on the right, so it is balanced.
Fifth Example
This fifth example represents the burning of propane. It has 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. |
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Figure
out how much CO2 and H2O will be made from the C3H8,
then figure out how much oxygen we will need to do it. |
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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. |
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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. |
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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. |
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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. |
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Now we should double check to see that everything is balanced. You have three carbons
on each side, eight hydrogens on each side, and ten oxygens on each side; so the balanced
equation becomes C3H8 + 5 O2 3 CO2 + 4 H2O.
Sixth Example
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. |
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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. |
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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. |
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We need 13 oxygen atoms, that would be the same as having 6 1/2 O2's. We can go ahead and write that down for the moment. |
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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 6½ O2 becomes 4 CO2 plus 5 H2O. We have the right ratio of numbers, we just don't have the right numbers. We can clear the fraction by doubling all of the numbers going
across. That gives us all whole number coefficients: 2 C4H10 + 13 O2 |
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For me, that is the easiest way to balance this type of equation. We can balance the equation by using a fractional number, so we do that first. Then clear the fraction. |
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Alternate Approach
There is another way of looking at that problem which is a little bit better conceptually, but not as easy mechanically.
That is to realize that if, indeed, 1 C4H10 is going
to take 6½ O2's--but we cannot have ½ of an O2--then we are
going to have to have two C4H10's. |
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We must have another one to take up the extra half-molecule of O2. |
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So, if we start off with a two in front of the C4H10,
then we have eight carbons and we need to have an eight in front of CO2. Also,
we have 20 hydrogens, so we need a ten in front of H2O. |
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Next add up the oxygens on the right. The 8 CO2's contain 16 oxygens, and the 10 H2O's which contain 10 oxygen. That gives us a total of 26 oxygens on the right. They come in pairs on the left so we need 13 pairs. |
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Note that whichever way you go about balancing the equation you get the same result.
Practice
What I think you need to do now is to practice balancing equations. There are a couple ways to get that practice. One way, which is the lab requirement for this lesson, is to work through a computer program, available in the lab, which gives you practice. You will need a record of the results of your practice session to submit in place of a lab report for this lesson. This program is not available in web format. In addition, there are also lots of equations in your textbook at the end of the chapter (or, use one of the textbooks in the lab or in the library on reserve). Get as much practice as you need.
I recommend that you work on balancing equations now. (If you plan to do the computer practice, first read through the "Categorizing Chemical Reactions" section of this lesson; the computer practice will ask you to balance and categorize equations.) When you resume working on the lesson, we will deal with how to use those balanced equations to get weight relationships from them, to figure out how many grams of "this" will react with how many grams of "that". But first I recommend that you stop and take some time to work on balancing equations, then continue with the lesson.