Mole recap, empirical, limiting, yield, simple titration

[00:00:06]Okay class, we are going to look at this chapter topic A stochometry skills guide. And um the one in your textbook looks like this.

[00:00:22]Let me scroll. Okay, the one in your notes looks like this. But we are not going to use this because I have condensed it in a another booklet which have I given out to you. So it looks like this the hard copy that I've given out to you. So we will revisit some of the more calculations as a revision before we go on to look at the titration calculations.

[00:00:44]So please use this set of notes. This set of notes is actually compiled from the the the chapter that was in your term one book. But I reorder the questions such that you know it flows better based on what I'm going to cover.

[00:01:04]Okay, first let's recap some of the formulas that give us the numbers of moles. So number of moles as we know we can obtain them by using mass over mr. So we usually take the mass divide by the total mr of the compound by summing up all the individual atomic mass from the periodic table. Number of moles is also equals to the number of particles over the avocado's constant.

[00:01:30]And avocado's constant is 6.02 * 10^ 23.

[00:01:34]So this avocado's constant can be found in the data booklet section 2 I think. So you do not really need to memorize the value but you need to be aware where to find the value.

[00:01:48]Okay. Now for gases, if we have gases, number of moles can also be calculated by volume over molar volume.

[00:02:00]And there are two different molar volume that we are looking at. It depends on the conditions. So we have STP and we have SATP.

[00:02:12]STP stands for standard temperature pressure and uh standard temperature pressure is referring to 0° C and at STP the molar volume is 22.7 dmQ. So the molar volume of a gas at STP is 22.7 dmq. SATP is something new to you. It is like your RTP room temperature pressure. It stands for standard ambience temperature pressure.

[00:02:47]So standard ambience temperature pressure is usually at about 298 Kelvin.

[00:02:53]And the value that we are using for the molar volume in the past you use 24 dnq but we are using 24.8 dnq for IB. Okay.

[00:03:03]So ATP is referring to 298 Kelvin 24.8 8 dm cq as the molar volume. So for gases the volume divide by molar volume will give us the number of moles because one mole of gas one mole of any gas occupies a certain volume which is the molar volume.

[00:03:26]So over here we also have a new definition for relative atomic mass and relative molecular mass. In the past, we only look at the atomic mass compared to 112 of the mass of carbon 12. But now we have to include the weighted average of uh all the isotopes. So atomic mass relative atomic mass is the weighted average of all the isotopes atomic mass compared to 112 of the carbon 12. So weight average means that you consider the isotopes present and the rate proportion that they are found. Okay. So isotopes and their abundance.

[00:04:08]So example if I have carbon for example if I have hydrogen maybe I should use chlorine. If I have chlorine 35 and I have a chlorine 37. So that means when I get a sample of chlorine I will have both of them and they exist in different ratio. This exist as 75% of uh the amount as chlorine 35 25% as chlorine 37. So when I get a bundle of chlorine what I'm getting is I'm getting an average of um containing 75% of chlorine 35 and 25% chlorine 37. So the weight will also be dependent on the isotope mass as well as the ratio that is present. So to factor that we will call it the weighted average of the isotopes.

[00:04:58]Okay, weighted average of the uh um atomic mass of the isotopes compared to 112 of carbon 12. So that's the definition of relative atomic mass.

[00:05:09]Okay. So we will also have one more formula that gives us the numbers of moles. Beside mass beside particles beside volume we also have number of moles is equals to concentration time volume. So concentration multiplied by volume will give us the number of moles and this is commonly used for titration.

[00:05:28]Okay. So let's look at some samples and then we will just quickly go through the questions.

[00:05:34]So calculate the mass of an atom in go 198. So in go 198 we are taking into consideration that the molar mass is 198 g per mole and one mole of particles contains 6.02 * 10 ^ of 23 particles of atoms. Right?

[00:06:02]So therefore the mass of one atom will be 98 / 6.02 * 10 ^ 23 to find the mass of per atom and that gives us about 3.29 * 10^ -2 g.

[00:06:29]Okay. So that's how we calculate the mass of atom. So in this formula we are using the idea of molar mass which is the mass of one mole and also the number of particles in one mole. Question two.

[00:06:43]An accurate value of a molar mass is 118.01528 g per mole. Calculate the mass of one molecule. So in this case uh we need one molecule and we need in kilog kilog.

[00:07:01]Okay. So one mole weighs this amount and one mole consists of avocado's number of molecules.

[00:07:09]So for two we will have to find out the mass of one molecule plus which is very similar 18.0152 01528 / by 6.02 * 10^ 23 and you will get the mass of one molecule as 2.9.

[00:07:33]Okay. So, so after you have gotten this, the answer is in grams. But the question says give the answers in kilogram. So, with this answer in gram, we will have to divide by a,000 to convert it into kilogram. And if you do that you will be getting 6.922 * 10 ^ -26 kilog per molecule of water.

[00:08:02]Okay. Question three. If the density of mercury is 13.534 g per cm cube and you have this amount of mercury, how many grams? How many moles and how many atoms of mercury do you have? Okay, so let's clean it up and then we put in question three.

[00:08:29]So question three, density is equals to mass over volume, right? And we know that that's 13.534.

[00:08:39]And we know the volume as 62.5 cm cub. So take note since this is cm cq and this is cm cq we can multiply directly and we get the mass.

[00:08:52]So that's how we answer the first part.

[00:08:55]How many gs? So the mass is equals to density time volume. And that is about 846 g.

[00:09:06]Okay. Next, they want to know how many moles is that equivalent to and how many atoms do we have. So, we will find the moles first using the formula. Moles equals to mass over Mr. and mass we have just calculated 846 and Mr is given as 200.659.

[00:09:27]So calculating the number of moles we'll get about 4.22 moles.

[00:09:40]So with 4.22 moles they want to know how many particles. So remember just now we have a formula number of moles is particles over avocado's constant. So moles will be 4.22.

[00:09:53]So we multiply the by avocado's constant.

[00:09:58]then you'll get the number of particles or number of atoms in this case particles refers to atoms. So if we press the calculator 4.22 * the avocado's constant you'll get about 2.54 * 10 ^ 24 atoms. Okay. So we are just revising what you have learned previously in year two and year three year three and year four calculation more calculation. Okay. We will continue with question um 4 5 6 7 going through the answer very quickly. Okay. So how many hydrogen atoms are required to prepare this number of moles of propane. So in chemistry uh in chemistry mass is is quite useless to us. So we don't usually deal with mass. So if we have questions in mass we will always convert it into moles. So we are not given the mr. So you but you are given the atomic mass find the mr and then find the moles. So for this question we will need to find the moles of C3 H8 which is mass over Mr. So mass will be 0.88 over mr will be 3 * 12.01 + 8 * 1.01.

[00:11:18]So a question that you may have is that can I use just 12 and one? Cannot. You cannot use just 12 and one. You must use the value in the data booklet. Okay. So the value in the data booklet is 12.01.

[00:11:34]So with that you will be able to get about 0.02 moles.

[00:11:40]Then the question is about hydrogen and it's about hydrogen atoms. It's not how many moles of hydrogen atom but how many hydrogen atoms. I repeat it's not about how many moles but it's how many hydrogen atoms. So before we can get to atoms we need to find moles of hydrogen.

[00:11:58]So we can see that in every formula one formula you seeing 8 moles of hydrogen.

[00:12:07]So therefore if this is 0.02 this should be eight times more than 0.02. So therefore the number of moles of hydrogen atoms will be 8 * of 0.2 uh sorry 8 * of 0.02 so will be 0.160 and that is moles but question is not about moles question is about number of hydrogen atoms. So moles to convert into particles will be moles time avocado's constant equals to number of particles.

[00:12:44]So moles will be 0.160.

[00:12:50]Avocados will be 6.02 * 10 ^ 23. And then you work out the best answer for question four. So it's about 9.6 6 * 10^ 22.

[00:13:06]Okay. Question five. All of the following statements are consistent with the concept of a mole asset. Okay. So this type of question a bit conf troublesome. You got to look at them each one at a time. One mole of sodium contains this amount of atoms which is correct because one mole of something always contains this number of particles. And in this case the particles is atom. So this statement is correct. One mole of carbon 12 has a mass of 12 g. So one mole will have a mass of the mr which is 12 for carbon.

[00:13:40]If you check the periodic table one mole of chlorine contains this number of molecules. So molecules referring to Cl2. Yes. One mole always contains this number of particles. But if it's asking how many chlorine atoms chlorine atoms then you need times two. So this is correct. One more of sodium chloride consists this number of ions. Now why is it wrong? Because in sodium chloride we get one sodium and one chloride. So the mole ratio is one sodium chloride to two ions. So therefore the number of ions should be time.

[00:14:20]Okay. So this is wrong D. Question five, calculate the mass of oxygen that can be obtained from decomposing this amount of water. So what we will do is that you we take it that that's the that's the amount that we have. Remember number of moles. Okay, mass is not useful in chemistry. So we always do whenever we get mass, the first thing is to find moles. So we will find moles. 1802 moles is equals to mass over mr. So we take the mr as 18.02.

[00:14:57]So that will give you about one oh sorry it's about 18.0.

[00:15:12]Yeah. So it's about 100 moles. So with 100 moles then the next stage after we find the moles of H2O the next stage we always apply mole ratio. So first always change mass into moles then apply mole ratio. So the mole ratio will be 2 H2O is to one oxygen.

[00:15:37]So therefore the moles of oxygen is lesser. The moles of oxygen will be 550 moles. And then but the question is not about moles, it's about mass. So to find mass, we will take the moles multiply by the mr which is 18.02.

[00:15:56]So we will have 50 * 18.02. But the question is asking in kg. So this mass is in g. So answer divide by a,000 to convert to kg. So that will give you around 1.6 kg.

[00:16:18]Okay. Last question. How many atoms are there in 18 g of water? So water is H2O.

[00:16:25]So in the formula we can see there is make up of two H and 1 O atom. So 2 H + 1 O. So therefore 3 atoms per molecule per 1 H2O it will be three atoms. So 18 gide by 18 right which is one mole but of H2O but in terms of atoms it will be three moles of atoms and with three moles of atoms we multiply by avocado's constant and you will work out the number of atoms. So 3 moles * 6.02 02 * 10^ 23 you get 1.81 * 10^ 24 atoms.

[00:17:09]Okay, that is for the first part.

[00:17:11]Recapping the formula of number of moles.

[00:17:24]Okay, after recapping on the number of moles formula, we are going to move next to the next calculation which is the empirical formula section. So as I repeat this set of notes is actually compiled from what you have in the term one book but I just pick up the questions and sort them according to the skills. So this part we're looking at the skills of doing molecular and empirical formula calculate determination.

[00:17:53]Okay. So after the earlier revision which is the foundation of the chapter we will move on and look at this part.

[00:18:00]So empirical formula is the simplest ratio of the atoms. So example if I have let's say C2 H4 that's not the simplest formula. The simplest formula will be CH2. So we will say that the empirical formula is CH2 while the molecular formula is C2H4.

[00:18:20]Okay. So molecular formula tells us the actual number of atoms present. So if it's C2H4 that means four hydrogen two oxygen in one molecule. So the usual procedure for empirical formula if you remember it will be a table whereby you are looking at in terms of the atoms. So sometimes you may have C H O something like and then you will write down the mass. Sometimes you're not given the mass then you write down the percentage.

[00:18:49]Okay? Because if we take it as 100 g then percentage is mass. So you put in the mass and then we usually find the number of moles from the mass.

[00:19:00]Okay. And then we always do since it's a relative ratio so we will divide by the smallest number of moles.

[00:19:09]And sometimes when we divide by smallest number of moles, we may just round to the nearest number. But it also depends, right? If I get 1.9, I may round it up to two. Okay? If I get 2.08, I may round it up to two. But there are some cases whereby you get 1.5, then you cannot be rounding up from 1.5 to 2 or you get 1.49 something like that. 489. You cannot be rounding up it to down to one and you also cannot round up to two. For this of cases, we will apply point number five.

[00:19:46]We multiply by a suitable integer such that we get a whole number. Okay. So if it's like 1 and a half, we will multiply by two. But let's say you get 1.333 which is 1 and 1/3. Then you may want to multiply by an integer of three to get a whole number. Okay? So usually we divide by the smallest um number almost we get the answer but in the event that the answers are some like fractions number like 1 and 1/3 or 1 and a half one and a quarter then we will usually have to multiply by an appropriate integer to get the whole number because we can't do a round up from 1.1 and 1/3 to become one. The difference is just too big.

[00:20:30]Okay. But if it's like one uh if it is like 1.99 yes we can put it as two.

[00:20:37]Okay. So after we get the molecular formula what's the re relationship between the molecular formula and the empirical formula. The the relationship between the molecular formula and empirical formula is that the molecular formula is n number of sets of uh of empirical formula.

[00:21:02]Molecular formula is n number of sets of empirical formula. Okay. So example if I have ch2 case again just now our CH2 case. So molecular formula is two sets of CH2. So that's why is C2H4.

[00:21:21]So therefore N will be 2 * C4.

[00:21:26]Okay, let's look at the examples which is already work out for you. So to recap the method. So first the questions give us the mass. We will put down the mass.

[00:21:35]Remember is oxygen not O2. Okay? It's O because we're looking at atoms not O2 not molecule. So do not put 32 over here. This is not 32. It's atoms. Okay.

[00:21:49]Then we'll take down the mass with write down the Mr. If you want to or write down the A R from the periodic table and you find moles. So after you got the moles, you divide by the smallest number. So the smallest number is about 0.399.

[00:22:02]So if you divide throughout and you get something like this. So remember if it's like very close, we can do a round off.

[00:22:09]It is 1 and a2. We cannot round half up to become two. So we understand that this is 1 and a half. So we will multiply by the factor of two to remove the denominator. So we 2 23. So we know the formula is Na2 S23.

[00:22:27]Okay, that's how we get the empirical formula and the molecular formula is how many sets of this basic formula. Okay.

[00:22:36]So let's look at an example here on finding molecular formula. So given empirical formula is C H2O and MR is 118 which is the molecular mass. So what is the molecular formula? So how many sets of C H2O will give me 18? How many sets of C H2O will give me 18. Right? So I will have to find the mass of one set of CH2O which is 30 and how many sets of it will give me 18 roughly six sets. So therefore we know the actual formula as to multiply by 6. So 6 * of C H2O will give me C6 H126.

[00:23:20]So that's how you work out from the empirical formula to molecular formula.

[00:23:24]So the key is we need to know the MR. If we do not know the MR then we cannot find the molecular formula from an empirical formula because we need the MR to help us to calculate. Okay. So question three over here is a practice for you to try out what we have just covered.

[00:23:50]Okay. So let's look at question three.

[00:23:54]Okay, I give you some time to do. You may want to pause the video if you want to try.

[00:24:07]So if we draw the table Okay. So you have a C H O 62.1 10.3 27.6.

[00:24:33]So we find by the number of moles. So be 62.1ide by 12.0 1 and then this will be around 10.3 / 1.01.

[00:24:50]oxygen I think is 27.6 / by 16. Okay. So you divide to find the number of moles and at the end you'll find that roughly roughly the number of moles is about 5.18.

[00:25:08]It's about 10.3 this about 1.73.

[00:25:18]So we need to divide by smallest number which is 1.73. So we divide by 1.73 and that will be approximately 3. So if you take a calculator 5.18ide by 1.73 you get about 2.994 which we will just approximate to three right. So therefore we will not write the 2.994 but we put it as three. Okay.

[00:25:45]So the same for the rest you get about 6 and 1. So the empirical formula is C3 H6O.

[00:25:58]Okay. The molecular formula is a bit difficult. So let's look at the molecular formula. The question says when 0.125 g of it evaporated in a strange at STP, this is the amount of vapor.

[00:26:14]So remember if I have empirical formula and I want to get molecular formula I need Mr. So I need to find M R. So I have a volume and I understand that number of moles is equals to volume over molar volume. So I can try that first 24.1 this is cm cq not dm cq. So you have to convert it to dm cq or you convert your m volume to dm cq. So at stp is 22.6. 7 dm cq. So 2270 0 22.7 dm cq is equals to 227 0 0 cm cub. Okay. So because is cm cub.

[00:26:58]So this has to be cn cub. So with that you will find the uh number of moles which is around 1.061 061 * 10 ^ -3 moles. But that's not mr, right? So what else do we learn? We learn that moles equal mass over mr. So moles will be 1.061 * 10 ^ of -3 moles. And if we need mr then the formula will be mr is equals to moles over uh sorry mr is mass over moles. Right? So mass is 0.125. So put in 0.125.

[00:27:43]That's the mass and moles was 1.061^ 3 minus 3. So put the most out. And then with that we'll be able to find Mr. is about 117.8 around that it is slightly different is by Ken because of the rounding off. So 117.8 it and that's the M R that's not the question the question is molecular formula the question is not find the MR right so with the MR then we can find molecular formula so what is the uh the mass of one set of C3 H6O the mass or the empirical formula MR is about 58 let me calculate so one uh carbon is 12.13 three of them. Last hydrogen is 1.01 * six of them. And then oxygen is 16. So it's about 58.09.

[00:28:54]So therefore, how many sets of 58.09 equals to this? So we can tell it's about two sets, right? So therefore 2 * C two sets of C3 H6 O two sets of them.

[00:29:09]So the molecular formula is C6 H12.

[00:29:16]Okay. So that is how we do the embracer formula and molecular formula that we have revised before. We are just doing revision of what we learned. Now this combustion data here is a bit new to you. So I will prefer to cover that in class. So what I'm going to do is we only do some revision of things that we learned before. Okay.

[00:29:41]So we will move on to bypass the combustion data and we will look at this part on page four. Skill four. Scale four is about our dilution, our number of moles, our mole calculation of concentration.

[00:30:02]So remember when we dilute something the numbers of moles do not change. So let's say this was the number of moles of uh uh chemical let's say number of moles of acid. So the number of mosa is so much right you added in more water.

[00:30:25]So the number of moles or the number of circles in my diagram did not change. So number of moles do not change. So for dilution we do not use mass over mr or volume over molar volume or particles over avocados constant. We use concentration times volume. So concentration time volume is number of moles. So at C1 V1 should equals to C2 V2 where one is before and one is after dilution.

[00:30:52]Okay. So understand this. Now the units for concentration is usually mole per dm cq. Sometimes they call it marity. Okay.

[00:31:01]But usually we use mole per dm cq. But marity means mole per dm cq. If you want to express the concentration in terms of grams that is also possible. We call gs per dm cq. is fine to express in this way from mole per dm cq to g per dm cq.

[00:31:19]How do we convert? Okay, how do we get moles per dmq into g per dmq? So to convert a moles into g, we multiply by m r. So the same for that for moles per dmq to g per dm cq multiply by mr. Okay. Another common units that we use for concentration will be something of ppm, parts per million. Parts per million means that if I have 1 million parts, 1 million of that thing, only one part is the chemical. So it's very diluted. You can tell like if 1 million part of water out of 1 million part of water, only one part is the chemical. So it's very very diluted. And how is it related to the previous concentration?

[00:32:06]We learned one part per billion a million is also known as 1 mg per dm cub.

[00:32:16]Right? So remember just now for mole per dm cq we want to find g per dm cq we multiply by mr. Then from grams per dm cube, if we want milligrams per dm cq, then 1 g is 1,00 mg. So we multiply by a,000. So that's how we convert something into ppm.

[00:32:45]Okay? And so therefore, if I give you ppm, you should be able to convert back.

[00:32:49]So if I'm given ppm, then you convert by dividing by, th000, right? The opposite.

[00:32:57]that over here instead of multiply by MR you divide by M R everything is the reverse. Okay. Now what we are doing so in the many of our calculations is that we also apply mole ratio. So mole ratios give us a comparison and allow us to find something. Okay. So sometimes when the equations a few equations are involved we may want to combine all the more ratios together rather than to apply one and calculate and then use the value and calculate again to the next one. Okay I'll give you an example later to show the combining of mole ratio.

[00:33:36]So most of the time when we solve a mole question the first step we will do is write an equation. You don't have an equation you cannot do a mole question.

[00:33:45]So you write an equation and mass is useless to us. So we will convert all the mass usually to moles.

[00:33:55]That's the minimum you can do is to convert the mass into moles. After you have the mass into moles or you use your other methods to find moles that we have covered earlier either volume or you concentration time volume or molar volume or particles. As soon as you get moles already the next step is to use mole ratio. get more ratio to find the moles. Okay, so it depends on the question. If the question say oh find the mass then you use your moles to continue to find the mass after you found the moles. If they say find percentage yield, then you use the moles to find percentage you also can or you use your mass to find percentage yield.

[00:34:34]Okay, so the idea is write equation convert the all the values to get moles and then apply your mole ratio.

[00:34:46]Okay. So let's look at some examples that apply what we have just covered. So if I have a dilution. So if this is a mount of diluted to a new volume, we don't know what is the new volume. So if we let the volume be V2. So we need this is the volume to give me this concentration. Can you see it's a different concentration. So if you call it C1 V1, then this must be the C2. So we apply the formula C1 V1= C2 V2. and we put that in to find the volume. Now this volume is not the volume of water to be added. So you if you so you read carefully. So if the question say what is the final volume then yes this is the answer. But sometimes the question say what is the volume of water to be added such that I can do that. So your the volume of water added is the final volume minus your initial 25 then you get u the volume of water to be added. So you read the question carefully see whether they asking for the final diluted volume or they are asking for the volume of water to be added. So if they asking for how much water should you add in order to make that solution then it will be you have to add this amount. So read carefully if the question say amount of water to be added then this will be the difference. If you say what's the final total volume of the diluted solution what is the final total volume like this question it will be the C2 is the V2 sorry.

[00:36:22]Okay. So this is a example of using mole ratio with uh with uh concentration. So again I have volume I have concentration. So usually volume and concentration. First step equation.

[00:36:36]Remember first step was equation. Second step find moles. So second step we find moles. First step we apply mole ratio.

[00:36:44]Okay. So since they say magnesium is excess. So we use our limiting region to find the hydrogen. So the mole ratio will be two hydrogen two acid to one hydrogen. 2 is to one. So the moles of hydrogen will be 0.05.

[00:37:02]Okay. That's how we do it right. Okay.

[00:37:07]Now remember I told you about combining mole ratio. If there are many moles many equation can I combine the mole ratio?

[00:37:16]Let me show you how to combine the mole ratio in this question. Okay. Question three.

[00:37:22]Calculate the mass of acid that can be produced from 5.6 g of nitrogen.

[00:37:29]5.6 g. And I need to find this one, right? So of course you can say, oh, I do 5.6 g first, then I find ammonia, how much? Then I take the ammonia and then I find how much N O. Then from the N O I calculate and find how much NO2 then for NO2 I find HNO3 which is very which is possible but very slow. So what we will do is that we will combine them together and I want to find the mole ratio of N2 is to HNO3 directly. I want to find the mole ratio of N2. How is it related to HNO3 directly? Okay by combining all the equations. So how let's look at the common items. So if we call this equation 1 2 3 4 the common thing that links one and two is ammonia. The common thing that links two and three is N O.

[00:38:28]And what is linking three and four is N O2. And then finally to what we want right and we started with N2. So we can see all these items that have highlighted are related. So for one mole of N_sub_2 I will get two ammonia.

[00:38:46]That's what equation one is telling me and apply the two ammonia into equation two. Now if I have two armonia what is the N O I'll be getting because I only have two ammonia I have four. So if I have two ammonia then I should get 2 N O because four to four. So 2 to two. So if I have 2 N O how many NO2 will I get? So 2 N O 2 NO2 so it will be two. Now if I have 2 N O2 apply to the last one. If I have 2 N O2 how much acid will I have? I have 2 NO2. This is four. So two four to four. So two to two.

[00:39:33]Right? So remember at the start I wanted the shortcut. I want to straight away link this to this and make make it easier bypass all the steps. So can you see a mole ratio of N2 to HNO3. So you can see N2 you can see HNO3. So the mole ratio is 1 N2 is to 2 HNO3.

[00:39:53]So instead of calculating all the intermediate steps we are all going to bypass all of them. We're going to bypass all of them and straight away link this to that. Okay. So with that let's try to solve question three.

[00:40:08]Okay, I give you some time to try and uh when you are ready then you unpause the video. You can pause the video to try and then when you're ready I pause to go through the answer. Okay.

[00:40:25]Okay. So we need to find the this is the mole ratio that we are going to work with N2 to 2 HNO3.

[00:40:35]So, so we find the number of moles of N2.

[00:40:39]Remember mass is useless to us. Mass is useless to us. So, we find the moles of N2. So, you look at periodic table. What is the nitrogen? Okay. Mr. So you get your mass 0.2 moles. So the 2.2 moles in moles of N2. So in mole ratio to HNO3 will be number of moles of HNO3 will be 0.4 4 double and but the question is asking for mass. So we end off the mass with 0.4 * the MR. So the MR will be um HNO3. So you add up the MR from your periodic table. Okay. The individual AR.

[00:41:25]Okay. I roughly roughly we should be quite close to 25.2 g something like that. Okay. we multiply by the MR. Okay, so this is an example of how we can combine multiple mole ratio so that we can bypass the intermediate step and skip the working such that we take from the start and we quickly relate to the end.

[00:41:47]Okay, question one, let's finish with question one and two. Question one is uh purchase standard solution of KOH as a concentration how much how would you prepare this to prepare one 100 CQ of the diluted one?

[00:42:06]So if this is C1 we need V1 okay and this is B2 and C2 right. So we can see that one remember cm cub me cube uh sorry cm cubq dmqr so I have to divide by a,000. So that's the volume in dm cq * 0.1.

[00:42:28]So this is c2 v2 is equals to c1 v1. Then you can rearrange your equation and you find V1.

[00:42:53]So if you let B1 be in CM cube, if you let it be then you will get the answer in CMQ. In this case I'm letting my V1 MQ.

[00:43:05]So I will get a value of 0.1 *.1 / by 1. So I'll get V1 is 0.01 d MM cq. But let's say if you want it cmq you multiply by,000. So it's about 10 cm cq.

[00:43:23]Okay remember to do this conversion is multiply by,000.

[00:43:34]Question two, calculate the marity of this.

[00:43:39]So remember concentration can be invarious for mq. So if you pay attention just now mill per dm cube is parts per million.

[00:43:52]It's also known as parts per million.

[00:43:54]Okay they want the marity. So what do you mean by marity? That means they want the concentration in moles. They don't want it in mig per dm cq but they want it in moles per dm cub. So what we will usually do is we will have our mill g per dm cq convert back into g per dm cq first and then grams is related to moles. So from grams converts to mole dm cub.

[00:44:22]So this is 6.2.

[00:44:30]So change into grams divide by 1,000.

[00:44:34]Millig to g is divide by 1,000. So I divide by a,000.

[00:44:40]So I'll get 0.62 g per dm cub. And then change to moles per dm cube. We multiply by mr. So the mr is oxygen. So be 32, right? So times 32. So I'll get about 0 984 moles per dm cq.

[00:45:14]Okay. But let's say if you want to follow SF and then you see this was 2 SF. So following the link SF you'll be 0.0 per EMQ.

[00:45:45]Okay, so we are done with this section.

[00:45:48]So we will go and revise skill five which is limiting region and percentage yield and then we will maybe take a stop here the skill five. So skill five limiting region and percentage yield.

[00:46:03]Let me see what is skill six. Skill six is direct titration. Okay, direct titration also can go through the examples. Okay, so we will stop after direct titration.

[00:46:15]Okay, so limiting region and percentage yield. Limiting region that is the one that is completely used up with nothing in left in excess. Okay, and it's the limiting region that determines the theoretical amount of product maximum that you can get. is a theoretical yield is determined by limiting reagent. It's the maximum that you can get. Okay.

[00:46:36]Excess region is the one that is left behind after the limiting reagent is used up. Okay. So, how do we identify limiting region? First, we must know how many moles of each reactant is present.

[00:46:49]And then we can divide the moles by the coefficient from the balance equation. Or we can just say oh if this number of moles is used up how much of the other do I need and then you check whether do you have enough to meet that need if you have enough to meet that need that means most probably that is in excess. If you have more than enough to meet the need then that means it's in excess. If you do not have enough to meet the need that is a limiting region. Okay.

[00:47:23]So that's how we determine limiting region yield. How do we know our yield?

[00:47:29]How much is our yield? Our yield will be the experiment what you get over what theoretically you should be getting. So theoretically what you should be getting is from the limiting region. If you remember this was the li the the theoretical yield is actually calculated based on limiting region in and mole ratio. So theoretical yield is calculated by the mole ratio to the limiting region.

[00:47:57]Okay. So after you get the theoretical yield, you have done an experiment, you compare it against it and you multiply by 100%. So that's percentage yield.

[00:48:11]Okay. So let's look at an example. So if this amount of calcium needs this amount of acid to react. So this is example of limiting reagent. Okay, one question I will ask or I always ask people is when do you know that the question needs you to determine limiting region?

[00:48:28]I repeat, when do you know that the question requires you to determine limiting region? So many people do not know when they just determine when the question ask or oh the question say this one is must do the limiting region then we will do it but it's not true. So you will have to determine the limiting reagent whenever all the reactance amounts are given to you. I repeat, you will have to determine limiting region whenever all your limit all your reagents amounts are given to you. And how do they give you all the amount or or all the reactants amount are given? How do they give you the amount? They may give you in terms of mass. They may give you in terms of volume for gases. They may give you in terms of concentration times volume.

[00:49:26]They may also give you in terms of articles. So as long as in some way they give you the amount of all the reactants present, you must determine limiting reagent. So you see this case, this one they tell me is 10 g. This one they tell me is 200 cm cube of 1 mole per len cube. So they gave me all the amount because in whichever way all the values that they give me allows me to find the number of moles of each one. So I know all the amounts. So that means this question must do limiting reagent. Okay. So take the 10 grams divide by the A R uh A R.

[00:50:04]So you get the number of moles of calcium. Then you take concentration plus volume you find the number of moles of acid. So then you ask yourself for for 0.2 moles of acid to be used up since the mole ratio is calcium to two acid right. So for 0.2 2 moles of acid to be used up. How much calcium I need?

[00:50:26]I need 0.1, right? M ratio Y is to two.

[00:50:29]So I know I need 0.1 calcium.

[00:50:33]And then I check do I have 0.1?

[00:50:36]Yeah, I have more than 0.1 more than right. So therefore this is excess. So this is limiting region.

[00:50:45]Okay. So the limiting region is HCl.

[00:50:50]So you can see here now next we look at one example of percentage U. So remember percentage U is over theoretical U and theoretical U comes from the limiting region.

[00:51:04]So in a way we have to determine limiting region and also if you read carefully the question give you 25 g of this and you have uh 28.5 g. So this 28.5 g is experiment.

[00:51:23]Okay. And if this question let's say I know it should be stated but it's missing. If you say excess oxygen, okay, oxidation of this in excess oxygen, can you add it in? Because I think this should be in the question in excess oxygen. So if it's in excess oxygen, we take that that's the limiting region already, right? So 25 divide by the mass. So remember my mass is not correct. You must use the data booklet mass. So data booklet mass will be 8 * 12.01 then + 10 * 1.01 01 you must use the value in the data of the fl which is the weighted average of all the isotopes compared to 112 of carbon 12 okay but so therefore you factor in the isotope that's why there's a 01 okay so you my example is not showing that but you during exam or during your practice you must use the data booklet one so 25 gide by mr you get the most so that's the limiting region so looking at C3 C8 H10 as the limiting wager. I need to find the the terra terapilic acid which is C H604 right so it's more ratio 1 is to 1 one is to 1 so therefore 0.236 is to 0.236 right and with that 0.236 236 I multiply by the MR of what of C8 H64 MR so this is assume this is the MR but by right you should have a two decimal place okay because we should use like values like 12.01 1.01 one. So you multiply this is a theoretical mass. But why? Because it is based on limiting region and based on mole ratio.

[00:53:22]So you did your calculation based on all this is theoretical one because you do just so the mole show is of the equation and the equation is a theoretical thing right. So we need supposed to get 39.2.

[00:53:36]So the formula is experimental over theoretical time 100%. Right? So experimental was 28.5 it was stated in the question here.

[00:53:52]Okay. And theoretical is what we have calculated using the limiting region 39.2. So 28.5ide by 39 sorry 39.2.

[00:54:02]So 28.5ide by 39.2.

[00:54:06]So that's the percentage yield that you're getting.

[00:54:14]Okay. So I think I have already uh done this part of the question 28.5. Right?

[00:54:22]So if you look at the example, we have already done this. This was the exact example. Okay. Just say that if 25 g limiting region. So which is what we have done just now. 25 g and we let it be the limiting region. Right? So we have gone through this example already.

[00:54:48]So we just find the other example.

[00:54:50]Calculate how much H2 will you get when this amount reacts with this amount. So the calcium is 0.623 623 and the HCl is 27.3 and 1.25 right moq so we will have to determine the number of moles of calcium and number of moles of HCl.

[00:55:21]So because both numbers of uh the amount are given since both amount are given we will have to determine limiting regions.

[00:55:28]So 0.623 / by 40.1 is about 0.0155 moles.

[00:55:36]HCl will be 1.25 time 27.3 over 1,000 to change to cm cq dm cq. So it's about 0.0341 more. So we can see from that is one calcium is to 2 hl. So if this is 0.0155, this should be 0.0155 * 2. So take your calculator 0155 * 2.

[00:56:08]So you get 0.031.

[00:56:10]This is what you need, right? If you want all to be reacted. So you check to have what you need. Sorry to HDL. You have what you need. Yeah, I have what I needed. I what I needed and I and and in fact this is an excess. So therefore this is limiting reagent. So using the limiting reagent amount 0.0155 we will do the mole ratio with limiting reagent calcium is to H2 is 1 is to 1.

[00:56:44]So number of moles of H2 will also be 0.0155 moles. Okay. And the question is asking for volume at SATP. So SATP molar volume is m volume we just recap on the first page 24.8 dm cq. So this moles number of volume will be equals to moles time molar volume. So 0.0155 times the molar volume 24.8 8 0155 * 4.8. So it's about 0 it's about 0.3844 uh dm cub. Now the problem is how many SF will we keep to write? What's the number of decimal places? So 0.3844 dn cub can or not. So we read the question usually remember what we are doing is we are doing multiplication and division. So we will follow the lease SF right. So this is 3 SF. So we will follow the lease SF 3 SF. So I cannot leave my answer like this by right. So I have to change to 3 SF 0.384 D MQ. Okay. This final answer should be 3 SF because the M SF in the data is three.

[00:58:25]Not because always must be 3 SF but you have to look at the data that you have used in the calculation. What is the least SF of the data then you will follow. Okay.

[00:58:37]Okay. So last part direct titration this all recapping. Okay. So direct titration in the titration we will have a titr titr is in the bureet. And then we have our sample sometime we call it analate which analyte which is in our conical flu. This is an light and then the the titrant is in the blue red.

[00:59:00]Okay. And we have a indicator and indicator give us the end point of the titration. So indicator give us n point and equivalence point is when based on the stochometric ratio the amount added is equal to neutralize it or or uh equivalence point is the stochometric amount has been added based on the equation to neutralize something okay or to write something. So usually titration calculation or any calculation start off with uh an equation. Start off with equation then find number of moles either using mass or using concentration times volume or using volume over m volume or using particle over avocado's constant. Okay. So find the moles and then apply your mole ratio.

[00:59:50]So let's look at this example that this amount of acid was neutralized by this amount of NaOH. The solution was made from this amount of uh acid in 250 g uh cmq things. But we do not know how to start. But at least we know we cannot take the 2.8 and try to divide by mr because we don't even know what is x. There's no mr for x. So but at least we see one set of complete data here.

[01:00:23]And this set of complete data is concentration times volume. Right? So first equation write an equation. Second find moles. So I'm going to find moles of using the concentration volume and get five moles. Then after you find moles that's the second step. Third step always mole ratio. So well mole ratio is with respect to H3X. So three sodium hydroxide to 1 H3X. So sodium hydroxide is this amount. So what is H3X? It will be divide by 3. So that's H3X.

[01:00:59]Okay. Question says that this is the number of moles of H3X and this is in 25 CNQ.

[01:01:09]But you have 250 CQ. So you read carefully here. You only took 25 to titrate, but actually you have 250.

[01:01:18]So that number of moles that you calculate was 425 okay is for 25 cm cub. So in terms of 250 cm cub then the number of moles will be 10 time the question is asking for the question is asking for um finding the mr. So of course if you do finding m there many way to do it. One way is concentration by using moles over 25 ml. Then take the concentration times the volume which is 250 to find moles and then you take um mass over moles to find mr. Of course also I can do it at this stage. Another way I use a proportion since 25 ml is 0.00 00286 mole. So 250 ml will be 10 times. So 0.0286 more. Right? And this is the moles and this moles equal to the um 2.8 g.

[01:02:35]This moles equals to 2.8 g. Right? So I can this equates to 2.8 8 g.

[01:02:47]So MR is equals to mass over mole.

[01:02:52]So 2.8 over 0.0286.

[01:02:56]So you will also get the MR in this manner.

[01:03:04]Okay.

[01:03:06]Okay. So I think we will not go through the questions here but we maybe stop around here at least we have gone through the method. So while school starts we will go through this practice questions since we already learned how to calculate okay we'll look at the practice question atom economy back titration and also the combustion which I skipped earlier because the combustion is new to you so I skip the combustion and breaker formula. Okay. So what we will do is that when school starts we will go straight to recap on this uh part.

[01:03:41]We will go straight to this practice and then go through them. Okay? If you want you can prepare first the practice questions so that when we go through in class you can follow better faster and you can check your answers quickly. So we will be going through the practice question when school starts and then continue with skill number seven and then skill number eight back titration.

[01:04:04]So once we finish all the titration then we'll go back to the combustion.

[01:04:11]Okay. So that will be your holiday homework. Holiday homework watch this video and if you want to please also try practice the question because you know all this well I'm the one doing the questions. So it will be good for you to try then when you come to class we will just go through you just a mark and then you just correct your mistakes in faster. Okay we stop here. Thank you.