Mohr's Salt

Transcript

0:00Introduction

Hello everyone! Let us begin with a small puzzle from the lab. Take two hundred and seventy eight grams of ferrous sulphate crystals, and one hundred and thirty two grams of ammonium sulphate. Together, that makes four hundred and ten grams. They combine mole for mole, into one mole of pale green crystals. But that one mole weighs only three hundred and ninety two grams. So where did eighteen grams go? We will answer that very soon. These crystals are Mohr's salt, and questions on its preparation and yield can come up in JEE practical chemistry. So, let's begin.

0:41What is a double salt?

So, what kind of substance is Mohr's salt? It is a double salt: a crystalline compound that contains two simple salts, in a fixed mole ratio. Okay? And here is the important condition: a double salt exists only in the solid state. So what do you think happens when you drop these crystals into water? Yes, exactly, it breaks up completely into the ions of the two salts. So its solution gives the test of every ion present. Right?

1:12Mohr's salt: ferrous ammonium sulphate

Now, Mohr's salt itself, ferrous ammonium sulphate. On the left: ferrous sulphate heptahydrate, F e S O four with seven water molecules, plus ammonium sulphate. The small numbers written low count the atoms. The bracket around N H four, with a small two outside, means two ammonium groups. And the raised dot joins the water of crystallisation to the salt. With me? The arrow means gives. On the right, both salts sit together in one crystal with six water molecules, plus one free water molecule at the end. So count the waters: seven on the left, but only six in the crystal. Now, can you solve our puzzle of the missing eighteen grams? Yes, exactly, one water molecule is released. Seven waters go in, six stay in the crystal, and one comes out as plain water. Clear? And one mole of water is eighteen grams. So two hundred and seventy eight plus one hundred and thirty two gives four hundred and ten, which equals three hundred and ninety two plus eighteen. Nothing disappeared; it left as water. Here is a topper's trick: remember three hundred and ninety two as the two salts together, minus one water. Got it?

2:27Preparing Mohr's salt in the lab

So how do we make it in the lab? Take equimolar amounts of the two salts, the same number of moles of each. The capital M is the molar mass: two hundred and seventy eight for ferrous sulphate heptahydrate, and one hundred and thirty two for ammonium sulphate. Okay? Dissolve them in water that contains a little dilute sulphuric acid, H two S O four. That acid has an important job, and it comes up next. Right? Then concentrate the solution to the crystallisation point, and cool it slowly. Clear? And you get pale green, monoclinic crystals, with a molar mass of three hundred and ninety two.

3:07Why the acid, and why no strong heating?

With me? Now, why the acid? First, it prevents the hydrolysis of the ferrous ion, F e two plus, where F e is the symbol for iron, and the small two plus written high is its charge. Okay? Second, it hinders the oxidation of F e two plus to F e three plus, the ferric ion. Got it? Now suppose you are in a hurry. You think: let me boil the solution hard, for a long time, and the crystals will come faster, right? That is the trap, so be careful here. The solution is not heated strongly or for long. If it turns yellow, that yellow colour means oxidation: some ferrous ions have become ferric ions. Clear?

3:47In water, and in titrations

Now dissolve Mohr's salt in water. It is a double salt, remember? So which ions will the solution give? Yes, exactly, all its simple ions: the ferrous ion, F e two plus, the ammonium ion, N H four plus, and the sulphate ion, S O four two minus. The small plus and two minus written high are the charges again. So the solution gives the tests of all three ions. Right? And here is why it is so useful. Mohr's salt is far more stable to air than ferrous sulphate itself. With me? And so it serves as a primary standard in redox titrations. Okay?

4:26Example 1: ammonium sulphate and yield for 7.0 g FeSO4.7H2O

Time for Example one. What mass of ammonium sulphate is needed for seven point zero grams of ferrous sulphate heptahydrate, and what is the theoretical yield of Mohr's salt? The small g on screen means grams. Pause the video and try it first. Done? Okay, let's check. Always go through moles. The small n is the number of moles, and the line in the fraction means over. Seven point zero over two hundred and seventy eight gives zero point zero two five two moles of ferrous sulphate heptahydrate, and the short unit after the number is read as moles. Clear? Now, the equation is one to one, so this needs the same number of moles of ammonium sulphate. The cross on screen means times. Zero point zero two five two times one hundred and thirty two gives three point three grams. Okay? Now the yield. Many students just add the two masses: seven point zero plus three point three gives ten point three grams. So the yield is ten point three grams, right? No, and our opening puzzle tells you why: one water molecule is released for each unit of Mohr's salt that forms. So use the molar mass of the product. Zero point zero two five two times three hundred and ninety two gives nine point nine grams. That is the theoretical yield. Got it?

5:50Example 2: percentage yield from 27.8 g of FeSO4.7H2O

Now a slightly harder one. Here, a student starts from twenty seven point eight grams of ferrous sulphate heptahydrate, with excess ammonium sulphate, and collects thirty one point three six grams of dry crystals. Find the percentage yield. Pause the video and try it first. Here is a hint: find the theoretical yield first. Done? Okay, let's check. The ammonium sulphate is in excess. So which salt decides how much Mohr's salt can form? Yes, exactly, the ferrous sulphate. There is more than enough ammonium sulphate, so the ferrous sulphate runs out first. Right? So, twenty seven point eight over two hundred and seventy eight gives zero point one zero zero moles of ferrous sulphate heptahydrate. With me? It is one to one again, so at most zero point one zero zero moles of Mohr's salt can form. Zero point one zero zero times three hundred and ninety two gives thirty nine point two grams, the theoretical yield. Clear? Now, a common mistake: comparing what you collected with what you started with. The slash in the box means over. Thirty one point three six over twenty seven point eight, and then times one hundred, gives about one hundred and thirteen percent. Is that the yield? Of course not, because a yield can never be more than one hundred percent. Always compare with the theoretical yield. Thirty one point three six over thirty nine point two, and then times one hundred, gives eighty percent. So the percentage yield is eighty percent. See how smoothly it goes through moles?

7:30Quick revision: double salt, acid, no strong heating, yield

So, quick revision. Mohr's salt is a double salt, pale green, with a molar mass of three hundred and ninety two. Make it from equimolar amounts of the two salts, in water with a little dilute sulphuric acid, and no strong heating. And for the theoretical yield, take the moles of the limiting salt, usually the ferrous sulphate heptahydrate, times three hundred and ninety two. Now go and try the practice problems on this page. All the best!