In our previous lectures we have determined the complete set of relative atomic masses for every element in the periodic table. In this particular lecture we're going to figure out how to use those relative atomic masses to count atoms. And in a particular to figure out what the molecular formula is for any compound that we might be interested in. By determining the relative numbers of atoms in each type in that compound. The material that we're going to cover here is from the second concept development study on atomic masses and molecular formula. You might want to review that material in conjunction with this particular lecture. What we're going to be attempting to do here is count atoms. To do this, we're only going to have to assume those relative atomic masses we've already figured out. And by counting atoms, keep in mind that we really are only going to care about the relative number of atoms, not the absolute numbers of atoms. We're going to be interested in, for every carbon atom, how many hydrogens are there in a particular hydrocarbon compound. We're going to do this in a particularly clever way. That is, we're going to show how we can count particles by measuring mass. That's the technique we use in chemistry. To do that, let's not start with something as complicated as atoms. Let's start with something more familiar, glass marbles. Imagine that you have a whole collection of glass marbles, and one glass marble weighs 1.5 grams. Now imagine that I have given you a sample in which there are 15 grams of marbles. It should be pretty easy to see that 15 grams is 10 times greater than 1.5 grams and as a consequence it's pretty clear that I have given you 10 marbles. You could easily have figured that out by simply counting from one to ten, but note that you could figure it out just by weighing the marbles rather than counting them. That becomes important when we go to larger masses. Let's ramp this up a bit. Let's talk about 1.5 kilograms, instead of 15 grams. This is much larger number. But it's still the case that 1.5 kilograms is exactly 1,000 times greater than 1.5 grams. So it must be the case that have now given you an example of which there are 1,000 marbles. Counting out those marbles one by one would take a while. Weighing them to determine that they have mass 1.5 kilograms would be very efficient and very quick. How have we done the calculation here? It's pretty clear that what we have done is to take the mass of the sample we have been given and divide that by the mass of a single marble. And the ratio between those two is equal to the number of marbles. Looking back, if I take 15 grams and divide by 1.5 I have ten marbles. If I take 1500 grams and divide by 1.5 grams per marble, I have 1000 grams. Let's illustrate how important this is by ramping up to a much larger mass. Let's say that we took a metric ton. Of marbles. That's 1000 kilograms. We clearly wouldn't to try to count out all of those marbles but we could simply say that the number or marbles in that sample is equal to 1000 kilograms, the mass, which is 1 times 10 to the 6th. Grams, divided by the mass of a single marb, marb, marble, which is 1.5 grams per marble, and if we just put those numbers in the calculator we've determined that there are essentially 667,000 Marbles in this sample. Far too large a number for us to want to count. But very simply, we've been able to figure out that number just by taking the mass. Note we figured out a formula here, and the formula is a simple one. The mass divided by the mass of a single marble Is equal to the number of marbles. We're going to use that formula when we start talking about atoms. Notice that the number is equal to the mass divided by the mass of one unit. Now often in chemistry what we want to do is compare numbers of particles of different types like carbon atoms and hydrogen atoms. Let's draw the analogy here by imagining that we have two different kinds of marbles. Let's imagine that we have some blue marbles, and let's imagine that we have some red marbles. And the blue marbles have mass 1.5 grams. And the red marbles have mass 2.3 grams. And then let's imagine that I've handed you two bags of marbles. One of the bags of marbles is 150 grams. That's the blue marbles. And the red marbles, the bag weighs 230 grams. It's quite clear that the ratio of the masses of the bags 230 to 150 is the same as the ratio of individual marbles 2.3 to 1.5. As a consequence, just by weighing the two bags of marbles I can actually tell you immediately that those two bags of marbles have exactly the same number of marbles in them. By comparing the relative masses of the samples, to the relative masses of the individual atoms, we've effectively determined the comparative number of atoms, or of marbles, of each type. Well, what if we now move over to some chemistry examples. Let's look instead blue marbles and red marbles, let's talk about carbon and sodium. Now we already know that one carbon atom is on a relative scale has a mass of 12 and one sodium atom has on the relative scale, a mass of 23. So if I had one carbon and one sodium, the mass ratio is 12 to 23. That's trivial. What if we ramp it up? Let's take 1,000 carbon atoms and 1,000 sodium atoms. Now it might be that we would think it would be complicated to figure out the mass ratio but it's not. Since each carbon atom weighs 12 then 1,000 carbon atoms weighs 12,000. 1,000 sodium atoms weighs 23,000 but notice that the mass ratio of the samples is still 12 to 23. It doesn't matter if I have one carbon atom and one sodium atom, or I have a thousand carbon atoms and a thousand sodium atoms. As long as I take the same number of atoms the mass ratio is exactly the same. 12 to 23. And therefore, any number of carbon atoms and that same number of sodium atoms will always have the mass ratio of 12 to 23. Let's turn this around then, and say instead, of giving you a fixed number of atoms, what if instead I gave you some masses. Let's say I handed you 12 grams of carbon, and I handed you 23 grams of sodium, and I asked you to compare the numbers of particles in those two samples. You could do that, because since one, carbon atom, and one sodium atom have ratio 12 to 23, then any number of carbon atoms and that same number of sodium atoms have ration 12 to 23. And since the ratio of the masses here is in fact, 12 to 23. Then it's straight forward to conclude that those two samples have in fact exactly the same number of atoms. Even though we could not possibly count that number, we can come to that conclusion. Well what is this number. Actually it doesn't matter to us too much how many atoms there are in 12 grams of carbon. What we really need to know is it's the same number of atoms in 23 grams of carbon. That actually gives us a rather simple way to determine the particles that we are interested in. By instead of worrying about exactly what those numbers are we're going to call that number a mole. It's really just like a dozen. It's a number but this number is in fact equal to the number of atoms which are in 12 grams of carbon. Why do we use that particular number? Well, an individual carbon atom weighs 12, so if I take 12 grams of carbon and call that a mole, then I can also say that a mole is the number of atoms in 23 grams of sodium, or 4 grams of helium, for that matter. Since an individual helium atom weighs 4 on the scale that a carbon atom weighs 12 That if I have 12 grams of carbon, and that's a certain number of particles, that same number of particles of helium will weigh 4 grams. Importantly actually down below, notice, that we can calculate the mass of a single mole, what we're going to call in a moment the molar mass, by simply looking at the relative atomic masses and assigning them a number in grams. So here's what we're going to do. We're going to count the number of moles of particles by determining the molar mass. Here's that quantity that we referred to on the previous slide. The molar mass is the mass of 1 mole, and it's in grams. And to determine what it's number is, we just take the mass in an atom, like 12 amu for carbon. Or in a molecule, let's say, 18 amu for water, and the mass of a mole will be that same number in grams. And then from the formula we worked out before, remember this formula up above. That the number of particles is the mass divided by the mass of sample one. Notice we have exactly the same formula here, expressed in chemical terms. That the number of moles is simply the mass of our sample divided by the mass of one mole. The equation which is expressed here is an extremely important equation because it gives us the opportunity to count the numbers of particles of each type just by doing calculations of the numbers of moles. Well we're going to do a couple of examples in a moment here. But before we do that let's think about why we would want to calculate the number of moles. Reason why is because the number of moles is a count of the number of particles, and counting particles is a very valuable thing to do. The molecular formula is the ratio of the number of particles of each type, so if we count those particles we can determine the molecular formula. The number of particles produced in a chemical reaction is determined by the number of particles we started off with in reactants. So if we count the particles we can do chemical reaction calculations. And likewise, if we are interested in the number of particles which are floating around in a solution, it will allow us to determine the concentration of the solution. Calculation of the number of moles are fairly ubiquitous in chemistry. We're going to do a lot of calculations of types of moles. Let's begin by illustrating one particular type of example. We're going to determine the empirical formula of the common compound called benzene. The empirical formula is the ratio of the number of atoms of each type in the molecule. Benzene consists only of carbon and hydrogen. And from the law of definite proportions, we've actually measured the mass composition of benzene. And it turns out it's 92.3 percent carbon and 7.7 percent hydrogen. How can we figure out the empirical formula? Well, remember, the empirical formula's the number of particles, so it's also going to be the ratio of the number of moles. That says that we ought to calculate the number of moles in these samples. Let's take a 100 grams of benzene. That's a nice easy number to work with. Since we have a 100 grams of benzene and benzene is 92.3% carbon by mass then in fact the mass of the carbon but in that sample must be 92.3 grams. Well, notice I can calculate them, the number of moles of carbon. From the formula that we've worked at before, which is, the mass of the carbon divided by the molar mass of the carbon. The mass of the carbon is 92.3 grams, the molar mass of the carbon, we already know, is 12 grams of carbon. And if we take that ratio, we can just using our calculator figure out that we have 7.68 moles of carbon atoms in our sample. In order to compare that the, the number of hydrogens, we need to figure out what the number of moles of hydrogen is. We need to count the number of hydrogen particles. Well let's see, the mass of the hydrogen, we know what that is, the 100 gram sample has 7.7 grams of hydrogen in it. So the mass of the hydrogen divided by the molar mass of the hydrogen is 7.7 grams divided by, let's see the molar mass of the hydrogen, hydrogen is on the relative mass ratio 1.008 grams per mole. And so, if we divide that out, what we discover is that we have 7.70 moles of hydrogen. And notice that the number of moles of hydrogen and the number of moles of carbon are to within our data here, essentially the same number. So where there's a one to one ratio between carbon and hydrogen. And therefore the formula, the empirical formula of the benzene is C1 H1. Other data would actually tell us that it's C6 H6, but we're only figuring out the ratios here. Let's do another example just to sort of drive this home and make sure that we know how to solve these kinds of problems. We're going to take a compound called chloroform that consists of carbon, hydrogen and chlorine. Once again, let's take a 100 gram sample. Here are the mass proportions associated with that. We have 10.06 grams of carbon. We have 0.84 grams of hydrogen and we have 89.00 grams of chlorine. From that we can calculate that the number of moles of carbon, which is 10.06 grams divided by 12.01 grams per mole, is actually equal to 0.833 moles. The number of moles of hydrogen, is 0.84 grams divided by the molar mass of hydrogen, 1.008 grams per mole, and that comes out to be 0.833 moles. And the number of moles of chlorine is the 89 grams of chlorine divided by the molar mass of chlorine which is 35.45 grams per mole. And that turns out to be 2.51 moles. And if we simply divide each one of those numbers by 0.833 we can show that the ratio is 1 to 1 to 3 so the molecular formula of chloroform turns out to be C1 H1 Cl3, and chloroform in fact turns out to be CHCl3. So we can do this with any kind of molecule as long as we knew what the relative masses are of the different elements in the compound. Let's illustrate one other way in which we can apply this concept of counting particles via moles, by relating the number of particles of reactants to the number of particles of products. In this particular case, we're going to imagine methane as being CH4. Simply we could have determined from the empirical formula calculation we did before. We're going to burn methane. And when we burn it we produce carbon dioxide. And here would be an interesting question. What if I started off with a kilogram of methane and I burned it, how much carbon dioxide am I going to produce from that reaction? How should I proceed? One thing I know is every time I burn a single methane molecule, there's only 1 carbon atom in that methane molecule, and therefore it can only produce 1 molecule of carbon dioxide. So the number of moles of carbon dioxide produced is exactly equal to the number of moles of methane molecules that I started off with. And I can determine the number of moles of methane molecules pretty straightforwardly by calculating the number of moles. Of methane. Well let's see. That's going to be 1 kilogram which is a 1000 grams divided by the mass of a mole of kilo of, of methane which is 16 grams per mole. And that number turns out then to be equal to, checking my notes here, 62.46 moles of methane. And if I then determine how many moles of carbon dioxide are produced by that remember for every mole of methane that burn for every particle of methane that burn I must produce a particle of carbon dioxide. So every mole of methane that burn I must produce a mole of carbon dioxide. So the number of moles the carbon dioxide produces is 62.46 moles and now what I have to do is figure out what the mass is corresponding to that. But you'll recall our formula before, the number of moles is equal to the mass divided by the molar mass. We know that the number of moles of carbon dioxide is 62.46. We don't know the mass, but we do know the molar mass is 44.01 grams per mole for carbon dioxide. And all we've gotta do then is multiply the mass per mole times the number of moles. And when we do that, the mass of carbon dioxide produced. Is actually equal to then 2.75 kilograms. So for every kilogram of methane we burn, we produce 2.75 kilograms of carbon dioxide. We'll do a few more calculations of this type in the coming lecture and over the course of the entire semester.