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Video transcript

- [Voiceover] We know what a chemical equation is, and we've learned how to balance it. Now we're ready to learn about stoichiometry. This is an ultra-fancy word that often makes people think it's difficult, but it really is just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives. "Stoichiometry is the calculation of quantitative, "or measurable, relationships of the reactants "and the products," and you're gonna see in chemistry sometimes people use the word "reagents". For most of really our purposes, you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you wanna throw a reagent in and see if something happens, and see if your belief about that substance is true or things like that. But for our purposes a reagent and a reactant is the same thing. So it's a relationship, the reactants and products in a balanced chemical equation. So if we're given an unbalanced one, we know how to get to the balanced point, balanced chemical equation. So let's do some stoichiometry. What I'm gonna do is I'm gonna start, just so we get practice balancing equations, I'm always gonna start with an unbalanced equation. Let's say we have iron(III) oxide. That's F... You have two iron atoms with three oxygen atoms. Plus aluminum, A-L, and it yields A-L two O three plus iron. So, remember, when we're doing stoichiometry, first of all, we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation, but I think it's good practice to actually balance the equations ourselves. Let's try to balance this one. We have two iron atoms here, right, in this iron(III) oxide. How many iron atoms do we have on the right hand side? We only have one, so let's multiply this. Let's multiply this by two, right here. All right. Oxygen, we have three on this side, we have three oxygens on that side, that looks good. Aluminum on the left hand side, we only have one aluminum atom. On the right hand side, we have two aluminum atoms, so we have to put a two here. And we have balanced this equation, so now we're ready to do some stoichiometry. So stoichiometry is essentially, look, if I give you, there's not just one type of stoichiometry problem, but they're all along the lines of, hey, if I give you x grams of this, how many grams of aluminum do I need to make this reaction happen? Or if I give you, you know, y grams of this molecule and z grams of this molecule, which one's gonna run out first? That's all stoichiometry, and we'll actually do those exact two types of problems in this video. So let's say that we were given, let's say we were given, 85 grams we were given 85 grams of the iron(III) oxide. So we're given 85 grams. My question to you is how many grams of aluminum do we need? How many grams of aluminum. Well, you look at the equation, you immediately see the mole ratio. For every mole of this, so for every one atom we use of iron(III) oxide, for every one of that, we need two aluminums, right? So what we need to do is figure out how many moles of this molecule there are in 85 grams, and then we need to have twice as many moles of aluminum, right, because for every mole of the iron(III) oxide, we have two moles of aluminum, and we're just looking at the coefficients. We're just looking at the numbers. One molecule of iron(III) oxide combines with two molecules of aluminum to make this reaction happen. So, let's first figure out how many moles 85 grams are. What's the atomic mass, or the mass number, of this entire molecule? Let me do it down here. So we have two irons and three oxygens. Let me go down and figure out the atomic masses of iron and oxygen. Iron is right here, 55.85. I dunno, I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron that has 30 neutrons. It has a mass number of 56, so iron has 56 atomic mass number, and then oxygen we already know is 16, so iron was 56. So we know that this mass is going to be two times 56 plus, plus three times 16, we can do that in our head. So one molecule of iron(III) oxide is going to have, is going to be 160 atomic mass units. One mole, one mole, one mole, or 6.02 times 10 to the 23, molecules of iron oxide... One of those pop-ups... Iron(III) oxide, is going to have a mass of 160 160 grams. In our reaction, we said we're starting off with 85 grams of iron oxide, how many moles is that? Well, 85 grams, 85 grams of iron(III) oxide is equal to 85 over 160 moles, is .53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron(III) oxide is. We said for every mole, and we figured out it's .53 moles, 'cause a full mole would have been 160 grams, but we only have 85. It's .53 moles. We know from this balanced equation that for every mole of iron(III) oxide we have we need to have two moles of aluminum. If we have .53 moles of the iron molecule, iron(III) oxide, then we're going to need twice as many aluminums, we're going to need 1.06 moles of aluminum, right? I just took .53 times two. 'Cause the ratio is one to two, for every molecule of this we need two molecules of that. For every mole of this, we need two moles of this. If we have .53 moles, you multiply that by two, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum, and then multiply that times 1.06 and we're done! So aluminum, or aluminium, as some of our friends across the pond might say. Aluminium, I enjoy that more. Aluminium has an atomic, well, the atomic weight or the weighted average is 26.98, but let's just say the aluminium that we're dealing with has a mass of 27 atomic mass units. One aluminium is 27 atomic mass units, so one mole of aluminium is going to be 27 grams, or 6.02 times 10 to the 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? 1.06 moles of aluminium is equal to 1.06 times 27 grams, and what is that, what is that? 1.06 times 27 equals 28.62. So we need 28.62 grams of aluminium. I won't write the whole thing there, in order to make, to essentially use up our 85 grams of the iron(III) oxide. If we had more than, what was the number, more than 28.62 grams of aluminium, then there'll be left over after this reaction happens, assuming we keep mixing it nice and the whole reaction happens all the way and we'll talk more about that in the future. If we have, and in that situation where we have more than 28.63 grams of aluminium, then this molecule would be the limiting reagent, because we had more than enough of this, so this is what's going to limit the amount of this process from happening. If we have less than 28.63 grams of, I'll start saying, aluminum, then the aluminum will be the limiting reagent, because then we wouldn't be able to use all of the 85 grams of our iron molecule our iron(III) oxide molecule. Anyway, I don't want to confuse you in the end with that limiting reagent. So the next video we'll do a whole problem devoted to limiting reagents.