Greetings. We've learned several different ways that we can use stoichiometry to solve chemical problems. This brings us to a skill that chemists use regularly in the laboratory, chemical reaction calculations. As usual, you're going to need a periodic table, a calculator, and a pencil and some paper to complete problem solving during this video. If you don't have those items available, please pause the video and get those items ready. This is the last of our set of stoichiometry topics. For reactions we need to convert from masses to moles, and then use the molar ratio stoichiometry in the chemical reaction to go from moles to moles. And then finally we need to convert back to mass. So we're going to go from mass to moles. And then from moles to moles and then from moles back to mass in all of these calculations. We need to use mole or mass to convert between mass and moles, so you need to be very adept at calculating mole or masses. And using those values to convert between moles and mass. Let's start with an example. This is good practice for a skill we've just learned. What mass of oxygen gas is needed for the complete combustion of 13 and a half grams of butane. Butane is a gaseous hydrocarbon that is used in these disposable lighters such as the one shown here on the right. This is a beautiful photograph. He managed to capture it just as the flint was igniting the butane. The first thing you need to do is write out the balance chemical reaction equation. To solve this problem, we first need to balance the chemical reaction equation. You've already done that. The next thing we need to do is determine the molar masses of the reactants. So, here, we have butane and looking on the periodic table, I can see the carbon has a mass of 12.011 grams per mole. So I'll say 4 times 12.011 grams per mole for the four carbons. Plus, there's 10 hydrogen's. So 10 times 1.0079 grams per mole which is the molar mass of hydrogen. I put that into the calculator and I find, that butane has a molar mass of 58.123 grams per mole. I do a similar calculation for oxygen that's a little simpler to do. I just say 2 times the molar mass of oxygen, which is about 16, my periodic table says that it's 15.999 grams per mole is the average atomic mass of oxygen. And that gives me 31.998 grams per mole of oxygen. When I solve these problems what I like to do is set them up in a very organized fashion. So I tend to put what I'm given right underneath that compound in the chemical reaction. I'm going to do this in a step wise fashion so we can talk about what I'm doing in each step. In the first step I don't need the coefficients and the reactions to stoichiometry at all. I'm only going to create the grams of butane into moles of butane. So I'll take my 13.5 grams of butane, and divide it by the molar mass that was up here, 58.123 grams per mole. My grams will cancel, and that results in 0.23227 moles of butane. But of course, not all of those digits are significant. So I'm going to remind myself that there was only three significant digits. If I put in a little line under that two. Alright, now I know how many moles of butane there are. But I need to know how many moles of oxygen there are, so what I can do is set up a proportion using the reactions to wikiometry, ratios. So looking at the reaction I'm going to change color. I see that two moles of butane reacts with 13 moles of oxygen. Do you see that? So I can write a calculation like this. I can say, two moles of butane reacts with 13 moles of oxygen. I don't have two moles of butane here. I have 0.2323, let's just round it that way, moles of butane. So I'm going to say how many moles of oxygen would that react with? So now I can cross multiply and divide, can't I? I can say X equals 13 times 0.2323 divided by 2. So, X equals 13 times 0.2323 moles; this is all going to be moles, divided by two moles, which equals 1.5097 Moles O2. Do you see what I did there? Finally I need to convert the moles of oxygen into grams of oxygen. So that many moles of oxygen, what do I need to do to convert it into grams of oxygen? 1.5097 moles of oxygen times the molar mass of oxygen, which I found up here is 31.998 grams per mole. Remind myself how many sig-figs g/ I have, and that works out to 48.3 grams of oxygen. And that's the answer to this question. What mass of oxygen, I'm going to re-read the questions to make sure I answered it correctly. What mass of oxygen is needed for the complete combustion of 13.5 grams of butane and the number I got is 48.3 grams of oxygen. The next thing I like to do when I'm solving a problem is ask myself if that answer makes sense. In this case, I know that mass has to be a positive number and it is, so that's good. It looks like I need more moles of oxygen than I had of butane. Just from the molar ratios and the chemical reaction equation and that is true I found that I need 0.2323 moles of butane reacting with 1.51 moles of oxygen. So my number makes sense and I feel pretty comfortable with it I'm going to check it off and say that I'm done with that problem. Here's one for you to try on your own. Here's practice. Are you ready? Heating zinc hydroxide results in its decomposition to give zinc oxide and water as products. Zinc oxide is used in a large number of products. It's a semi-conductor and is also widely used as a white pigment. I fondly remember zinc oxide from my days at the beach. When I was a child, my mother used to put a sunscreen that contained the zinc oxide powder in it, to protect my nose and cheeks and ears from the sun, just like these boys have done, looks like they're getting ready to go surfing. So, the first thing you need to do is write the balanced chemical reaction equation. That means you need to remember how to figure out the formulas of zinc hydroxide and zinc oxide, doesn't it? So, we're using the naming rules that we learned last week to write equations this week. Go ahead and try that now. Well done. It was easy to figure out that zinc hydroxide needed hydroxide anions, which have a negative one charge. Zinc has a plus two charge. So, to balance those charges, I needed one zinc and two hydroxides. Now it's a solid. That is going to decompose in the presence of heat. So we'll put a little triangle there for the heat symbol to zinc oxide. Zinc has a charge of plus two, and oxygen typically the oxide has a charge of minus two. So zinc oxide's just ZNO. That's also a solid, plus water. Which is a liquid here. So that's how we write the balanced chemical reaction equation. That was good review and practice. The next thing we need to do is the reaction calculation, which is the topic of this video. Great. Now that we've got the balanced chemical reaction we can go ahead and perform the calculations with the numbers. The first thing we need to do is determine the molar masses of all the compounds in this reaction. I like to do this in a very organized way so that I can follow my work later, just in case I make a small mistake. I'm going to write the molar masses over here, all in one spot. So, zinc hydroxide, molar mass, is comprised of one zinc, which has a mass of 65.39 grams per mole. Plus two oxygens which each have masses of 15.999 grams per mole plus the two hydrogens which are each 1.0079 grams per mole. And I can look up all of those molar masses from the periodic table. Once I perform that calculation, I see that that gives a molar mass of the compound of 99.4038 grams per mole. And I'm going to underline the hundredths place because that is the last significant digit of this number if I use my rules for significant figures. I can do a similar calculation for zinc oxide. I'll use a different color to keep things straight. Zinc oxide is comprised of one zinc atom, which has a mass of 65.39 grams in one mole. Plus one oxygen, which has a mass of 15 I'm sorry, 15.999 grams per mole. So zinc oxide's molar mass is 81.389 grams per mole. I can also calculate the molar mass of water. Now, I don't need to do that for this particular problem the way it's phrased, but we could do that just as easily. Water is two hydrogrens, so two times the molar mass of hydrogen, which is 1.0079 grams per mole, plus one oxygen atom, 15.999 grams per mole. So, water's more mass is 18.016 grams per mole. So, now I have the molar masses of all three compounds in this reaction. I'm interested in determining the amount of zinc oxide that is formed from the decomposition of five grams of zinc hydroxide; that's what I'm given in this problem. So, I'm going to start over here with five grams of zinc hydroxide [SOUND]. And in order to get the number of moles of zinc hydroxide, in addition, divide that by the molar mass of zinc hydroxide which I think I got over here. Was 99.4038. Remember, we're carrying extra significant figures to avoid rounding errors. 4038 grams per mole of zinc hydroxide. So that gives me a total number of moles of 0.0503 moles of zinc hydroxide. The next thing I do is look at the mole ratio and the balanced chemical reaction equation. Here, one mole of zinc hydroxide decomposes to produce one mole of zinc oxide. Well that's really handy and simple, isn't it? So I have 0.0503 moles of zinc hydroxide that gives me 0.0503 moles of zinc oxide. And remember, I don't have four significant figures here. I have fewer than that but I'm writing some extra ones out. I'll go ahead and round it at the end. The next step is to convert. So I've gone from grams, if we just think about what I've done, I've gone from grams to moles to moles. The last thing I need to do is convert those moles into grams. So what do I do to do that? Well, 0.0503 moles of zinc oxide times the molar mass of zinc oxide, which is 81.389 grams per mole. Then, those are going to cancel, and that gives me 4.1 grams of zinc oxide, and that is the answer to this problem. 4.1 grams. Because I have conservation of mass, the total weight of both products together should be five grams. So I'm saying I had five grams of zinc hydroxide up here. And these two things together should weigh five grams. We'll see if that's true. If I take the number of moles of water that should form. Let's go ahead and do that in red. So I should also be forming 0.0503 moles of water, correct? Because I also have a one to one ratio there. And I multiply that times the molar mass of water, which is 18.016 grams per mole. And see how much water I form. Let me use my calculator, because I haven't done that one out in advance. So I'm going to say 0.0503 times 18.016. And that gives me 0.91 grams [SOUND] of water. So, if I add those things together [SOUND], and I look at the significant figures, I see that I do, in fact, make five grams of product. Well you've worked hard today doing these reaction calculations, and I know that you're going to enjoy practicing some more reaction calculations on this week's quiz.