Halogens: Carius' Method

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You take zero point two five grams of an organic compound that contains chlorine. You do one experiment, and at the end you weigh the solid you collect. Zero point two eight seven grams. Wait. Zero point two five grams went in, and zero point two eight seven grams came out? More mass out than in? So where did the extra mass come from? And how can this one weight tell us how much chlorine the compound had? Today we will find out, with Carius' method. And believe me, this is one of the easiest numericals to score in JEE, if you avoid one trap. So, let's begin.

0:45Carius' method: trap the halogen as AgX

So, what is Carius' method? First, we weigh the organic compound. Call its mass w, the small w you see on screen, okay? Then we heat it with fuming nitric acid and silver nitrate, the two formulas on screen. And where do we heat it? Not in an open test tube, but in a sealed hard glass tube. Sealed, okay? So remember the setup in one breath: fuming nitric acid, silver nitrate, sealed hard glass tube. Clear? Now think. Our compound has a halogen in it: chlorine, bromine or iodine. And we have put silver into the tube, through silver nitrate. So where does the halogen end up? Yes, exactly! It is precipitated as silver halide, written AGX on screen, where X stands for the halogen. So we get silver chloride, silver bromide or silver iodide, as a precipitate, okay? Then we filter this precipitate, dry it and weigh it. We call its mass m sub AGX, right? That's it! Two weights, w and m: that is all the experiment gives us. Everything after this is pure calculation.

1:54One mole of AgX, one mole of halogen

Now the key idea of this whole topic. Look at the formula, AGX: one silver, one halogen. So one mole of silver halide contains exactly one mole of halogen atoms, okay? Every halogen atom of the compound lands, atom for atom, in this precipitate. So you see, Carius' method is really a mole calculation in disguise. Got it? Next, the three molar masses you must know. Silver chloride, one hundred forty three point five. Silver bromide, one hundred eighty eight. Silver iodide, two hundred thirty five. Too many numbers to remember? Here is a topper's trick. Each one is just one hundred and eight, for silver, plus the atomic mass of the halogen. Chlorine thirty five point five, bromine eighty, iodine one hundred twenty seven. Add one hundred and eight to each, and all three molar masses come out. Simple, isn't it?

2:53JEE trap: the precipitate is not the halogen

Now, back to our puzzle. Zero point two five grams of compound gave zero point two eight seven grams of silver chloride. Many students say: the chlorine is zero point two eight seven grams. Divide by zero point two five, multiply by one hundred. Is that right? What answer do you get? One hundred fourteen point eight percent! What, more than one hundred percent chlorine? No, no, careful! This is the JEE trap. The weight of the precipitate is not the weight of the halogen. Why not? Because the precipitate is silver plus halogen, and the silver came from silver nitrate, not from our compound. So that is where the extra mass in our puzzle came from: the silver! So only a part of the precipitate is halogen. Which part? In one mole of silver halide, the halogen weighs its atomic mass, A sub X on screen, out of the molar mass of the silver halide, capital M sub AGX, okay? So the halogen is the fraction A sub X by M sub AGX, of the mass of the precipitate. With me?

3:56The Carius formula

So, let's write it properly. Mass of halogen equals m sub AGX, the mass of the precipitate, times the fraction A sub X by M sub AGX, okay? And the percentage? Divide by w, the mass of the compound, and multiply by one hundred. Percent of X equals A sub X by M sub AGX, times m sub AGX by w, times one hundred. This is the Carius formula. Write it down. Clear? Now a quick question. Suppose the precipitate is silver bromide. Which fraction will you use? Eighty by one eighty eight, yes! Bromine with silver bromide. Always match the pair: thirty five point five by one forty three point five for chlorine, eighty by one eighty eight for bromine, and one twenty seven by two thirty five for iodine. Putting one forty three point five into a bromine question is a classic way to lose easy marks. Don't do it!

4:49Example 1: percentage of chlorine from AgCl

Alright, now let's solve our puzzle properly. In a Carius estimation, zero point two five grams of a compound gives zero point two eight seven grams of silver chloride. Find the percentage of chlorine. Pause the video, and try it first. Done? Okay, let's check. The precipitate is silver chloride, so the pair is thirty five point five by one forty three point five. Chlorine equals zero point two eight seven, times thirty five point five, divided by one forty three point five. Zero point zero seven one grams. Now divide by the mass of the compound. Zero point zero seven one by zero point two five, times one hundred. Twenty eight point four percent. Not one hundred fourteen point eight, but twenty eight point four. The extra weight was the silver all along. Mystery solved!

5:45Example 2: molar mass of chlorobenzene from AgCl

Now a harder one, JEE level. Zero point two two five grams of a compound, with one chlorine atom per molecule, gives zero point two eight seven grams of silver chloride. Find its molar mass, and suggest a formula if it is aromatic. Pause the video, and try it. Hint: count moles, not grams. Done? Okay, let's check. Look, this time nobody wants a percentage. They want the molar mass, and for that we need the moles of the compound. So think: how are the moles of silver chloride linked to the moles of the compound? Step one: moles of silver chloride. Zero point two eight seven by one forty three point five, that is zero point zero zero two moles. Same precipitate as in example one, right? Now the key idea. One mole of silver chloride means one mole of chlorine atoms. And each molecule has exactly one chlorine atom. So the compound is also zero point zero zero two moles. Simple as that! Molar mass, capital M on screen, is mass divided by moles. Zero point two two five by zero point zero zero two. One hundred twelve point five. Now, aromatic, with one chlorine. Let's try C six H five C L, okay? Six carbons give seventy two, five hydrogens give five, and chlorine gives thirty five point five. Total, one hundred twelve point five. Exactly our answer! So the compound is chlorobenzene. See? One weighing gave us the molar mass, and even the formula.

7:23Quick revision: the Carius formula and matching pairs

So, quick revision. In Carius' method, the halogen is caught as silver halide, filtered, dried and weighed. Percentage of halogen: A sub X by M sub AGX, times m sub AGX by w, times one hundred, and always match the halogen with its own silver halide. And one mole of silver halide means one mole of halogen atoms. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!