Stoichiometry example problem 2

We're told that glucose reacts with oxygen to give carbon dioxide and water. What mass of oxygen, in grams, is required to complete the reaction of 25 grams of glucose? And they also want to know what masses of carbon dioxide and water are formed? Well let's first just write the reaction. So they're saying glucose reacts with oxygen to give carbon dioxide and water. So glucose, C6H12O6, reacts with oxygen in its molecular form-- there's two atoms there. The products are carbon dioxide and water. And you might already recognize this. This is a stoichiometry problem. They're saying, hey, we have 25 grams of glucose. How much oxygen is required to completely react with that glucose? And then how much carbon dioxide and how much water is going to be produced in grams? That's what stoichiometry problems are all about. And if you remember from the last video, the first thing you should always do is make sure that your equation is balanced. So let's make sure it's balanced. On the left hand side-- and you always want to do the most complicated molecules first, and then do the simplest molecules last, because those are the easiest ones to balance out. So on the left hand side, here, I have a 6 carbons. On the entire left hand side. On the right hand side, I only have 1 carbon. So let me multiply this over here by 6. And now I have 6 carbons on both sides of this equation. Let's move to the hydrogens. I have 12 hydrogens on the left hand side of this equation. The 12 are all sitting right there in the glucose. How many do I have on the right hand side? Well I only have 2 hydrogens. So let me multiply that times 6. And I didn't mess with the carbons at all, so that shouldn't change anything-- so now I have 12 hydrogens on both sides of this equation. 12 here, 12 there. And now I can just balance out the oxygen. I saved that for last because I just have the oxygen molecule here. That's the easiest one to balance. So how may oxygens do I have on the right hand side? I have 6 times 2 here. I have 12 oxygens there. And then I have another 6 oxygens over here. So plus 6 oxygens. I have 18 oxygens on the right hand side of my equation. So I need to have 18 oxygens on the left hand side of my equation. How many do I have right now? I have 6 oxygens over here. So I'm going to need 12 over here. Right? This is the last thing I want to mess with. I don't want to put a coefficient out here. That'll change everything else. I just want to put a coefficient here. That'll make everything balance out. I have 18 on the right hand side. I already have 6 here. I want to have 12 right here. So let me multiply this times 6. And now everything should work out. I have 6 carbons on both sides. I have 12 hydrogens on both sides. And I have 18 oxygens. Six here, 12 here. 12 here, 6 here. I have 18 oxygens on both sides of this equation. Now, the next thing we're going to want to do is figure out how many moles of the reactants that they're telling us about that we have. So they're telling us that we have 25.0 grams of glucose. So let's figure out how many moles per gram, or how many grams per mole, there are of a glucose molecule. And since everything here is dealing with carbons, and oxygens, and hydrogens, let's look up the atomic weights of all of them. So carbons, hydrogens, and oxygens are pretty common. So at some point you might want to memorize their atomic weights. And I want to give proper credit to the person whose periodic table I'm using. Le Van Han Cedric. I got this off of Wikimedia. It's a creative commons attribution license. So I want to make sure I attribute the person who made the periodic table. But we have oxygen. It has an atomic weight of 15.9999. Usually it's given as 16. But I'll just write it like that. So oxygen is 15.999. And we're going to have to figure out carbon, and hydrogen. Let's go back to the periodic table. We have carbon, has an atomic weight of 12.011. And hydrogen is 1.0079. So let's do the carbon, 12.011. So the carbon is 12.011. And then the hydrogen-- remember atomic weights are all the weighted average of all of the isotopes on earth-- is 1.0079. 1.00-- what was it? Was it one zero or two zeroes? It was two zeroes. 1.0079. So what's the atomic weight of glucose, given all of that? So what's the atomic weight of glucose? Let me scroll down a little bit. So the atomic weight of glucose, C6H12O6, it's going to be equal to 12.011 times 6. 6 times 12.011. Plus 12 times hydrogen. Just to keep things simple, actually, let me just make this into 16. let me make this into 12, and let me make this into 1. That's going to make our math a lot easier. So let me just do it like that. Let me clear this. So the atomic weight of glucose is 6 times the atomic weight of carbon. Six times 12. Six times 12. Plus 12 times the atomic weight of hydrogen. Plus one times 12, or maybe I have to write 12 times one, just to be consistent. 12 times one. Plus 6 times the atomic weight of oxygen. Plus 6 times 16. So what is this? This is equal to 72 plus 12. Plus, 6 times 16 is 60 plus 36. It's 96. And that is-- I'll get the calculator out-- 72. Plus 12, plus 96, is equal to 180. So the atomic weight of glucose is equal to 180. Which tells us, that let me start doing the problem. We have 25.0 grams of glucose. So I'll just write grams of C6H12O6. We want to write this in terms of moles of glucose. So we want to cancel out the grams. So we want the grams in the denominator. Right? Here it's in the numerator. If we divide by that unit, we're going to cancel it out. So we want the grams of C6H12O6, or the grams of glucose in the denominator. And we want the moles of glucose in the numerator. Because then when we perform this calculation, that will cancel with that, and we will be left with moles. So how many grams per mole? Well why don't we just figure it out? The atomic weight is a 180. So it's 180. If we have Avagadro's number of these molecules, it's going to have a mass, I should say, of 180 grams per mole. That's the information we got by figuring out glucose's atomic weight. So let's just perform this calculation. These units cancel out with those units. And so we just take 25 times one, divided by 180. So this is equal to 25 over 180. And all were left with in units is moles C6H12O6, or glucose. And what is 25 five divided by 180? I should say 25.0 just so we know that we have three significant digits here. And I don't want to be too particular about significant digits. Sometimes I'm a little bit loosey goosey about it. And I was already loosey goosey about the topic weight. But we'll try to be someplace in the ballpark. So 25 divided by 180 is equal to 0.139. So we have 0.139 moles of-- I'm tired of writing, I'm just going to write-- moles of glucose. Now, we want to figure out how many moles of oxygen-- or, well, the first step we want to figure out is for every mole of glucose, how many moles on oxygen? And we know that. It's 1 to 6. So let's write that down. And that's going to tell us how many moles of oxygen we're going to need. And, remember, we want to get rid of these moles of glucose. So we want to write that in the denominator. So moles of glucose required. For every 1 mole of glucose required, we figured out when we balanced out the equation, for every mole of glucose, we need 6 moles of oxygen. We need 6 moles of O2 required. And then what do we get here? Well we have the moles of glucose canceling out with the moles of glucose. And then we multiply the 6 times 0.139. So let's just multiply this times 6 is equal to 0.833. So we need 0-- that's too fat-- 0.833 moles of oxygen required. Now, we're almost there, we know how many moles of oxygen are required, molecular oxygen. Now we just have to figure out how many grams that is. So let me just write it over here. We have 0.833 moles of molecular oxygen are required. We're going to multiply that. Remember, we want to get this into grams now. So let's try the moles of O2 in the denomenator. And we're going to put grams of O2 in the numerator. So how many grams of O2 are there per mole? Well, we know the atomic weight oxygen is 16. So the atomic weight of molecular oxygen, where you have two of these, is equal to 16 times two. Which is equal to 32. So it's 32 grams per mole. So when we perform the calculation, the moles of O2 in the numerator cancel out with moles of O2 in the denominator. And we're just going to have 0.833 times 32 is equal to 26-- I'll just say, seven. So this is equal to 26.7 grams of O2 required. Grams of O2, and I'll write required in that same exact color. So we finished one part of the question. We have two more parts left. If we go back to the problem, it asks us, we already figured out, what mass of oxygen is required for the complete reaction? We did that part, so we can write a little check there. We did that first part. But they also ask, what masses of carbon dioxide and water are formed? So now we need to figure out carbon dioxide and water. And so we go back here. We know we have 0.139 moles of glucose. We figured that out in the first part. So 0.139 moles of glucose. And let's do the carbon dioxide first. We know for every mole of glucose, we're going to produce 6 moles of carbon dioxide. We see that right there. There's a one out here implicitly. So for every mole of glucose, you're going to produce 6 moles of carbon dioxide. So let's write that. So for every one mole of glucose, you're going to have 6 moles of CO2 produced. We just got that directly from the equation. And I wrote it this way, instead of one over the 6, because I wanted it to cancel out with the moles of glucose. Moles of glucose in the numerator, moles of glucose in the denominator. So that canceled out with that. And this is going to be equal to 6 times 0.139 moles of CO2 produced. And we know what that is. We already multiplied 6 times 0.139 before. This is going to be equal to 0.833 moles of CO2 produced. We've already done that calculation. And actually we could even-- well, I won't skip too many steps-- but we see over here, for every one mole of glucose, 6 moles of this, 6 moles of this, and 6 moles of that. So that's why we had the 0.833 moles of oxygen. And we're also getting the 0.833 moles of carbon dioxide. And, if we did the exact same thing, we've performed the exact same calculation. For every mole of this used, you have a mole of that produced. Or every 6 of this, you have 6 of that, 6 of that. So you're also going to have, by the very same logic, I mean, you could do this again with the water. You're also going to have 0.833 moles of water produced. Right? For every mole of carbon dioxide, you have a mole of water. Here it's 6 for 6, but it's a 1:1 ratio. You could think about it like that. Now let's figure out how many grams this is. We have the same number of moles of carbon dioxide and the same number of moles of water. But they're going to have different number of grams. So the actual number of molecules are the same, but the actual mass of those molecules are going to be different. So what's carbon dioxide's atomic weight? Just to remind ourselves. Carbon has atomic weight of 12, oxygen of 16. So it's going to be 12 plus 16 times two, which is equal to 12 plus 32, which is equal to 44. So if we are starting off with 0.833 moles of carbon dioxide that are produced, we want to figure out how many grams per mole-- we're going to put the moles of carbon dioxide in the denominator because we want it to cancel with that. And then we have the grams of carbon dioxide. We know carbon dioxide's atomic weight is 44. So that means there are 44 grams. If we have a mole of carbon dioxide, if we have 6.022 times 10 to the 23 carbon dioxide molecules. So this is going to be 44 times 0.833 is-- what do we get? 0.833 times 44 is equal to 36 point-- let's just say, 7. 36.7 grams of CO2 are going to be produced. And then we're almost done. We just have to figure out how many grams of water are going to be produced. We know it's the same number of moles, but how many grams is that? So water's atomic weight is going to be 2 times 1, right? Because we have to 2 hydrogens. Plus 16. Which is equal to 18. So if we're starting off with 0.833 moles, or if we're producing 0.833 moles of water, we just multiply that times how many grams per mole of water? So 1 mole of H2O-- once again we want this in the denominator so it cancels out with this in the numerator. 1 mole of H20 has a mass of 18 grams. Or 18 grams of H20 for every mole of H20. So this cancels with that and we get-- get the calculator out-- 18 times 0.833 is equal to 14.994. So we can just round that to 15.0. So we have 15.0 grams of water produced. And we're done! We are done. We figured out that if we start off with that 25 grams of glucose, like they told us at the beginning of the problem, we're going to require 26.7 grams of oxygen to react with it, we're going to produce 36.7 grams of carbon dioxide, and 15 grams of water.