 Hi students, in today's lecture I have come up with four small topics starting off with percentage yield because this is kind of related to stoichiometry like we will be using the concepts of stoichiometry and limiting reagent here. So starting off what is percentage yield? We need to know a formula. So percentage yield of any reaction is equal to actual yield, actual yield means like we have actually have done the experiment and we have got this much amount of yield let's say divided by theoretical yield, theoretical yield and theoretical yield is nothing but what we were calculating using those stoichiometry and limiting reagent concepts. Now in place of actual yield we will be basically writing what actual moles, what are the parameters that we will be using? So actual moles upon theoretical moles or we can also write actual mass upon theoretical mass and as because it is a percentage we need to multiply it with 100. So using this formula we will be solving the questions. Now straight away we will understand this concept using numericals. So first numerical is initially two moles of each A and B are taken and 0.5 moles of A to B2 is formed. So here I am writing the equation 2A plus 2B is giving A to B2. I have already balanced it. Now we are also given with two moles of A and two moles of B. So apparently we also know according to the stoichiometry two moles of A and two moles of B should react and give how many moles of A to B to one mole. So normally in the question also two moles only is given. So definitely according to the question also one mole of A to B2 should be formed. But they have told that 0.5 moles is formed. Now we will be finding out the percentage yield of this reaction. So what formula did we write? Percentage yield is equal to actual moles. So this whatever is given in the question is the actual moles that was obtained divided by theoretical moles. Theoretical moles. So using stoichiometry we found that out. This is our theoretical value and this is our actual value. So now into 100 will also be there as it is a percentage. So it will be 0.5 upon 1 into 100. That will be 50 percent. So the yield of this percentage yield of this particular reaction has been found to be 50 percent. Coming on to the next question. On heating 50 grams of CaCO3 20 gram of residue was obtained. So when we heat CaCO3 first what do we get? CaO plus CaO2. So this is our solid. This is solid and this is gaseous. Now what they are saying 50 grams. On heating 50 grams we are getting 20 gram of residue. Residue means this calcium oxide they are talking about because this will go out as gas. So now 20 grams of residue was obtained. Find the percentage yield of the reaction. So this was the actual value. We need to find out according to stoichiometry like theoretical value we need to find out. So we know that according to the equation 100 gram should give 56 grams of CaO. 100 grams of what? Calcium carbonate. Should give 56 grams of calcium oxide. You can see one mole here, one mole here and one mole of calcium carbonate means 100 grams. One mole of calcium oxide means 56 grams. So 50 grams should give how much? 56 upon 100 into 50. So it should give 28 grams. So this is our theoretical value. Now we will simply apply the formula. Percentage yield is equal to now actual mass, actual mass what was obtained? 20, 20 and divided by theoretical mass that was 28 into 100. So if we solve this what we will get 20 upon 28 into 100 that will come around 71.4 percent. So this is the percentage yield of this particular reaction. I hope this particular topic is clear this is a very small topic but you need to know the stoichiometry and limiting reagent concepts for solving the percentage yield questions because you need to find out the theoretical expected moles or mass whatever. Now coming on to the next topic of this lecture that is our percentage purity. What is the now here also we will have certain one particular formula. Percentage purity is equal to mass of pure compound in the sample, pure compound in the sample divided by total mass of impure sample, impure sample of course into 100. So whenever they will say sample we need to understand one thing that inside that sample sample we have the compound also and some impurities also impurities also. So if they ask percentage purity in the numerator we will put mass of pure compound in the sample. Suppose they can ask percentage impurity also right. So percentage impurity will be equal to mass of impurities in the sample upon total mass of the impure, impure sample into 100. So by using these formula we will be solving the numericals of this particular topic. Now the first question for this type is read the question and try to find it out on your own because it is quite similar to percentage yield only. A 150 gram sample of copper ore contains 75 grams of pure copper, find percentage impurity very easy just you have to put the formula percentage purity was mass of pure compound in the sample. So this is the mass of pure compound in the sample in the copper ore. So we will put 75 upon the total mass of the impure sample that is nothing but 150 gram. So 150 into 100. So the percentage purity of the particular substance will be 50 percent here. This will get cancelled with 2, so the sample is 50 percent pure. Now coming on to the next question the percentage purity of calcium carbonate sample if 150 grams of sample gives 44 grams of CO2 on heating. So here 150 grams of sample is giving 44 grams of CO2. So I can say that according to the stoichiometry I know that CaCO3 is giving CaO plus CO2. I already know that 44 grams is given by how much grams of CaCO3 100 grams. So percentage purity will be, so that means what in this 150 grams of sample the pure substance present is only 100 grams because how much of CO2 are we obtaining 44 grams. So percentage purity will be the pure compound that is present in the sample that is our 100 upon the total mass of the impure sample that is nothing but 150 here into 100. So we will get it as 2 upon 3 into 100 that will give us a value of 66.67 percent. So I hope you have completely understood percentage yield and percentage purity. Next topic is our POAC. What is POAC? What is POAC? Principle of atom conservation. So basically this particular thing is like based on the law of conservation of mass. Law of conservation of mass. So here I can say that for example I can take one equation. Let's say CaO3 is giving CaO2. So according to the principle of atom conservation we need to say that like it is not like we need to say it is always like that the atoms of potassium on the left side will be equal to the atoms of potassium on the right side. Only the concept on which balancing is dependent on. Okay the way we balance that is only principle of atom conservation. So I can say that the total number of moles of potassium should be on the left side should be equal to the total number of moles of potassium on the right side. So according to this POAC we can express it the POAC for potassium. I can express how many number of potassium is present on the left side one. So one into moles of CaO3. So how many moles are present in the mole of CaO3? One is present. So that this particular amount should be equal to one into here also one because only one case K is present this side also moles of CaO3. This is the POAC expression for potassium in this particular reaction. Now similarly we can do the same thing for oxygen as well let's see. So moles of oxygen on the left side should be equal to moles of oxygen on the right side. So here what we will do? The number of oxygen present in the compound that is 3 into the moles of the compound in which it is present. This amount should be equal to the number of oxygen here 2 into the moles of the compound that is nO2 this time. So this should be our equal. So this is our POAC expression for oxygen and chlorine will be you can understand it will be the same as this. It will be same as this because both contents potassium and chlorine both on the left and right it is present like in one one moles. Now one question we will be doing based on this this is nothing this is very simple we studied law of conservation of mass in the first lecture that is completely based on this. So now one question we will be solving read the question 27.6 grams of K2 CaO3 was heated by a series of reagents so as to convert all of its carbon to this compound. So we can say that this carbon has not gone any not gone to any other product it has only gone here. So I can say that this particular compound and this particular compound may this carbon has moved. So that means if I make the carbon conserved in these two compounds I will get whatever the question is asking. So they have asked the weight of the product the weight of this product when I take 27.6 grams of K2 CaO3 and you can see there is a series of reagents involved but we don't need to see any reagent anything what is forming in in between we just know that all the carbon has moved from this compound to this compound. So we will just make that conservation. So carbon here is conserved so if I say K2 CaO3 series of reagents we have put and it has gone to K2 Zn3 FeCn6 whole twice okay. Now I can say that if I if I conserve this carbon I can say that how many carbon are there in this compound one. So 1 into the moles of K2 CaO3 should be equal to how many carbon are there here Cn6 and 6 also that too 6 into 12 6 into 2 is 12 12 into moles of this product not writing the entire thing this product. Now we know this mass of K2 CaO3 and we know the molecular weight of this as well. So we already know moles is equal to weight upon molecular weight. So here I can write as 1 into 27.6 divided by 138 will be equal to 12 into the weight that we need to find out weight of the product we need to find out we don't know that W by MW of the product molecular weight of the product that is given as 698. So from here we need to find W so W will be equal to what W will be equal to 27.6 into 698 upon 138 into 12 when you calculate this entire thing you will get the answer as 11.699 grams. So when we start with 27.6 grams of K2 CaO3 and we do N number of reactions we put N number of series of reagents we don't need to know this reagents also in this case we know that carbon is conserved means this equation we need to follow. So this POAC is making our work easier to find out the weight of the product right. So this was all about POAC is a very small concept so you can expect questions like this itself right our next topic is our empirical formula and molecular formula. So we will be understanding this particular concept with the help of one question itself but before that what exactly is empirical formula now molecular formula we already have known molecular formula it expresses the actual number the actual number of each atom in a molecule that is our molecular formula but when we say empirical formula it expresses the simplest whole number ratio ratio of atoms in a molecule right. So that means for example molecular formula let's say is H2O2 so how many atoms of hydrogen is there 2 how many atoms of oxygen is there 2 so 2 is to 2 that is nothing but 1 is to 1 so empirical formula will be equal to HO. So we are representing a formula where the simplest whole number ratio of the atoms are taken one more example we will see for example C6H12O6 this is the molecular formula if I take the ratio of 6 is to 12 is to 6 we will get 1 is to 2 is to 1 so I can express the empirical formula as CH2O okay CH2O now we need to know one thing for example we have a substance like C686 that is our benzene and C2H2 that is our ethane or acetylene right. So here in both the cases the empirical formula will be CH here also it will be CH so are you saying one thing that there are two different compounds different substances but they both have same empirical formula so this is possible okay so two different substances can have same empirical formula this you have to note down right now we will understand this particular concept how to find out the empirical formula and molecular formula also from that we will be understanding with the help of a question. Now what is the question so elements are given and their percentage composition is given to us so this particular thing is given in the question. Now we need to make a table like this whenever we are trying to find out the empirical formula we need to make a table like this what is this this is our atomic weight of the element atomic weight of the element for carbon what is the atomic weight 12 for hydrogen it is 1 for oxygen it is 16 now next column is we need to find percentage composition upon atomic weight we need to find this ratio so in case of carbon it will be 40.6 divided by 12 that will be around somewhere around 3.38 for hydrogen it will be 5 upon 1 that is 5 itself and for oxygen it will be 54.4 divided by 16 that will be around 3.4 so these are the values this is the value of what let's say we are calling this as x this is the value of x that we have got. Now next column is for simple ratio how to find out the simple ratio whichever is the smallest among these let's say this is a this is b this is c which one is the smallest one a so all of them a b c I have to divide it with a the smallest one so here I will do a upon a we will get what 3.38 divided by 3.38 that we will get as 1 next we will do b upon a so we will get 5 upon 3.38 that will be nearly equal to 1.5 then this one that is c upon a that will be 3.4 upon 3.38 this will also be almost equal to 1 now simple whole number ratio next column is our simple whole number ratio so here we can see that one of them is in fraction we can take it as 3 upon 2 so we have to make it a whole number ratio so if I multiply this 3 upon 2 with 2 then it will become 3 but if I am multiplying this particular factor with 2 then I have to multiply this and this also with 2 okay so 1 into 2 it will give 2 3 upon 2 into 2 it will give 3 and 1 into 2 again it will give 2 so the simplest whole number ratio of carbon we got as 2 for hydrogen we got it as 3 and for oxygen also we got it as 2 so what will be the empirical formula the empirical formula will be equal to c 2 H 3 O 2 O 2 now there is one formula molecular formula is equal to empirical formula whole n and what is this n natural number natural number it can be 1 2 3 dot dot dot okay now how to find out this n how to find out this n n is nothing but molecular formula weight which is given already to us where is it given 118 grams you can see the molecular weight of the compound is 118 grams so n is equal to molecular formula weight upon empirical formula weight and what will be our empirical formula weight empirical formula weight you can easily find out from this particular formula that we have got so it will be 2 into 12 plus 3 plus 2 into 16 so calculate this how much it will be 24 plus 3 plus 32 that will be 59 so we are just placing the values 118 upon 59 that will be an exact value of 2 now we can easily calculate the formula so the molecular formula will be equal to empirical formula that we got around c 2 H 3 O 2 whole n and what was n value 2 so the molecular formula that we are getting is c 4 8 6 O 4 so you can again check it check it with the mass molecular weight that is given here okay and cross check it so this is how we solve empirical formula questions how to find out empirical formula and molecular formula so today's lecture we have covered four small topics our next lecture will be completely we will be starting with concentration terms which is again very important part of this particular chapter I hope so far all the basic concepts are clear to you thank you so much see you in the next class