In this lecture, I'll be discussing how mole ratios can be used to determine elemental compositions of compounds. You'll want to have both a periodic table and a calculator at your finger tips. There are some in-video questions where you'll need those. So, please pause the video now to get these items ready. I have my calculator right here, ready to go. Recall from a previous lecture, that the Stoichiometric Ratio or mole ratio for a compound is simply the moles of one element in that compound compared to the moles of another element in a compound. Here is a example compound, Calcium Nitrate. Calcium Nitrate's molecular formula is shown here. Remember that the subscript numbers, the 2 and the 3 and the implied 1s, simply are used to compare the moles of one element in this compound with the moles of another element in this compound. We can write a whole bunch of mole ratios using this information. For example, I can see that there's one mole of Calcium for every two moles of nitrogen. Because the two here would be multiplied times the 1 for the nitrogen. Similarly I could say that there's one mole of calcium for six moles of oxygen. And I can also say that there's two moles of nitrogen for six moles of oxygen. Or I could say, one mole of nitrogen for three moles of oxygen. All of these are simply ratios that I've gotten from looking at the compound's formula. Let's do a calculation. If we have 14 moles of nitrogen, how many moles of calcium and how many moles of oxygen would we need to make calcium nitrate, that consumed all 14 moles of the nitrogen? The calculation would be set up like this. We start with what we're given, 14 moles of nitrogen, and then we multiply that times the mole ratio that's relevant. In order to determine how much calcium we need, we would multiply it times the mole ratio one mole of calcium to two moles of nitrogen. So really we would just need to take 14 moles of nitrogen and divide it by 2. The moles of nitrogen cancel out and leave us with seven moles of calcium. We would also need oxygen to build the calcium nitrate. We're starting with 14 moles of nitrogen and we need six moles of oxygen for every two moles of nitrogen. Again our moles of nitrogen cancel out. We could have also used a 3:1 ratio here and that also works. And we find that we need 42 moles of oxygen to build all the calcium nitrate that we could possibly build, starting with our 14 moles of nitrogen. All of those calculations were simply examples of using compound stoichiometry. [BLANK_AUDIO] We can use the compound stoichiometry to do even more complicated calculations where we have to convert between grams and moles. This is where we need our periodic table because we need to know the masses of each of the elements in the compound. Let's start with this problem. A calcium nitrate sample contains 2.04 grams of nitrogen. What is the mass of that calcium nitrate sample? Which is another way of asking how much does that sample weigh? The strategy for solving this problem relies on using several stoichiometric ratios. We need to convert the grams of nitrogen into moles of nitrogen. And in order to do that, we need a ratio, of the type that we just showed. Then we need to convert the number of moles of nitrogen into the number of moles of calcium nitrate, and again, there's a ratio for that. And finally, we need to convert the moles of calcium nitrate into the grams of calcium nitrate. There's also a ratio for that. Just like there's an app for everything, right? In fact, there's probably an app that will do this calculation for you but I'd like you to know how to do it yourself with just a piece of paper and a calculator. These ratios are chosen so that the units cancel out. This is dimensional analysis, right. Cancelling out the units. So in the numerator we would use the units that we're trying to have as our result, and in the denominator we would use the units to the left. What we're starting with. Of course we can write all kinds of stoichiometric ratios for our calcium nitrate. From the periodic table we could look up the mass of each of these types of atoms. So these just came from the periodic table. And of course, I could write all of these ratios upside down as well. So I know that there's 40.078 grams of calcium in one mole of calcium, but I can also write one mole of calcium has a mass of 40.078 grams. I could write calcium there also if I wanted to. And I could write these other two upside down also, couldn't I. [BLANK_AUDIO] There's also several stoichiometric ratios contained within the chemical formula. This is review. For example, for calcium nitrate, there's a ratio between the moles of calcium atoms and the moles of calcium nitrate molecules. Similarly, there's ratios for the moles of nitrogen atoms to moles of molecules and the moles of oxygen atoms to moles of molecules. We can write all of these ratios upside down, or we can also write ratios comparing atoms to atoms, as I showed before. We know the ratio of calcium to nitrogen, calcium to oxygen, and nitrogen to oxygen. We can also look at the periodic table and determine the molar mass of calcium nitrate. In order to do that, we would need to add together the atomic masses of all the atoms that would be needed. If we do that calculation, we find that we have 164.086 grams of calcium nitrate in a sample that contains one mole of calcium nitrate. I'll show you this calculation later in another problem. Now that we've listed a bunch of ratios that we can determine just from inspection of the molecular formula and also from looking at the periodic table, let's return to the original problem. First, I'm going to do the problem the way that I like to do it. And I like to think through problems in a step-by-step logical fashion. I've got my calculator ready to go, and I'm just going to write down the calculations that we would do for each step, using the mole ratios. First of all, I'm starting with 2.04 grams of nitrogen. And I need to convert that into moles of nitrogen. Well, I can look at the periodic table and see that one mole of nitrogen has a mass of 14.007 grams. I got that just from looking on the periodic table. 14.007 is the average atomic mass of nitrogen. If I multiply those two numbers together, the grams of nitrogen is going to cancel out and leave me with moles of nitrogen. And that is 0.14564 moles of nitrogen. Now I've used too many significant figures here, haven't I? I always like to conclude a couple of extra sig, a couple of extra digits so that I don't create a rounding error when I do a problem in a step by step fashion. But what I'm going to do here is underline the last digit that's significant just to remind myself of that. The next ratio I need to use is to convert from moles of nitrogen to moles of calcium nitrate molecules. So let me start with my moles of nitrogen. 0.014564 moles of nitrogen. And let's see, I want to have moles of calcium nitrate so I'm going to write that on the top, moles calcium nitrate. And I'm going to write moles of nitrogen on the bottom. And then I'm going to look for the mole ratio, and it looks to me like one mole of calcium nitrate contains two moles of nitrogen. So basically I need to take that number, 0.014564 and divide it by 2, and that gives me an answer that I'm going to write over here so that I'm kind of, going over of 0.072821. And again, I only have three significant figures so I'm going to underline the eight. Finally, I need a ratio of going from moles of calcium nitrate to grams of calcium nitrate. The ratio there would be the molar mass of calcium nitrate. So, up at the top, I'll calculate the molar mass of calcium nitrate. What's that equal? Well I have one calcium, to look on the periodic table and see that, that weighs 40.078 grams per mole. So there's my calcium, 40.078 grams per mole. Plus I have two nitrogens. And those weigh each 14.007, grams per mole. And I have six oxygen, so 6 times the average molar mass of oxygen, which is 15.999 grams per mole. And all of that adds up to 164.086 grams per mole. Which was on a previous slide. I'm just showing you where we got that number. So I can take this number here, sorry I'm scribbling out because I'm being messy. This number is the moles of calcium nitrate isn't it? 0.0728. And I'm going to take that times, I want to have moles of calcium nitrate in the denominator. And I know that I have 164.086 grams and one mole of calcium nitrate. So I do that multiplication and I find that I have 11.9 grams. So this sample has a mass of 11.9 grams. I use three significant figures. I always check that at the end. Now it's your turn to try one. So I just did that problem in a step by step fashion. But it's actually also convenient to do the problem as one gigantic equation. And if you like to do that, it would look like this. I start with 2.04 grams of nitrogen. I want to convert that to moles of nitrogen, so I use this ratio, that there is one mole of nitrogen in 14.007 grams. Instead of doing the intermediate calculation, I can then just multiply the moles of nitrogen, cancel my grams, with the ratio that one mole of calcium nitrate has two moles of nitrogen atoms. Now I can cancel out the moles of nitrogen. And ultimately, I would like to have grams of calcium nitrate, so I can multiply that quantity by the molar mass of calcium nitrate. This is all of the calculations I did on the previous slide, wrapped up into one. And it should give me the same result. The final unit is going to be grams of calcium nitrate and the answer is still 11.9 grams of calcium nitrate. So, it worked out just as I expected. Now it's your turn to try one. Again, make sure that you have your calculator and your periodic table handy, so that you can answer this in-video question. This time our calcium nitrate sample contains 14.03 grams of oxygen. And I want to ask you, how many grams of nitrogen does it contain? So I've twisted the problem a little bit. I'm not asking about grams of calcium nitrate this time. I'm asking you start with grams of oxygen and calculate how many grams of nitrogen there are. Go ahead and try this. To answer this in-video question, I can do a bunch of calculations using mole ratios, just as I did on the previous example. In this case, I'm starting with grams of oxygen. Starting with 14.03 grams of oxygen, I want to convert that to moles of oxygen, so I'm going to multiply that by a ratio. There's one mole of oxygen that weighs 15.999 grams, and that just came straight from the periodic table. Then I want to convert that number of moles of oxygen. Because remember, my grams of oxygen has cancelled out into moles of nitrogen. I can inspect the compound formula and determine that there are two moles of nitrogen for every six moles of oxygen. You could also use one to three here if you'd like. So now, my moles of oxygen is cancelling out, and I have moles of nitrogen. But ultimately, I'd like to find grams of nitrogen. Well, that's easy enough. I know the ratio of moles of nitrogen to grams of nitrogen because I can just look that up on the periodic table. One mole of nitrogen contains 14.007 grams. So my moles of nitrogen has canceled out and the answer to this problem then becomes 4.094 grams of nitrogen. As a quick check, I'm going to make sure that I express the answer with the correct number of significant figures. The quantity I was given initially had four significant figures. I did all multiplication and division in this problem, so the answer should also have four significant figures, and it does. I have one more practice example to work together. Here's how we did the last problem. Here's the last practice problem. How many grams of oxygen are present in 12 moles of sodium nitrite? The first thing we need to do to answer this problem, is determine the formula of the ionic compound, sodium nitrite. We need to be careful because nitrite, that ends with ite, is not the same as nitrate that we've been using. Nitrite is a polyatomic ion, NO2 1 minus. Sodium of course we can look up on the periodic table. It's an alkaline metal, which means it tends to lose one valence electron to become the sodium cation. So the two species that combine to make sodium nitrite are the sodium cation and the nitrogen anion. So the compound here has this molecular formula, NaNO2. So we're starting with 12 moles of sodium nitrate, 12.0 moles of sodium nitrate. And we're trying to find grams of oxygen, so we need to convert from moles of sodium nitrite into moles of oxygen, that needs to be the first thing we do. There's two moles of oxygen in one mole of sodium nitrite. Then we need to convert those moles of oxygen into grams of oxygen, and I find that by looking up the mass of oxygen on the periodic table. It's average molar mass is 19, I'm sorry, its average molar mass is 15.999 grams of oxygen in one mole of oxygen. So I just take my calculator out, plug those numbers in quickly, and I see that all this equals 383.976 grams of oxygen. But I do my double check on the significant figures and I see I should only have three significant figures. So I need to round that, it'll round up to 384 grams of oxygen. So this concludes our lecture on calculations using mole ratios and compound stoichiometry to convert between grams and moles. This is a skill that you'll use frequently in chemistry. So I encourage you to continue practicing this as you take the quizzes for the class and also work on the advanced problem sets.
