Stoichiometry: mole-to-mole and percent yield | Chemistry | Khan Academy

As a chemist, your goal is to produce some ammonia, and you decide to use this chemical reaction to do that. Ammonia is useful in making fertilizers, for example, to improve the crop yields.

Anyways, suppose you react 4.43 moles of hydrogen with excess of nitrogen, meaning you have a lot of nitrogen, but you have only limited quantities of hydrogen—only 4.43 moles of hydrogen. So, you react them. Then the question is, how many moles of ammonia can be theoretically produced? This is called stoichiometry, which deals with figuring out the amount of products if you are given a certain amount of reactants, or figuring out how much reactants you need to get a certain amount of products.

So, how do we figure this out? Well, our first step is to ensure that we have a balanced equation. Okay, so over here, you can see it is balanced. You have two atoms of nitrogen on the left-hand side, two atoms of nitrogen on the right. You have six atoms of hydrogen over here, six atoms of hydrogen over here.

Next, let's see what this balanced equation is telling us. It's telling us that one molecule of nitrogen will react with three molecules of hydrogen to give two molecules of ammonia. But it's also saying that if I had one dozen molecules of nitrogen, well then it will react with three dozen molecules of hydrogen to give me two dozen molecules of ammonia. Makes sense, right? Or in other words, if I had one Avogadro number of nitrogen molecules, meaning one mole of nitrogen, it would react with three moles of hydrogen to give me two moles of ammonia.

So, look, the balanced equation is telling me the ratios in which the reactants and products must be for the reaction to work. And the same applies to recipes when you're cooking something. For example, let's say you want to bake cookies. I don’t bake cookies, but let’s just say. Let’s just say that you can take one cup of sugar, mix it with three cups of flour, and I'm probably missing a lot of ingredients over here, but I'm just trying to make this analogous to the problem that we have at hand.

Anyway, if you mix one cup of sugar with three cups of flour, let’s say you can bake two cookies. All right, then if this recipe is accurate, which I don’t think it is, but if it was, then this ratio gets set. Meaning if I now double the amount of sugar, meaning I have two cups of sugar, and I double the amount of flour, and so I mix it with six cups of flour, then I’ll get double. I get double the amount of cookies, meaning I’ll get four cookies, and so on.

So now imagine the question was, what if I had lots of sugar, excess of sugar, but I only had, for some reason, 4.43 cups of flour? How many cookies would I get? The question is exactly the same mathematically. So if you can answer that question, you can answer our original question as well.

So it would be a great idea to pause the video and just explore this problem, see if you can answer this question yourself. You can answer the cookie question, and then see if you can answer this question yourself, and then let’s figure it out together.

All right, let’s do this. So we have excess of nitrogen; we don’t have to worry about it; we have plenty of it. Okay, but we have hydrogen in limited quantity, so we have only 4.43 moles of hydrogen, and from that, we need to figure out how many moles of ammonia I would get.

So, in some sense, I need to be able to convert from the amount of hydrogen to the amount of ammonia, right? So for that, I just need to build a conversion factor, and I can do that by using the ratios I have built over here. I know that for this reaction to work, I need three moles of hydrogen for 2 moles of ammonia.

So here’s how I like to think about it: let’s first write down what’s given to us. We have 4.43 moles of hydrogen, and now, to convert from the moles of hydrogen to moles of ammonia, let me build a conversion factor. My conversion factor is I will get 2 moles of ammonia per 3 moles of hydrogen.

Notice the moment I write it this way: what happens? The moles of hydrogen here and the moles of hydrogen here will cancel out, and so what I'm left with is with moles of ammonia. So this will give me the amount of ammonia I would get for 4.43 moles of hydrogen.

Now, the key thing to ensure that we are on the right track is to ensure we write the full units, not only just write down moles, but also write down which molecule we're dealing with.

Over here, for example, let’s say I did not write the conversion factor correctly. Let’s say I wrote the conversion factor; I got confused, and let’s say I wrote the conversion factor this way. I wrote it down as, you know, I need three moles of hydrogen for 2 moles of ammonia. Let’s say I wrote it down this way.

Now when I look at the units, I say, “Wait, wait, wait a second! Moles of hydrogen, they're both in the numerator; they're not cancelling out.” Ah, that's how I realized that, oh no, no, this should have been in the denominator, and this should have been in the numerator. So this is how, just by looking at the units, I can be sure whether I'm on the right track or not. This is called dimensional analysis, and it's a powerful tool to get our stoichiometry right.

Anyways, once we have this, we can just plug the numbers in the calculator. We just have to do 4.43 * 2 / 3. Once we do that, we will get 2.95 moles of NH3, and that's our result.

Now, a quick thing before we move on to the next problem is that this result that we obtained is called the theoretical yield because this is the number that we get on paper, it’s in theory. But the actual amount of ammonia that you might really get in the lab if you perform this experiment could be quite different from this, and there are a few reasons for that.

One could be that not all the 4.43 moles of hydrogen would probably react with nitrogen; there could be some leftover hydrogen that just didn't react, so incomplete reaction could be one of the reasons. But another reason could be that, you know, maybe there was some impurity in the hydrogen gas that you had. Maybe, you know, it was not 4.43 moles of pure hydrogen to begin with, so you actually had less hydrogen, and therefore you'll get less amount of ammonia.

So whatever reason it is, the actual yield that you might get in a lab could be quite different. So for example, let’s say you perform this in a lab and you get the yield to be—the actual yield to be about 1.89 moles of ammonia.

Now we can define something called the percent yield, which is basically the actual yield divided by the theoretical yield multiplied by 100. So in our example, the percent yield would be the actual yield that you got in the lab divided by the theoretical yield that we got from our theoretical calculation, multiplied by 100.

Notice that the moles of NH3 cancel out, so the percent yield will not have a unit; it will be a unitless number. But if you multiply this, you will get about 64.1%. This number tells us that in the lab, we achieved about 64.1% of the maximum yield that we could theoretically get.

It’s kind of like the test results; that number tells us what percent of the maximum score have we achieved.

Anyways, our calculations on the paper will only give us the theoretical yield. Of course, for the actual yield, we have to do the experiment in the lab.

So let's move on to the next problem. In this case, we are reacting propane with oxygen to give us water and carbon dioxide. And the question is, how many moles of oxygen are needed to burn an excess of propane and produce 5.73 moles of carbon dioxide?

Great! You should pause the video and see if you can try this on your own.

All right, let’s do this. Our first step is always to check for whether our equations are balanced, and the equation is balanced. You can check that over here; for example, you have four atoms of oxygen and six atoms of oxygen, so total 10 atoms of oxygen on the right-hand side, 10 atoms of oxygen on the left-hand side. You can check for each atom, and it is balanced.

And then what does this balanced equation tell us in terms of moles? It says that one mole of propane needs to react with five moles of oxygen, giving us four moles of water and three moles of carbon dioxide. This is the mole ratio for this particular equation.

And what are we given? We are given that there’s excess of propane, so I don’t have to worry about propane. I have to now figure out how many moles of oxygen have I used up in this particular reaction to get 5.73 moles of carbon dioxide.

So this time I’m given the amount of products, and I need to figure out the amount of reactant. Does that change anything? No, because it doesn’t matter whether we are given the reactants or the products; the whole idea is to come up with a conversion factor, and you convert from one thing that’s given to another thing that we need to find out.

Okay, so here we are given the moles of carbon dioxide, and from that, we need to convert to moles of oxygen using this ratio 3:5 over here.

So again, I start by writing down what’s given to us. We’re given 5.73 moles of carbon dioxide. Now, to convert from there to oxygen, I'm going to use a conversion factor where the moles of oxygen are in the numerator and the moles of carbon dioxide are in the denominator.

So I can, you know, cancel carbon dioxide. So I write it as 5 moles of oxygen used up to give me 3 moles of carbon dioxide. That is my conversion factor over here. And I check with my dimensional analysis, and I see that, hey, moles of carbon dioxide cancel out.

Moles of carbon dioxide cancel out over here, and I get moles of oxygen, so I’m on the right track. So the dimensional analysis will always come to my rescue to make sure I'm on the right track.

Now, since this is giving me the moles of oxygen, I just have to put the numbers in the calculator: 5.73 * 5 / 3. And once we do that, we get about 9.55 moles of oxygen.

We can also always do a sense check. If we had used up five moles of oxygen, we would have gotten three moles of carbon dioxide, right? But what we actually achieved was slightly less than the double of that number, slightly less than six.

So that means we must have also used up slightly less than the double of this number, slightly less than 10, and that's exactly what we get. The ratios will always stay the same, and therefore you can see this is also slightly less than 10, and that's pretty much what we got.

Yes, if you’re wondering, again, this represents the theoretical yield. When we do stoichiometry on paper, the numbers we get for the product will always be the theoretical yield.