 Okay, in this video we are going to be covering a different type of problem than the type we covered in the previous video. This type of problem is going to be called a mole-to-gram problem, or at least some people call it that. Before we actually go to the problem itself, you can ignore that. You should try to balance this equation, so you should spend a minute or so working on that. Pause the video and then I'll go over it. Okay, unpausing. We do, or at least I do, the typical make a table. Left side, right side of the arrow. So and then we systematically go through each element. How many carbons on the left? Oh, let's put the numbers in first. One, one, one, okay. How many carbons on the left? Two, because we're doing one times two. How many carbons on the right? Only one, because we're doing one times one. So we've got our first unbalanced atom. We have to fix the side that's short, so the easiest way to fix that is to double this. Change this to a two, and then we have two carbons on the left and the right. How many hydrogens on the left? Well, one times two, also two hydrogens. How many hydrogens on the right? One times two, two hydrogens. So we've got two things balanced. Last up is oxygen. We're going to have a problem here. It's not going to be the end of the world, but you're going to have to live with this for at least a little while. How many oxygens on the left? Two. How many oxygens on the right? Remember, if you look carefully, the oxygens are distributed over two molecules. How many oxygens on this guy here? Well, two times two, that's four. How many oxygens here? Well, one times one is one. So we have a total on the right side of five oxygen atoms. Okay, so it's not balanced, right? Our equation, as it stands, with this one, one, a two, and a one, not balanced. The left side doesn't have enough oxygens. You need to turn this two into a five. The way to do it, you can struggle with this if you want, but we can put a 2.5 here, or two and a half. Two and a half times two is equal to five. At this point, a lot of times, some of the students basically cry out and say, you can't do that, or I didn't know you could do that. Sometimes they say that. And what I'm going to say is, you bet you can. The reason you can is, if we're thinking about this in moles, then we can certainly have 2.5 moles of something. The reason students fight it is either they have been told in the past that they can't stuff a fraction in here, but I'm telling you you can. We're all consenting adults here, so as long as we're not breaking any rules, we can do it. What they've also been told is this means one molecule, and you can't have 2.5 molecules of something. You can't split a molecule down the middle, which is true, but what I'm telling you is that we're talking about moles. Now I will tell you that usually if you can, people will try to adjust equations like this to get rid of the fraction. There is a way to adjust it, and what you can do in this case is you can double all of the front numbers, and it will stay balanced, and if you double this, it will turn into a whole number. So if we double it, let's just check. We turn this 1 into a 2, so it's 2C2H2. If we double this 2.5, that's 5, 5O2. If we double this 2 here, that's 4CO2, and if we double this 1, that's 2H2Os. Let's just check to see if we still have a balanced equation. How many carbons? Four on the left. How many carbons? Four on the right. How many hydrogens? Four on the left. This is carbon, this is hydrogen. How many hydrogens on the right? Four. How many oxygens? Ten. How many oxygens here? Eight, because we're doing four times two. And how many oxygens here? Two times one is two. Eight plus two is ten. So we have ten oxygens on the left, ten on the right, four hydrogens on the left and the right, four carbons on the left and the right. So what I am going to tell you is that it's perfectly okay to use fractions of non-whole numbers in front of your formulas in a balanced equation. Either way is correct. So you can write one C2H2 plus 2.502 makes two CO2 plus one H2O. Or you can say two C2H2 plus 502 makes four CO2 plus two H2O. Either equation is correct. They're both balanced. So whatever makes you happier. So now let's go on to the actual problem. So let me rewrite the numbers in for the balanced equation that I believe is balanced. Now, and don't worry about the energy again. That was just me trying to faithfully reproduce the question from the book. Here's the question. How many grams of CO2 can be produced when 54.6 grams of C2H2 is burned? So here's our C2H2. Burned basically means we heated it up. So I'm going to put the triangle in the presence of oxygen. The first question I usually ask the on-ground students is what is the problem with this problem? And the problem with it is that these numbers here, the one, the two and a half, the two and the one, those are moles. Or at least I want you to think of them as being moles. The units that they're asking about are grams. So there's an extra step that we need to do. We need to somehow convert moles to grams or we need to convert grams to moles, which is why this is called a mole-to-gram problem. Now I will tell you there are about 10 zillion different ways of solving this problem. Each has its good points and its bad points. I'm going to show you the way that I would do it. If you have a better way that makes you happier, you can do it your way. So what I'm going to do is I'm going to take these numbers, one mole of C2H2, 2.5 moles of O2, etc. I'm going to convert them to grams. If you remember at the beginning of this unit, I said if you know number of moles and you know the formula, you can figure out how much something weighs, which is basically grams. And we know the number of moles and we know the formula. So we can convert the grams, but this requires using the idea of molar mass. So I want to know how much one mole of C2H2 weighs, or more formally the molar mass of C2H2. And we're only going to round to the nearest whole number to keep things simple in these calculations. If you look up the molar mass of carbon, it's 12 grams, but we have two carbons, so it's 12 times 2. That's 24 grams. If you look up the molar mass of hydrogen, that's 1 gram, but we have two hydrogen, so it's 1 times 2 is 2 grams. So the molar mass of C2H2 is these two numbers added together, 26 grams of C2H2 per mole. So instead of saying one mole of C2H2 like we've written in this equation, I could replace it, I could say 26 grams of C2H2. And up here, what I'm going to do is I'm going to write the balanced equation, but I'm going to write it in grams instead of moles. So we're going to do this type of calculation here. We're going to do it for every molecule in our equation. So let's get to work. That's the molar mass of O2, because that's the next guy in our equation. Molar mass of O2, well, the molar mass of O is about 16 grams, but O2 has two of those, so 16 times 2, 32 grams of O2 per mole. So if I had one mole of O2, you would weigh 32 grams. But in the balanced equation, we have two and a half moles of O2. So it's going to be 32 times 2.5. And what is that? That's 64 plus another 16, 74, that's 80 grams. So instead of saying two and a half moles of O2, I can say 80 grams of O2. So see how I've written the front part of this equation. Instead of saying one mole of C2H2 plus 2.5 moles of O2, I can say 26 grams of C2H2 needs to mix with 80 grams of O2 to make, well, let's keep going. What's the molar mass of CO2? Because CO2 is the next molecule in our equation. Molar mass of carbon is 12 grams. We only have one carbon in our formula, so it's 12 grams, 12 times 1. Molar mass of oxygen is about 16 grams. We have two oxygens, 16 times 2 is 32 grams. If we add these two numbers together, 44 grams of CO2 per one mole of CO2. So if I had one mole of CO2, it would weigh 44 grams. But in my balanced equation, how many moles do I have? I have two. So two moles of CO2 must weigh 88 grams because I just doubled this number. Let's go on to the final molecule. Final molecule is water, molar mass of H2O. Molar mass of hydrogen is about one gram, and we have two hydrogens, so one gram times two. Two grams molar mass of oxygen, about 16 grams. We only have one, so 16 times one, 16 grams. Add them all together, 18 grams of water is one mole of water. How many moles do we have in our equation that's balanced? One mole. So this is 18 grams, 88 grams of CO2 plus 18 grams H2O. So we have rewritten this bottom balanced equation in a completely different way. The bottom equation says one mole of C2H2 needs to mix with 2.5 moles of O2. And I rewrote it as 26 grams of C2H2 needs to mix with 80 grams of O2. And it says that can make two moles of CO2, but I said that just is another way of saying 88 grams of CO2. And it also makes one mole of H2O, and I said that's just another way of saying 18 grams of H2O. So this top equation is just the bottom equation rewritten in grams. Now before we go on to the problem, I want to ask a question about the top equation. The question is, and you should pause the video and answer, I guess two questions. How much does everything on the left side of the arrow weigh together? How much does everything on the right side of the arrow weigh together? And if you pause and figure that out, and then on pause, everything on the left 80 plus 26 grams, it's 106 grams on the left, 88 plus 18 grams, also 106 grams on the right. The reason they match is because the equation is balanced. In other words, we have not gained or lost any material. We didn't lose any weight, we didn't gain any weight. We started with 106 grams worth of stuff. We ended up with 106 grams worth of stuff. This is another crude test to see if you've balanced your equation correctly, if the weights all match up on the left and the right. So now let's go on to the problem. The problem is how many grams of CO2 can you make when you use 54.6 grams of C2H2? Well, according to this equation, 26 grams of C2H2 can make 88 grams of CO2. That's just coming from this, can make this much CO2. Assuming that we have enough oxygen, but that's why the word can is in there, because we're just saying, under the best possible conditions, how much CO2 could you make, and if I had 26 grams of C2H2 under the best possible conditions, I could make 88 grams of CO2. But the question doesn't say that we have 26 grams of C2H2. It says we have 54.6 grams of C2H2, can make, I don't know how many grams of CO2. So how do we do this? Well, we can write these numbers as a ratio. I can say 26 grams of C2H2 can make 88 grams of CO2. And then I make an equal fraction off to the right, and I say, but I don't have 26 grams of C2H2, I have 54.6 of C2H2. How many grams of CO2 can I make? And then we can cross-multiply and solve for x. If you want to at least get a crude estimate of what x is going to be, you might notice that this number is a little bit more than twice as large as this number. This is 2x plus bigger than 26. So whatever x is, it better be a little bit more than 2 times bigger than 88. So it should be in the 180 gram range, around 180 grams. Let's see if we are correct. We're going to cross-multiply, so these two numbers, multiplied together, 26 grams C2H2 times x grams CO2 equals these numbers, multiplied together, 88 grams CO2 times 54.6 grams C2H2. And we want to get the x alone, which means we want to get rid of this 26 grams of C2H2. To do that, we divide both sides by 26 grams C2H2, 26 grams C2H2. And on the left side, they reduce down to 1. On the right side, grams of C2H2 cancel. The only unit we're left with on both sides of the equation are grams of CO2, grams of CO2. And what is the question asking? How many grams of CO2 can we make? So we've probably set it up correctly. X grams CO2 is equal to 88 grams CO2 times 54.6 divided by 26. And what is that equal? 184.8 grams of CO2 can be made, which is close to what our estimate was, right? I said it was going to be about 180 grams, so we probably did the calculation correctly. And don't worry about significant digits. So this is a typical example of a mole to gram problem. It's one of the more difficult problems in the course. Again, there will probably be one question like this that shows up on the second exam. Probably be one question like this that shows up on the final exam. But usually, students feel like they're being tortured with this. So this is a tiny bit of the course, but it's somewhat difficult, so it gets its own video. So that is it for mole to gram problems. Again, there are probably about 10 billion practice ones in your book and on the internet. So I don't know, practice if you have time.