[BLANK_AUDIO] Previously we learned how to determine mass percent composition. Data of that type can be determined through an experimental technique called elemental analysis. In this lecture, we'll learn how to use that data to convert the mass percents into empirical formula. As usual, you will find it handy to have a periodic table and a calculator ready to work through these problems. The empirical formula is the simplest formula for a compound. It gives the mole ratios of the different elements that compose that compound. For example, here are two compounds with different molecular formulas. The molecular formula shows exactly how many of each type of element are present in the compound. So the compound I just circled, which is acetic acid, a component of vinegar, has two carbons, six hydrogens, and two oxygens. Another compound, shown here, is a simple sugar. It could be glucose or fructose. It has the formula C6H12O6. But this is not the simplest formula. It doesn't just show the mole ratios, it shows the actual number of atoms present. We can divide all of the numbers here, by the greatest common factor. So, for example, if I look at my example of acetic acid, the greatest common factor for the 2 and the 4 would be 2 so I could divide all of the numbers by 2 and that would leave you with a formula CH2O. I can do a similar process for the sugar. But in this case, the greatest common factor is 6. So if I divide each number by 6, 6 divided by 6 is 1, 12 divided by 6 is 2, and 6 divided by 6 is 1. I get the same empirical formula for both of these compounds. Both of these molecules have the same empirical formula, CH2O, which happens to be the molecular formula for formaldehyde. So again, the empirical formula is the simplest formula, and it only conveys the mole ratio of elements in the compound. In that previous example, there was one mole of carbon, two moles of hydrogen and one mole on, of oxygen. In the molecule, that was the mole ratio. For example, perhaps we've prepared these little crystals. They look like little rocks, don't they? And we may have prepared them by heating iron in the presence of oxygen. So this sample, which contains 10.36 grams of iron oxide, was prepared by heating 7.5 grams of the metal, iron, which is a solid. All metals are solids, except for mercury. At room temperature and pressure, anyway. What is the empirical formula of this product? So we're given the masses. We're given some masses, but for the empirical formula, we need moles, so we can determine mole ratio. The iron oxide, we don't know what the empirical formula is. We could say, we know it contains iron and we know it contains oxygen, but we don't know how how much of each type of element is present. We were given the mass of the iron directly, weren't we? But we were not given the mass of the oxygen directly. What is the mass of the oxygen in this product? What we need to do to determine the mass of the oxygen in this product, is take the total mass, which was 10.36 grams, and subtract the mass of the iron, which is 7.50 grams. If we do that, we see that we're left with 2.86 grams of oxygen, in this sample. Is there any other information that we would need to do this problem that is not given in the problem? We're given the masses in grams. What do we need to figure out, for the number of moles of each type of element? That's correct, we need to look up the atomic masses. If you pull out your periodic table, you see that iron, which you can just look up on the periodic table, has a molar mass of 55.847 grams in one mole of iron. And oxygen has a molar mass of 15.999 grams per mole. So now we have the masses and we have the molar masses, we should be able to determine the number of moles of each type of element. And once we have the moles of iron and the moles of oxygen, then we can write the empirical formula. To determine the moles of iron, we're going to take the mass of iron in the sample and divide it by the molar mass. You might remember, perhaps from your algebra classes, that we can do something like this. We can take a over b, and divide by c over d, and that equals a over d divided by b over c. So, we could, instead of showing it the way I've shown it, which is with a ratio where the grams cancel out, like that, we could also do it in this fashion, and this is how I do it. I could write it like this. 7.5 grams divided by 55.847 grams per mole. So this is really like grams over 1. So I end up with 0.134, the grams is times the moles, right? If I'm doing it with the units. And then I have moles, oops. Then I have grams in the denominator, so grams cancel. And that gives me the same answer. Just showing you a slightly different way to do it. Okay, for moles of oxygen, we would take the amount of oxygen that we calculated by doing that subtraction on the previous slide and divide that by the molar mass of oxygen, which is 15.999, and we see that we have 0.179 moles of oxygen. So now I have the mole ratio, right? I have the moles of iron and I have the moles of oxygen. And I can actually write a formula with those moles, like this. So I just put the number of moles of each as the subscript. That's not exactly how we would like the formula to look though, isn't it? We would like to have whole numbers, as these subscripts. So we need to do something to convert these decimal place numbers to whole numbers. One of the things we can do to turn one of them into a whole number is divide both of them by the same, smaller number. Now, because this is already a fixed ratio, 0.134 to 0.179, anything we need to do to one number, we also must do to the other number. Otherwise, we'll mess up the ratio. So I'm going to divide both of these numbers by 0.134, to keep the ratio constant, and that'll give me, that I have 1 iron to 1.33 moles of oxygen. So, if I have 1 mole of iron to 1.33 moles of oxygen, I'm making progress. I've turned one of the numbers into a whole number. Now I need to get both of the numbers into whole numbers. And I can do that by multiplying the numbers by different whole numbers. For example, in this case, if I multiply both the 1 and the 1.33 times 3, I get 3 and about 4, don't I? If you get this close, you can go ahead and round. So in this case, the iron oxide that was formed is Fe3O4, that is the empirical formula. Again, the empirical formula is the simplest formula. It may or may not be the true, or actual, molecular formula. But it does give us information about the ratios of moles of each type of element. The reason that the empirical formula is so useful is that we can determine the empirical formula given data of mass percent from that elemental analysis technique that we talked about before. The molecular formula, remember, is the actual number of each type of element in a compound. So here's an example. The empirical formula for benzene, which is an organic solvent, is CH. There's one carbon for every one hydrogen, but the molecular formula is C6H6. The compound is actually a hexagons of carbon, where each of the carbons has one hydrogen attached. So the molecular formi, formula tells me more about the structure of the molecule. So to review, what is empirical formula? Well remember, the empirical formula contains mole ratios. Let's suppose that we do a calculation from some mass percent data and it gives us this information, this type of mole ratio. Again, we need to divide everything by the smallest of these numbers, first. And if we do that, in this case, we don't end up having to do any additional steps. We get whole numbers simply by dividing all of those subscripts by the smallest of them. And we see that this particular compound has the empirical formula CH2O. So the empirical formula is only telling us the ratio of the elements. One carbon for every two hydrogens, for example. The molecular formula tells us the actual number of atoms. In order to determine the molecular formula, we need the molar mass of the compound. The actual molar mass of the compound allows us to convert the empirical formula into a molecular formula. For example, if we're given the empirical formula that we just calculated on the previous slide, and we know the molecular weight, then we can determine the molecular formula. And usually when someone says, it's the chemical formula, usually they mean the molecular formula. 'Kay, so those two numbers, or those two expressions, excuse me, equate molecular formula and chemical formula. And again, I use the word, molecular weight, sometimes, when I mean molar mass. Those are also interchangeable terms. So, if the molar mass is 30 grams per mole, then the chemical formula needs to be CH2O. And that's because, if we added up the masses of these elements. Carbon weighs 12, right? Hydrogen weighs 1 each, so say, 2 times 1. Oxygen weighs 16. We see that that all weighs 30 grams per mole. So if the actual compound weighs 30, has a molar mass of 30 grams per mole, then the formula of the molecular formula is the same as the empirical formula. If the compound weighed 60 grams per mole, that would mean that the actual molecule was twice as heavy as the empirical formula. So, therefore, we would multiply each of the subscripts by 2. If the molar mass was much greater, for example, here, 180 grams per mole, well, 180 divided by 30 equals 6, so we need to take each of the subscripts and multiply them times 6, and that tells us that we have C6H12O6, for the molecule. Are you ready to try one on your own? Okay, here's a review example. If you have your calculator and your periodic table ready, you can go ahead and try and do this problem now, just by pausing the video. If you'd like to watch me work through the problem, you can do that as well. But I really think that it's a good idea to try to do the problem yourself, first. That's how you learn how to do things the most effectively. Let me start by giving you a hint. The hint is, you're given mass percents of the three different elements, here, carbon, hydrogen and oxygen. But you're not given an amount, of the compound. But you can assume that you have 100 grams, if you like. Of course, we're going to need a periodic table to solve this one, unless, of course, you have the masses of these elements memorized, which you might, because I do. What do you think is the first step in solving this problem? You might have said that the first step is looking at the molar masses of carbon, oxygen, and hydrogen. You might have said that the first step is that you need to assume a sample size, as I gave you in the hint. Or, that you need to find the number of moles of each type of element. All of these are correct ideas. Let's go ahead and work through the problem, here. So I said we're going to assume a total mass of 100 grams. So I'm going to say, assume 100 gram sample. Therefore, my 52.14% of carbon is 52.14 grams of carbon. And if I look on my periodic table that I have on the side, here, I know that there are 12.011 grams of carbon in one mole of carbon. So if I take 52, I'm going to use my calculator, take 52.14 divided by 12.011, and I see that that's 4.341. And I'm going to use one more digit, even though I know there's only four significant figures here, I'll put a little line underneath the 1 to remind myself that that's the last significant digit. So there is a number of moles of carbon that are present in my 100 gram sample. I do a similar calculation with the hydrogen. In this case, I look on the periodic table, and I see that one mole of hydrogen weighs 1.0079 grams. That's, that's just something I'm looking up. So, if I take my calculator again, and I say, 13.13 divided by 1.0079, I get 13.027. 13.027 moles of hydrogen. And, finally, for oxygen, I have 34.73 grams of oxygen. Oxygen has a molar mass of 15.999 grams per mole. This is getting kind of messy, draw a little line to make it neater. And using my calculator again, 34.73 divided by 15.999 equals 2.1707. So in all these cases, I just decided to carry an extra digit so that I don't create any rounding errors. Alright, now, I have the mole ratios, already calculated. So I could write the formula, if I'd like, like this, 4.341. Hydrogen has 13.027, and oxygen has 2.1707. What is the next step in solving this problem to find the empirical formula? That's correct, the next step is dividing by the smallest of those numbers. So if I divide each of those numbers by 2.1707, 2.1707 and I'm going to divide the oxygen by 2.1707. Of course, the oxygen number ends up being 1. What about the carbon number? That's 1.9998, so it's about 2. For the hydrogen, I'm going to take 13.027 and divide it by 2.1707, and that equals six. So this number equals 2, this number equals 6. So, to the answer to that problem was, C2H6O, and that is the empirical formula. In order to find out if that is actually the molecular formula of this compound or the real chemical formula, we would need to have the molar mass, of the substance. And we'll talk about ways that we can calculate molar masses from experimental data, later in the course. Thank you for joining me for these calculations of empirical formulas. You'll be able to do some practice on this if you do the quizzes this week. There will also be some practice on the advanced problem set that will be available and due next week.
