Thank you. Hey and welcome to IB stoichiometry revision. This is a revision session on IB topic one which is all about moles and calculations and ratios and it's a topic that's suitable for both HL and SL so no matter what level of chemistry you're studying this is a great video for you. I've kind of designed this so that it's a perfect video to kind of start your IB chemistry revision for this topic It goes through the equations in the unit and then different types of questions that you might be asked in the unit. So you can use it as a way to find out what your strengths and your weaknesses are in the unit and then kind of seek out more questions or more information if it doesn't make sense to you here. So let's get started. The first thing that you can see here is that I've got open my revision worksheet which you can download from the information under this video. As well as that I've got the data bookcloth open which you should really have access to if you're doing any form of revision for chemistry. And then thirdly I have a version of the iB Periodic Table just as a separate tab to kind of save me scrolling through the data booklet to find the periodic table every time I want it. So that's it. because I inevitably want it so often. So let's get started by looking at the equations that I kind of prepared here for us already to speed up this video a little bit. So the first one, hopefully you feel familiar with. It's used on... exclusively for solids. And what we've got is we've got N, which is the moles measured in moles. And then that's equal to the mass in grams divided by the molar mass or relative molecular mass, depending on what you're doing, in grams per mole. Now, what we're going to do here is we're going to highlight the molar mass in yellow, because we can find this in our data booklet. And it's really important in this unit that you know what's available to you in your data booklet, and that you're not just memorizing a ton of stuff for the sake of it. But this, the formula masses, you can definitely find from your data booklet. You'll find them in the periodic table. You've got relative atomic masses. And so you can use these numbers in order to calculate the overall relative molecular mass and chuck it into that equation there. So one equation, done. The second one is using Avogadro's number to calculate the number of particles that you have in a sample. I use the word particles kind of generically because I could mean atoms or ions or molecules, just like things, the number of things that you have. And we calculate this by taking the moles times Avogadro's. number and the units for Avogadro's number is in per mole. So Avogadro's number, this 6.02 times 10 to the 23, you can definitely find in the data booklet. So if you scroll along here to table one, you've got all the equations and then underneath in table two, we've got Avogadro's constant. which is given in our unit in our section on constants. So that is super useful to know that that exists and to know that that is there for us. So we're going to highlight that in yellow too. So thirdly, we've got percentage yield. So that's the actual amount of stuff we make divided by the theoretical amount of stuff that we could have made. Notice again I'm being super vague. This word amount normally isn't in my chemistry vocabulary but here it could mean mass in grams, it could mean volume in centimeters cubed or decimeters cubed, it could mean a multitude of things. So we use the word amount but what you need to remember is that both of those amounts need to be in the same units. So if your actual amount is in grams your theoretical amount also needs to be in grams to give you the right number overall. Okay, so that is percentage yield. The next equation we use just for gases. So this here is how we work out the one way of working out the moles of a gas. And you take the volume of that gas in this unit here called decimetres cubed, and we divide it by something called the molar volume of that gas, which is really like how many decimetres cubed are there per mole of gas at a certain temperature and pressure. Now this... exists in the data booklet at standard temperature and pressure. They tell us the standard temperature and pressure, one mole of any gas will occupy 22.7 decimetres cubed. And that can be a really, really easy way to answer certain questions. If they're asking about standard temperature and pressure and volumes of gases, then this can give you a really quick shortcut to the answer. So let's keep going. We've then got this equation here, PV equals nRT. This is actually the only one that exists in the data booklet, the only one from this unit that is there for you. PV equals nRT, so we'll make that nice and yellow. And this is called the ideal gas equation. If there is anything that is going to screw you over in this set of equations, it's this one and its units. We really, really need to nail the units and make sure we know what units are required. So the P stands for pressure. It is measured in pascals. The V, which is volume, is measured in meters cubed. So a really like key thing to remember is that to go from centimeters cubed to decimeters cubed to meters cubed, all of these you're dividing by a thousand. You're a thousand times bigger in units than each other. So there's like you're going in order. This, again, is in the data booklet. They tell you here. Let me show you. and the data booklet, they tell you here, one decimetre cubed equals one metre, which is one times ten to the minus three metres cubed, which is one times ten to the three centimetres cubed. So they're trying to help you with that conversion there, but make sure those conversions are really, really strong for you. The n is most. The r is our gas constant. I feel like you're getting fed up with me telling you things are in the data booklet, but this, guess what? It's in the data booklet. It's 8.31 and it's measured in joules per Kelvin per mole. This is in the data booklet. It's a constant, so unsurprisingly we're finding it. in the constant section where we've been working most of this time but actually a lot of the stuff that you need here is available in the data booklet and I think knowing that can be really helpful Lastly, we've got the T, which is temperature, which is measured in Kelvin. So making sure that you also know that to get from degrees C to Kelvin, we're going to add 273. That is not given to you. That's something that you're just expected to know. So make sure that you do know that. OK, last equation. This is our equation for solutions. So for solutions, you've got... your moles, n, equals the concentration in moles per decimetres cubed, and then times by the volume, which is measured in decimetres cubed. So again, making sure that if your volume's in centimetres cubed, then you're doing the conversions between those units. And that is all of the equations that you need for this unit, with the units of all of those equations. for you there. So let's have a go at some questions. The first thing that we're going to do is we're going to look at how we calculate molar mass. So what we've got here is some multiple choice questions. Remembering that if questions are multiple choice, that means they are non-calculator. So I'm also going to help you to do some of the mental arithmetic and how to think about these questions to get the answers when the calculation looks almost impossible. So what we've got here says what amount in moles, so lots of our equations have moles in them, that's not particularly helpful to me just yet. What amount in moles is present in 2 grams of sodium hydroxide? If it's in grams, that tells me that this is definitely a solid. So I know that I should be using my moles equals mass divided by formula mass. In your paper one, you do get given a periodic table. And so what we've got here is we can take the formula mass for sodium, which is the roughly 23. and then you've got the oxygen over here, which is 16, and then you've got the hydrogen at the top, which is around 1. So really, all that they're asking you to do is they gave you the mass, they gave you the 2.0, divided by the formula mass, which is going to be roughly your 23 plus 16 plus 1. Okay? So this is roughly 2.0 divided by 40. So mental arithmetic wise, even if you're not very good at working out like how many decimal points it can be, you can do some tricks to kind of work out which one is most likely. If you're doing 2 divided by 40, you know that number is definitely less than 1. And so therefore, there's no way the answer can be C or D. Also, 2 divided by 4 is a half. And so the answer here is A. because what you've got is you're definitely going to have 5 in that number, because it's got to be 0.5 something, 0.05, 0.005, depending on where those numbers fall. There are different ways to solve this. There's no right way, so whatever makes sense to you is good to go. Let's go on. The second question here is about working out the mass of one mole. And this statement here actually, this statement, the one that says mass of one mole, really just means like the molar mass, the mass of one mole of it, of hydrated copper sulfate. And people get confused here because you don't realize that the water of crystallization, this dot 5 H2O, contributes to the formula mass. Like it's part of the mass of the thing, so you have to take it into consideration. So using your periodic table, or using information that they've given you, you can literally just add these together, but you have to include those 5H2O. So you end up with 64 plus 32 plus your 4 times 16 plus 5 and then times your 2 plus 16. I should probably put extra brackets there. Okay. Overall. This comes to 250 grams per mole. So the answer here is D. Okay, so let's move on to what we do when we're calculating number of particles in multiple choice questions. Because often people find this quite hard because the calculations look really tricky from the outside. So it says, what amount of oxygen in moles contains 1.8 times 10 to the 22 molecules? Now that sounds like a huge amount of information to process. The way that I solve these is that I always think first about what one mole of particles look like. So if I've got one mole of oxygen molecules, then that means that I've got 6 times 10 to the 23. molecules. Okay, so what they're asking is not for 6 times 10 to the 23 molecules, but they're saying, oh, how many moles is your 1.8 times 10 to the 22? So what I'm thinking is I'm thinking, oh, is that a bigger number or is it a smaller number than 6 times 10 to the 23? So this is definitely smaller. than one mole. So we know for sure that it can't be D because it's definitely smaller than one mole. So really the question is like, how much smaller than one mole is it? So if I took this down to six times 10 to the 22 molecules, so by a factor of 10, that would make this 0.1 moles. 0.1 moles. So what we've got here is something that's still smaller than this. If we go down again, we're going to 6 times 10 to the 21 molecules. And that becomes 0.01 moles. So the way I think about this is that the number that they have given me, the number they've given me, this 1.8 times 10 to the 22, is between these two numbers. It's between the two. So therefore the moles must also be between these two numbers here, between 0.1 and 0.01. So therefore the only thing that satisfies that criteria is B. You're 0.03. moles it's the only one that's between the two so often when you're solving these it makes sense to think about the 6 times 10 to the 23 and how it relates to that let's have a look at another one because these are really like you need a lot of practice to get good at these and to get good at solving them without a calculator so the next one what is the total number of oxygen atoms in your 0.2 moles of glucose So firstly, you should identify that there are six oxygen atoms within this formula. So if I've got 0.2 moles of glucose, like 0.2 glucoses, then that means that within that I need to times this number by six to work out how many oxygens I've got. So overall, this means that you would have 1.2 moles of oxygen atoms. Okay, notice that we're not doing particles, I don't need the two, anything like that. They're just asking about the oxygen atoms. So if they want the number of oxygen atoms, then that means that I can use my formula. So I can do like the number of particles equals the moles times Avogadro's number. So this becomes 1.2 times 6 times 10 to the 23, Avogadro's number here. Now, what I really want is I'm looking for something, a number that's a bit bigger than 6 times 10 to the 23. I'm approximating. There's no way I'm going to try and do this calculation in the exam, because often in the exam you can just pull out and destroy answers because they don't make any sense. The only one that is bigger than 6 times 10 to the 23 is this one here, D. 7.22. times 10 to the 23. So that has to be the answer. And if you worked out the calculation and had a calculator, then of course you do that in two seconds. Let's have a quick look at empirical formula. So the definition of an empirical formula is the simplest ratio of atoms within a compound. So we're looking for that simplest ratio overall. Multiple choice questions often look like this, which says which compound has the empirical formula with the greatest mass? So they're asking me to work out the empirical formula, the simplest ratio for each of these, And then which one is going to have the greatest mass or really like the greatest relative molecular mass in this case? So what we're going to do is we're going to simplify all of these down to the lowest. ratio that they can possibly be. I can't talk and think about this at the same time. So let's just see H. So this is the simplest ratio for each of these compounds, each of these hydrocarbons. And then which one is the heaviest? I know carbon is way heavier than hydrogen, so it's going to matter a lot more. So the answer is definitely going to be B because it has two carbons and each of them are 12, so that's a lot. Okay, let's look at how we work out the empirical formula when we're given some information about a compound. So this compound is a compound containing carbon, hydrogen and oxygen. That's what I tell you in the question. They give you the percentage of carbon and hydrogen. And then we can work out the percentage of oxygen by doing 100 minus these two. And that gives me 36.4%. You can convince yourself that if you're a calculator. The way we work out empirical formula is the ratio of the moles of each element within that compound. So we need to work out the ratio of the moles. Like if I've got one mole of one, how many moles by comparison do I have of the other one? In this case, though, I can't work out moles from a percentage. So what I need to do is I need to pretend that the compound weighs 100 grams. And then what the percentages mean is just a mass of that 100 grams. Because it doesn't matter how much it weighs. It's the ratio of one to the other. So for the purposes of this, we're imagining that the sample that we've got weighs 100 grams. So therefore, our percentages are just masses themselves. To calculate this, I often will set up a little table to help me to organize the information that you're given in a question. Because often in these questions, you're given a ton of numbers. And it's really hard to work out which numbers go with which elements and all of that kind of stuff. So if you've got space, then I really recommend you do this to work out what's going on. So in the first... row, we're going to put the mass in grams. And remember, this is going to be the same as the percentages because we're going to assume that the sample weighs 100 grams. And so therefore, you are all good. Then what we're going to do is we're going to work out the moles of each of these. So to work out the moles, you're going to need to take the formula mass of each one. So carbon is around 12. 12.01, hydrogen is 1.01, and oxygen is 16.00. That information I'm pulling straight from the periodic table, it's coming straight from these numbers here. Oh, straight from these numbers here. Okay. So to work out the moles of each of these, we're going to do moles equals mass divided by formula mass. So I'm going to do 54.5 divided by 12, which gives me 4.54. Then 9.1 divided by 1, which gives me 9.1 still. And then 36.4 divided by that 16, which gives me 2.28. So these are the moles. So all we need to do now is look at the ratio between these. Now sometimes it might be really clear what the ratio is just from the numbers that you're given. But if you're not sure, then I really recommend you divide by the smallest of these numbers. So in this case, between the 4.54, the 9.1 and the 2.28, the smallest one is your 2.28. So what that means is I'm going to divide all of them by 2.28. So what that means is I'm essentially making this one equal one, so that I can see what the ratio is by comparison to it. So if I do 4.54 divided by the 2.28, I get roughly two, and then the 9.1 becomes roughly four. So this is the ratio of the moles. The last step, the really important step to get these three marks, is to actually write the empirical formula itself, which will be C2H4. Oh, we don't need the one. Another different type of calculation that is kind of similar to the ones on empirical formula is how you work out the number of moles of water of crystallization in a hydrated compound. So in this question, they're asking about hydrated copper chloride and what this X is, like how many waters are there per copper chloride? molecule. So you're really working out the ratio of the moles of CuCl2 to H2O, and you want the ratio of those two things. They've given you some experimental data here, which is actually really useful. But let's have a think about what we actually are doing first. So I can actually write an equation for what they're doing in this experiment. So they've got the hydrated copper. and what they're doing is they're heating it a lot to drive off all of the water and leaving anhydrous copper chloride so copper chloride without the water and this has got the x here so what we're really doing is trying to work out what that x is In this, they tell us the mass of the crucible and then the initial mass of crucible and hydrated copper chloride. So if I subtract the 16.221 from this, I get 2.139, which gives me this number here, 2.139 grams of that hydrated copper chloride. Afterwards, they say the final mass, again, minus the mass of the crucible, and this time, that number, the 1.696, is going to be the mass of the copper chloride on its own. Remember, the water here is gaseous, it's left as a vapor, so it's no longer with inside your crucible. So this is 1.696 grams. What's cool about this is because we have this thing called the conservation of mass, we can just do a subtraction because we know that whatever mass we started with has to be the mass at the end. So the water has to be 0.443 grams. So now you're kind of at a point where you can do a similar calculation to your empirical formula calculation, because you're doing the ratio of the copper chloride to the water. So you'll notice I'm going to set this up in a really, really similar way. But at the top this time, rather than having individual elements, what we're going to have is we're going to have just the copper. chloride versus the water. And I can do the same process overall. So this time I do have a mass in grams, which is your 1.696 and your 0.443 overall. What we're going to do is we're going to divide through by the formula mass for the copper chloride that's 134.45 just getting that information again from your periodic table so taking the copper and then two of the chlorines overall to give you 134.45 and then the water which is 18.02 Next step, just like before, we're going to work out the moles. So in this case, we get 0.126 moles of the copper chloride. And then we get 0.0246 of the water. Now, for me, I'm looking at this and I'm like, oh, it's pretty obvious that the number for the water is double that of the copper chloride. Like, to get from here to here, you're timesing by two. Okay. So I'm not going to do the divide by the smallest because I don't need to, to help me work out what the ratio is. So what this means is that the formula of this compound here is actually CuCl2.2H2O. For every copper chloride you have, you have two water of crystallization. And that's how you solve the problem. So X equals 2 for this example. Let's move on to reacting mass calculations. So what we've got here is we've got an equation of lithium hydroxide reacting with carbon dioxide. And then they've said what mass in grams of lithium hydroxide is needed to react with 11 grams of carbon dioxide. So whenever you're given equations like this, what you're doing is you're working out the moles of each and then you're trying to work out the mass overall. So again, I really recommend trying to set up kind of tables to work out what's going on, which numbers relate to what and what you're trying to work out. So we're going to go mass because this is all about. solids here, so you've got grams and grams on both sides, and then your formula mass, and then your moles overall. In the question, they say what mass in grams of lithium hydroxide is needed to react with 11 grams of CO2? So carbon dioxide has a formula mass or a relative molecular mass of 44. So what we're going to do is we're going to take that 11 divided by 44, which gives us 0.25. It's a quarter of it. We look at the mole ratio to get the moles of lithium hydroxide. Now this 2 here means that for every 1 mole of carbon dioxide, I have 2 moles of lithium hydroxide. But in this case, I only have 0.25. So I'm going to times it by 2, and that will give me 0.5 moles of lithium hydroxide. The formula mass, to go back up again, of that lithium hydroxide is 24. Overall, convince yourself that with your periodic tables. And then if we use the equation mass equals moles times formula mass, we can then times these two together to give me the answer, which is 12 grams overall. So in every single case, we're working out the moles, the moles of the other thing, and then trying to work out the mass again. So let's look at a different question on reacting mass calculations before we try something else. So here we've got, as it says, an excess of calcium carbonate is added to. a solution containing 0.10 moles of HCl. So immediately what I want to do is I'm trying to write 0.10 moles here. I want to make sure that I'm attaching numbers to the compounds that they relate to. What mass of calcium carbonate, so they want the mass of calcium carbonate reacts, and what mass of carbon dioxide is formed. So they want the mass of these two things. So what we need to do is we first need to look at the ratio here. So if I've got 0. 0.10 moles of HCl, then looking at the molar ratio, that must mean that I have 0.05 moles of the calcium carbonate. I would also have 0.05 moles of calcium chloride, 0.05 moles of this, and 0.05 moles of the CO2. So using that ratio of the big numbers, the coefficients, in front of each species. So now let's try and answer the question. So they want to know the mass of calcium carbonate and carbon dioxide formed. So we're going to use that first equation again. The mass equals the... moles times the formula mass. So you've got your 0.05 times your formula mass. So for calcium carbonate, that's 100. So if I do 0.05 times 100, I'm going to get 5. So I know it's between C and D at this point. And then you are looking at CO2. CO2, again, is still 0.05 moles. Remember, we worked out here. times 44, because they give it to me here, and that's going to be 2.2 grams overall. So that means that my answer must be C. Okay, and that's how you're doing your like reacting mass calculations. These ones are pretty simple numbers, but practice more of these if they're challenging for you. So let's go on to... percentage yield. Now percentage yield is something that people often find kind of okay and so I haven't actually got any questions on it in this video here. Just making sure that you're doing lots of questions and you will come across them. They're kind of like scattered amongst everything but I couldn't find any individual questions on this. So we're just gonna put a smiley face and assume everything's okay. Okay, so let's have a look at how we do calculations and how we work out what's going on when we have to deal with limiting reagents. These can make things a bit more complicated. So this type of question, you can kind of tell that you have one when it gives you a lot of information about both of the starting materials. If it's a normal reacting mass calculation, they'll often say like one is in excess or they'll tell you which one's limiting. But if they give you the moles of both, then that means that you're going to have to do some extra work in this calculation. So in this case, I've got 0.600 moles of my aluminium hydroxide and also 0.600 moles of my sulfuric acid. And in the first question here, they say determine the limiting reagent. And what that means is it means which one is going to run out first. So for each reaction here, you need two of the... aluminium hydroxides and you need three of the sulfuric acid. So for every one I need like I need more of the sulfuric acid. So logically it makes sense to me that the sulfuric acid is gonna run out first. But how do you show that? The way we show it is we divide the numbers of moles by the coefficients. And if you divide by the coefficients then that tells you you've got no point. 3.00 here and then you've got 0.200 here and whichever one is lowest of these two numbers will always be the limiting reagent so if it's not quite as easy to see which one's going to run out first you should do this so in my case the sulfuric acid the H2SO4 is limiting Okay, so this calculation that we did here was really just to work out which one was limiting. It's actually, we can't use it, like it's not useful to us in our like further calculations. It was just like a little trick to help us to work it out. Okay, so the next part of the question says calculate the mass of aluminium sulfate produced. So that's this one here, the aluminium sulfate that we care about. Remember that the... sulfuric acid is limiting. So I'm going to put an L there to kind of remind me that that's limiting, which means that the aluminium hydroxide is in excess. If it's in excess, I can't really use the number of moles there because it's going to be too many. Like, there's no point in using that number. So we're going to stick with using the moles of our limiting. Then, just like normal, we're going to take the mole ratio, which is a 3 to 1 ratio here, and the moles of aluminium. Hydroxide is going to be 0.200 moles. The question wants the mass though for the second mark, so we're going to times this by the formula mass of the aluminium sulphate. So to do that you're going to need to do your 22.9 times 2 from your periodic table plus your 3 times 32.07 plus 16 times 4. brackets, and overall you should get 66.8 grams. That's a point there, 66.8 grams overall. Cool. So let's have a look at our final question. It says, determine the amount in moles of excess reactant that remains. So my excess reactant is my aluminium hydroxide. So how much aluminium hydroxide is left over? So if I use my limiting reagent moles here, my sulfuric acid, and I go back this way, these are now in a 3 to 2 ratio. So that means the moles that would actually react is 0.400, would actually react. So to get the overall leftover stuff, what I'm going to do is I'm going to take 0.600, which is what we started with, minus the 0.4, which is what would have actually reacted, to give me 0.2. You want a lower pressure overall, because PV has to equal a constant. If you're changing nothing else. So if I want the volume to go up, that means the pressure has to come down. So that means I'm kind of down to B or D as an option. And then the second variable they're asking you about is the temperature. So if I want the volume to increase, then that means the other side of the equation I also want to increase. So therefore I need to increase the temperature and therefore my answer is D overall. So what we're trying to do is we're looking at those relationships and making sure that you have a really strong grasp of what causes what in the equation. Make sure you've practiced these, they almost certainly come up in some capacity, either like this or like a graph. Let's have a look at how we use PV equals nRT in paper two style questions. So once we've got your calculator and you can actually use the equation itself and how that might become a bit confusing. So this here, we're going to use PV equals nRT still. And this is the second half of question 6, which was an empirical formula question that we did earlier in this video. And we worked out that the empirical formula was C4H4O. That's what part A told us. So let's see what they want us to work out here. They've given us the mass of the sample. They've given us its volume, its temperature, its pressure, and then asked us to work out the relative molecular mass. So the only formula that I know that's got relative molecular mass in is the one that's got MR equals the mass of it divided by the moles of it. So currently I have the mass, I have the 0.230 grams. but I don't know what the moles are. And I can't work it out using this equation because this formula here is an empirical formula. It's not the molecular formula. It could be C4H8O2. It could be C200H400O100. I don't know how far it needs to be factored up, so I need to work that out first. So what we're trying to do is we're going to use PV equals NRT to work out the moles. So let's have a look at how that looks. Not with that pen. PV equals nRT. So the pressure was this 102 kilopascals. Remember, the pressure was in pascals. So I need to change that into pascals. We're going to make that 102,000 pascals is the pressure. times the volume. So the volume they gave me it in decimetres cubed. So decimetres cubed is a factor of a thousand smaller than a metre cubed. So what I'm going to need to do is take my 0.0785 and then times it by a factor of 10 to the minus 3. I know that's not in correct standard form but you get the idea. and then what we're going to do is we're going to say that equals to the other side of the equation which is the moles which is the thing that I'm trying to work out this time times r which is the gas constant remember we find that in the data booklet here I'm going to take that 8.31 8.31. And then T, which is the temperature, and they gave me 95, but it's in degrees centigrade, so I'm going to need to add 273 to get that into Kelvin. Okay? If you put all of that into your calculator, you'll get a value for the moles of 2.62 times 10 to the minus 3 moles. Okay, I really recommend you check that you get that number on your calculator. Otherwise, it might indicate that you have problems with inputting numbers into the calculator you're using. Okay, so now I can put that number back into my equation up here. The 2.62 times 10 to the minus 3 moles to get the relative molecular mass, which is 87.8 grams per mole. Okay, so that is the answer to this question. Three marks in total. So making sure that you can get all of the stages of that, it's really worth it to get those marks overall. It's all about the unit changes. Part C to this question says, determine the molecular formula of A using your answers from part A and B. So remember, from part A, we deduced that the empirical formula is C2H4O. And from part B, I deduced that the relative molecular mass was 87.8 grams per mole. Okay, so the two pieces of information that I have. So what I need to do first is to work out the relative formula mass of this empirical formula unit. And if you work that out, you get a value of 44 grams per mole, roughly. So what that means is I know that if I just had the empirical unit was the molecular formula, this number here would actually be 44, but it's not. The relationship between these two numbers is that I need to times by... 2. So that means that I've got two of the empirical formula units, I don't have one. So I'm going to do the same thing with the empirical formula, and I'm going to say it's the C2H4O, and I also need to times that by 2, and I get C4H8O2. This also has a relative molecular mass of the 87.8, so I know that this is the correct formula for this compound. So in IB chemistry, this section here that I'm doing now, I would classify as something that is level seven content. You need to make sure that you've gone through these tiny syllabus points to make sure that you really know all of your stuff. And there's this one syllabus point that's all about ideal gases and what conditions, what an ideal gas means, what we assume when we're saying that PV equals NRT and why it's not so great all the time, like why the numbers aren't great. So, you see questions like this, where it says, suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled, but nitrogen doesn't really deviate. And it's a really, really tricky question. So, things that you should know is that when we say ideal gases, the assumptions that I'm making is that the particles themselves have no volume at all. Have no volume. And what I mean by that is that when I'm looking at the space that it occupies, what we're saying is that we're just looking at the space between particles. I don't care about how big the particle is itself, which is kind of a fair assumption, because in gases, the particles are so far apart from each other that the size of the particle itself doesn't matter so much. You also have to assume that there's no intermolecular forces between... So we're assuming that there's no attraction between those molecules because that attraction, again, could influence how they're moving and how close they are together. So in this graph here, what you will see is kind of the main content of this, which is that you've got the ideal gas across here. You remember that. But you can see that. really low temperatures and really high pressures that gases deviate significantly from the ideal gas law. Please read about that more if you want some more information. So the question here talking about water vapor versus nitrogen. In terms of the size of these particles, they're pretty similar. Water has a relative molecular mass of around 18. Nitrogen is around 28. It's probably not the most significant difference between them. They're both pretty small. But the intermolecular forces are super different between these two. And that's actually what's causing the problem. So... To get the marks in this question, you need to talk about how nitrogen molecules form London dispersion forces only, dispersion forces only, whereas water molecules form hydrogen bonds. So that will get you one mark. For the second mark, you really want to be talking about hydrogen bonds. are much stronger. And then if you really want to go into more detail with that, that means that the water is going to occupy a smaller volume than predicted. So if we're looking at the ideal gas equation, I would expect water's volume to be smaller because they have those intermolecular forces, which is really pulling everything together. Okay, so let's move on to titration calculations. This is the last section of this video. It's all about moles and concentrations in solutions. So the equation we're going to be using for this section is moles equals concentration times volume. and that will help us out throughout. This first one is a fairly simple level of titration calculation. They give you some information about the reactants and ask you to work out the missing thing. So what we've been given is we've been given determine the volume. So the volume of HCl is what we're trying to work out. and we've been given that it's 1.5 moles per dm cubed, which is a concentration. So I can see here that this is definitely going to be utilizing my equation at the top, my n equals cv. I just need to work out the moles first. I've also been given that calcium carbonate is 1.25 grams. If something's in grams that should be telling you that it's definitely a solid so we're going to need to use that moles equals mass divided by formula mass to work out the moles of that. So before we get started, I've just noticed that this equation is not balanced. There should be a 2 in front of this HCl. I'm assuming it's just a typo. The IB don't tend to be sneaky and horrible like that and just like sneak in that it's not balanced. But it is always worth a check. So I'm glad that we rescued that today. The first thing I'm going to do is to work out the moles of calcium carbonate, though, because I've got the mass of it. And I can work out the formula mass, remember, from the periodic table. So the moles of calcium carbonate equals the mass, which is 1.25 grams. And we're going to divide that by 100 grams per mole, which is the relative formula mass from the periodic table. And that's going to give us 0.0125 moles overall. So using the ratio, I can now work out the moles. of hydrochloric acid. So the ratio is 1 to 2, so I'm going to need to times my moles by 2. So I'm going to take my 0.0125, times that by 2, and that will give me 0.0250 moles. So that's how many moles of hydrochloric acid that we have. And now we need to just work out the volume equals the... moles divided by the concentration. So I've got my 0.0250 moles. And we're going to divide that by the concentration that they gave us, which was the 1.50 moles per dm cubed. And that equals 0.0167 decimeters cubed. Ideally, that would pick you up the three marks. If you're reading this question properly, you will have noticed that there's this little thing here saying that they want it in centimeters cubed. So to get from dm cubed to centimeters cubed, remember we're timesing by 1000, is the change between those. So overall, that gives me 16.7 centimeters cubed for the full three marks here. Okay, let's do the last question in this worksheet. This is a bit of a trickier titration question. Overall, so we're going to try and break it down into smaller pieces so that you feel a bit more comfortable breaking down this type of question in the exam. So they're saying an unknown metal carbonate. So I know it's a carbonate, so I know it's got the CO3 2-, but I don't know what the positive ion is with it. I don't know what that metal is that's combining with it. An unknown metal carbonate reacts with hydrochloric acid according to the following equation. So this is metal carbonate plus acid gives salt plus CO2 plus water. Pretty standard reaction overall. So it says... A 3.44 gram sample of it was dissolved in water, they made a solution, 25 cm3 portion of this solution required, and my head is already exploding. This is a ton of information, loads of numbers, this really requires us to think about what relates to what. The first thing that I'm going to do is I'm going to draw my best and kind of what's going on to help us work out what we're seeing here. And I really recommend that if you're stuck in an exam that you try and do this. So they told us we've got 3.44 grams of our mystery M2CO3. I don't know anything about that. I don't know its formula mass. There's no way I can do anything with that mass because I don't know what the M stands for. Like there's no element for it. I don't know what it is yet. They said, so I did that, was dissolved in distilled water. Okay, so normally when we're dissolving in distilled water, that means putting it in a volumetric flask. It could be a conical flask, whatever you want to do, but I'm going to draw it like this. So they put it into water up to 250 centimetres cubed. So I'm envisioning that I've now got this solution that contains that 3.44 grams of powder that we started with. Cool, done. a 25 cm cubed portion. So that means I'm going to take out 25 cm cubed. It's a dodgy conical flask. This is 25 cm cubed portion of this. So I like took a tenth out and then they titrated it. with the HCl. So I can draw this in, and it says it needed 33.2 cm cubed, and of 0.150 moles per tm cubed. of the HCl. Okay, so this is kind of what you've got going on in this reaction overall, and what they're doing to try and wrap your head around what they're asking you in the question. But the IB does a pretty good job of helping you to work out what's going on with the questions. So it says, calculate the amount in moles of HCl in the 33.2 cm cubed bit. So here we're going to use n equals cv, And so we've got the concentration, the 0.150 moles per diem cubed. And we're going to times that by the 33.2 cm. Oh wait, since cm cubed, I need to make sure that I convert that into dm cubed. So we can leave it in cm cubed actually and just divide it by 1000 to make sure that it's converted into decimeters cubed. That's the thing that's going to come up a lot where you need to be really, really careful of these units in topic one in particular. Overall, that gives us 0.00. 4, 9, 8 moles. Making sure that you give your answer to three significant figures. So counting those significant figures when you get to the first number that is not a 0. So this is the right answer for part 1. It then says, calculate the amount in moles of the mystery carbonate that would react with this amount. So in this I've worked out that I've got 4, 9, 8 moles of the hydrochloric acid. These are in a 2 to 1 ratio. So to get the moles of the carbonate, what we're going to do is we're going to divide by 2 because the moles are in that same ratio. So if we do that, divide by 2, then we get 0.00249 moles. This is a one mark question, so if you're doing any work that's more than one step, you're definitely making a mistake and you want to have a real look at what you're doing. OK, three significant figures, done. All good, no problems. It then says calculate the amount in moles. of the sample in the 3.44 grams sample. So what we've got here is you've worked out the moles that were in this 25 cm cubed in the conical flask. This is your 0.0249 moles were in there. So that must mean, is that the same as how many moles were in this powder at the beginning? Did all of this powder end up in this conical flask? No. So what we need to look at is if we look at these stages, we can see that because we took out Because we took out 25 cm3 from the 250, I actually took out a tenth of it. So this 0.0249 is actually only like a tenth of the powder. So to get back again, we need to times it by 10 to get back to that big volumetric flask. So here in this step, all that we need to do is times by 10, which gives you 0.0249 moles. Because whatever moles were inside my volumetric flask were also inside the powder, because all the powder went into the volumetric flask, and then I took a tenth of it out. So this is the part that people often get a bit stuck on. You really need to think about what was happening in the process. Okay, cool. So I've got the moles now. It says calculate the relative formula mass of M2CO3 to one decimal place. So for this part of the calculation, we're going to use the formula mass equals the mass divided by the moles. So the mass is 3.44 grams. That was given to us in the question. The moles of it are 0.0249 moles we just looked at. So overall, this gives me 138.2. They want it to one decimal place, so that's what they get. And that's your final answer for the formula mass. Okay, last part of the question. Hence determine the relative atomic mass of the metal M and deduce its identity. So if I know that I've got M2CO3, I can work out the formula mass of the CO3. and then I can subtract it to give me the formula mass of the M2. So the M2 overall gives me a 78.14, but that's not the formula mass because that's two of them. It's not one of them. So with that 78.14, I'm going to divide it by two, and that gives me 39.1 as the relative atomic mass of that metal. It then says what is M? So if I go to my periodic table looking for a metal and you can see here that potassium has a relative atomic mass of 39.1. I kind of had a clue that it was group 1 because it formed M2CO3 so the identity of M is K or potassium. And that is all for topic one. This video has been a lot. I hope that you've got something out of it. If you have any questions post them in the comments. This is a new format for me so if it's been particularly useful or if you've got any tips then please let me know or if you've just got a topic that you want me to have a go at next. Alright see you next time!