Stoichiometry of Balanced Equations

Now that you know how to balance equations, you need to know what to do with them and how you can use them. The way that you will use them is to come up with the mass relationships that exist between various chemicals involved in chemical reactions.

Information from the Balanced Equation

In the following table, the first line gives the word equation for the reaction of magnesium and oxygen, and the second line gives the balanced equation. The coefficients in front of the formulas - the "2" in front of magnesium, the "1" that's not shown in front of oxygen, and the "2" that's shown in front of magnesium oxide - can be interpreted in terms of how many atoms and molecules react or in terms of how many moles of these chemicals will react with one another.

Because of the way a mole is defined, if two magnesium atoms react with one oxygen molecule, then two moles of magnesium atoms will react with one mole of oxygen molecules. When two moles of Mg react with one mole of O₂, two moles of MgO will be formed. Once you have the balanced equation, the coefficients immediately give you the mole relationship. Two moles of magnesium reacts with one mole of oxygen to give two moles of magnesium oxide.

Since you already know how to figure the formula masses of compounds, you can use that along with the number of moles to find out the massrelationships that are involved in the chemical reaction. One mole of magnesium weighs 24.3 grams; therefore, two moles of magnesium weighs 48.6 grams. Oxygen has an atomic mass of 16.0, so the formula mass for O₂ is 32.0 grams. Going over to the right side of the formula mass of magnesium oxide is the sum of the atomic mass of magnesium and the atomic mass of oxygen, which is 40.3. That is the mass in grams of one mole of magnesium oxide. Thus the mass of two moles of magnesium oxide is 80.6 grams.

Note that we can determine the mass relationship in a chemical reaction just from knowing what the chemicals are. If we know the formulas of all the chemicals involved, we can write a balanced equation. From the balanced equation, we can figure out the mole relationship. From the number of moles of each chemical that are involved in the reaction and their formula masses, we can figure out the mass of each one of these.

So from the balanced equation, you can get the mass relationships. These are the same kind of mass relationship that you were using earlier in the lesson when you were working with calculations that involved the mass relationships in chemical reactions. Now, you can do that starting with a balanced equation or actually even starting with an unbalanced equation because you can balance the equation. You don't have to carry out the reaction and weigh the chemicals or be told what the mass relationship is. You can figure that out doing the very kind of thing that we just went through in this example.

The next five examples show how to set up calculations to answer a variety of questions using information from a balanced equation. All of these examples use the equation for the formation of magnesium oxide. The questions are listed in example 20 in your workbook. If you wish, try your hand at answering these questions and then check your answers by looking at the following examples. Or you can begin by following through the worked out examples to see how they are done. Remember, the mole and mass relationships in chemical reactions based on balanced equations are called stoichiometry. After you have worked through these examples, try your hand with the practice problems.

Example 1

How much oxygen would be needed for 20.0 g of magnesium to react?

Before reading through my walkthrough of this example, I encourage you to revisit the Stoichiometry section earlier in the lesson and look at the table for this reaction above. See if you can piece together how to approach the problem on your own

Example 2

How much magnesium oxide (in grams) could be made from 30.0 g of magnesium?

This problem is very similar to the previous one. See if you can set it up and solve it before reading my walkthrough.

This time we want the mass of MgO from our table on top, with the mass of Mg still going on the bottom. Working through the multiplication and division gives us a mass of 49.8 g MgO that would be produced.

Example 3

What mass of oxygen would be needed to make 50.0 g of MgO?

This time you should definitely take a good hard stab at setting the problem up on your own before continuing to read.

Because we are going from magnesium oxide to oxygen, we place the mass relationship for oxygen on top and for MgO on the bottom. By now you should be getting a good feel for these types of problems.

Example 4

How many moles of magnesium would it take to react with 3.00 moles of oxygen?

This one should bring you up a little short; it is quite different from the previous examples. Those asked about masses, whereas this one asks about moles. However, if anything, this question is even simpler than the first.

The coefficients in our balanced equation (2 Mg and 1 O₂) give us the mole relationship we need. They tell us that for every mole of oxygen we will need two moles of magnesium. Thus, we get the values and units for our dimensional analysis, plug them in, and solve to get 6.00 mol O₂.

Example 5

How many grams of magnesium would it take to make with 3.00 moles of magnesium oxide?

Now we have an example that mixes grams and moles. Fortunately, the relationships we worked out above apply even when mixing the two types of measurement. That is, looking at our table we find that 48.6 g of Mg is equivalent (stoichiometrically) to 2 moles of MgO. We can put these values together into a single conversion factor.

Working through the math, we come up with an answer f 72.9 g Mg required to make 3.00 mol MgO.

You will definitely want to practice with these skills. All these examples are derived from the same equation, so it will be valuable to get practice with a variety of chemical equations. The practice problems in your lab workbook will be very useful in digging into these skills.