Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages

How to Solve AP Chemistry Stoichiometry Problems

Stoichiometry is one of the most important topics on the AP Chemistry exam, so it’s vital that you understand it and all of its applications. Stoichiometry derives from the Greek stoicheionmetron, meaning “element measure”. In other words, stoichiometry is the practice of using a chemical reaction equation to predict the results of the reaction. We’ll go into more detail about that later.

Of course, that means that we need to start with a chemical reaction. When we look at the two sides of the reaction, we have to make sure that the number of atoms of each element on each side is the same because of the principle of conservation of mass. This is known as balancing a reaction. In this section of the AP Chemistry Crash Course, we’ll start by looking at the basic concepts of stoichiometry, and then we’ll cover five applications for stoichiometry on the AP Chemistry exam.

First, let’s talk about the basic concepts of stoichiometry. First of all, we have to start with a balanced reaction. Let’s say that we have this equation, which is the combustion of methane.

CH_4 + O_2 \rightarrow CO_2 + H_2O

Now, this equation is not yet balanced. How can I tell? I add up the number of atoms of each element on the left side: one carbon, four hydrogens, and two oxygens. Then I do the same on the right side: one carbon, two hydrogens, and three oxygens. Uh-oh! How do we balance this equation? First, we look at the left side and see that we need four hydrogens. The only way to get that four hydrogens is to double the amount of H_2O on the right side. So now we have:

CH_4 + O_2 \rightarrow CO_2 + 2H_2O

However, now we still have more oxygen on the right side. In fact, it’s even worse than before; the ratio is now 2:4. So, we double the number of O_2 on the left side, yielding:

CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O

…and this equation is now balanced. You’ve probably learned much of this even in a non-AP chemistry class, so bear with me; it’ll get more advanced.

This balanced equation tells us a lot about how this reaction works. The core concept we can take away from it is that one molecule of methane combines with two molecules of diatomic oxygen to form one molecule of carbon dioxide and two molecules of water, or correspondingly, that one mole of methane combines with two moles of diatomic oxygen to form one mole of carbon dioxide and two moles of water. This concept is very important to the applications of stoichiometry.

Application 1: Expected Value

Let’s get to it! First of all, let’s look at the expected product.

Here’s a simple problem: If 34.7 \text{ g} of O_2 is reacted with excess methane, how much CO_2 (in grams) is produced?

The first step to this problem is to convert the grams into moles. We know that in the combustion of methane, 2 \text{ moles} of O_2 are necessary to make 1 \text{ mole} of CO_2, but we don’t know anything about how many grams of O_2 are necessary to make 1 \text{ gram} of CO_2. So, our first step is to find out how many moles of oxygen we have.

The molar mass of O_2 is just about 32 \text{ g/mol} (since oxygen has a molar mass of 16 \text{ g/mol}, and obviously there are two oxygens in O_2). So, we divide 34.7 \text{ g} by the molar mass of O_2, yielding 1.08 \text{ moles}.

Next, we need to figure out how many moles of CO_2 that will yield. We know from the balanced equation above that 2 \text{ moles} of O_2 are necessary to make 1 \text{ mole} of CO_2, so 1.08 \text{ moles} of O_2 would yield half as much CO_2, or 0.54\text{ moles}.

Now, we have the number of moles of CO_2, but the questions ask for the answer to be in grams. So, we have to multiply the number of moles (0.54) by the molar mass (12+16 \times 2) = 44 \text{ g/mol}, yielding 23.8 \text{ grams} of CO_2.

Application 2: Limiting Reactant/Reagent

This one is another very important application of stoichiometry for the AP Chemistry exam. A note about nomenclature: saying reagent makes me feel like a mysterious alchemist, but I’m going to favor reactant because I think it’s a little bit clearer.

For the sake of simplicity, we’re going to continue to use the combustion of methane as our reaction in question.

Let’s say that we have 12.4 \text{ grams} of methane and 39.8 \text{ grams} of oxygen. Which is the limiting reactant in this reaction?

First, as before, we’ll convert the two quantities to moles. The molar mass of methane is (12.0+1.0 \times 4) = 16.0 \text{ g/mol}, and the molar mass of diatomic oxygen is (16.0 \times 2) = 32.0 \text{ g/mol}. You may already be able to tell which of these will be the limiting reactant, but let’s continue on anyways. We find that we have 0.775 \text{ moles} of methane and 1.24 \text{ moles} of diatomic oxygen.

We have more moles of oxygen, so methane is the limiting reactant, right? No, because in the reaction, we can see that we need twice as many moles of oxygen as of methane. The standard practice to eliminate this problem is to simply carry out both calculations as expected value calculations.

So, let’s pretend that we’re calculating the expected value of CO_2 in moles. We have 0.775 \text{ moles} of methane, so if we had enough oxygen, we could make 0.775 \text{ moles} of CO_2. So far, so good. Then, we have 1.24 \text{ moles} of diatomic oxygen, so if we had enough methane, we could make 0.62 moles of CO_2.

0.62 is clearly less than 0.775, so oxygen is the limiting reactant here because its expected value in the reaction was less.

Application 3: Percent Yield

In the real world, no reaction ever goes to completion. These expected values that we calculate are just the ideals of what would happen if the reaction went to completion. Percent yield is a way of quantifying the difference between the theoretical yield (the expected value) and the experimental yield.

Using the same reaction yet again, let’s say that you were given a problem like this.

7.8 \text{ grams} of methane is reacted with excess oxygen, yielding 16 \text{ grams } H_2O. What was the percent yield of the reaction?

By now, you should know the first step: convert to moles. Methane’s molar mass is 16 \text{ g/mol}, so we have 0.49 \text{ moles} of methane.

Next step: figure out how many moles of the product are produced. In the equation, 2 \text{ moles} of H_2O are produced for every mole of methane, so we get (0.49 \times 2) = 0.98 \text{ moles} of H_2O.

How many grams is that? The molar mass of H_2O is (1 \times 2+16) = 18, so we should get (0.98 \times 18) = 17.6 \text{ grams} of water.

But the problem says that we only got 16 \text{ grams} of water. What percentage of 17.6 is 16?

\dfrac{16}{17.6} = 0.91 = 91\%

So this particular reaction had a 91\% yield.

Application 4: Expected Values in Gas

A lot of this will be very familiar because, fortunately, it’s almost exactly the same as the expected value! Let’s use our favorite reaction, methane combustion, yet again.

14.0 \text{ L} of O_2 is reacted with excess methane at STP (standard temperature and pressure). How many liters of CO_2 are produced?

What’s the first step? Convert to moles! However, this part is even simpler than it was in grams. At STP, 22.4 \text{ L} of any gas will be 1 \text{ mole} of that gas. So 14.0 \text{ L} of O_2 is:

\dfrac{14.0}{22.4} = 0.625\text{ moles}

Next, we figure out how many moles of the product are produced. From the reaction equation, we know that 2 \text{ moles} of O_2 are necessary to make 1 \text{ mole} of CO_2, so 0.625 \text{ moles} of O_2 would make:

\dfrac{0.625}{2} = 0.313 \text{ moles } CO_2.

The final step, of course, is simply to convert back to moles. 1 \text{ mole} of CO_2 is, of course, equal to 22.4 \text{ L} of CO_2 at STP, so 0.313 \text{ moles} of CO_2 would be 7.01 \text{ L} of CO_2.

Actually, the answer is 7 \text{ L}; the 0.01 is due to rounding up the 0.3125 \text{ moles} of CO_2 to 0.313. In fact, there is an easier way to do this calculation. You see, the conversions from liters to moles and back to liters are the same for both gases, so the only number that really mattered, in this case, was the mole ratio between the reactant and the product (in other words, the fact that it took two O_2 to make one CO_2). So, knowing that the mole ratio was 2:1, we could have simply said that the liter ratio would also be 2:1, or in this case, 14:7.

Why didn’t we do that the first time? Well, first of all, it’s good practice to do it the long way, and it helps to understand the fundamentals. And on the AP Chemistry exam, it’s often worth the extra two minutes to write out your entire thought process on a stoichiometry problem, especially if it’s a long answer. However, on a multiple choice stoichiometry problem, you may want to use that little trick.

Why can’t we use that trick for gram-based stoichiometry problems? You can, but it’s more complicated and not usually worth the effort. With a gram-based question, you have to take into account the different molar masses of the reactant and the product. That being said, if you’re pressed for time, you can try it.

Recall this question:

Let’s say that we have 12.4 \text{ grams} of methane and 39.8 \text{ grams} of oxygen. Which is the limiting reactant in this reaction?

The molar mass of methane is 16 \text{ g/mol}, and the molar mass of diatomic oxygen is 32 \text{ g/mol}. In the reaction, their mole ratio is 1:2, meaning that we need twice as many moles of oxygen as we do methane. So, because oxygen has twice the molar mass and we need twice as much of it, we need four times as much oxygen in grams as we need methane.

Is 12.4 \times 4 more or less than 39.8? It’s more, so we can tell that oxygen is the limiting reactant in this reaction, as we discovered through the more involved calculation.

If this isn’t immediately intuitive, don’t waste too much time thinking about it; it’s almost always more useful to just carry out the two expected value calculations. However, if this seems obvious to you, keep it in mind; it may come in handy at some point.

Application 5: Titration

Titration may seem like an intimidating topic, but while it is a little involved, it simply builds on the concepts we already know from expected values and limiting reactants. Sadly, for this example, we have to move on from our beloved methane combustion equation (sniff).

The equation we’ll use here is this one:

2NaOH + H_2C_2O_4 \rightarrow Na_2C_2O_4 + 2H_2O

It’s already balanced because I’m nice like that.

Given that the NaOH solution is at a 1 \text{ molar} concentration, how much oxalic acid (in moles) is in the titration?

Pretty intimidating if you don’t know what to do, right? Guess what the first step is.

That’s right — it’s to convert to moles!

A little explanation: titration is basically just adding one compound to another in solution (in this case, adding a base to an acid) and seeing how the pH changes. In this case, we can see that the pH passes through the equivalence point (where the acid and the base cancel each other out and the pH is 7) at 30 \text{ mL} of NaOH added. So of course, our first step will be to find out how many moles of NaOH that is.

30 \text{ mL} is by definition 0.030 \text{ L}, and a 1 \text{ molar} solution of NaOH will contain 1 \text{ mole} of NaOH per liter, so we have 0.030 \text{ moles} of NaOH in the solution.

Now, as I said earlier, 30 \text{ mL} of NaOH (or as we know now, 0.030 \text{ moles} of NaOH) is the equivalence point, when the acid and the base cancel each other out. But as you can see, the molar ratio of NaOH to oxalic acid in the reaction equation is 2:1. So, to find out how many moles of oxalic acid we have, we need to use the same ratio. 2:1 means that there are:

\dfrac{0.030}{2} = 0.015 \text{ moles of oxalic acid in the solution}

If the problem asked for the answer in grams instead, what would you do? You’d simply multiply the number of moles by the molar mass, as usual. The molar mass of oxalic acid is (1 \times 2+12 \times 2+16 \times 4) = 90 \text{ g/mol}, so the answer in grams would be (0.015 \times 90) = 1.35 \text{ grams} of oxalic acid.

Yay! You’re done! Those are the five main applications of stoichiometry on the AP Chemistry exam. You’ll notice that the common thread running through all of them is the expected value. Essentially, if you understand expected value really well, you should be able to figure out most of the rest. Stoichiometry can be confusing, but hopefully, you can see that it all boils down to that one relatively simple idea. Best of luck on your AP exam!

Looking for AP Chemistry practice?

Kickstart your AP Chemistry prep with Albert. Start your AP exam prep today.

Bring alert to your class

Bring Alert To Your Class

Teachers get free, limited access to Albert's code academic areas to use with their students.

Get Started
Learn about school wide solutions

Learn About School-Wide Solutions

Deploy Albert to all classes and unify your school on a single practice and assesment platform.

Exploring Prices

Article written by The Albert.io Team

Learn anything through interactive practice with Albert.io. Thousands of practice questions in college math and science, Advanced Placement, SAT, ACT, GRE, GMAT, literature, social science, history, and more.