Stoichiometry

Let's look at an example of a chemical reaction and the way it obeys conservation of mass, as discussed in the previous section. As you can see below, the masses of methane and chlorine in this reaction add up to the masses of methyl chloride and hydrogen chloride.

As you can also see, when the mass of methane involved in the reaction is doubled, the masses of all the other components of the reaction double as well. This illustrates an important fact about chemical reactions: in any reaction, the relationships between the masses of the chemicals that react and form are fixed. For example, when methane reacts with chlorine, the mass of chlorine that reacts is always just a little less than four times the mass of the methane. There is a constant ratio between the amounts of reactants and/or products.

methane + chlorine → methyl chloride + hydrogen chloride

16.0 g + 61.0 g → 40.5 g + 36.5 g

32.0 g + 122.0 g → 81.0 g + 73.0 g

The amount of methane that reacts with a given amount of chlorine, and the amounts of methyl chloride and hydrogen chloride that form, are determined by the details of the reaction. Later in this lesson you will learn how to calculate these amounts from the formulas of the chemicals involved and a balanced equation that describes the reaction. You will also see why mass is conserved in ordinary chemical reactions. In this section, we'll apply this idea of fixed relationships to problems of stoichiometry: the measure of amounts of chemicals involved in reactions.

Stoichiometry Basics

Let's take one of our examples from the previous section.

methane + chlorine → methyl chloride + hydrogen chloride

16.0 g + 61.0 g → 40.5 g + 36.5 g

Once we know that 61.0 g of chlorine are required to react with 16.0 g of methane, we can use that information to determine how much chlorine would be required to react with any other mass of methane, or how much methane would react with any other mass of chlorine.

Keep in mind that these numbers are not arbitrary. They are like the amounts of ingredients in a recipe. To get, say, six servings of a particular dish (the “product”) you might need 2 eggs, 4 cups of flour, as well as specified mounts of a number of other “reactants.” To get nine servings, half again as many, you would use half again as much of each ingredient – 3 eggs, 6 cups of flour, and so on. Note that, like the reactants and products in a chemical reaction, the ratio of ingredients to one another and to servings stays constant: the ratio of flour to eggs stays at two to one; the ratio of servings to eggs stays at three to one, and so on.

Using this concept, we can do some simple stoichiometry in our heads. For example: how much methyl chloride would be produced from the reaction of 8.0 g of methane? To answer this problem, we note that we have halved the amount of methane in the reaction, from 16.0 g to 8.0 g. So we will also halve the output of the reaction, meaning the amount of methyl chloride produced will be 20.25 g.

The Dimensional Analysis Approach

As you can imagine, not all stoichiometry problems will be neat and tidy in this way, with values being cut in half. For example, let's say you wanted to know how much chlorine would react with 22.9 g of methane. We won't be able to do this one in our heads. Instead, we will use the proportions of the masses from the equation above as conversion factors. What does that mean? Well, we start with our 22.9 g methane. We know from our balanced equation that the ratio of chlorine to methane is 61.0:16.0, so by dividing by 16.0 and multiplying by 61.0, we get the mass of chlorine needed in the reaction. This approach is shown below as a dimensional analysis, using the conversion factor derived from the mass data above.

This conversion factor method can be used any time the two quantities are related to each other by a constant ratio – that is whenever they are directly proportional to one another. A convenient test of such a relationship is to ask “if one is doubled will the other necessarily double?” Problems solved this way are relatively easy to set up and including the units in the conversion factor helps to insure the ratio is applied correctly.

Constructing the conversion factor in this way provides a convenient way to make sure we have the ratio correct. Since we’re converting from grams of methane, the units “g methane” must cancel, so that the units that remain on the left, “g chlorine,” are the same as the units on the right. It is also easy to determine how much methane might be needed to react with a given mass of chlorine. We just turn the conversion factor upside down, allowing us to determine that we would need 2.62 g of methane to react with 10.0 g of chlorine.

Notice again how the units on the left side of the equation cancel. The units “g chlorine” cancel, leaving only “g methane” on the left, the same units that are on the right side of the equation.

Stoichiometry is a powerful tool: we can use these ratios to calculate an amount of any component of a reaction, given any other component. For example: how many grams of methane are needed to produce 17.8 g of methyl chloride? We can set the problem up in the same way. Note that the setup is identical, even though the problem is now seeking a mass of a reactant, given a mass of a product. It turns out that 7.03 g of methane are needed to make 17.8 g of methyl chloride.

The chief difficulty students seem to have with these problems is keeping straight which set of numbers are part of the “recipe,” that is, describe the actual ratio between masses of reactants and products, and which number is the arbitrary mass of some reactant or product from which the mass of some other reactant or products is to be calculated. This difficulty disappears with practice and the habit of reading the problem carefully.

The Cross-Multiplying Approach

Here is another method you may have learned for solving this kind of problem: proportionality. To find out how many grams of chlorine are needed to react with 2.2 grams of methane, we use the fact that the ratio of that number of grams of chlorine to 2.2 grams of methane is the same as the ratio of 61.0 grams of chlorine to 16.0 grams of methane. This is the same idea (constant proportions) that we applied above, just presented a little differently.

Because we know the proportion of chlorine to methane is the same in every case, we are able to set up this equality between the two ratios. Then, using the technique of cross-multiplication, we are able to solve for X, which comes out to 8.4 g of chlorine. The important thing to keep straight, once again, is that the 2.2 g of methane are an arbitrary amount, provided as part of the question, whereas the 61.0 and 16.0 are more fundamental to the reaction - or, at least, the ratio between them is.

It makes no difference which method you use to solve these problems, so use the one that seems most natural to you. They are mathematically equivalent. Which ever method you choose, however, get into the habit of always including the units in your problem solving. There are a series of exercises in your workbook that will give you practice in working this type of problem. Be sure an complete them before moving on to the next part of the lesson.