Limiting Reactants and Percent Yield


Mr. Andersen explains the concept of a limiting reactant (or a limiting reagent) in a chemical reaction. He also shows you how to calculate the limiting reactant and the percent yield in a chemical reaction.


Transcript Provided by YouTube:

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Hi. It’s Mr. Andersen. And today I’m going to be talking about limiting reactants and
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percent yield. Sometimes we call limiting reactant limiting reagents. But those are
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essentially the same thing. And this is always going to occur in a chemical reaction. So
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before we talk about chemical reactions, let’s actually talk about food. And so if we were
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making cookies, let’s say we were making Nestle’s Chocolate Chip Cookies. This is the recipe
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for it. And if you were to go down your ingredients list before you start making the cookies and
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you see, yeah I’ve got the flour. I’ve got the teaspoon of baking soda. I’ve got 3/4
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cup of sugar. But then you get down here and you look in the fridge and you realize, oh
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I only have one egg. And I need to two eggs for this recipe. Well chemical reactions work
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the same way. You’ve got reactants. And you can think of those as like the ingredients
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before the actual reaction. And if you don’t have all of them, then there’s going to be
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one thing that’s going to limit that reaction. We call that the limiting reactant. In this
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case it would be eggs. Now can I not make cookies anymore? No way. All you do is just
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cut all of these down. So if they only have one large egg instead of two, then I just
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cut everything in half. So it’s a half of teaspoon of salt. A half a cup of butter.
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And so it still works. I’m still able to make cookies. But I just cut everything in half.
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And so those coefficients that you see in front of equations, if you think of those
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as like the amount is a good way to think about it. So let me kind of start by showing
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this in a visual sense. So if we’re doing hydrogen combustion and an example of this
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would be, this is in the main rocket chamber in the space shuttle. What we have in here
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is liquid hydrogen and liquid oxygen. And those are combining and we’re getting this
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huge exothermic reaction which is creating a lot of the lift. So this would be the chemical
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reaction. Hydrogen plus oxygen yields H2O. If we actually draw that out then, here would
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be the reactants on the left side. This would be hydrogen gas and this would be oxygen gas.
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And then the products I haven’t shown you yet. And so if you look at this, you know
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that H2 and O are going to combine together to make H2O. And so you should, if you pause
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the video right now, you’d be able to take a second and probably figure out what the
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limiting reactant is. Looks like we’ve got a little more hydrogen then oxygen. So what’s
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the limiting reactant? Let’s do the actual reaction. So what do we get? We get 4 water
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molecules. And then we have 1 oxygen left over. And so what was the limiting reactant?
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The limiting reactant in this case was the hydrogen gas. Because how many atoms of hydrogen
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gas did we have? We’ve got 2, 4, 6, 8. And so we only had enough to make four water molecules.
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Now we had enough oxygen to make 2 more water molecules. But the hydrogen in this case were
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like the eggs in our recipe. They were limiting the reaction. Let’s go to the next one. This
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is combustion of methane. Combustion of methane, what you’re doing in that is you’re actually
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combining natural gas. And that’s, methane is a hydrocarbon. So it’s 1 carbon and 4 hydrogens.
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With oxygen just found in the room and we’re making carbon dioxide and water. That’s how
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a bunsen burner works. So if we show you the reactants that I happen to have in this really
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small area. I’ve got 1 methane molecule. And then we’ve got, it looks like 4 oxygen molecules.
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And those are going to make water and carbon dioxide. So if you actually do the reaction,
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there we go. I get 2 water molecules and I’ve got 1 carbon dioxide molecule. So what was
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the limiting reagent this time? The same thing. Again we’ve got plenty of oxygen right here,
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but we’ve only got 1 methane molecule. And so what was limiting us was the methane. So
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we call that the limiting reactant or the limiting reagent. So what would be left over?
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Well everything that didn’t’ react. In other words that oxygen that was left over was just
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going to stay as oxygen. It’s not used. It’s like the other half a cup of butter that we
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used in our recipe. And so those are graphically kind of showing you what limiting reagents
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are. Unfortunately atoms aren’t this big in the laboratory. If they were this big we could
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just look at how much we have and we’d be able to figure out which is going to be the
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limiting reactants. And so we have to use our friend the mole to actually figure out
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how much there is. And so let’s say we have this reaction down here. We’ve got methane
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plus oxygen yields carbon dioxide plus water. And let’s say together I’m going to mix two
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things. I’m going to take 10 grams of oxygen and I’m going to mix that with 10 grams of
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methane. So the question that you could be posed with is which of those is the limiting
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reagent? Which of these is the limiting reactant? Is it the methane or is it the oxygen? Well
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you can’t tell by looking at a mass. So you have to do some calculations. And we can’t
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see the atoms so we’re kind of stuck. And so let me talk you through this process. First
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thing we do is we convert them to moles. We then compare the moles. In other words we’re
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going to compare the amount of methane and oxygen to one of the products on the other
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side. In this case I’ve chosen carbon dioxide. We’re next going to convert that back to grams.
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And then we’re going to figure out all the possible products. And so if these steps are
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confusing to you, you may want to look at the video on converting grams to moles. So
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first thing we do is convert to moles. And so I’m going to put my grams on the bottom
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and then my moles on the top. These numbers right here are the formula masses. And so
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we’re getting that from the periodic table. In other words to get methane we’re taking
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the atomic mass of 1 carbon plus 4 hydrogens to get that. Okay. So if we look at it what
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we’ve converted to, if I were to cancel out my grams of methane, is I’ve converted to
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moles of methane and moles of oxygen. If we do our next conversion I can then compare
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the moles. And so on the bottom here I have 1 mole of methane. I’m grabbing that from
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the top over here. And one mole of oxygen to two moles of oxygen. Now where am I getting
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2 on the bottom and 1 mole on the top? I’m doing that from my actual equation here. And
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so there’s an implied 1 in front of the methane right here. And so that’s why there’s 1 mole
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here. And there’s 2 moles of oxygen. Why did I only put 1 mole of carbon dioxide on the
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top? Because on the other side, on my product side, there’s only 1 mole. Now we could have
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chosen instead on this comparing moles, we could have put one mole of methane on the
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bottom and 2 moles of water. In other words we could have compared both of these reactants
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with water. And we still would have figured out which is the limiting reactant. Next I
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could convert it to grams. And I do that just by now taking the moles of carbon dioxide
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and figuring out the formula mass of carbon dioxide. And then I finally get my possible
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results. And so if I had 10 grams of methane. That’s enough methane to make 27.4 grams of
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carbon dioxide. So that’s a lot. That 10 grams of oxygen however could only make 6.88 grams
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of carbon dioxide. And so what’s the limiting reactant? Well this equation or these equations
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or these conversions show me that I’m going to run out of oxygen long before I’m gong
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to run out of methane. And so what’s the limiting reactant? The limiting reactant is the amount
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of oxygen. In other words 10 grams of oxygen only gets me 6.88 grams of carbon dioxide.
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And so that’s the limiting reagent, the amount of carbon dioxide. And I can even predict
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how much grams of carbon dioxide I’m going to get. So if you think of yourself as a chemist,
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what you could now do is put in a 6.88 grams of carbon dioxide. You could work backwards.
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And you could figure out exactly how much methane you have to add to get 6.88 grams
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of carbon dioxide. And if you’re a smart chemist you’re always in any reaction that you do,
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you’re always going to make your most expensive chemical as a reactant be your limiting reactant.
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So you’re not going to send, you’re not going to use any bit more of that. Okay. The other
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thing I mentioned I would talk about is the percent yield. And so in this theoretical
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experiment we’re going to get 6.88 grams of carbon dioxide. That’s with all the oxygen
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reacting with all of the methane. But that never happens. So this is a what we call,
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6.88 would be a predicted yield. If everything went perfect. In other words if every molecule
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of methane reacted with every molecule of oxygen we’d get 6.88 grams of carbon dioxide.
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Now you can think about it. That’s never going to happen if you think about the billions
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and billions and billions of molecules. The chances that they’re all going to react and
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some aren’t going to be lost is really, really rare. And so lots of times you’re actual yield
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will be much less than your predicted yield. In other words I might be hoping to get the
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6.88 but I only get 6.19. And so those 2 values can tell me the percent yield. And so the
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percent yield is essentially the actual yield. In other words how much I get divided by the
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predicted yield, how much I was hoping to get. And so if I throw those values in here,
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you can see the smaller value is going to be on the top. That’s how much I actually
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get back. And the precent yield is never 100 percent. And if I divide the top by the bottom
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and then multiply it by 100 percent I get 90.0 percent, percent yield. And that show
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me that this is a pretty good reaction. In other words I’m getting most of the product
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that I predicted that I would be able to get. And so that’s limiting reactants. That’s percent
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yield. And I hope that’s helpful.


This post was previously published on YouTube.

Photo credit: Screenshot from video