Hello, students, let us continue our discussion on mole and equivalent concept. First we will take an example on percentage purity and then we will discuss, what is the percentage yield of a reaction.

Let us take an example. An impure 6g of NaCl is dissolved in water and then treated with excess of silver nitrate. The mass of AgCl precipitate is found to be 14.35g. What is the percentage purity of NaCl? Now, in this case first of all we will see the reaction what is happening. It is double displacementry. It is a double displacement action. NaCl plus silver nitrate it gives you AgCl plus NaNO3. And now we will first develop the strategy here. What should be our strategy here? That any AgCl will be precipitated here it will be precipitated by the actual AgCl only. So, using the mass of AgCl precipitate, we will first calculate the mass of NaCl that is present and then using this information we will find the percentage purity. Whatever is actual, 6g is theoretically given and we can calculate. So, we write down the reaction and then we first do the stoichiometric analysis. 1 mole NaCl gives you 1 mole AgCl. So, using the molar mass data, now they have given atomic rates, for silver it is 108, sodium 23 and chlorine 35.5. So, from here we can calculate the molar mass of NaCl as 58.5 and that of AgCl as 143.5. Now, from here converting the moles into mass, we can write down 58.5 g of NaCl is producing 143.5 g of AgCl. Now, we will do reverse analysis. 143.5 g of AgCl is going to be produced by 58.5 g of NaCl. So, 1g of AgCl will be produced by 58.5 divided by 143.5 g of NaCl. Now, how much we are getting, 14.35 g of AgCl so that is going to produced, rather that is going to be produced by 58.5 by 143.5 into 14.35 g of NaCl. Which comes out as 5.85 g of NaCl, this was the actual amount. Theoretically how much it is given? 6 g. so, percentage purity simply what, 5.85 divided by 6 times 100 that is equal to 97.5.

Now, let us discuss a very important term percentage yield. What is percentage yield? Sometimes the reaction may not get completed according to the initial amount taken. Means whatever amount you have taken, sometimes you predict that it will happen but that doesn’t happen only. What are the possible reasons? The first possible reason is establishment of equilibrium. Now, this chapter you will read later in the portion of chemistry, but let me tell you, the amount reactant taken by you after a certain time it stabilises, it doesn’t proceed further. So, that is called as equilibrium. And therefore the reaction that you expect for the corresponding product doesn’t occur. Now, for the second reason, some of the reactants are not able to react due to formation of a protective layer. Suppose there is one solid and if you pour liquid on top of it, that reacts with the surface and then covers it up and the remaining solid cannot react and therefore we will not get the actual amount that we are thinking. Now, the third possible case would be when we take gaseous reactants they may simply escape out of the reaction chamber and will not give you the products. And the fourth possible reason can be, other reactions happening in parallel for the same reactants. Like till now you have studied that, N2 and H2 can give you ammonia NH3 but they can also combine to give you N2H4 which is called as hydrazine. So, you may not get ammonia in total that you are thinking. Now, in such cases the efficiency of the process is measured using percentage yield. What is percentage yield? It is a product mass that you actually get divided by the product mass that you theoretically predict times 100.

Now, let us take an example for this. A on controlled oxidation gives X according to the reaction. We have got stoichiometric. They are saying some of the reactant A oxidises and only 70% of maximum yield is obtained. What mass of X is produced if 200 g of A is taken? Now, what should be our strategy? First of all we will calculate normally, we will think 100% yield is there and let us find out how much mass of A that we can produce. And then using the mass of X that we have calculated, we will calculate the actual mass by using the percentage yield 70%. So, let us see, how we do it. We write down stoichiometry of the reaction. 2A plus 9B gives you 2X plus 4Y plus 5Z but what is the interesting part for me, 2A and 2X. So, we write down 2 mole A is going to produce 2 mole X. So, 1 mole A is going to produce 1 mole X and we first convert the moles into mass. So, 128 g of A will give you 128 g of X based on a molecular mass data that was given to us. So, from here we can say 128 g of A produces 148 g of X and therefore 1 g of A is going to produce 148 by 128 g of X. Assuming 100% yield, 200 g of A is going to produce 148 by 128 times 200 g of X. Now, since percentage yield is equal to product mass actual divided by product mass theoretical times 100. We will rearrange this equation because we want the actual mass. So, we can write it down as mass actual equal to percentage yield by 100 times mass theoretical. Now, putting in the value, value of yield is 70%, so, 70 by 100 times this much mass and that is going to give us 161.875 g. So, first we calculate the actual amount or rather the theoretical amount and then we calculate the actual amount using yield.

Thank you.

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