We're going to begin today with a so-called review of the concepts that we developed in the previous lectures. Remember, our fundamental goal here is to attempt to count numbers of particles, at least in relationship to one another, because when we can count those particles, we could do things like determine the formulas of individual molecules or do a chemical reaction calculations. We figured out a rather systematic and clever way to count the numbers of particles by counting the number of moles of particular particles. Where a mole, you'll recall is a specific number, and it is the number of atoms in 12 grams of carbon. So, it's just some specific number. And every time you hear us talk about moles, what you want to think is, that's a specific number of particles just like a dozen, is a specific number of particles, or a gross is a specific number of particles. And if I know that I have one mole of carbon and I have one mole of helium, that I know I have exactly the same number of atoms of carbon than I have of helium. Furthermore, we can figure out the mass of a mole of any particles. We know what the mass of a mole of carbon is. By definition, it's 12 grams. But since the ratio of the mass of a carbon atom to a helium atom is 12 to 4 , than the mass of a mole of carbon atoms to a mole of helium atoms is 12 to 4. And therefore, a helium mole weighs 4 grams. So, it's straightforward to then figure out what the mass of a single mole is and the mass of a single mole then is something that we will call the molar mass and it is equal in grams to the mass of a single atom or a single molecule weighed in the original atomic mass units. Once we know the mass of one mole, then it's a relatively straightforward thing for us to figure out how many moles we have from this ratio here. We'll just take the ratio of the mass of the sample that we have weighed to the mass of one mole and that ratio, is simply going to then be the number of moles. And remember, the number of moles is a count of particles. So, by weighing the mass of the particles, we've actually figured out how many particles there are. This formula here, this simple ratio, is an equation that we will use often. And why do we calculate the number of moles? Because it's a way of counting the number of particles. And counting particles is important to us, as we saw in the previous lecture, to determine molecular formula. As we will see in this lecture, to do calculations about chemical reactions. And as we'll see in the next lecture, is a means of determining the concentrations of solutions. Let's illustrate this with a particular example. Let's say, we have substance, acetylene. And we don't actually know what acetylene is at this point, other than it contains hydrogen and carbon. We'll burn it in oxygen and when we burn it, we're going to produce the products, carbon dioxide and water. And we'll weigh the mass of those products that come out of burning, 10 grams of acetylene. Well, how are we going to use these data? Maybe the best way to do this is to attempt to count the atoms of carbon and hydrogen. And we can do that because we can actually count the number of particles of carbon dioxide by counting the number of moles of carbon dioxide. And how do we do that? We do it by measuring the mass of the carbon dioxide in our sample and dividing by the molar mass of the carbon dioxide which is the mass of a single mole of CO2. How do we work that out? Remember, from what we just said, we will take the mass of a single particle of carbon dioxide, which is 12 for the carbon plus 16 for each oxygen, so that totals up to 44. And then, that is grams per mole because the mass of a single mole is the mass in grams of a single particle in AMUs. Well, let's see then. The mass of CO2 was determined to be 33.8 grams, the molar mass is 44.0 grams per mole, and if we just plug that into the calculator to find out how many moles we have, we'll discover that we have 0.76 moles of carbon dioxide. Now notice, we weren't actually trying to determine the number of moles of carbon dioxide. What we wanted to know was the number of moles of carbon because we're trying to figure out how many carbon atoms there are in acetylene. But, in each mole of carbon dioxide, since each molecule of carbon dioxide contains a single carbon, then the number of carbon atoms is simply equal to the number of moles of carbon dioxide. The number of moles of carbon atoms that are in the reactants must be equal to the number of carbon atoms in the products, and the number of carbon atoms in the products is simply equal to the number of moles of carbon dioxide. So, we have actually counted the number of atoms of carbon that we started off with in the acetylene. We can do the same trick now with the water. The number of moles of water is equal to the mass of the water divided by the molar mass of the water. The molar mass of the water is easy to find because the oxygen is 16 and each hydrogen is 1, so it's 18.0 grams per mole. And if we do the calculation, let's see the mass of the water according to the problem here is 6.9 grams and the molar mass of the water is 18.0 grams per mole. And if we just do the calculator math there and plug that in, we wind up with 0.38 moles of water. Again, that wasn't what we wanted. We wanted to know the number of moles of hydrogen, but it's straightforward to figure out the number of hydrogen, moles of hydrogen once we know the number of moles of water, because every molecule of water contains 2 hydrogens. So, we can then figure out that the number of moles of hydrogen must be equal to double the number of moles of water because every water molecule has 2 hydrogen atoms in it. So, the number of moles of hydrogen is then equal to 0.76 mole. And we notice that the number of moles of carbon and the number of moles of hydrogen are exactly the same, and as a consequence, we can say that the molar ratio of carbon to hydrogen is equal to 1 to 1, which means that the atomic ratio of carbon to hydrogen is 1 to 1. As a consequence, we can answer the question, and say the empirical formula of acetylene is, we can say it this way, C1H1 or just CH. And it turns out acetylene ultimately is C2H2, so the ratio is, in fact, 1 to 1 just as we have calculated. We can actually push this a little bit further and understand even more about chemical reactions by considering what's called a balanced chemical equation. Remember that according to the atomic molecular theory, the atoms in chemical reaction are always conserved. We have exactly the same atoms that we started off with that we started off with when we wind up the chemical reaction. All of the atoms that were present at the outset, must be present in the products. And there can be no atoms in the products that were not present in the reactants. So, what that tells us is we can just count the number of atoms in the reactants and that number of atoms must also be present in the products. Let's illustrate that with a simple example. We're going to burn propane. Using the kinds of calculations we've described here, propane can be determined to be C3H8. When we burn something, we're reacting it in oxygen and we wind up with our products being carbon dioxide and water, as shown by the equation here on the screen. But there is something wrong with the equation on the screen, as you can immediately tell. Because if I count the number of carbon atoms, which are in the reactants, there are 3. If I count the number of carbon atoms which are in the products, there is only 1. So, we can't have carbon atoms disappearing so we're going to have to do something to adjust this reaction. One way that we could do that is to simply say maybe, we should have 3 carbon dioxides. That's supposed to be a 3. If that's the case, then I would have simply 3 carbons on both sides of the equation. And now the carbons are accounted for, but how about the hydrogens? We started off with 8 hydrogens here and we wind up with only 2 hydrogens. One way to simply solve that would be, if we were to put a 4 in front of the water, and now we would have 8 hydrogens on both sides of the equation and 3 carbons on both sides of the equation. The only thing left to look at is the oxygens. Right now, we've got 6 oxgens from the carbon dioxide and 4 oxygens from the water for a total of 10 oxygens in the product, but there's only 2 oxygens in the reactants. Well, we can fix that easily enough by simply putting a 5 in front of the O2. The result, written rather more neatly over here on the page, is that a single propane reacts with 5 oxygen atoms to produce 3 carbon dioxides plus 4 H20 molecules and if we look carefully at the, all of the 3 types of atoms, carbon, hydrogen, and oxygen have the same number of atoms of each type on both sides of the equations. We call this a balanced chemical equation. Let's do another example here. What if we took copper metal and dissolved it in nitric acid? The products of that reaction turn out to be copper nitrate, CuNO32, nitrogen, nitrogen dioxide and water. If we write that down symbolically, then what we've got here is, of course, the copper with the nitric acid reacting to form the copper nitrate, nitrogen dioxide, and water. So, that expresses a list of the reactants in the products, but it's clearly not an equation. I can look at the copper and immediately see that there's one copper on each side of the equation but as soon as I look at, say, the nitrogens, there's only one nitrogen listed there in the reactants and there are clearly 3 nitrogens listed in the products. And if I look at the hydrogens, there's only 1 hydrogen in the reactants, but 2 in the products, and the oxygens were all over the place. We have 6 oxygens from the copper nitrate, 2 from the nitrogen dioxide, and 1 from the water, so that's 9. But there are only 3 that we started off with in the original reactants. In order to rectify this, we actually need to do as we did in the previous reaction and what we'll actually see is that we can balance this equation by putting actually a 4 in front the nitric acid. We don't need to put anything in front of the cooper nitrate because the coppers are already balanced. But now, let's see, I'm going to need a 2 in front of the nitrogen dioxide so that there are 4 nitrogens on each side of the equation, and I will have a 2 in front of the H2O so that there are 4 hydrogens on each side of the equation. Slide that over. Let's see that equation on the next page here. That's the balanced equation that we just worked out and if you take a look at this, you'll easily see that there's 1 copper, 4 nitrogens, 12 oxygens, and 4 hydrogens on each side of the reaction. We can actually work with this very systematically with some fairly interesting , calculations which can be done. Let's imagine we started off with a piece of copper metal which was 10 grams and ask, what mass can be produce from the product? So, we are trying to relate the amount of the copper we started off with to the amount of the nitrogen dioxide we wind up with. How should we proceed? Well, if I could determine the number of moles of copper, then I would know the number of moles of carbon dioxide, because note I'm sorry, of nitrogen dioxide. Because you'll notice that there's a 2 to 1 ratio for every 1 copper, there are 2 nitrogen dioxides produced. As a consequence, one mole of copper will produce 2 moles of nitrogen dioxide. That means, I should figure out how many moles of copper did we actually start off with. And let's see, we've got 10 grams of copper. And if we divide that by the molar mass of copper, which is just equal to the mass in grams of the mass of a single atom of copper in atomic mass units, which is, from the periodic table, 63.55 grams per mole. And if I just plug that into the calculator, I'll discover that is 0.76 moles, I'm sorry, that's the wrong number, it is equal to 0.157 moles of copper that we started off with. But if there's one mole of copper, we would produce 2 moles of nitrogen dioxide so the number of moles of nitrogen dioxide produced is always num 2 times the number of moles of copper that we started off with, and therefore, it must be the case that we have produced 0.314 moles of nitrogen dioxide. And if I wanted to know what mass of the nitrogen dioxide that was equal to, then I could go back to the formula that we reviewed earlier in this lecture, which is that the mass of the sample divided by the molar mass of the substance is equal to the number of moles. Notice we know this number. It's 0.314. We've determined that. We know what the molar mass of nitrogen dioxide is, because each nitrogen is 14 and each oxygen is 16, so that's 44 grams per mole. Notice that we know this number and we know this number, then it's fairly straightforward to calculate that the mass of the nitrogen dioxide is the number of moles of the nitrogen dioxide multiplied by the molar mass of the nitrogen dioxide. And if I multiply those numbers together, I determine that the mass of the nitrogen dioxide I've produced is 14.5 grams. So, this is an illustration of what is called chemist, chemical stoichiometry. Stoichiometry is kind of a complicated word here, but I defined it for you to simply mean chemical algebra. It's basically just doing algebra involving the numbers of atoms of each type and their associated masses used in a balanced chemical equation. Now, there's something else interesting that can show up here. Notice that all we calculated was the mass of the copper that we started off with assuming essentially that we have plenty of nitric acid around to dissolve that amount of copper. But we don't always have that. We might actually have limited amounts of different materials. So, if I take a look at that particular reaction and say, what if we don't have an unlimited amount of nitric acid, would we actually produce as much nitrogen dioxide as we have just predicted from the previous calculation? Let's do an example. Let's imagine that we have 0.20 moles of nitric acid and let's ask the question, how much copper can I dissolve in 0.2 moles of nitric acid? Notice that the ratio of the number of moles of copper to the number of moles of nitric acid, in the chemical equation, is 1 to 4. Here they are, 1 to 4. So, what that means is I need to have 4 times as many molecules of nitric acid as I have atoms of copper or, stated differently, I need to have 4 times the number of moles of nitric acid that I have of the number of moles of copper. Well, let's see, what we had at the outset you'll recall, is that the number of moles of copper is 0.157 so the number of moles of nitric acid, which are needed in order to react with 0 I'm sorry, 10 grams of copper, must be equal to 4 times 0.157 moles which is equal to 0.628 moles. That's how many moles of, of nitric acid are needed in order for me to dissolve that mass of copper. Do I have that? The answer is no. Because if we go back over here and take a look, we only gave 0.2 moles of nitric acid. Consequently, there's not enough nitric acid to dissolve all of the copper. The result is that the production of the product nitrogen dioxide is now limited by the nitric acid. We now call the nitric acid the limiting reactant because it limits the amount of product that we can produce. It's the thing that we will run out of before we run out of anything else. So, if I now ask the question, how many moles of nitrogen dioxide can I produce, it's determined by the number of moles of nitric acid, not the number of moles of copper because I've got plenty of copper, but I'm going to run out of nitric acid first. Notice that the ratio between the number of molels between nitric acid that we start off with, and the number of moles of nitrogen dioxide we produced is in ratio 4 to 2, which is also the ratio 2 to 1. Therefore, for every 2 moles of nitric acid I start off with, I produce 1 mole of the nitrogen dioxide. As a consequence then, since the number of moles of nitric oxide sorry, of nitrogen, nitric acid that we started off with was given to be 0.20 moles, then the number of moles of nitrogen dioxide that we can produce is simply equal to 0.10 moles, much smaller than what we were able to produce before. And now, if we ask what is the mass of that number of moles of nitrogen dioxide, it's simply equal to the number of moles of the nitrogen dioxide multiplied by the mass of a single mole of nitrogen dioxide. And again, since the mass of a single mole of nitrogen dioxide is 46 grams per mole, I have an error up here, 46 grams per mole. Then, if I multiply those two numbers together, I wind up with the mass being 4.6 grams of nitrogen dioxide. So, we can actually do predictions of the masses of materials produce from the masses of the materials that we started off with. But only, if we do calculations of the numbers of moles of reactants and use that to determine the number of moles of products, and take into account this idea that we might run out of one of the reactants before we run out of the other type of reactant. We're going to continue these kinds of calculations in the next lecture.