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Worked example: Calculating amounts of reactants and products

AP.Chem:
SPQ‑4 (EU)
,
SPQ‑4.A (LO)
,
SPQ‑4.A.1 (EK)
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SPQ‑4.A.2 (EK)

Video transcript

- [Instructor] We're told that glucose, C6H12O6, reacts with oxygen to give carbon dioxide and water. What mass of oxygen, in grams, is required for complete reaction of 25.0 grams of glucose? What masses of carbon dioxide and water, in grams, are formed? So, pause this video and see if you can have a go at this and then we'll work through this together. All right, now first let's just set up the reaction. So, this is going to be, we have glucose, so that is C6H12O6, is going to react with oxygen. Now, oxygen in its molecular form is going to be O2. And what it gives is carbon dioxide and water. Plus water. Now the next question is are we balanced. Do we have a conservation of mass here? And let's just go element by element. So first let's focus on the carbons. So, we have six carbons on the left-hand side of this reaction. How many carbons do we have on the right-hand side? Well, right now we only have one carbon in this carbon dioxide molecule. So if we want the carbons to be conserved we need to have six carbons on the right-hand side as well. So let me see what'll happen when I throw a six there. So now my carbons are balanced: six on the left, six on the right. Now let's look at the other elements. So on the left-hand side I have 12 hydrogens in total. On the right-hand side I only have two hydrogens. So if I want to balance that I could multiply the water molecule. Instead of just one here, I could have six, and now this would be 12 hydrogen atoms so both the carbons and the hydrogens are now balanced. By putting that six there I haven't messed with the carbons. And now last, but not least, let's think about the oxygens here. So on the left, we have six oxygens there and then another two oxygens there. So that's a total of eight oxygens. And on the right, I have six times two, I have 12 plus another six oxygens. So I have 18 oxygens on the right-hand side, and I only have eight oxygens on the left-hand side. So I have to increase the number of oxygens on the left-hand side. So let's see. This six you can imagine matches up with this six. So if I could somehow make this into 12 oxygens, I'll be good. So the best way to make this into 12 oxygens is to multiply this by six, so let me do that. So if I put a six right over here, I think I'm all balanced. I have six carbons on both sides, I have 12 hydrogens on both sides, and it looks like I have 18 oxygens on both sides. So I'm a fully balanced equation here. So let me get a periodic table of elements and I only need to think about carbon, hydrogen, and oxygen. So for the sake of space, let me scroll down a little bit. I just need to look at this stuff over here. And so, let's first, let's see hydrogen's right up here. And we can see in this periodic table of elements, it gives the average atomic mass, but you can also view that number 1.008, as its molar mass. So it's 1.008 grams per mole of hydrogen. Now we can move on to carbon. Carbon is 12.01, 12.01 grams per mole of carbon. And then last but not least, oxygen over here. That is 16.00 grams per mole of oxygen. And now we can use that information. I can now remove our periodic table of elements. We can use this information to figure out the molar masses of each of these molecules. So for example, glucose right over here, if we're talking about C6H12O6, how many grams per mole is that going to be? How many grams is a mole of glucose going to be? Well, it's six carbons, 12 hydrogens, and six oxygens. So one way to think about is going to be six times this, so it's going to be six times 12.01 plus we have 12 times 1.008 plus six times 16.00, and then this is going to be grams per mole. And this is going to be equal to, let's see, 72.06 plus 12.096 plus six times 16 is 96.00. Of course all of this is in grams per mole. And so this is going to be equal to, let's see, 72 plus 12 is 84 plus 96 is 180, 180 point, and let's see, we have 60 thousandths plus 96 thousandths, which would be 156 thousandths, so 156 thousandths grams per mole. Let me make sure I got the significant figures right. The six, the 12 and six are pure numbers, so I'm still good with this. And then when I add these numbers together, I need to round to as much precision as I have in the least one. So here I go up to the hundredths, up to the hundredths, up to the thousandths. So I need to round this, actually, to the hundredths place. So this is going to be 180.16 grams per mole of glucose. And now let's think about the other molecules. If we think about oxygen, I'll do that over here to save space, oxygen, that's pretty straightforward. A molecular oxygen molecule just has two oxygen atoms, so it's going to be two times this, so it's going to be 32.00 grams per mole. And then carbon dioxide. Carbon dioxide is going to be, it's two oxygens plus one carbon. So it's going to be this plus one carbon, so that's this plus 12.01, so that's 44.01 grams per mole. And then last but not least we have the water. And so, H2O. This is going to be two hydrogens, so that's two times 1.008, so that's 2.016 plus an oxygen. So that's going to be 2.016 plus 16 is going to be 18.016. Once again, we only go to the hundredths place here, so I'm going to round to the hundredths place here. So 18.02 grams per mole. Now the next step is to think about, all right we're reacting with 25 grams of glucose. How many moles of glucose is that? So we have 25.0, 25.0 grams of glucose. And we want to turn this into moles of glucose. And we know the molar mass is 180.16 grams per mole, or we could multiply and say that this is for every one mole, per mole, it is 180.16 grams. All I did is take the reciprocal of this over here. And notice, the grams will cancel with the grams and I'll be left with moles of glucose. That's going to be equal to 25.0 divided by 180.16 moles of glucose. 25 divided by 180.16 is equal to this. And let's see, I have three significant figures divided by five significant figures. So I'm going to round to three significant figures. So 0.139. So this is approximately equal to 0.139 mole of glucose. Now, they say the first question is what mass of oxygen is required for a complete reaction of this. Well, for every mole of glucose we need six moles of molecular oxygen. And so, we're going to need, let me just multiply that times six. So times six is equal to that. And remember, I had three significant figures. So, 0.833. So we're going to need 0.833 moles of molecular oxygen. And then I just multiply that times the molar mass of molecular oxygen. So, times 32.00 grams per mole of molecular oxygen. 0.833 times 32 is equal to that. If you go three significant figures, it's 26.7. 26.7 grams of oxygen, of molecular oxygen. And so we've answered the first part of the question. What mass of oxygen is required for complete reaction? So, that's this right over here. And then the next question is what masses of carbon dioxide and water in grams are formed. Remember, for every one mole of glucose, we needed six moles of molecular oxygen and we produce six moles of carbon dioxide and we produce six moles of water. So this is the number of moles of glucose we input into the reaction. Six times that was this. So this is actually also the number of moles of carbon dioxide or water that we're going to produce. So, if we take our original 0.833 mole of carbon dioxide and I multiply that times carbon dioxide's molar mass, 44.01 grams per mole, this is going to be approximately equal to, .833 times 44.01 is equal to three significant figures, 36.7 grams. 36.7 grams of carbon dioxide. And once again, you can see that the moles cancel with the moles, just as they did before, to give us grams of carbon dioxide. And then last but not least, we can do that with the water. 0.833 mole of H2O times its molar mass, times 18.02 grams per mole of water is going to give us approximately .833 times 18.02, gives us, three significant figures is going to be 15.0. 15.0 grams of water. And once again, those moles canceled out to give us the right units. And we're done, which is pretty cool. This is really useful in chemistry to be able to understand, based on a balanced chemical equation, to be able to understand, hey, if I have a certain mass of one of the inputs, one of the things that are one of the reactants, how much do I need of the other? And then how much mass of the products am I actually going to produce? All of which we just figured out.
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