Ch 7

69 Chapter 7 STOICHIOMETRY, Part 1 ©2004, 2008, 2011, 2022 Mark E. Noble We continue the quantitative features of reactions. In the last Chapter, our quantitative aspects were based on balance and the conservation of atoms. This was based on a simple counting of the atoms in the various chemical units which were on the left and on the right of the chemical equation. But, alas, when it comes time to actually do a reaction, we can't count the number of atoms or of molecules or of formula units. They're too tiny and there're too many. Remember? What good is an equation based on a count if we can't do a count? No, we cannot count actual atoms or chemical units directly but, as I told you in the beginning of Chapter 5, we can measure how many. In that Chapter, we used the mass and molar mass to measure how many formula units were in a sample. We will do so again in this Chapter. Later in Chapter 9, we will see how volume can also measure a number of formula units. In Chapter 16 when we talk about gases, we will even see how pressure can be used. Since mass and volume are measurable things, they can provide the necessary connection to the number of formula units in a real sample. Then, when it comes time to do a reaction with some number of formula units, we can calculate the mass or the volume which is needed to provide the desired number for each reactant. We can also calculate the mass or volume for each product in the process. These mass and/or volume relationships between the reactants and products of a reaction are called "stoichiometry". I'm going to briefly return to the notion of cooking as a follow-up on its mention in Chapter 5. Many people have at least attempted to cook something and they at least know what a recipe is. There are several parallels to note between cooking and chemical reactions. First, both the recipe and the chemical equation relate what goes in and what comes out. Second, the recipe gives measurable masses (e.g., oz, Ib) and volumes (e.g., tsp, tbsp, c) directly for all ingredients. This part is not true for the chemical equation. The balanced chemical equation gives all reagents on a numbers-of-particles basis, but we can't use this directly. Stoichiometry takes this information and transforms it into measurables such as mass (e.g., g) and volume (e.g., L). In other words, the combination of {balanced equation + stoichiometry} is equivalent to the recipe. When you cook and when you do a chemical reaction, you want to know how much you want to make and you want to know how much of each ingredient or reactant to throw into the pot. And you need to know this in terms of something measurable, not just numbers of particles. You also need to know how to adjust the recipe. If the recipe is for 48 brownies and you need 82, then you need to change the amounts of the ingredients. How would you calculate the amount of ingredients for 82 brownies? (OK, OK, I know what some of you are thinking. Double the recipe and eat 14 by yourself. That's not fair.) There is one more difference between a recipe and a chemical equation: the recipe usually comes with directions. When we actually do chemical reactions in the laboratory, we do follow directions or, like some cooking, we make them up as we go. Needless to say, one has to be very careful with that. The consequences of a screw-up in a laboratory can be far more dire than a screw-up in a kitchen. Let's just say that I've used a fire extinguisher. Several times, actually. 7.1 We begin stoichiometry with mass. Let's start with a basic example: we will consider the reaction of elemental hydrogen and elemental chlorine to produce hydrogen chloride. All of these are gases. All are diatomic molecules with covalent bonding. The balanced equation is the following. Hx(g) + Cl(g9) - 2 HCI(g) From Chapter 6, the balanced equation relates the numbers of the chemical units involved. H, + Cl, - 2 HCI one molecule ] one molecule two molecules of of hydrogen, H, reacts with of chlorine, Cl, to produce hydrogen chloride, HCI 00 &) E)

70 Chapter 7: Stoichiometry, Part 1 We can also say the following. H, + Cl, - 2 HCI one mole . one mole two moles of of hydrogen, H, reacts with of chlorine, Cl, to produce hydrogen chloride, HCI Remember from Chapter 5 that whenever I say a mole of some compound, I mean a mole of formula units of that compound. Everything here is composed of diatomic molecules so all the formula units are the molecular units. H, + Cl, — 2 HCI one mole . one mole two moles of of hydrogen, H,, reacts with of chlorine, Cl,, to produce hydrogen chloride, HCI, molecules molecules molecules Now, let's bring in the molar masses for each reagent. What's the mass of one mol of H, molecules? Of Cl, molecules? Of HCI molecules? The molar masses are 2.016 g, 70.90 g and 36.46 g. Put these in. Don't forget that you're dealing with two moles of HCl in the equation. H, + Cl, - 2 HCI 2.016 g H, reacts with 70.90 g Cl, to produce 2 X 36.46 g HCI = 72.92 g HCI Notice that the masses of the reactants, 2.016 g H, and 70.90 g Cl,, add up to 72.92 g, which is the mass of the products. This is a direct consequence of the conservation of mass: a properly balanced chemical equation will always reflect the conservation of atoms, which is the same as the conservation of matter and the conservation of mass. These numbers (2.016 g H,, 70.90 g Cl, and 72.92 g HCI) are the "recipe amounts": they are the masses for the reagents in the balanced equation as given. If you want to make 72.92 g HCI, it will take 2.016 g H, and 70.90 g Cl, (and not a smidgeon more or less). This is the simplest stoichiometry set-up. Unfortunately, life's not always simple. You have to be able to adjust the recipe. To any amount. For any reactant or product. What if you have 4.032 g H, to react? Well, that's just a doubling of the recipe. You double the Cl, (141.8 g) and then you can make double the HCI (145.8 g). What if you want to react 35.45 g Cl,? That's one-half the recipe amounts. You will need 1.008 g H, and you can make 36.46 g HCI. What if you want to react 24.77 g H,? How much Cl, do you need? How much HCI can be made? OK, that's not so simple. Before I show you how to do that, I want you to think about how you actually go about adjusting a recipe. What calculation goes through your head? Can you visualize your calculation? Could you write it down to show somebody? Whether you are consciously aware of it or not, the calculations amount to a ratio between the desired amount of ingredient (or product) and the recipe amount. Let's consider A and B as some ingredient or product of a recipe. The calculation ratios are desired amount of A desired amount of B recipe amount of A recipe amount of B Some people use an alternate form. desired amount of A recipe amount of A desired amount of B recipe amount of B They're equal by rearrangement, so I'll stick with the first set. Go back to the brownies example earlier. Let's say that the recipe calls for 1% c flour for 48 brownies, but you need 82 brownies. How much flour do you need for 82 brownies? You could do the calculation based on the following ratios. 82 brownies _ desired amount of flour 48 brownies 1%2 ¢ flour This gives an answer of "2.5625" ¢ flour which is {22 ¢ + 1 tbsp} flour.

Chapter 7: Stoichiometry, Part 1 71 OK, now let's return to the reaction using the 24.77 g H,. How much Cl, do you need? How much HCl can be made? TI'll start with the CI, part. We can handle this in a manner similar to the recipe approach. We need our ratios. desired amount of H, _ desired amount of Cl, recipe amount of H, recipe amount of Cl, The recipe amounts are those corresponding to what is given by the balanced equation directly. desired amount of H, _ desired amount of Cl, 2.016 g H, 70.90 g Cl, We desire to use 24.77 g H,. 24.77gH, _ desired amount of Cl, 2.016 g H, 70.90 g Cl, Take your calculator, plug in the numbers and punch out the answer. Round off to the correct sigfigs and you get 871.1 g Cl, for the desired amount. That means, in order to do this reaction with 24.77 g H,, you need 871.1 g Cl,. No more, no less. Now go to the HCI part. Calculate how much HCI can be made. Set up those ratios. 24.77gH, _ desired amount of HCI 2.016 g H, 72.92 g HCI Plug it in, punch it out, round it off. The answer is 895.9 g HCI. In summary, 24.77 g H, would react with 871.1 g Cl, in order to make 895.9 g HCI. We didn't really need to do ratios for HCl. We could have used the conservation of mass. The reactant masses of 24.77 g and 871.1 g add to 895.9 g, just as calculated by the ratios. Unfortunately, this addition works when there's only one product, so this method is very limited. Be sure to understand what we just did. Beginning with the balanced chemical equation, we determined the "recipe amounts” for each reagent. The recipe amounts were the masses for the moles of each reagent as given by the balanced equation; in other words, the molar mass times the coefficient in the balanced equation. We then adjusted the recipe for the actual amount we wanted. 7.2 Using dimensional analysis Another way of doing stoichiometry calculations is with dimensional analysis. This incorporates all the separate steps above, although in a different manner. The two methods are mathematically identical, which means they're both right and they'll both give you the right answer. They're two different tools in the toolbox. When I'm working on my car and I have to remove some hex bolt, sometimes I use a closed- end wrench, sometimes I use a socket and sometimes I use an adjustable. Sometimes one works better than the other. Sometimes it really doesn't matter. (And when you're under the hood and it's a hot day and you ran the engine until it got up to temperature and your arm is wedged in front of the engine block right next to the exhaust manifold and you can't see because it's so crowded and because they-don't- make-cars-like-they-used-to, then you just may not care what tool is in your hand.) Regardless, it is very useful to have different tools for different uses. We'll redo the same two problems as above with the 24.77 g H,, now using dimensional analysis. We will calculate how many grams chlorine we need and how many grams hydrogen chloride we can make. Dimensional analysis can appear confusing at first to students because it strings all steps together. I'll break it down so that we can see what's going on. First question: how many grams of Cl, are needed to react with 24.77 g H,? Keep in mind that we're starting over here. We're not setting up the mass relationships between all the reagents like we did in the recipe method. Dimensional analysis does not use mass relationships directly. Instead, it uses mole relationships directly. The masses are then derived from those. The mole relationships are the coefficients of the balanced equation. These become conversion factors. Go back to the original equation. H, + Cl, - 2 HCI If it's easier to follow, you can show the ones.