Lecture 22. Midterm II Review
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Potma, Eric
UCI Chem 1P is a preparation go General Chemistry that covers: units of measurement, dimensional analysis, significant figures; elementary concepts of volume, mass, force, pressure, energy, density, temperature, heat, work; fundamentals of atomic and molecular structure; the mole concept, stoichiometry; properties of the states of matter; gas laws; solutions concentrations. Course may be offered online. Slides:
 All right. He is unaware of the fact that there's a midterm Friday. You don't know? Guess what? There's a midterm. So today is a good day to walk through a couple of topics that will be on midterm. So I'll go through the eggs. You can ask questions. Not right through, of course. In addition, tonight is a new session organized by the peer tours of the Department of Chemistry. At 7 o'clock, we're going to call Room 101, so you are welcome to join that session. They have a review package online that they prepared. So they will talk about those questions and you can ask questions to them. So it's pretty useful. In addition, tomorrow night, at least Thursday night, there is another review session for those folks who really want to talk about this kind of stuff. There's an opportunity at 7 o'clock in Don Grand Park, 11-0-0. So that is probably the final review session that you can attend. And it's all there for you. So many, many opportunities for us to actually improve your skills. Okay, so what's the deal? Friday, we do a midterm. Where? Right here? What do you bring? Same things as before. A calculator and you can bring the e by 5 flash card. Okay? What we do not like is cell phones that go off. So, turning off, really do that because when they go off, it's just upwardly annoying. Alright, so what will be on the midterm? Basically, it's everything that's covered in Lectures 11-18. Let me quickly go through that. Mold and mass, that is something we talked about. Mass to scientists, talked about the mass to scientists and how to use it to determine the empirical form of reference. So empirical in the likelihood of formulas, you have to know what they are. The difference is between the two and how you can determine your empirical form of reference and your likelihood of formulas. Long-lead and perfect formulas. Now, balancing counter equations. Identity is a very important thing that we should be able to do to balance an equation. Because if you do stoichiometric calculations, we have to know the stoichiometric numbers and that comes from equation. So balancing equations is very important. Limiting reagents, something we talked about, that enables you to determine how much product you generate and also how much excess reagent you have left after the reaction. Equal solutions, which one do well and which one do not. Polarity, that is a measure of concentration, number of moles per volume of the solution. And then we talked about precipitation reaction, know where they are and understand the solubility table. But solubility table is very important. There's a lot of detail on that table. So you're welcome to put it on your index card. That's a typical thing that I would put on there. And then finally, acid-base reaction. The final topic we talked about in lecture 18. And it is part of the return. Okay, so this is the, these are the ingredients of the return. Not bad, not bad. Alright, so let me, let me go through a couple of these, these items. So we'll not go through all of these, but there's a couple of topics that I think are very relevant. So, the mole, you know what the mole is then, right? The mole is just a number. What number is it? This number. That's it. That's the definition of the mole. And we use it exactly the same way as if we used the word dozen times. The word dozen in calculations. It doesn't mean it's trouble, and mole means it's not. That's it. Nothing to be serious about. What is very convenient for us is that one mole of an atomic mass unit, so atomic mass unit, if you have one mole of that, you have one gram. Okay? And that enables us to do calculations of molar masses in terms of grams. For instance, one mole of any compound, let's take beta-carotene. You've never seen this, let me burn beta-carotene, but this is a thermo-lock, C40H56. Very strange numbers, doesn't matter. This is a real compound. You should be able to calculate the molar mass of this. Because 40 times this thing here, which is the molar mass of carbon, this is, if you have one mole of carbon, carbon contributes 12.01 gram. That's because of this relation right here. 56 times the molar mass of hydrogen, okay, that's this guy. You just have these contributions up to come to the molar mass. We've done it many times, and it's the worst of every particular compound. No matter how strange they look. So, whenever you see a strange compound, a double matter, you can calculate it, okay? All the elements are in here at table, and you should be able to determine the molar mass of each and every compound that is served to you. Okay, so then, once you know the molar mass, you can do a couple of conversions, which were very important to us. For instance, you can convert from moles to mass. If you know the amount of moles you have this relation here, you multiply with the molar mass, you get the amount in grams, and conversion can also be done from mass to moles. If you have a certain amount of material, give it grams, amount of grams divided by the molar mass gives you the amount of moles. These are useful approaches, okay? So, and use them many, many times, okay? Okay, so, what else? Mass percentage, something we talked about. What is the definition? The mass of a particular atom, a particular element in a compound, okay, so let's take one mole of the compound, then the mass that that single atom contributes, that the element contributes to one mole of the compound, goes here in the numerator, and in the denominator goes the mass of one mole of the compound, which is nothing but the molar mass of the compound. So here, molar mass of the compound, here the mass contributed by that element, times 100%. We've done this several times. Let me show you an example that appeared on a midterm two years ago, or a couple of years ago. What are the mass percentages of the different elements in sorbitol? Three elements, carbon, hydrogen, and oxygen, and we can determine the mass percentages of each through this formula. So let's do that, you see how that works. Carbon, this is six moles of carbon per one mole of the compound. Six times the molar mass of carbon is the contribution of carbon to the total. This is the total of one mole of the compound, that's basically the molar mass, 182.17, is the molar mass of sorbitol. This ratio, six times this, divided by the molar mass, times 100 gives us 39.56% of this particular case. What are these six things that you need in this calculation? Yeah, you need four, because this number here is given as four. Let's say you grab this from the pure table, and that's four smithing figures, this is a limiting factor, that means this should be four. Some people ask, should percentage be in some kind. If the number is smithing figure, no, it's a calculation, you just stick to the rules. People ask, when do I need to apply significant figures? Yes, the very quick answer is, if you simply always stick to the rules, you can never go wrong. If you decide, well, maybe not when necessary, you actually risk a chance of doing it wrong. So just applying those rules actually will teach you to get into the habit, it's not really that difficult, you just have to do it several times, and then you can assure yourself that you at least won't lose points in that department. So the different figures, just stick to it, even though it's not explicitly said. Okay, mass of hydrogen, how do you do that? Same way, 14 times the molar mass of hydrogen, this guy here is the contribution of the mass of hydrogen, contribution of hydrogen to the mass, and this is the molar mass of the total compound once again, that ratio times 100 gives us 7.75%. That is the percentage, the mass percentage of hydrogen and sorbitol, and finally, the last element is oxygen, same deal, 6, that's a subscript that means 6 moles of oxygen per one mole of the compound, 6 times the molar mass, 16 grams per mole of oxygen, divided by the molar mass of sorbitol, that ratio times 100%, gives us 52.70%. Okay, if you're not sure that you did it correctly, you can always do a final check, and that is adding these guys up. Should add up to 100 if you include all elements. We have included all elements here, so they should add up to 100%. In this case, 100.01. This 01 is within experimental uncertainty, so this is perfect. Question? Why does hydrogen have 3.60? Say again, why does hydrogen have 3.60? This one here, this one should be in 4, that's a very good point, this one should have one more 6.00. That's right, absolutely. So this answer is not completely correct, but it's a more 6.00. Okay, another one. This is balancing equations. Balance the following molecular equations. This is a funny reaction. We haven't seen this one before, but it doesn't matter. We should be able to balance it, nonetheless, right? Barium oxide, this is reacting with hydrochloric acid, this is hydrogen peroxide, and this is barium chloride. So we haven't really worked a lot with this hydrogen peroxide, but it doesn't matter in this question, we just have to determine the correct stoichiometric coefficients in this equation. We can just try and balance. Okay? So rule of thumb is, we first want to look at those elements that are not hydrogen peroxide, so they are more exotic elements, so to speak. Barium is one of those. Barium has one here, one there. That seems to be fine. Chlorine, that's the next one I'm going to look at, because chlorine is not hydrogen peroxide. So take that one, I have two here, I have one there, so I need at least two there. So I'm going to put that there for starters. So I'm going to put it with another two, make it four, who knows, but at this point I have taken care of this chlorine, relative to this contribution right there. So there's a two here, that means I also have two hydrogens. Does that make sense? Hey, look at that, there's two there. That's good. Two oxygens and two oxygens, that means that putting a two here was sufficient for balance in the equation. So at this stage I'm done with this guy. Okay, another one, a typical question where everything is spelled out in words, and you have to first of all find the compounds and then balance them in the equation. Okay? Iron III sulfide is one reactant. The other one is hydrogen chloride. Gas is hydrogen chloride. Okay? So if you put hydrogen chloride in water, it becomes hydrochloric acid, HCl. Right? In the gas phase it's called hydrogen chloride. Those are the two reactants. The products are iron III chloride and hydrogen sulfide, which is a gas as well. So let's write down those reactants and products. Iron III is iron III plus, right? It's an ion. Sulfide has as its charge two minus. To make them equal and charge anionic side and cationic side, you need to have the lowers multiplier. That does that for you. So this is III plus, this is II minus. I need to multiply the anion with III and the cation with II to make a neutral compound. Thus, I have iron II and then SIII. That is iron III sulfide. Make sure you do that, okay? This is very important. We've done it some of the while back, but being able to write these compounds correctly is of course very important. If you make a mistake here, everything else is going to be different. If you want to balance the equation. Because you get a different number of elements that you have to balance out if you make a mistake. Okay, the other reactant is not so difficult. It's hydrogen chloride, which is HCl. In the gas phase, it's called hydrogen chloride. Okay, on the product side, I have iron III chloride, chlorine, the chlorine ion, as charge one minus. I've used only three of them to make a neutral compound. So, I have FECL III, which is a solid. Let's give it a year in question. And then the other product is hydrogen sulfide. Sulfide, again, is an anion with charge two minus O vice. That means I need two hydrogens to make a neutral compound. So, hydrogen sulfide is H2S in the gas phase. If you put this in water, it's called iron sulfuric acid. Okay. So, now we have to balance this. So, first let's look at the iron here. So, iron is two here, there is only one there. So, I'm going to put it two there. So, that's iron. And then let's look at the next element, which is sulfur. So, here, there's only one there, so I put a three there. Okay, so at least this part is balanced with respect to this. Now, I'm going to worry about the second reaction here. It says I have three times two is six hydrogens. So, that means I need a six there. Okay. And then, luckily for me, two times three, that means also six fluorines. So, I'm done. All right. So, this one is also balanced. These are real questions from exams. Okay. So, if you are comfortable with this, you should be fine on the exam. You feel like I am very jittery on this, then you know what to focus on. Okay. So, here we have a reaction. And the question is, write this reaction and balance it, but do that in a molecular equation way, the complete ionic equation, and the net ionic equation. Okay, so I have to write the same equation basically, three times. Molecular, complete ionic, and net ionic. So, let's do that. This seems like an acid-base reaction. Okay. So, this is called hydrosyanic acid. Hydrosyanic acid. We've seen it several times. That is sodium hydroxide. Those are the two reactants. It's an acid-base reaction. What happens? This H is going to meet up with this OH- to form water. Okay. So, let's write that. Here they are. These are the specimens, the reactants, in form of molecular compounds. This H plus here is going to interact with this OH- to form water, which is a liquid. Okay. And then the other two ions are sodium plus and CN minus, and they can be written as one compound in the 8-base space. Once again, 8-base space means this thing is fully dissolved, and either it is an ionic compound, or it means that they are dissolved into ions. That is not a specific set of equations. In the complete A-ion equation, we do explicitly write out the ions. Okay. We write out everything that exists in a ionic form in solution. That's what the complete A-ion equation does for you. So, how do you write this? H plus N. And CN minus. That's right. H plus, CN minus, both are dissolved. Okay. I have seen folks do the following, and please do not do this. They write H plus C4 plus N3 plus. Okay. I have no idea where this comes from, but it happens every year on exams. Every year, a couple of folks will do that. So, somebody must have told them that, but it was not me. I tell you this. And since I am the one that is taking your exam, you better listen to me. Okay. I am the joke. This is the truth. Okay. Okay. Sodium hydroxide, right there, and you have hydroxide, any ion. Sodium, any hydroxide, any ion. Okay. So, they are both stable ions in solution. On the other side, does water split? No. Let's say compound, that's already in solution, it's very happy, is non-ionic species. Okay. So, water stays water. This, however, is the ionic species, fully dissolved into sodium and the CN minus Na. Okay. Last step, net ionic. What we do here, of course, is cross out spectator ions. Spectator ions are those ions that appear on both sides of the equation in the same form. Okay. For instance, the CN minus here in the aqueous form, and there, it's exactly the same, I can cross that out. And the sodium in the same form here, as it appears on the other side, can also be crossed out. And what I am left with is simply the proton interacting with the hydroxide and the ion to form our good friend, the water one. Some people have asked me, in the acid-base reactions, is the net ionic equation always this? The answer is not always. Okay. For instance, I can have an acid-base reaction where the base is NH3, which is ammonia, which can accept the proton, then the product is NH4 plus. You do not form water explicitly. So be aware, it's not always this. You don't always have to form water in acid-base reactions. Acid-base reaction is defined in this particular case as the reaction where the proton is exchanged between two components. Okay. Another example. We have to write, again, the molecular, completely ionic and net ionic equation of this reaction. This is not an acid-base, but a precipitation reaction. So that means we have to recognize what is in precipitate. That means you have to know your solubility state. Okay. Aluminum nitrate bearing hydroxide are the two reactants. So let's write them down again for all intents and purposes. These are the reactants. The products are, well, let's figure out which is the one that is the precipitate. I can make two more combinations. Barium with nitrate and aluminum with hydroxide. Which one do you think is the precipitate? Aluminum hydroxide or barium nitrate? It's not barium nitrate because nitrates always dissolve well. That means it must be aluminum hydroxide. Okay. So the precipitate is aluminum hydroxide. Let's write that down. Aluminum hydroxide. Aluminum is 3 plus, OH minus, 1 minus, of course. That means I need three of these guys. Right? So it's Al OH 3. And then barium nitrate, two nitrates per barium because barium, of course, is 2 plus, always 2 plus. So these are the compounds. Okay? I belt this already for you. There's a 3 there, a 2 there, a 3 there, and a 2 there. Okay, so let's now split all these things into ions if they exist. This one is ionic and dissolves well. So I get aluminum and nitrate, barium and hydroxide. Okay? There's a 3 there, there's a 2 there. That means I get six hydroxides. Please take note, folks. Don't make a mistake. Don't write three hydroxides because you have really six. There's a 2 there. So look at the subscripts very carefully. Same here. There's a 2 there. The 2 multiplies with this 3. You have six nitrates. Okay? On the other side, I have my precipitate, which, of course, does not split up into ions. It is a solid. And then this guy is fully dissolved, which means, as a younger compound, it exists in a form of cation and anion. Again, 3 times 2 makes nitrate and anion in storage. Okay, so let's figure out which ones are the spectators. They need to appear on both sides in the same form. Okay? Nitrate, I see here, six nitrates right there. Beautifully crosses out. Three bariums and three bariums right there. Also very nicely crosses out. This hydroxide does not cross out because even though it's over here, this is part of a solid. Okay? Well, here it is fully dissolved. So this hydroxide does not cross out. So what I'm left with is the aluminum, the hydroxide, and the precipitate. That is the nettinger equation. One note is there is a 2 there, a 6 there, and a 2 there. That means 2, 6, 2. In the nettinger equation, what I want to do is to show you only the lowest stoichiometric coefficient. I can divide the whole equation by 2 to have 1, 3, 1. Okay? Do that. In the nettinger equation, you want to show the lowest stoichiometric coefficient. If you can design the whole thing by 2 or 3 or 1, whatever, do it. Okay? Then we talk about concentration, in terms of molarity. So the definition is mol of the compound over the volume in the air of the solution. Actually, this one should not be there. Okay? It's not the definition. It is mol of the compound in the volume of the solution, in terms of the air. So this one is incorrect. So if you dilute the solution, and you solve a stormalac, which is very convenient to use in several calculations, we realize that if you dilute a solution, the number of moles of solute do not change. Okay? So if I put sugar in a cup, let's put it even more simply. I have a cup of coffee. I put sugar in there, 1 cube. Then I put more coffee on top. Does the amount of sugar change? No, still 1 cube remains. But the volume has changed. That means the concentration has changed. The number of moles of sugar before and after I put extra coffee on top is the same. So before and after I have the same number of moles. Mold equals volume times molarity. So that means M times V before equals M times V after. That's where the formula comes from. This formula is indeed very convenient. You can use it when doing calculations in the volume of the solution. However, use it in terms of understanding the problem. Okay? Don't blindly apply this to anyone. Because you're going to run yourself into the ground at some point. Okay? So let me give you an example. This is a question which is very similar to the one in the practice measure. It says the following. I have a 0.675 liter solution of sodium chloride. It also says how many moles sits in there. Okay? How many liters of water needs to be added to make 0.15 molar of the solution of NaCl? Okay, so let's have a look. I can use this equation. Okay? I can use this equation. So what I have, I have a situation before which is this many liters. And I can calculate the molarity too. This amount of mol divided by this number is the amount of liters. So is the molarity. Okay? So I have a molarity and I have volume. And I can also determine the... I also have the molarity after. That means I can calculate the volume after. So this is V2. So let's plug in those things. Okay? So before times V. So m times V. m times V really is the number of moles. Okay? The number of moles is actually given right there. I don't even have to calculate this step explicitly. It is 0.34 moles. Okay? I'll show you this example because I wanted to realize that just m times V simply means number of moles. If the number of moles is already given, you don't have to make a conversion back to molarity and then multiply the volume again. Okay? So this is the numerator. This is the denominator, which is the concentration that you want to have. Okay? And 2, that's the one that you want to have. That means V2 can be calculated. V2 is 2.67. So that means initially I had this, 0.675. My final situation says the volume is 2.67. Okay? The volume is larger than before. I have added more water to it. That means the volume has gone off. However, the question is this. How many liters do you need to add? Okay? I don't need to add this much. This is the final volume. I already had something before I started. So what I have to add is the difference between the two. Okay? This is the final situation. This is what you started with, so what you have to add to get to the new volume is 1.59 liters. That is the answer. The answer is not this. Okay? And here's an example of understanding the question rather than just plugging in a formula and hoping that you're doing it right. Okay? I can very easily see whether you understand the question or not. If you forget this step, you probably just blinded your by the formula and everything will turn out well. Okay? But if you do this step, you actually have a whole of the problem. You know what the problem means. This is the final answer. So read carefully and visualize the problem. Try to understand what is going on in this particular situation. Okay. What are the reactions? You know I have a problem for red and blue sears. Okay? So I'm going to go through this once again. Hope on it. I have 1, 2, 3, 4, 5 red guys and 1, 2, 3 blue guys. Okay? That's what I have. Now these guys can react. We've got another. Namely, a blue guy can interact with a red guy in a form of a bomb. Two friends, arm in arm, walk forever. Okay? The question is, what is the limit in radiation? Blue or red? Blue. Yeah, you can see that very quickly. Blue. You don't even have to do anything special. However, we can do this also in the brain of these dumbbells. We have molecules. And this situation is so easy, you can just look at it and see the answer right in front of us. Okay? If you do molecular calculation with molecular compounds, you don't see that right in front of you like that. Okay? But I wanted to show you a similar area in two situations. Okay? So the way we do it is this. You look at the ratio between red and blue. They react one to one. Okay? What I have here actually is not one to one. I have five to three. Okay? That is larger than one. And then the definition is that I have a limiting reagent, which in this case is the denominator. Okay? The limiting reagent is the denominator if this is larger than one. Okay? So blue is the limiting reagent. How much excess reagent is left? Two. I can see that directly. No calculation needed. Again, look at the similarity here. Okay? The answer should be two. What do we do? We start with the limiting reagent, relay that to how much of the other guy has reacted. Okay? Through the mole ratio. The mole ratio is one to one. So I convert blue to red. That means I have reacted three red guys. What is left is what you started with minus what has reacted, which is two. We do the calculation as we do it in chemistry. And it is really the same kind of calculation as we do intuitively, but to spell it out in such a way that if you follow this track, you won't make a mistake. When things get more difficult, your intuition will leave you at some point. Let me give you another example. This is situation number two. Three guys in blue. One, two, three, four, five, six, seven, eight guys in red. Okay? The only reaction is four. Not one to one, but one to two. Okay? Each blue guy likes to have two reference. Okay? So which one is the limiting reagent? Is it blue or red? Let's calculate that. I mean, you can maybe say blue, but since it is a record two to one, you have to take into account that you have to multiply this with two first to see how much you have left. Okay? So let's look at a little bit more involved. You can imagine when the numbers are very different, and the ratios are completely different, if this is three to fifteen, you lose track. Okay? So this is when your intuition is escaping you, and you better stick to this way of applying it. It's the same way as if your intuition spelled out mathematically. Okay? The equation says this reacts one to two. Okay? Two red for each one blue. Actually, the actual situation is eight over three. That is larger than two. That means blue again, once again, is the limiting reagent. Okay? To calculate how much you have left, you take the limiting reagent, convert that to the other guy. Through the ratio two to one, that means six hundred reactors, you have eight minus six to the left. Two is the normal axis. Right? So the calculation here with the sears is down in exactly the same manner with molecules. No difference. Absolutely no difference. The ratios are ratios of number molecules, or moles of molecules. It doesn't matter. The ratio is a ratio. Okay, so let's apply that in the following way. Oh, this is actually the recipe that we're going to follow. This is what we have done so far. Okay? If you've got such a question where you have to determine, for instance, how much water do you pour, or how much excess reagent is left, you always want to start with writing out the equation and be correct. So write a balance equation. Then you want to determine which one is the limiting reagent. The recipe that always works is to identify the mole ratio of the reactants according to the equation. Then you cover the actual numbers, determine the ratio, and compare it to. From that, you can determine the limiting reagent. The ratio is more, the denominator is limiting reagent, and it's less, the numerator is the limiting reagent. And then finally, once you know the limiting reagent, the limiting reagent determines how much product will be poured and how much of the excess reagent will react. You always start using the amount of moles of the limiting reagent to calculate how much of the other two guys you have. So here's an example. This is a precipitation reaction. I have barium chloride, and I have iron sulfide. The question is, how much of the precipitate, in this case, it's given what the precipitate is, it is barium sulfide. So we have to determine how much in grams of the precipitate do we pour. First things first, I want to write down the balanced equation. Let's write down the reagent, barium chloride, that is 1, and then iron 3 sulfide. You've seen it before. Iron 3 sulfide is Fe2S3. Barium 2+, chlorine 1-, that means barium Cl2. Barium sulfide is this compound right here, BASS. Barium is 2+, sulfide is 2-, that means one sulfur for each barium. And iron 3 chloride, iron 3+, chlorine minus of the rest ions, and there's three chlorines for each one iron 3+. If you balance it, there's going to be two there, three there, and a three there. Okay, so now I know this equation. To simplify matters, you don't have to do this. Some people ask, when do I need to use the net end equation? I mean, it's up to you. The net end equation is very convenient in the case of precipitation of the actin strength, it tells you directly what are the things to look at. Okay? The net end equation of this situation is the following. Barium needs one ion of sulfur to form barium sulfide. Sulfide, sorry, barium sulfide. So this simplifies things dramatically. I don't have to worry about all these things, the stoichiometric coefficients are all different now, because the only thing I have to worry about in the net end equation is, one barium ion to one sulfide ion gives me barium sulfide. That's it. Okay? I have to determine which one is the linear reagent, the barium ion or the sulfide. And then I have to determine how many moles of these guys do I have. The equation says it reacts one to one. So let's see how much we actually have. How do you find it out? Well, look at this. I can calculate the number of moles of barium chloride. Each barium chloride unit has one barium. So if I multiply this with that, the molarity with the volume, I get the number of moles. Okay? The number of moles is 0.1 liter times 0.2 molarity is the number of moles of barium 2+. That's over here in the numerator. In the denominator, I do exactly the same to calculate the number of moles of the sulfide anions. Okay? There's three sulfide anions for each one mole of the compound. So this liter times the molarity times three. Okay? Because there's three of these guys for each one mole of the compound. So this denominator here is the number of moles of the sulfide anion. To complete this, you'll find the following 0.67 is the ratio. That means the numerator is the limiting reagent, which is the barium ion. Okay? The barium ion is the limiting reagent. How many barium ions do I have? Well, I've just calculated. It's right here. 0.02. 0.02 is the number of moles of barium ions. Is this a solution? Yes. Okay. Okay. Who can explain to me why do you simply multiply this number here with three? Why is that? Yes? It's three, I think, and the number is three. Yeah, three is because 0.1 liter times 0.1 molar to do the local technique, you get the number of moles of iron-3 sulfide. Number of moles of iron-3 sulfide. Each mole of iron-3 sulfide has three sulfides. Right? That's the three right there. Each one mole of iron-3 sulfide has three sulfides. If you throw in one mole of iron-3 sulfide in a solution, you will have three moles of sulfide anion. That's why there has to be a three. This is an important step, folks. If you don't get this, you should really ask the credit for me to explain it to you one more time. Because it's important to realize that this compound gives you three sulfide anions in solution if you dissolve one mole of it, three moles of sulfide. Okay. Barium is limited in reagent, you conclude it. So let's start with barium sulfide. So let's start with barium to calculate how much barium sulfide will produce. Okay? One to one. I saw in a previous page, one barium produces one barium sulfide. That's the product, so one to one. That means that it generates the same amount in terms of moles of barium sulfide. 0.02 times one is 0.02 moles of barium sulfide. The amount of moles of barium sulfide multiplied by the molar mass gives me the amount in grams, 3.4 grams of barium sulfide. Okay, let's do one more or not. Let's do a precipitation reaction. Okay? This is a table of condensed form that we already saw. So be sure you can work with this. Okay? And in the following way. This is precipitation reaction. Potassium sulfide is reactant with barium chloride. Again, to calculate the amount of product in precipitation. Same kind of question. Always starts with writing down the equation. Nobody's talking about it. Okay? So potassium sulfide, that's one reactant. Barium chloride is the other guy. Here they are. Okay? Potassium sulfide. This is barium chloride. The precipitate is going to be barium sulfate. Because potassium chloride is not a precipitate. Potassium dissolves with almost everything else. So it will be no precipitate with potassium. Barium chloride is not precipitate either. So the only option left is barium sulfate. And indeed, sulfate forms a precipitate with barium 2+. That's an exception. It's in the table. Okay? Now, this equation is already balanced for you. Determine the limiting reagents. We do it, of course, the same thing, because we always do. Look at the ratios in the equation. We have these two reactants. Okay? This is one mole to one mole of barium chloride. One to one. That means this ratio of the equation is one. And the actual numbers are here. One mole of potassium sulfate, four moles of barium chloride. So one to four is 0.25. That means this guy here is the limiting reagent. Potassium sulfate is the limiting reagent. Okay? In this case, I've not converted it to a net-eonic equation. We don't have to. Okay? Whatever works best for you. Okay. So I have to calculate how many grams of the precipitate. This is the precipitate. This is the limiting reagent. Okay? So what I'm going to do is I'm going to start with the number of moles of the limiting reagent, potassium sulfate, convert that to the number of moles of the precipitate. Let me go back very quick. One to one. Okay? Number of moles of the limiting reagent, converting that into the number of moles of the product, barium sulfate, one to one. That means I generate one mole of barium sulfate. Multiply that by the molar mass of barium sulfate. Gives me the amount of grams of barium sulfate. The nice thing about this example is that I have already everything written out in moles. Okay? All right. Very last example, two seconds, is an acid-base reaction. I want to show you that not every reaction has to form water. And here's an example. This is an acid. This H splits off. Ammonia is a base, a base, an acid. This H goes right there. Okay? Forming NH4 plus and the rest NIN. So take note of that because acid-base, in our definition, needs a proton to be transferred from one compound to the next. Often times, forming water in the counter ion is always minus. But if it's something else, like ammonia, people count the notes. Okay, folks. See you on Friday.
 All right.  He is unaware of the fact that there's a midterm Friday.  You don't know?  Guess what?  There's a midterm.  So today is a good day to walk through a couple of topics that will be on midterm.  So I'll go through the eggs.  You can ask questions.  Not right through, of course.  In addition, tonight is a new session organized by the peer tours of the Department of Chemistry.  At 7 o'clock, we're going to call Room 101, so you are welcome to join that session.  They have a review package online that they prepared.  So they will talk about those questions and you can ask questions to them.  So it's pretty useful.  In addition, tomorrow night, at least Thursday night, there is another review session  for those folks who really want to talk about this kind of stuff.  There's an opportunity at 7 o'clock in Don Grand Park, 11-0-0.  So that is probably the final review session that you can attend.  And it's all there for you.  So many, many opportunities for us to actually improve your skills.  Okay, so what's the deal?  Friday, we do a midterm.  Where? Right here?  What do you bring?  Same things as before.  A calculator and you can bring the e by 5 flash card.  Okay?  What we do not like is cell phones that go off.  So, turning off, really do that because when they go off, it's just upwardly annoying.  Alright, so what will be on the midterm?  Basically, it's everything that's covered in Lectures 11-18.  Let me quickly go through that.  Mold and mass, that is something we talked about.  Mass to scientists, talked about the mass to scientists and how to use it to determine the empirical form of reference.  So empirical in the likelihood of formulas, you have to know what they are.  The difference is between the two and how you can determine your empirical form of reference and your likelihood of formulas.  Long-lead and perfect formulas.  Now, balancing counter equations.  Identity is a very important thing that we should be able to do to balance an equation.  Because if you do stoichiometric calculations, we have to know the stoichiometric numbers and that comes from equation.  So balancing equations is very important.  Limiting reagents, something we talked about, that enables you to determine how much product you generate and also how much excess reagent you have left after the reaction.  Equal solutions, which one do well and which one do not.  Polarity, that is a measure of concentration, number of moles per volume of the solution.  And then we talked about precipitation reaction, know where they are and understand the solubility table.  But solubility table is very important.  There's a lot of detail on that table.  So you're welcome to put it on your index card.  That's a typical thing that I would put on there.  And then finally, acid-base reaction.  The final topic we talked about in lecture 18.  And it is part of the return.  Okay, so this is the, these are the ingredients of the return.  Not bad, not bad.  Alright, so let me, let me go through a couple of these, these items.  So we'll not go through all of these, but there's a couple of topics that I think are very relevant.  So, the mole, you know what the mole is then, right?  The mole is just a number.  What number is it?  This number.  That's it.  That's the definition of the mole.  And we use it exactly the same way as if we used the word dozen times.  The word dozen in calculations.  It doesn't mean it's trouble, and mole means it's not.  That's it.  Nothing to be serious about.  What is very convenient for us is that one mole of an atomic mass unit,  so atomic mass unit, if you have one mole of that, you have one gram.  Okay?  And that enables us to do calculations of molar masses in terms of grams.  For instance, one mole of any compound, let's take beta-carotene.  You've never seen this, let me burn beta-carotene, but this is a thermo-lock, C40H56.  Very strange numbers, doesn't matter.  This is a real compound.  You should be able to calculate the molar mass of this.  Because 40 times this thing here, which is the molar mass of carbon,  this is, if you have one mole of carbon, carbon contributes 12.01 gram.  That's because of this relation right here.  56 times the molar mass of hydrogen, okay, that's this guy.  You just have these contributions up to come to the molar mass.  We've done it many times, and it's the worst of every particular compound.  No matter how strange they look.  So, whenever you see a strange compound,  a double matter, you can calculate it, okay?  All the elements are in here at table,  and you should be able to determine the molar mass of each and every compound  that is served to you.  Okay, so then, once you know the molar mass,  you can do a couple of conversions, which were very important to us.  For instance, you can convert from moles to mass.  If you know the amount of moles you have this relation here,  you multiply with the molar mass, you get the amount in grams,  and conversion can also be done from mass to moles.  If you have a certain amount of material, give it grams,  amount of grams divided by the molar mass gives you the amount of moles.  These are useful approaches, okay?  So, and use them many, many times, okay?  Okay, so, what else?  Mass percentage, something we talked about.  What is the definition?  The mass of a particular atom, a particular element in a compound,  okay, so let's take one mole of the compound,  then the mass that that single atom contributes,  that the element contributes to one mole of the compound,  goes here in the numerator, and in the denominator goes the mass  of one mole of the compound, which is nothing but the molar mass  of the compound.  So here, molar mass of the compound, here the mass contributed  by that element, times 100%.  We've done this several times.  Let me show you an example that appeared on a midterm two years ago,  or a couple of years ago.  What are the mass percentages of the different elements in sorbitol?  Three elements, carbon, hydrogen, and oxygen,  and we can determine the mass percentages of each through this formula.  So let's do that, you see how that works.  Carbon, this is six moles of carbon per one mole of the compound.  Six times the molar mass of carbon is the contribution of carbon  to the total.  This is the total of one mole of the compound,  that's basically the molar mass, 182.17,  is the molar mass of sorbitol.  This ratio, six times this, divided by the molar mass,  times 100 gives us 39.56% of this particular case.  What are these six things that you need in this calculation?  Yeah, you need four, because this number here is given as four.  Let's say you grab this from the pure table,  and that's four smithing figures,  this is a limiting factor, that means this should be four.  Some people ask, should percentage be in some kind.  If the number is smithing figure, no, it's a calculation,  you just stick to the rules.  People ask, when do I need to apply significant figures?  Yes, the very quick answer is,  if you simply always stick to the rules,  you can never go wrong.  If you decide, well, maybe not when necessary,  you actually risk a chance of doing it wrong.  So just applying those rules actually will teach you  to get into the habit, it's not really that difficult,  you just have to do it several times,  and then you can assure yourself that you at least won't lose points  in that department.  So the different figures, just stick to it,  even though it's not explicitly said.  Okay, mass of hydrogen, how do you do that?  Same way, 14 times the molar mass of hydrogen,  this guy here is the contribution of the mass of hydrogen,  contribution of hydrogen to the mass,  and this is the molar mass of the total compound once again,  that ratio times 100 gives us 7.75%.  That is the percentage, the mass percentage of hydrogen and sorbitol,  and finally, the last element is oxygen,  same deal, 6, that's a subscript that means 6 moles of oxygen  per one mole of the compound,  6 times the molar mass, 16 grams per mole of oxygen,  divided by the molar mass of sorbitol,  that ratio times 100%,  gives us 52.70%.  Okay, if you're not sure that you did it correctly,  you can always do a final check,  and that is adding these guys up.  Should add up to 100 if you include all elements.  We have included all elements here,  so they should add up to 100%.  In this case, 100.01.  This 01 is within experimental uncertainty,  so this is perfect.  Question?  Why does hydrogen have 3.60?  Say again, why does hydrogen have 3.60?  This one here, this one should be in 4,  that's a very good point,  this one should have one more 6.00.  That's right, absolutely.  So this answer is not completely correct,  but it's a more 6.00.  Okay, another one.  This is balancing equations.  Balance the following molecular equations.  This is a funny reaction.  We haven't seen this one before,  but it doesn't matter.  We should be able to balance it,  nonetheless, right?  Barium oxide, this is reacting with hydrochloric acid,  this is hydrogen peroxide,  and this is barium chloride.  So we haven't really worked a lot with this hydrogen peroxide,  but it doesn't matter in this question,  we just have to determine the correct  stoichiometric coefficients in this equation.  We can just try and balance.  Okay?  So rule of thumb is,  we first want to look at those elements  that are not hydrogen peroxide,  so they are more exotic elements, so to speak.  Barium is one of those.  Barium has one here, one there.  That seems to be fine.  Chlorine, that's the next one I'm going to look at,  because chlorine is not hydrogen peroxide.  So take that one, I have two here,  I have one there, so I need at least two there.  So I'm going to put that there for starters.  So I'm going to put it with another two,  make it four, who knows,  but at this point I have taken care of this chlorine,  relative to this contribution right there.  So there's a two here, that means I also have two hydrogens.  Does that make sense?  Hey, look at that, there's two there.  That's good.  Two oxygens and two oxygens,  that means that putting a two here was sufficient  for balance in the equation.  So at this stage I'm done with this guy.  Okay, another one,  a typical question where everything is spelled out in words,  and you have to first of all find the compounds  and then balance them in the equation.  Okay?  Iron III sulfide is one reactant.  The other one is hydrogen chloride.  Gas is hydrogen chloride.  Okay?  So if you put hydrogen chloride in water,  it becomes hydrochloric acid, HCl.  Right?  In the gas phase it's called hydrogen chloride.  Those are the two reactants.  The products are iron III chloride and hydrogen sulfide,  which is a gas as well.  So let's write down those reactants and products.  Iron III is iron III plus, right?  It's an ion.  Sulfide has as its charge two minus.  To make them equal and charge anionic side and cationic side,  you need to have the lowers multiplier.  That does that for you.  So this is III plus, this is II minus.  I need to multiply the anion with III  and the cation with II to make a neutral compound.  Thus, I have iron II and then SIII.  That is iron III sulfide.  Make sure you do that, okay?  This is very important.  We've done it some of the while back,  but being able to write these compounds correctly  is of course very important.  If you make a mistake here,  everything else is going to be different.  If you want to balance the equation.  Because you get a different number of elements that you have to balance out  if you make a mistake.  Okay, the other reactant is not so difficult.  It's hydrogen chloride, which is HCl.  In the gas phase, it's called hydrogen chloride.  Okay, on the product side,  I have iron III chloride, chlorine, the chlorine ion,  as charge one minus.  I've used only three of them to make a neutral compound.  So, I have FECL III, which is a solid.  Let's give it a year in question.  And then the other product is hydrogen sulfide.  Sulfide, again, is an anion with charge two minus O vice.  That means I need two hydrogens to make a neutral compound.  So, hydrogen sulfide is H2S in the gas phase.  If you put this in water, it's called iron sulfuric acid.  Okay.  So, now we have to balance this.  So, first let's look at the iron here.  So, iron is two here, there is only one there.  So, I'm going to put it two there.  So, that's iron.  And then let's look at the next element, which is sulfur.  So, here, there's only one there, so I put a three there.  Okay, so at least this part is balanced with respect to this.  Now, I'm going to worry about the second reaction here.  It says I have three times two is six hydrogens.  So, that means I need a six there.  Okay.  And then, luckily for me, two times three, that means also six fluorines.  So, I'm done.  All right.  So, this one is also balanced.  These are real questions from exams.  Okay.  So, if you are comfortable with this, you should be fine on the exam.  You feel like I am very jittery on this, then you know what to focus on.  Okay.  So, here we have a reaction.  And the question is, write this reaction and balance it,  but do that in a molecular equation way, the complete ionic equation,  and the net ionic equation.  Okay, so I have to write the same equation basically, three times.  Molecular, complete ionic, and net ionic.  So, let's do that.  This seems like an acid-base reaction.  Okay.  So, this is called hydrosyanic acid.  Hydrosyanic acid.  We've seen it several times.  That is sodium hydroxide.  Those are the two reactants.  It's an acid-base reaction.  What happens?  This H is going to meet up with this OH- to form water.  Okay.  So, let's write that.  Here they are.  These are the specimens, the reactants, in form of molecular compounds.  This H plus here is going to interact with this OH-  to form water, which is a liquid.  Okay.  And then the other two ions are sodium plus and CN minus,  and they can be written as one compound in the 8-base space.  Once again, 8-base space means this thing is fully dissolved,  and either it is an ionic compound,  or it means that they are dissolved into ions.  That is not a specific set of equations.  In the complete A-ion equation,  we do explicitly write out the ions.  Okay.  We write out everything that exists in a ionic form in solution.  That's what the complete A-ion equation does for you.  So, how do you write this?  H plus N.  And CN minus.  That's right.  H plus, CN minus, both are dissolved.  Okay.  I have seen folks do the following,  and please do not do this.  They write H plus C4 plus N3 plus.  Okay.  I have no idea where this comes from,  but it happens every year on exams.  Every year, a couple of folks will do that.  So, somebody must have told them that,  but it was not me.  I tell you this.  And since I am the one that is taking your exam,  you better listen to me.  Okay.  I am the joke.  This is the truth.  Okay.  Okay.  Sodium hydroxide, right there,  and you have hydroxide, any ion.  Sodium, any hydroxide, any ion.  Okay.  So, they are both stable ions in solution.  On the other side, does water split?  No.  Let's say compound, that's already in solution,  it's very happy, is non-ionic species.  Okay.  So, water stays water.  This, however, is the ionic species,  fully dissolved into sodium and the CN minus Na.  Okay.  Last step, net ionic.  What we do here, of course, is cross out spectator ions.  Spectator ions are those ions that appear  on both sides of the equation in the same form.  Okay.  For instance, the CN minus here in the aqueous form,  and there, it's exactly the same,  I can cross that out.  And the sodium in the same form here,  as it appears on the other side,  can also be crossed out.  And what I am left with is simply the proton  interacting with the hydroxide and the ion  to form our good friend, the water one.  Some people have asked me,  in the acid-base reactions,  is the net ionic equation always this?  The answer is not always.  Okay.  For instance, I can have an acid-base reaction  where the base is NH3, which is ammonia,  which can accept the proton, then the product is NH4 plus.  You do not form water explicitly.  So be aware, it's not always this.  You don't always have to form water in acid-base reactions.  Acid-base reaction is defined in this particular case  as the reaction where the proton is exchanged  between two components.  Okay.  Another example.  We have to write, again, the molecular,  completely ionic and net ionic equation of this reaction.  This is not an acid-base, but a precipitation reaction.  So that means we have to recognize what is in precipitate.  That means you have to know your solubility state.  Okay.  Aluminum nitrate bearing hydroxide are the two reactants.  So let's write them down again for all intents and purposes.  These are the reactants.  The products are, well, let's figure out  which is the one that is the precipitate.  I can make two more combinations.  Barium with nitrate and aluminum with hydroxide.  Which one do you think is the precipitate?  Aluminum hydroxide or barium nitrate?  It's not barium nitrate because nitrates always dissolve well.  That means it must be aluminum hydroxide.  Okay.  So the precipitate is aluminum hydroxide.  Let's write that down.  Aluminum hydroxide.  Aluminum is 3 plus, OH minus, 1 minus, of course.  That means I need three of these guys.  Right?  So it's Al OH 3.  And then barium nitrate, two nitrates per barium  because barium, of course, is 2 plus, always 2 plus.  So these are the compounds.  Okay?  I belt this already for you.  There's a 3 there, a 2 there, a 3 there, and a 2 there.  Okay, so let's now split all these things into ions if they exist.  This one is ionic and dissolves well.  So I get aluminum and nitrate, barium and hydroxide.  Okay?  There's a 3 there, there's a 2 there.  That means I get six hydroxides.  Please take note, folks.  Don't make a mistake.  Don't write three hydroxides because you have really six.  There's a 2 there.  So look at the subscripts very carefully.  Same here.  There's a 2 there.  The 2 multiplies with this 3.  You have six nitrates.  Okay?  On the other side, I have my precipitate,  which, of course, does not split up into ions.  It is a solid.  And then this guy is fully dissolved,  which means, as a younger compound,  it exists in a form of cation and anion.  Again, 3 times 2 makes nitrate and anion in storage.  Okay, so let's figure out which ones are the spectators.  They need to appear on both sides in the same form.  Okay?  Nitrate, I see here, six nitrates right there.  Beautifully crosses out.  Three bariums and three bariums right there.  Also very nicely crosses out.  This hydroxide does not cross out  because even though it's over here,  this is part of a solid.  Okay?  Well, here it is fully dissolved.  So this hydroxide does not cross out.  So what I'm left with is the aluminum,  the hydroxide, and the precipitate.  That is the nettinger equation.  One note is there is a 2 there,  a 6 there, and a 2 there.  That means 2, 6, 2.  In the nettinger equation, what I want to do  is to show you only the lowest stoichiometric coefficient.  I can divide the whole equation by 2 to have 1, 3, 1.  Okay?  Do that.  In the nettinger equation,  you want to show the lowest stoichiometric coefficient.  If you can design the whole thing by 2 or 3 or 1, whatever, do it.  Okay?  Then we talk about concentration, in terms of molarity.  So the definition is mol of the compound  over the volume in the air of the solution.  Actually, this one should not be there.  Okay?  It's not the definition.  It is mol of the compound in the volume of the solution,  in terms of the air.  So this one is incorrect.  So if you dilute the solution,  and you solve a stormalac,  which is very convenient to use in several calculations,  we realize that if you dilute a solution,  the number of moles of solute do not change.  Okay?  So if I put sugar in a cup,  let's put it even more simply.  I have a cup of coffee.  I put sugar in there, 1 cube.  Then I put more coffee on top.  Does the amount of sugar change?  No, still 1 cube remains.  But the volume has changed.  That means the concentration has changed.  The number of moles of sugar before and after  I put extra coffee on top is the same.  So before and after I have the same number of moles.  Mold equals volume times molarity.  So that means M times V before equals M times V after.  That's where the formula comes from.  This formula is indeed very convenient.  You can use it when doing calculations in the volume of the solution.  However, use it in terms of understanding the problem.  Okay?  Don't blindly apply this to anyone.  Because you're going to run yourself into the ground at some point.  Okay?  So let me give you an example.  This is a question which is very similar to the one in the practice measure.  It says the following.  I have a 0.675 liter solution of sodium chloride.  It also says how many moles sits in there.  Okay?  How many liters of water needs to be added to make 0.15 molar of the solution of NaCl?  Okay, so let's have a look.  I can use this equation.  Okay?  I can use this equation.  So what I have, I have a situation before which is this many liters.  And I can calculate the molarity too.  This amount of mol divided by this number is the amount of liters.  So is the molarity.  Okay?  So I have a molarity and I have volume.  And I can also determine the...  I also have the molarity after.  That means I can calculate the volume after.  So this is V2.  So let's plug in those things.  Okay?  So before times V.  So m times V.  m times V really is the number of moles.  Okay?  The number of moles is actually given right there.  I don't even have to calculate this step explicitly.  It is 0.34 moles.  Okay?  I'll show you this example because I wanted to realize that just m times V simply means number of moles.  If the number of moles is already given, you don't have to make a conversion back to molarity  and then multiply the volume again.  Okay?  So this is the numerator.  This is the denominator, which is the concentration that you want to have.  Okay?  And 2, that's the one that you want to have.  That means V2 can be calculated.  V2 is 2.67.  So that means initially I had this, 0.675.  My final situation says the volume is 2.67.  Okay?  The volume is larger than before.  I have added more water to it.  That means the volume has gone off.  However, the question is this.  How many liters do you need to add?  Okay?  I don't need to add this much.  This is the final volume.  I already had something before I started.  So what I have to add is the difference between the two.  Okay?  This is the final situation.  This is what you started with, so what you have to add to get to the new volume is 1.59 liters.  That is the answer.  The answer is not this.  Okay?  And here's an example of understanding the question rather than just plugging in a formula  and hoping that you're doing it right.  Okay?  I can very easily see whether you understand the question or not.  If you forget this step, you probably just blinded your by the formula  and everything will turn out well.  Okay?  But if you do this step, you actually have a whole of the problem.  You know what the problem means.  This is the final answer.  So read carefully and visualize the problem.  Try to understand what is going on in this particular situation.  Okay.  What are the reactions?  You know I have a problem for red and blue sears.  Okay?  So I'm going to go through this once again.  Hope on it.  I have 1, 2, 3, 4, 5 red guys and 1, 2, 3 blue guys.  Okay?  That's what I have.  Now these guys can react.  We've got another.  Namely, a blue guy can interact with a red guy in a form of a bomb.  Two friends, arm in arm, walk forever.  Okay?  The question is, what is the limit in radiation?  Blue or red?  Blue.  Yeah, you can see that very quickly.  Blue.  You don't even have to do anything special.  However, we can do this also in the brain of these dumbbells.  We have molecules.  And this situation is so easy, you can just look at it and see the answer right in front of us.  Okay?  If you do molecular calculation with molecular compounds, you don't see that right in front of you like that.  Okay?  But I wanted to show you a similar area in two situations.  Okay?  So the way we do it is this.  You look at the ratio between red and blue.  They react one to one.  Okay?  What I have here actually is not one to one.  I have five to three.  Okay?  That is larger than one.  And then the definition is that I have a limiting reagent, which in this case is the denominator.  Okay?  The limiting reagent is the denominator if this is larger than one.  Okay?  So blue is the limiting reagent.  How much excess reagent is left?  Two.  I can see that directly.  No calculation needed.  Again, look at the similarity here.  Okay?  The answer should be two.  What do we do?  We start with the limiting reagent,  relay that to how much of the other guy has reacted.  Okay?  Through the mole ratio.  The mole ratio is one to one.  So I convert blue to red.  That means I have reacted three red guys.  What is left is what you started with minus what has reacted, which is two.  We do the calculation as we do it in chemistry.  And it is really the same kind of calculation as we do intuitively,  but to spell it out in such a way that if you follow this track, you won't make a mistake.  When things get more difficult, your intuition will leave you at some point.  Let me give you another example.  This is situation number two.  Three guys in blue.  One, two, three, four, five, six, seven, eight guys in red.  Okay?  The only reaction is four.  Not one to one, but one to two.  Okay?  Each blue guy likes to have two reference.  Okay?  So which one is the limiting reagent?  Is it blue or red?  Let's calculate that.  I mean, you can maybe say blue,  but since it is a record two to one,  you have to take into account that you have to multiply this with two first  to see how much you have left.  Okay? So let's look at a little bit more involved.  You can imagine when the numbers are very different,  and the ratios are completely different,  if this is three to fifteen, you lose track.  Okay?  So this is when your intuition is escaping you,  and you better stick to this way of applying it.  It's the same way as if your intuition spelled out mathematically.  Okay?  The equation says this reacts one to two.  Okay?  Two red for each one blue.  Actually, the actual situation is eight over three.  That is larger than two.  That means blue again, once again, is the limiting reagent.  Okay?  To calculate how much you have left,  you take the limiting reagent,  convert that to the other guy.  Through the ratio two to one,  that means six hundred reactors,  you have eight minus six to the left.  Two is the normal axis.  Right?  So the calculation here with the sears  is down in exactly the same manner with molecules.  No difference.  Absolutely no difference.  The ratios are ratios of number molecules,  or moles of molecules.  It doesn't matter.  The ratio is a ratio.  Okay, so let's apply that in the following way.  Oh, this is actually the recipe that we're going to follow.  This is what we have done so far.  Okay?  If you've got such a question where you have to determine,  for instance, how much water do you pour,  or how much excess reagent is left,  you always want to start with writing out the equation  and be correct.  So write a balance equation.  Then you want to determine which one is the limiting reagent.  The recipe that always works  is to identify the mole ratio of the reactants  according to the equation.  Then you cover the actual numbers,  determine the ratio, and compare it to.  From that, you can determine the limiting reagent.  The ratio is more, the denominator is limiting reagent,  and it's less, the numerator is the limiting reagent.  And then finally, once you know the limiting reagent,  the limiting reagent determines how much product will be poured  and how much of the excess reagent will react.  You always start using the amount of moles of the limiting reagent  to calculate how much of the other two guys you have.  So here's an example.  This is a precipitation reaction.  I have barium chloride, and I have iron sulfide.  The question is, how much of the precipitate,  in this case, it's given what the precipitate is,  it is barium sulfide.  So we have to determine how much in grams of the precipitate  do we pour.  First things first, I want to write down the balanced equation.  Let's write down the reagent, barium chloride,  that is 1, and then iron 3 sulfide.  You've seen it before. Iron 3 sulfide is Fe2S3.  Barium 2+, chlorine 1-, that means barium Cl2.  Barium sulfide is this compound right here, BASS.  Barium is 2+, sulfide is 2-, that means one sulfur for each barium.  And iron 3 chloride, iron 3+, chlorine minus of the rest ions,  and there's three chlorines for each one iron 3+.  If you balance it, there's going to be two there,  three there, and a three there.  Okay, so now I know this equation.  To simplify matters, you don't have to do this.  Some people ask, when do I need to use the net end equation?  I mean, it's up to you.  The net end equation is very convenient in the case of precipitation of the actin strength,  it tells you directly what are the things to look at.  Okay? The net end equation of this situation is the following.  Barium needs one ion of sulfur to form barium sulfide.  Sulfide, sorry, barium sulfide.  So this simplifies things dramatically.  I don't have to worry about all these things,  the stoichiometric coefficients are all different now,  because the only thing I have to worry about in the net end equation is,  one barium ion to one sulfide ion gives me barium sulfide.  That's it.  Okay? I have to determine which one is the linear reagent,  the barium ion or the sulfide.  And then I have to determine how many moles of these guys do I have.  The equation says it reacts one to one.  So let's see how much we actually have.  How do you find it out?  Well, look at this.  I can calculate the number of moles of barium chloride.  Each barium chloride unit has one barium.  So if I multiply this with that, the molarity with the volume,  I get the number of moles.  Okay?  The number of moles is 0.1 liter times 0.2 molarity  is the number of moles of barium 2+.  That's over here in the numerator.  In the denominator, I do exactly the same  to calculate the number of moles of the sulfide  anions.  Okay?  There's three sulfide anions for each one mole of the compound.  So this liter times the molarity times three.  Okay?  Because there's three of these guys for each one mole of the compound.  So this denominator here is the number of moles of the sulfide anion.  To complete this, you'll find the following 0.67 is the ratio.  That means the numerator is the limiting reagent,  which is the barium ion.  Okay?  The barium ion is the limiting reagent.  How many barium ions do I have?  Well, I've just calculated.  It's right here.  0.02.  0.02 is the number of moles of barium ions.  Is this a solution?  Yes.  Okay.  Okay.  Who can explain to me why do you simply multiply this number here with three?  Why is that?  Yes?  It's three, I think, and the number is three.  Yeah, three is because 0.1 liter times 0.1 molar  to do the local technique, you get the number of moles of iron-3 sulfide.  Number of moles of iron-3 sulfide.  Each mole of iron-3 sulfide has three sulfides.  Right?  That's the three right there.  Each one mole of iron-3 sulfide has three sulfides.  If you throw in one mole of iron-3 sulfide in a solution,  you will have three moles of sulfide anion.  That's why there has to be a three.  This is an important step, folks.  If you don't get this, you should really ask the credit for me to explain it to you one more time.  Because it's important to realize that this compound gives you three sulfide anions in solution  if you dissolve one mole of it, three moles of sulfide.  Okay.  Barium is limited in reagent, you conclude it.  So let's start with barium sulfide.  So let's start with barium to calculate how much barium sulfide will produce.  Okay?  One to one.  I saw in a previous page, one barium produces one barium sulfide.  That's the product, so one to one.  That means that it generates the same amount in terms of moles of barium sulfide.  0.02 times one is 0.02 moles of barium sulfide.  The amount of moles of barium sulfide multiplied by the molar mass  gives me the amount in grams, 3.4 grams of barium sulfide.  Okay, let's do one more or not.  Let's do a precipitation reaction.  Okay?  This is a table of condensed form that we already saw.  So be sure you can work with this.  Okay?  And in the following way.  This is precipitation reaction.  Potassium sulfide is reactant with barium chloride.  Again, to calculate the amount of product in precipitation.  Same kind of question.  Always starts with writing down the equation.  Nobody's talking about it.  Okay?  So potassium sulfide, that's one reactant.  Barium chloride is the other guy.  Here they are.  Okay?  Potassium sulfide.  This is barium chloride.  The precipitate is going to be barium sulfate.  Because potassium chloride is not a precipitate.  Potassium dissolves with almost everything else.  So it will be no precipitate with potassium.  Barium chloride is not precipitate either.  So the only option left is barium sulfate.  And indeed, sulfate forms a precipitate with barium 2+.  That's an exception.  It's in the table.  Okay?  Now, this equation is already balanced for you.  Determine the limiting reagents.  We do it, of course, the same thing, because we always do.  Look at the ratios in the equation.  We have these two reactants.  Okay?  This is one mole to one mole of barium chloride.  One to one.  That means this ratio of the equation is one.  And the actual numbers are here.  One mole of potassium sulfate, four moles of barium chloride.  So one to four is 0.25.  That means this guy here is the limiting reagent.  Potassium sulfate is the limiting reagent.  Okay?  In this case, I've not converted it to a net-eonic equation.  We don't have to.  Okay?  Whatever works best for you.  Okay.  So I have to calculate how many grams of the precipitate.  This is the precipitate.  This is the limiting reagent.  Okay?  So what I'm going to do is I'm going to start with the number of moles  of the limiting reagent, potassium sulfate,  convert that to the number of moles of the precipitate.  Let me go back very quick.  One to one.  Okay?  Number of moles of the limiting reagent,  converting that into the number of moles of the product,  barium sulfate, one to one.  That means I generate one mole of barium sulfate.  Multiply that by the molar mass of barium sulfate.  Gives me the amount of grams of barium sulfate.  The nice thing about this example is that I have already everything  written out in moles.  Okay?  All right.  Very last example, two seconds, is an acid-base reaction.  I want to show you that not every reaction has to form water.  And here's an example.  This is an acid.  This H splits off.  Ammonia is a base, a base, an acid.  This H goes right there.  Okay?  Forming NH4 plus and the rest NIN.  So take note of that because acid-base, in our definition,  needs a proton to be transferred from one compound to the next.  Often times, forming water in the counter ion is always minus.  But if it's something else, like ammonia, people count the notes.  Okay, folks.  See you on Friday.
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