Yeah. >> Alright. First of all, a status report out your exams. As we told you on exam day, what we do is we separate the exam pages for grading purposes. Each page goes to a different greater. So five babies on the exam, five different graders, as things currently stand, four out of five raters have actually picked up their pages degrade, and one of those betas is actually greater than returns. >> I'm hoping that the fifth grader will show up today and I'm hoping to have all of them back to you by sometime next week, I will say, and everybody in email to let them know what day they can come to class and pick up their exams. >> I will bring to this route. >> Now in the meantime, and certainly after you get your exams back, one thing you might want to do is stop over at my office, Brown lab to 33 and check out the bulletin board that's thick and hereafter. When I refer to my bulletin board, I do not mean an electronic bulletin board. >> I mean a thing made of cork that I stick push pins into to attach pieces of paper to it. >> If you were to go over there this morning, you would find that the answer key for the exam you took on tuesday has been posted. So you can stop by there anytime you want. >> If you want to take pictures of what's up there, that's up to you. You want to stop by and look at the answers to the questions, in a sense, in your own mind, how you think you might have done that's up to you to. >> But we would certainly hope that you would check out the bulletin board after you get your exams back so you can see where you made your mistakes. Hopefully learn the process, hopefully that that sort of thing again, And speaking of learning from the process and not doing that sort of thing. >> Again, let me call your attention to the fact that your second big thing is less than four weeks away. And that may seem like it's a decent distance away, but I'm going to promise you you will want as much time as you have to prepare for that. One thing, I got a lot, but this happens every time. >> In the days leading up to the first exam is a lot of email requests for clarification on how to print out the practice exams. >> I'm hoping that everybody understands how to do that and you could do that anytime, and I strongly recommend that you do so more. Exam number two. So you'll have a print copy and available to you and you won't have to worry about what happens when the system crashes the day before. >> The next big thing, what I mostly want to talk about today is the rest of chapter three, or at least as much of it. >> And as we'll have time to get to today, we had briefly touched on section 3.5 and the title of which is combustion analysis. >> But please recall that the main focus of chapter three in your textbook is the concept of the mole and everything that you can do surrounding that concept. I'll come back to that picture in just a moment. >> But one thing we took a brief look at last time was this slide. >> And one of the reasons I'm suggesting that you start preparing now for your second exam. The second exam involves a lot of quantitative analysis, crunching lots of questions that will be answered using numbers. And generally people find that they need some extra time, fair, for things like that. For those of you who are not paid back, they don't worry too much. We will get back to more qualitative topics before the semester is over. >> But between now and October 20th, it's going to be pretty heavy on quantitative analysis. >> One of the things I want to talk about today is how to solve the kinds of problems in combustion analysis that you will see in your textbook. >> And more importantly, a more general method that you could use to solve them and other similar problems. I hope what I mean by that, let me clear by the end of class today. >> Here is fundamentally what happens in a combustion analysis experiment. Some new compound gets discovery, happens to be composed of just three elements, carbon, hydrogen, and oxygen. >> And by the way, there are millions of compounds containing only those three elements that are known to exist. And to find out more about the gaba, like what its molecular formula is. >> We can subject it to combustion. >> That is, we can expose it to excess oxygen at high temperatures. >> And when that happens, all of the carbon atoms get converted into carbon dioxide molecule. All the hydrogen atoms windup with water molecules. Here is a schematic diagram that appears in your textbook for a combustion analysis apparatus. There's a lot of it over here where the sample for analysis is placed. >> We blowing oxygen at the high temperatures. >> And then the chambers over here contain other substances whose mission in life is to absorb water or absorb carbon dioxide. Carbon dioxide, of course, is normally a gas. >> Water at the high temperatures of the combustion oven is also a gas. >> So those two gases get absorbed by these things. >> And the point is, if you weigh these names before >> And then after the experiment, you can find out the mass of CO2 and the mass of H2O there was absorbed in the course of the combustion analysis experiment. If you also weighed your original comment on it and if you know the compounds molecular weight. And again, as being the 21st century, we have many very effective methods of determining compounds molecular weights. You should be able to determine the molecular formula of the combat. >> Now for those of you following along in the lecture notes, there are a couple of different approaches that you can take the solving problems like this. >> One of them appears on page so this method is actually pretty similar to what is shown in your textbook for determining the empirical formula. We've talked about the difference between empirical formulas and molecular formulas before. The point is, yes, you should be able to determine the empirical formula using this technique. >> But if you also know the compounds molecular weight, finding out the molecular formula should not be that count within the method on page 27 at the method of the textbook do in fact work. >> But they're a bit tedious and that they involve a lot of rammed them all back to Graham and ran back them all again, kind of calculations. There is an alternative method presented on page 28 in the lecture notes, which was actually suggested some years ago by one of my former camp 101 students. That is a bit more straightforward, doesn't involve quake so much. Number-crunching. And I think might be a bit more efficient if you're called upon to solve a problem like this during an exam, which you will almost certainly willing. So let me just share with you as we go through and solve this problem. >> And it's the same problem that appears on both pages 2728 in the lecture notes. >> But let me just share with you as we go through solving this problem, what this technique is MY focus, let's look at the problem first. >> Let's suppose just for a moment that you are the first person to discover this interesting compound called vitamin C, which can be used to prevent scurvy, has other health benefits as well. Vitamin C, as it turns out, is composed only of carbon, hydrogen, and oxygen. So it's formula could be represented as C. >> Next HY Oozie. >> That using various techniques for determining molecular weights, we find that the molecular weight of vitamin C is a 176 grams per mole. If we subject this compound to combustion analysis using an apparatus sort of like the picture shown in the textbook. That picture appears on page 99. >> For those of you following on that if we carefully weigh everything, we find that combustion of 5.1100 milligrams of vitamin C produced 7.66 milligrams of carbon dioxide and 2.09 milligrams of one. >> Armed with the information that you currently have on the screen, can you figure out the molecular formula? >> Can you figure out the values of x, y, and z in vitamins? >> Now, one thing that is kind of nice about combustion analysis is that you don't have to have large amounts of material. What to find out the formula of to make this work. If you just have a few milligrams, that's all it takes. Now, if you're comfortable with the concepts we've already discussed in this course. Looking at the information that you see here, what can you determine that might be helpful in solving this problem? >> Yes, sir. >> The oxygen? >> Yes. >> In other words, we're given the masses of everything else. >> The only thing we don't know is the mass of oxygen that was conceived. >> And that's hard to measure right off the bat because you're blowing oxygen through the sample as it goes. >> But what you can do is take advantage of the law of conservation of mass and figure out the mass of oxygen that was consumed. >> So as a general technique for solving problems like this, step one involves using the law of conservation of mass to find out how much oxygen it was consumed in the course of this particular experiment. And that should be pretty straightforward. >> That's just adding and subtracting. >> If you know, the products of this reaction are carbon dioxide and what? Here's the mass of carbon dioxide, 7.66 milligrams. Here's the mass of water, 2.09 milligrams. >> The reactants are the vitamin C and the oxygen. >> The mass of the vitamin C was 5.1 milligrams. >> Subtract that from the sum of the masses of the other things. And what we find by taking advantage of the law of conservation of mass is that 4.64 milligrams of oxygen must have reacted. In the course of this experiment. Are there any questions about that much? Okay, so now that we know the mass of oxygen, suggestions for how we might proceed to figure out what the molecular formula is. But a day this When I hope it's obvious that X, Y, and Z represent the number of atoms of each element in the molecule, right? But the quantities that we have on screen are gram quantities, actually milligram quantities. But the point is they are masses. How do you get from a ratio of masses to a ratio of atoms? >> Say again. >> Okay. >> Using what you're trying to convert what to what you give her milligrams to grams. >> That'll turn out to be useful. >> But yes, the magic word here is moles, because it turns out that the gram ratios here don't really telling you anything. But if you invert them into the corresponding mole ratios, that will tell you a lot, because it turns out that the mole ratios between the different atoms that exist is the same basic thing is the Adam ratios in the formula of the compound. So step two in our general method is to convert all of the masses that appear on the screen into low quantities by dividing by the appropriate molecular weights. >> And we can do that because we know the molecular weight for vitamin C that's given in the problem. >> And since we know the formulas of everything else, we can just look up the atomic weights on the periodic table and figure out the molecular weights too big. >> So we do those things. >> Here's what we get. >> In the experiment, we used 5.1 milligrams of vitamin C. He suggested converting milligram, milligrams to grams per day. That's valid. We can divide by 1000 to convert milligrams to grams. >> And that's reasonable because we have molecular weights in grams per mole. >> Also, they convert grams into moles. So we take 5.1 milligrams divided by 1000 grams and divide by the molecular weight of vitamin C, which was given in the problem as a 176 grams per mole. >> We find that 2 times ten to the negative five moles of Vitamin C was consumed in this experiment. >> And then we do the same thing where everything else, oxygen has the formula O2. From our previous calculation, we found that 4.64 milligrams of it were consumed divide by fasting and grams divided by the molecular weight of O2, which is 32-zero grams per mole. 1.45 times ten to the fourth moles of O2 consume in this reaction, the products. Carbon dioxide, 7.67.66 milligrams divided by 1000 per grams divided by the molecular weight of CO, 244.01 grams per mole, 1.74 for the moles of CO2. >> And finally water, 2.09 milligrams divided by 1000 divided by 18.02 grams per mole, 1.6 times ten to the negative fourth pulls and one. Any questions about step two? >> All right. So far, hopefully these all represent skills that you were supposed to be responsible for for your first exam. >> Now let me go back to the previous slide for a moment and take a look at the chemical equation up here. Now the point is, this is obviously not a balanced equation as written at the moment. >> We really wouldn't have any idea how to make it a balanced equation because we don't know the numerical values of x, y, and z. >> But hopefully you are comfortable with the idea that if we're going to try to balance this equation, we would insert various coefficients in front of the formulas that we're looking at. >> Now here's the thing with those coefficients. >> And that turns out to be one of the most important things we're gonna talk about today. The coefficients in front of the various species in a balanced chemical equation can refer to you how many molecules or atoms of those things that you then also refer to? >> How many moles of things that you have them. >> So in principle, having just calculated at these mole quantities, we could use these things as the coefficients in the balanced equation. >> However, most of the time when we write balanced equations, we like for our coefficients to be small whole numbers whenever possible. Can anybody think of a way that we can keep these four numbers in the same ratio, but represent them as small whole numbers. >> What's the little trick for doing that? >> Receive, Yes, divide by the smallest number. >> Which of these four numbers is the smallest? >> Say again, the one with the exponent negative five. >> Yes, 2.90 times negative five is smaller than all of the others because anything times ten to the negative five is smaller than anything times ten to the negative four. >> So step three of our general process is going to be to convert these into whole numbers of moles by dividing all of them by the smallest number. >> The smallest number in this case is 2.90 times negative five. >> So we know that when we divide 2 times that they unify by itself, we're going to gift one. >> Now with any luck, if we do that for the others will also get whole numbers. >> And it turns out that we do 2.9 times ten to the negative five divided by itself is of course, one coefficient for oxygen. >> We had one boy, 4-5 times negative four moles of oxygen. >> Divide that by 2.5, you get five. >> Here's the number of moles of carbon dioxide, or the number of moles of water. Dividing each of those by dou y 9-0 times negative five, you get 64 respectively. >> So the point is at this point, we can use these as the coefficients in our balanced equation. >> In other words, what we can do at this point is rewriting the equation that we showed you a few moments ago. >> Go ahead and insert these numbers as the coefficients, so we don't put one in front of the formula for vitamin C. >> He doesn't just one, but we do put a five in front of the oxygen, six in front of the CO2, and a four in front of the H2O brings us to the final step in the process. >> Normally, when we balance the equation, we know that the formulas are at a time and then we just fiddle with the coefficients. In this case, we determined what the coefficients we're using the results of the experiment. >> The only thing we know is the values of x, y, and z. >> But that's what we're trying to figure out anyway. So the last step in the process here is to simply insert the correct values for x, y, and z to make this a balanced equation. >> So for what we're looking at right now, what should be the value of x six? >> Because we have six carbon atoms on the right side. So we need six apartment applicable inside. >> What should be the value of y? >> Not 68, because we have four water molecules over here. >> Four times two is eight. Therefore, why should be eight? >> What should be the value of z? >> Six does turn out to be the right answer. >> But just in case that's not obvious, how many total oxygens do we have? >> On the right-hand side, Douglass 166 times to dwell right here, plus four more from the water. >> That's 16. >> So 16 oxygens on the right side. We need 16 oxygens on the left side, but we have ten oxygens right ear, from the five moles of O2, the reactants. >> So five times two is ten. >> You subtract that from 16 limits it. So if you do those things, you should get that x is six, y is eight, z is six, and therefore the correct molecular formula for vitamin C should be C6, age eight. >> Those things, how can you check yourself when all of that is done to convince yourself that you didn't directly Yep. >> Check to see if it matches the molecular weight or the data for this problem is the fact that the molecular weight for this compound is a 176 grams per mole. So if you've done it correctly, whatever formula you get should match up with the molecular weight. >> If you determine the molecular weight based on the formula C6H12O6 does come out to be pretty close to 176. >> So the method that I'm going to suggest that you use for solving problems like this. >> Looks like this. >> Step one, the law of conservation of mass, to find out the mass of oxygen that was consumed with the experiment, which is probably not going to be given it has data but shouldn't be too hard to figure out. >> Step to convert everything to moles. Step three, take a look at those bull on these. Figure out with honest, divide everything by the smallest number of moles that you have, the IDB. And try to get whole number coefficients where your balanced equation. >> And then finally balance that equation by inserting subscripts due to form. >> Now I'm going to do is assign you a couple of problems that appear at the end of chapter three that are similar except for one thing. >> And that is that those problems, they don't happen to give you the molecular weights. So what I'm gonna do right here, right now is give you the molecular weights. >> And then presumably you can solve those problems using the same technique, found one of them, and then looking for the other one. >> Well, I'll find that later. >> I'll give you the one that I like that. But what I'm going to assigned problem 3.63 on page 121 in your textbook. Problem 3.63 has to do with a compound called menthol that you might recognize. And the point is, they give you enough information to that problem to figure out the empirical formula, which is what they're asking you to do. >> But we live in a day and age in which it's very reasonable to expect to be able to determine molecular weights of compounds that might tell you white life. That the molecular weight of menthol is a 156 grams per mole. >> Try solving problem 3.63 on that basis, using the technique we just described, people totally get the molecular formula as well as the empirical formula. >> Let me know how it goes. >> And you could also find a simpler problem like this on the second practice exam if you go ahead and print that out. Now the point is there's a reason this technique works. And the reason the technique works Has to do with a word that I've come to the realization I don't know what's going on out in the high schools these days when people, they chemistry classes. But there's this word and it just scares the **** out of. So I'm just going to say the word. The word is stoichiometry. >> And it doesn't need to be a scary word of the title of chapter three in your textbook is stoichiometry ratios of combination. But the section that follows, section 3.5, which is merely, just weren't talking about combustion analysis. >> Is Section 3.6, the title of which is calculations with balanced chemical equations. Well, folks, I'm here to tell you that calculations with balanced chemical equations is stoichiometry. Here's what we mean by that. All we mean by the stoichiometry of a chemical reaction. >> The relative amounts of the substances present in moles in that chemical orange. >> That's it. >> And the point is you can figure that out simply by looking at the balanced equation. >> For example, they can look at this equation which happens to be a balanced equate. Pcl5 is phosphorus pentachloride. >> One mole of phosphorus penta chloride reacts with four moles of water to form one mole of phosphoric acid H3PO4 at LHC. >> Yeah, that's it. That's a statement of the stoichiometry of that grant. >> In other words, when you look at these coefficients and you realize that this is a balanced equation, you can say to yourself, OK, those coefficients could be used to refer to you members of molecules of each of these things. >> When we balance that equation, we usually think about how many atoms of each element we have on each side of the arrow. >> That's fine. >> So certainly one reasonable way to interpret these coefficients. >> This, I think in terms of the number of molecules. >> One thing this balanced equation tells you is that one molecule of PCl5 and four molecules of water react to form one molecule of phosphoric acid and five molecules of HCl. >> That's fine. >> But one thing we pointed out earlier, we were talking about writing balanced equation. A multiple of a balanced equation is still a balanced equation. If we multiply all of these numbers by 22 molecules, eight molecules do molecules, molecules is still a balanced equation. >> I'm all like 1010401050 or a 100 hundred, four hundred, five hundred, or any number we want. >> As long as you multiply everything by the same number, it's still a balanced equation. >> Okay? >> So we'll take that line of reasoning one step further. What happens if we multiply everything by Avogadro's number? >> Well, the answer is, if we do, we get the same thing, but in moles. >> So the coefficients also tell you that one mole of B, C alpha reacts with four moles of water to form one mole of us boric acid, five moles of HCl. >> And that's useful because we now how to get from bulk quantities, the gram quantities, by multiplying by the appropriate molecular weights. So here's the thing about the stoichiometry leverage. >> Stoichiometry expresses the ratios and the materials present in moles. The problem is that when you're working in the laboratory, you can't measure out mole quantities directly. The only thing you can do is measure out gram chlorines. But hopefully by now you're comfortable enough with converting random quantities, them all quantities, and vice versa, that you can take advantage of that. >> But I'm going to show you on the bottom half of the slide is the fundamental three-step process by which in principle, any stoichiometry problem can be solved. >> Because stoichiometry problems generally take this form. Given a gram, a quantity of substance a that's involved in a chemical reaction. >> What is the gram quantity of some other substance B that's also involved in that reaction. >> And the problem is you can't go directly from rams, they'd brands of b just by looking at the balanced equation. >> Because me It's in the balanced equation. >> I'll tell you about brands, but they do tell you about molecules or moles. So we can go from moles of a to moles of B by looking in the balanced equation. >> So to solve the problem, we just have to be able to convert gram quantities, interval quantities or vice versa. And David RAM1, and he's involved quantities. We just divide by the appropriate molecular weight and you go in the opposite direction, we just multiply by the appropriate pointed away. >> So the usual process that we go through to solve stoichiometry problems. >> The step one, get Bergson Graham quantity. >> If you sum all quantity by dividing by the appropriate molecular weight. Step two, convert bolt quantity of one thing and do bulk quantity of something else using the coefficients in the balanced equation. >> Step three, that recalls. And that second thing back in grams by multiplying by its molecular weight. >> Hopefully that's pretty straightforward, but let's see how it applies to a particular chemical reaction. If you are following along in the lecture notes >> This reaction appears on page 31. This is the balanced equation of a chemical process known as the Ostwald process, or I should say it's the balanced equation of one reaction in the Ostwald process. The Ostwald process isn't industrial process whose mission is to convert ammonia, NH3 eventually into nitric acid. Now you won't see the formula for nitric acid, which is HNO3 in this equation anywhere. That's because this is only one reaction in a sequence of reactions. >> But it's a good example for doing something like they still camera, because the balanced equation tells us that four moles of ammonia react with firewalls. The form four moles of this compound, n. >> O, is known by the name nitric oxide and six moles of water. >> And the question that we're asking here is, suppose we have 56.8 grams of ammonia. How many grams of oxygen do we need to react with that much opinion? How many grams of NO will be formed in the process? How many grams of water will be formed in the process? >> Bearing in mind what we just showed you on the previous slide about the basic three-step process for solving stoichiometry problems. >> I want to recommend we get started solving for the mass of oxygen involved in this range. >> Let's do this part first. >> Where do we begin? >> Yes. Okay. >> In other words, start with the only information you have, which is the 56.8 grams of ammonia. And step one is to convert that into a mole quantity by dividing by the molecular weight of ammonia. >> They go to the periodic table and you look up the atomic weights for nitrogen and hydrogen. >> And you calculate the molecular weight of ammonia is approximately 17.03 grams per mol. >> Divide 56.8 by 17.03. >> Take the three sig figs you're allowed. We find that we have 3.34 moles of ammonia reacting under these conditions. >> Again at step one. >> What's step two? >> Remember, we want to get from here to grams of oxygen. >> Okay? >> Okay, exactly. >> What we do next is we come up here to the balanced equation and we'll look at the coefficients for ammonia and oxygen. >> And the five tells us that when four moles of ammonia react, five moles of oxygen react. So all we do is we set up a conversion factor that converts moles of ammonia into moles of oxygen. Now that, what main piece of advice I will give you for solving stoichiometry problems is that when you do a calculation like this, to get started, they'll just write down 3.3 for moulds, right down on what its poles, 3.34 moles of ammonia. >> Because it, the next step we're going to set up a conversion factor that looks like this, which allows you to convert moles of ammonia and moles of oxygen, multiplying by five over four. But the main point is when we do that, moles of ammonia cancels out, leaving us with moles of oxygen. So we take the 3.34 moles of ammonia and multiply by five divided by four, we find that 4.17 moles of oxygen reacts with any questions about stiff to. >> Okay, finally, step three, we get from where we are to grams of O2. >> Finish it off. You multiply by the molecular weight of votive? >> Correct? And the molecular weight of O2 is approximately 32 grams per mole. So put this all into the calculator. Take that three sig figs that you're loud here because you started with three sig figs over here. And we find that we need a 133 grams of oxygen to react with this many grams of ammonia in this particular process that before we go any further, are there any questions about any aspect of solving this particular stoichiometry problem. >> Yes, that's the molecular weight of O2. >> Will go to the periodic table, look up the atomic weight of oxygen and multiply it by two. >> Because the formula says, go to any other questions about how this works? >> Alright, let's try the next part. How many grams of NO can we expect to form again, starting from 56.8 grams of ammonia? >> What's the first step saying that? >> And I don't know you, but not you eat. >> And the last, somebody else is gonna get involved there. >> What's step one? >> Yes. Okay. >> In other words, step one here is exactly the same as step one of the previous problem. >> Take that 56.8 grams of a login, divide by the molecular weight of ammonia, which is about 17 grams per mole, get this many moles hugging that step one. >> Now we're trying to find out about it. >> So let's step two. >> Yeah, we will eventually is the molecular weight of it. We have to do something else first. What's step two? >> In general, once you convert something to moles of O2 is, I'm sorry, you set up a conversion factor which converts What the, what mols of NH3 to moles of N up. >> And in this case, the arithmetic is going to be very easy because it's a four to four mole ratio, which is the same thing as a one to one mole ratio. >> So starting from our 3.34 moles of ammonia as before, and set up a mole ratio that looks like this, which converts moles of ammonia into moles of NO. >> No surprise, the answer comes out 3.34 moles of NO. >> And then finally, to get back to grams, we do want close, not divide here turns out to be multiplied by, but if you set it up as a conversion factor, you probably won't make that mistake. So the final step in here is to multiply by the molecular weight of n, turns out to be 30 grams per mole. >> Now you're allowed three sig figs here, because the original mass had three sig figs. Punch this up on the calculator. K36 phase. The answer terms out to be a 100. I suppose if you wrote it this way, you might get away with it on the exam graders in a generous mood. I think if you write it in scientific notation, like we're showing at the bottom line right here, that 100 written as 1 times ten to the negative two, much more clearly shows that there are three sig, figs. But the bottom line is the right answer is a 100 grams of it. >> I'm still making sense, getting more comfortable with the basic idea. Grams to moles, moles to moles, moles back brains, because they have one part of this left to figure out, and that's the mass of water that would be involved here. >> But the point is I've been taking it very slowly, going out one step at a time just to make sure everybody's clear on where all this is coming from. >> But once you get the hang of it, you don't have to do that. And if you want to stream to everything together on one line, we can do that. So solving for the water in this thing, let me put the balanced equation back of solving for the water. >> This thing is also very straightforward. >> Once again, we start with our 56.8 grams of ammonia step one, divide by the molecular weight of ammonia step to look at the mole ratio from the balanced equation. >> The mole ratio says, but six moles of water, four moles of ammonia react. So multiply by six over four, and then finally multiplied by the molecular weight of water, which is about 18 grams per mole. Again, your lab three sig figs, we find that 90.1 grams of water should be formed during this 3M. So point is, once you're comfortable with the basics of stoichiometry problems. >> If you want to just write everything out on one line and save a little time. That's fine. >> You can do to make sure you're comfortable. >> How might we check ourselves to make sure we did this entire problem? In other words, what the problem was asking was starting from 56.8 branching alone, yet grams of oxygen, how many grams of grams of water? >> We calculated all those things. >> How can we check ourselves to be sure we did it right? In other words, if we did it right, the results should obey the law of conservation of mass. The sum of the masses of the reactants should be equal to the sum of the masses of the Bronx. We started with at the 6.8 grams of ammonia calculated that we needed and a 133 grams of oxygen to react with it. And when we're done, we should have a 100 grams of NO, 90.1 grams of water. >> If we add these up, are these two numbers the same? >> It's a yes or no question. >> Are these two numbers the same? >> I'll make it even easier to those two numbers look the same now. >> But here's the thing. These are the numbers that come out of the calculated. What do calculators do? Remembers when calculators do. Okay. I'm going to refresh your memory. Calculators lie to you about sig figs. We did say that engineering. So here's a thing, these numbers are coming out of the calculator. But how many sig figs Are we allowed for each of these numbers? >> The answer turns out to be three, because in 56 E, the uncertain digit is the eight in the tens column. >> But at a 133, the uncertain digit is the three and the units column. And the rule for adding and subtracting is you're not allowed to go beyond whichever column contains an uncertain digit. >> So you shouldn't go beyond the units column. >> Year should go beyond the units column here either because the uncertain digit in the 100 is the 0 and the units don't. So if you rewrite these numbers, taking into account the fact that you really are only allowed three sig figs at each one, and they are the same. So the point is, yes, the law of conservation of mass is obey just because the calculator appears to not say so, doesn't mean that the calculator is right, because calculators lie to you about saying things. So just make sure that when you have a number coming out of the calculator, you actually write it down to the correct composing things. But in general, you can check yourself on a stoichiometry problem using the law of conservation of mass? >> Questions, yes. Well, let's put it this way. If you did all the calculations correctly, the numbers should come out to be the same. >> But that's why you check yourself because the numbers don't come out the same. Then you could say, oops, made a mistake somewhere. Then you go back and see if you can find me a major mistake. >> Any other questions about this? >> There are plenty of examples of stoichiometry problems in chapter three in your textbook. >> So please feel free to explore and see what you can find. >> Let me wrap up for the day by appealing to my nutrition majors. >> Because on page 29 on the lecture notes, I include a recipe for short read. >> It's a very simple minded recipe that I found in a very simple-minded Cookbook, which is good because I'm a very simple-minded. The recipe says in a bowl cream, one cup of softened butter at 1.5 cup of sifted powdered sugar, a tail light and fluffy stirring, and two cups of all-purpose flour. It blend well. Gather up the dough at present evenly over the bottom of a likely Grease 19 square, making bake for 45 minutes and 325 degrees Fahrenheit transfer pan to wire Act and cut 18-25 squares. >> Let Julien Pain that simple. >> Those are the kinds of step-by-step instructions that you would like to have if you're in your pitch and write. >> A limb of short-run. >> Here, of course, is how a chemist would mangle the same thing. Don't go looking on your periodic table for V USE, or SQ stands for butter, sugar, and short red squares. And I should probably change my notation here. Fl is supposed to represent flour, although if you have an up to date Periodic Table, one of the most recently discovered elements, fluorophore, you actually has the symbol FL. But for our purposes right now, that means flower point is, here's the quote, unquote balanced equation for this chemical reaction. >> And let's face it, cooking is a category. >> When you cook, you transform substances, two other substances which will hopefully be a bit more. >> Bowel syllabi formwork. >> Them 101 students that actually whipped up some short brand based on this recipe and brought an aid for sharing and its rich but good. Here's the difference though between this equation and a typical chemical equation. Because the numbers a year, one for the butter, 1.5 for the sugar, to, for the flower represent cups on each of those ingredients. And the 25 represents short red squares. And the point is you can directly measure cups, you can directly count short red squares in a chemical equation, the coefficients represent balls which you cannot measure directly. So I have to measure the gram quantities and then convert, which is what we're doing. But here's what I want to point out about this recipe. Now, if I'm correct, before much longer we have parents and family weekend coming isn't right. >> So before too much longer, the family might come down. >> Okay, let's suppose you want to impress them by whipping up a batch of short read using this recipe that you found out about neurochemistry glance. But here's the thing. >> You check out your supplies. If you find that you have ten cups of butter or seven cup of sugar, two cups of flour. That being a four-star BE acknowledged. You know, money with which to go buy anything more that's close enough to the truth, right? Given those quantities, how many short red squares can you make? Here's the recipe slash balanced equation up here. >> So I'm going to short red squares. Can you make from what you have say again, more than 50? >> Yes. >> I'm sorry. Say that 150 is the correct answer. >> How did you get a 155 day are okay with other words, You do that? >> Yes. Whatever that but that now that's important because that's what goes on and they do this event as well. I mean, that's good, but here's one that means, here's the recipe which basically says one cup of butter, 25 short red squares, have a cup of sugar, 25 certain bread squared, two cups of flour, 25-year grids where it's, it's very much like a mole ratio again. So here's one way of solving this problem from your ten cups of, but suppose you had ten cups of butter and all the sugar and flour, you possibly one. How many short red squares could you make? Well, the answer to that question is simple, because one cup and lettering, if your 20th, I actually read pairs. >> So ten cups gave you ten times as much. >> And you can set up a conversion factor that looks like this. And you very easily conclude that based on the butter, you could make 250 shortcuts. Same line of reasoning for the sugar. You need a half a cup of sugar to make your batch of shorter read. >> And you have seven cups of sugar. >> So starting from seven cups multiplied by 25 for red squares over Africa, pressure turns out to be 350 for Fred's whoops. >> And finally, the same thing for the flow. >> Two cups of flour per batch, we have cups of flour. >> So talk UPS flour times 25 over 250 short Brits wins. In other words, this is what we can make from ten cups of butter with all the sugar and flour we could possibly want. >> Here's a, we can make from seven cups of sugar with all the butter and flower we could possibly one. >> Here's what we can make from 12 cups of flour with all the butter and sugar would you possibly want in the actual situation? By the time we get done making a 150 short red squares, we've used up all our flour. And so it doesn't matter that we could have made more from the literature that we have because we run out of, well, so by the time they went out of flour, we can't make any more short grid squares that we've done things well, 150 short red squares that we get you. >> Makes sense? Oh yeah, it makes sense. Now it's food, newsflash dog food is chemicals, works with chemicals do are going to be talking about in more detail. Next steps are the following concepts, which you will find out all about in section 3.7 in your textbook, which is limiting reactants. >> The limiting reactant in a chemical reaction is simply the reactant that is completely consumed. Or you want to think about it that way, the reactant that is not present in excess in the short-run situation, flower is our limiting reactant because the flower gets used up first. It turns out we have access other and excess sugar, which simply means that by the time we're done using up the flower, we have butter and sugar leftover >> But the point is we used up the flower. >> So flower is the limiting reactant. It's called the limiting reactant because it's the thing that determines how much of your desired product you can make. The theoretical yield from a chemical reaction is the amount of product that we expect to be able to make, assuming that we can completely convert our limiting reactant into the desired product. So when I said how many short red squares we made, the answer turns out to be a 150. That's the theoretical yield of short read. >> In this situation, the yield of a chemical reaction is the amount of the product that we actually get, which is usually less than the theoretical yield. >> Just to stick with the short read situation for a little bit. I don't know. It baking day is like in your house? At our house, we fire up the oven, which is that superbly well calibrated. Maybe we set it to the right temperature or maybe we don't maybe with bird the first batch before we had the dial the oven back a little bit. >> And even if we get lucky, we didn't have to have a couple of English breaker spaniels at home. >> Maybe a few of those short breads where's accidentally land on the floor, wind up inside the dog, for whatever reason, you'll probably get less than what you were really expected. >> That's the yield. >> So you could use something called the percentage yield to describe the efficiency with which the reaction takes place. The percentage yield is just the yield divided by the theoretical yield and multiplied by a 100 to express it as a percentage. >> The point is just like we can calculate it based on food. >> We can calculated based on any other chemical reaction. The only difference with the other the three reactions is that the coefficients represent moles. So we have to get out at basic three-step stoichiometry technique. Grams to moles, moles to moles, moles back to grams. Be able to do it the exact same things in the short run, but we'll look at some more specific examples of these concepts. >> Next time enjoying it, we end up, we'll have your exams back where you next week, but watch your email for more information. Yeah. Oh, Mm-hm. Yeah. Oh, yeah. Yeah, yeah. Yeah. Yeah. Okay. Yeah. Yeah.
Recording from Sep 22, 2016
From Dana Chatellier March 03, 2020
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