Hydrates, Metal Compounds and Non-stoichiometric Oxides

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Iron oxide, F e O: F e is iron, so one iron for every one oxygen, right? But a sample of iron oxide can have the composition F e zero point nine four O, okay? Zero point nine four iron atoms for every oxygen? Wait, some iron is simply missing! Then how is the charge still balanced? Today we will solve this mystery. On the way, we will count the water inside crystals, and find the atomic mass of an unknown metal. And look, these three set-ups come again and again in JEE numericals. So, let's begin.

0:44Hydrates: count the water in moles

So, first set-up: hydrates. Here is a salt, capital A, with a little dot, and then x H two O, okay? The dot says: along with every unit of the salt, there are x molecules of water sitting in the crystal. This water is called water of crystallisation. Now, what happens when we heat it? The water of crystallisation is driven off, and only the dry salt is left behind. We call it the anhydrous salt. Anhydrous simply means without water. Clear? So, the question gives you two masses: the water lost, and the anhydrous salt left behind. How do you get x from these two? Just divide one mass by the other? Hmm, wait. Careful. Think about what x really is. It counts water molecules per unit of salt. So it is a ratio of counts, and counts means moles, right? So, x equals moles of water lost, divided by moles of anhydrous salt. Dividing the grams directly is the classic JEE trap. Don't do that! For the water, the molar mass is eighteen grams per mole, every time. For the salt, use its own molar mass. Then just change both masses to moles, divide, and x is done.

2:06Atomic mass of a metal from its compound MXk

Alright, second set-up. We have an unknown metal, capital M, joined with a partner, capital X, and the formula is M X k, with a small k below. So for every one atom of the metal, there are k atoms of the partner, okay? Quick question. Take a metal chloride, M C l two. In one mole of it, which has more moles: the metal, or the chlorine? Chlorine, of course! Two moles of chlorine for every one mole of the metal. So, moles of M equals one over k, times moles of X, okay? For a chloride with two chlorines, you simply halve the moles of chlorine. And the trap? Students take the moles of metal equal to the moles of chlorine. Then the atomic mass comes out half of the right value. Don't fall for it! Once you have the moles of the metal, the last step is easy. Capital A sub M is the atomic mass of the metal. It equals the mass of the metal, divided by the moles of the metal. Grams divided by moles. As simple as that. And just read k from the formula. For an oxide, M O, k is one, so moles of metal equal moles of oxygen. For the chloride M C l two, k is two. Got it?

3:24Topper's trick: percentages? Take 100 g

Now, a trick that toppers use. Sometimes the question gives only the percentage of each element. So what mass should you start with? One hundred grams, always! Why? Because in one hundred grams, every percentage is directly a mass in grams. Twenty two point seven six percent zinc? That is simply twenty two point seven six grams of zinc. No extra multiplication, right? Then follow the arrows on screen: change grams to moles, and compare the moles. This works for hydrates and for metal compounds alike. Nice and fast.

4:02Non-stoichiometric oxides and charge balance

Now, back to our mystery from the start. Some oxides do not follow a neat whole number formula. We write them as M sub a O, where a can be a decimal, like zero point nine four. These are called non-stoichiometric oxides. And the secret? The metal is present in two forms: some as M two plus, and some as M three plus. So, let a fraction f of the metal be plus three. Then the rest, one minus f, is plus two, okay? Now think. Each oxide ion carries a charge of minus two. So, for every oxygen, how much positive charge must the metal ions carry? Exactly two, yes! The compound is neutral, right? So, here is our balance condition on screen. Neutral compound, so the positive charge per oxygen equals two. Everything else is just counting that positive charge carefully. Clear? Now count that positive charge. Each plus two ion gives two, each plus three ion gives three. So the average charge per metal atom is two into one minus f, plus three f. There are a metal atoms per oxygen, so multiply by a. So, a into the bracket, two into one minus f, plus three f, equals two. Write this one down. Got it?

5:20Example 1: water of crystallisation in washing soda

Let's do Example one. Two point eight six grams of washing soda, N a two C O three dot x H two O, is heated. It leaves one point zero six grams of anhydrous sodium carbonate. Find x, okay? Pause the video, and try it first. Done? Okay, let's check. First, how much water was lost? Many students see one point zero six grams and take that as the water. No, no, careful! One point zero six grams is what is left, not what is lost. So, water lost is two point eight six minus one point zero six: one point eight zero grams. With me? Now moles. Sodium carbonate has a molar mass of one hundred and six. One point zero six divided by one hundred and six: zero point zero one zero mole. Water: one point eight zero divided by eighteen. Zero point one zero mole. So, x equals zero point one zero divided by zero point zero one zero. Ten! Washing soda is N a two C O three dot ten H two O, okay? See? Moles, not grams, and the answer is a clean whole number.

6:42Example 2: fraction of iron as Fe3+ in Fe0.94O

Now the harder one, and it is our puzzle from the start. Here it is: a sample of iron oxide has the composition F e zero point nine four O, okay? What fraction of the iron is present as F e three plus? Pause the video, and use the charge balance. Alright, first a quick guess. One minus zero point nine four is zero point zero six. So six percent of the iron is plus three, right? No, no, careful! This is the JEE trap. Zero point zero six only tells you how much iron is missing, not how much is plus three. Let the equation decide. Here, a is zero point nine four. So, zero point nine four into the bracket, two into one minus f, plus three f, equals two. Divide two by zero point nine four: about two point one two eight. So two plus f is two point one two eight, and f is zero point one two eight. Divide two by zero point nine four: two point one two eight. So two plus f is two point one two eight, and f is zero point one two eight. That means twelve point eight percent of the iron is F e three plus. Mystery solved! Some iron ions carry three charges instead of two, so fewer iron ions are enough to balance each oxygen. Not six percent, but twelve point eight percent. All clear?

8:13Quick revision: hydrates, metal from MXk, Fe0.94O

So, quick revision. Hydrates: x is moles of water divided by moles of anhydrous salt. Moles, never grams. Metal compounds: moles of metal is one over k, times moles of the partner, and then atomic mass is mass divided by moles. Non-stoichiometric oxides: balance the charge, a into the bracket, two into one minus f, plus three f, equals two. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!