Lesson 3: Stoichiometry

You may have noticed in the last two examples of Conservation of Mass that when we doubled the mass of methane, the mass of chlorine that reacted with it also doubled, and so did the masses of methyl chloride and hydrogen chloride that were formed.

methane	+	chlorine		methyl chloride	+	hydrogen chloride
16.0g	+	61.0 g	=	40.5 g	+	36.5 g
32.0g	+	122.0 g	=	81.0 g	+	73.0 g

 

This is because in every chemical 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.

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 more stoichiometry problems.

Stoichiometry Basics | The Conversion Factor Approach

The Ratio-Proportion Approach

Stoichiometry Basics

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

methane	+	chlorine		methyl chloride	+	hydrogen chloride
16.0g	+	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.

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The Conversion Factor Approach

How much chlorine is needed to react with 10.0 g of methane?

We use the original numbers to construct a “conversion factor” that converts from the mass of methane to the mass of chlorine.

 

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. In fact, no matter how a problem is solved, keeping track of the units – a process called “dimensional analysis” – can prevent many errors.

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.

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

And in fact we can use our original mass data to compute how much of any reactant will be needed or how much of any product will be formed when we are given the mass of any other reactant or product.

How much methyl chloride can be produced from 10.0 of methane?

methane	+	chlorine		methyl chloride	+	hydrogen chloride
16.0g	+	61.0 g	=	40.5 g	+	36.5 g

 

 

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.

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The Ratio-Proportion 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 10.0 grams of methane, we use the fact that the ratio of that number of grams of chlorine to 10.0 grams of methane is the same as the ratio of 61.0 grams of chlorine to 16.0 grams of methane.

 

In words, the problem is set up by saying, “X grams of chlorine are to 10 grams of methane as 61 grams of chlorine are to 16 grams of methane.” You can solve the resulting equation by multiplying both sides by 10.0 g methane, or by cross multiplying, then dividing through by 16.0 g of methane. You must keep straight that the 10.0 g methane are an arbitrary amount, while the 61.0 g chlorine and the 16.0 g methane are part of the “recipe.”

Solving this equation for X gives us the same answer we got before.

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. Exercises 10, 11, and 12 show you some sample problems worked out. Exercise 13 gives you some problems to work on your own (the answers are given below). Be sure you can do these practice problems before you move on to the next part of the lesson.

Answers to Ex. 13:

a. 280 g magnesium chloride

b. 110 g sulfur

c. 39 g iron

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