Crystallisation and Yield

Transcript

0:00Introduction

Hello everyone! Let us begin with a small lab puzzle. Two students have the same hot, concentrated solution of a salt. The first student leaves it to cool slowly on the bench. The second is in a hurry and puts the dish straight into ice. Both get crystals. So tell me, whose crystals are purer? We will check it in a few minutes. Today we see how to get a pure, dry product, and how to work out its yield. Crystallisation and yield are part of practical chemistry in the JEE syllabus, so let's begin.

0:35Crystallisation point and slow cooling

So, first we concentrate the solution to its crystallisation point, the stage where it is ready to crystallise as it cools. But how do we know that we have reached it? Here is the check. Put a drop of the solution on a glass rod. If the drop crusts on cooling, the crystallisation point is reached. Now, back to our two students. Should we cool it fast in ice, or slowly? Yes, exactly, we cool it slowly, because slow cooling gives large, pure crystals. So the student who left the dish on the bench gets the purer crystals. Clear? Now let us see what rapid cooling does. Many nuclei, the tiny first crystals, form at once, and they grow quickly. Okay? Growing so fast, they trap mother liquor, the solution left around the crystals, and impurities between them. So rapid cooling gives small, less pure crystals. That answers our puzzle.

1:31Choosing a solvent for recrystallisation

Got it? Next, recrystallisation: crystallising the crude product again from a solvent to purify it. First, the solvent must dissolve the compound well when hot, okay? Then why not a solvent that dissolves it well even when cold? That sounds even better, right? That would be a trap, because if the compound dissolves well even when cold, it stays in the solution and we lose it. So a good solvent dissolves the compound poorly when cold, and the crystals come back out as it cools. With me? Now look at the impurities. The solvent either keeps them dissolved, so they stay behind in the mother liquor, or it does not dissolve them at all. Either way, they stay apart from our crystals.

2:15Washing and drying the crystals

Right? After filtering, we wash the crystals with a small volume of cold solvent. Let me ask you this. Why only a little solvent, and why cold? Yes, because the wash only has to remove the mother liquor clinging to the crystals. Cold solvent, used sparingly, does this without dissolving an appreciable amount of our product. Clear? Then we dry the crystals by pressing them between filter papers. Now, why are crystals of Mohr's salt or alum not dried by heating in an oven? It would be much faster, right? These crystals hold water of crystallisation, water that is part of the crystal itself. In an oven they would lose this water, and Mohr's salt would also be oxidised. So crystals with water of crystallisation are never dried by strong heating. Got it?

3:09Percentage yield

Now let us come to the yield itself. Percentage yield equals the actual yield over the theoretical yield, times one hundred. Actual yield is what we really collect, and theoretical yield is what the equation predicts. Okay? But when two reagents react, which one decides the theoretical yield? It is the limiting reagent, the one that gets used up first. Once it is gone, no more product can form. Right? Now suppose your actual yield comes out below the theoretical yield. Did you make a mistake? That alone does not mean a mistake, because the actual yield is always below the theoretical yield. Clear? Let us see why that happens. Some product stays dissolved in the mother liquor, and some is lost in transfers and filtration. With me? Also, side reactions occur, and the reaction may be incomplete. So do not panic over a yield below one hundred percent. Okay?

4:07Checking purity: the melting point

One more idea before the examples. Say you have made acetanilide and recrystallised it. How do we check that it is pure? We check its melting point. Pure compounds melt sharply at the recorded value, and for acetanilide that value is one hundred and fourteen degrees Celsius. The small raised circle and the C on screen mean degrees Celsius. Right? For a second check, mix it with an authentic sample. If our product is pure, the mixture melts at the same temperature.

4:37Example 1: limiting reagent and yield of acetanilide

Got it? Alright, example one, from the preparation of acetanilide. Five point zero grams of aniline is treated with five point zero grams of acetic anhydride. Which is the limiting reagent, and what is the theoretical yield of acetanilide? Pause the video and try it first. Done? Okay, let's check. The equation: aniline plus acetic anhydride, and the arrow gives acetanilide plus acetic acid. The small numbers written low count the atoms, and the two after the bracket means two acetyl groups. One mole reacts with one mole. Now, five grams of each, equal masses, so neither is limiting, right? That is a common trap, because we must compare moles, never grams. Capital M on screen is the molar mass, in grams per mole. For aniline it is ninety three, so five point zero over ninety three gives zero point zero five three eight moles. Okay? Acetic anhydride has a molar mass of one hundred and two. Five point zero over one hundred and two gives zero point zero four nine zero moles, the smaller number. Clear? They react one to one, so the anhydride runs out first. Acetic anhydride is the limiting reagent, and aniline is in excess. Right? One mole of anhydride gives one mole of acetanilide, and its molar mass is one hundred and thirty five. Zero point zero four nine zero times one hundred and thirty five gives six point six grams. That is the theoretical yield. Got it?

6:12Example 2: Mohr's salt yield, which reagent is in excess

Now a slightly harder one. Twenty grams of ferrous sulphate heptahydrate is mixed with ten grams of ammonium sulphate. On screen, the raised dot joins seven water molecules to the ferrous sulphate, and the bracket with a small two means two ammonium groups. Which is in excess, and what is the theoretical yield of Mohr's salt? Pause the video and try it first. Here is a hint: count the water. Done? Okay, let's check. The equation: ferrous sulphate heptahydrate plus ammonium sulphate gives Mohr's salt, which keeps six waters, and one water molecule comes out. Once again, one mole reacts with one mole. Now, for the ferrous sulphate, do the seven waters count in the molar mass? Yes, absolutely, because the twenty grams we weighed includes that water. So its molar mass is two hundred and seventy eight, and twenty over two hundred and seventy eight gives zero point zero seven one nine moles. Okay? Ammonium sulphate has a molar mass of one hundred and thirty two. Ten over one hundred and thirty two gives zero point zero seven five eight moles, the larger number. Right? Look at this. Ferrous sulphate had twice the mass, twenty grams against ten, but fewer moles. So ammonium sulphate is in excess, and ferrous sulphate is the limiting reagent. Clear? Mohr's salt has a molar mass of three hundred and ninety two. Zero point zero seven one nine times three hundred and ninety two gives twenty eight point two grams. That is the theoretical yield, and what you actually collect will always be less. Well done if you got it!

7:51Quick revision: cooling, solvent, limiting reagent

So, quick revision. Cool slowly for large, pure crystals, wash them with a little cold solvent, and dry them between filter papers. Choose a solvent that dissolves the compound well when hot and poorly when cold. And find the theoretical yield from the limiting reagent, comparing moles, not grams. Now go and try the practice problems on this page. All the best!