Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a little mystery from the lab. One yellow compound, and two methods for finding its nitrogen. Dumas' method says twenty point three per cent nitrogen. Kjeldahl's method, on the very same compound, says only ten point one per cent. So one of the two methods must be wrong, right? Hmm, not so fast, let us think about it. We will solve it today, in this unit problem bank, where one question often needs tools from the whole unit. Questions like this come up often in JEE, so let's begin.

0:37How to use this chapter

So, first, how to use this chapter. It has three parts. Part A is at the JEE Main level. Attempt it as a thirty minute test, after you finish Chapters one to six. Okay? Part B is at the JEE Advanced level. Here, the same compound goes through several analyses in one question. So you must join the tools together. Right? And Part C checks the statements most often tested as true or false, or as assertion and reason items; the slash on screen means or. They look easy, but one word can change the answer. Clear?

1:16Part C warm-up: true or false?

Now, let us warm up with three statements from Part C, okay? The first statement says that in steam distillation, the mixture boils below one hundred degrees Celsius; the small circle and the C mean degrees Celsius. True or false? Yes, it is true, well done! The vapour pressures of water and of the liquid add up. So together they reach atmospheric pressure while the temperature is still below one hundred degrees. Right? The second statement is about TLC, thin layer chromatography, on silica. It says the least polar compound has the lowest R f; that is a capital R with a small f written low. What do you say, true or false? No, no, it is false, and many students fall for it! The least polar compound is held least by the silica, so it moves furthest up the plate. That gives it the highest R f, not the lowest. Got it? And the third statement: in Kjeldahl's method, K two S O four, potassium sulphate, is the catalyst. What do you think? That is false as well, so be careful here! Potassium sulphate is added to raise the boiling point, the b p on screen. The catalyst is C u S O four, copper sulphate. With me?

2:34Nitrogen tools for Parts A and B

Parts A and B have many nitrogen problems, so let us sharpen two tools. In Dumas' method, the compound is heated with C u O, copper two oxide, in an atmosphere of carbon dioxide, C O two. The nitrogen comes out as N two gas, and we measure its volume. This works for all organic nitrogen. Okay? In Kjeldahl's method, the nitrogen first becomes ammonium sulphate; on screen, the N H four group sits in brackets with a small two, because there are two of them. The arrows mean becomes. This then gives ammonia, N H three, which is caught in standard acid. Right? Then comes the formula. Per cent N equals one point four, times the milliequivalents of acid used by the ammonia, over w, the mass of the compound in grams. The little sign at the start is the per cent sign, the cross means times, the long line is a fraction bar, and m e q means milliequivalents. Clear? Now let me ask you a quick question. For sulphuric acid, H two S O four, are the milliequivalents the same as the millimoles? No, they are not, and this is a classic trap! Sulphuric acid has two acidic hydrogens, so its milliequivalents are two times its millimoles, m m o l on screen. Got it? And now comes the big limitation of this method. Kjeldahl's method fails for nitrogen in nitro and azo groups, or in a ring, as in pyridine. That nitrogen is not converted into ammonium sulphate. Remember this line; it will solve our mystery soon. With me?

4:13Example 1: Kjeldahl percent N with 25 mL of 1 M HCl

Now Example one, an easy one to start, from Part A, okay? The ammonia from zero point seven grams of a compound neutralises twenty five millilitres of one molar H C l, hydrochloric acid. Find the per cent of nitrogen. On screen, g means grams, m L means millilitres, and capital M means molar. Pause the video and try it first. Done? Okay, let's check. Twenty five times one gives twenty five millimoles of acid. Now, we just learnt to double, so that makes fifty milliequivalents, right? Then the nitrogen comes out as one hundred per cent. Is that possible? Oh no, careful, a compound that is all nitrogen, with no carbon at all, is not organic! The doubling was only for sulphuric acid. H C l has one acidic hydrogen, so its milliequivalents equal its millimoles, which is twenty five. Okay? Now we use the formula. So one point four times twenty five gives thirty five, and thirty five over zero point seven gives fifty. So the compound has fifty per cent nitrogen. Always ask how many acidic hydrogens the acid has.

5:29Example 2: Dumas 20.3% vs Kjeldahl 10.1%, find the compound

Right? Now Example two, from Part B, and this is our opening mystery. The yellow nitrogen compound gives twenty point three per cent nitrogen by Dumas' method, but only ten point one per cent by Kjeldahl's method. Explain this, and suggest the compound if its molar mass is one hundred and thirty eight. Pause the video and try it first. Done? Okay, let's check. First, which of the two methods could have missed some of the nitrogen, and why? It is Kjeldahl's method, yes, exactly! Dumas' method counts all the organic nitrogen. But Kjeldahl's method misses nitrogen in nitro and azo groups, remember? So both results are right; they simply count different nitrogen. Okay? Now let us compare the numbers. Ten point one is about half of twenty point three; the wavy equals sign means approximately equal. So half of the nitrogen escapes Kjeldahl's method. It must be nitro, azo or ring nitrogen; we will see which one soon. Right? Now let us find the compound. The molar mass is one hundred and thirty eight, and each nitrogen atom weighs fourteen. So how many nitrogen atoms give twenty point three per cent? It is two atoms, very good! Two nitrogens weigh twenty eight, and twenty eight over one hundred and thirty eight, times one hundred, gives twenty point three per cent. That matches Dumas' method perfectly. Clear? And Kjeldahl's method counted just one of them. Fourteen over one hundred and thirty eight, times one hundred, gives ten point one per cent. There it is, exactly the Kjeldahl value. Got it? So one nitrogen is an amino nitrogen, which Kjeldahl's method converts, and the other one is missed. Can the missed one be azo nitrogen? No, because an azo group holds two nitrogens side by side, and here only one is missed. Also, the yellow colour fits a nitro group, so a compound that fits is para nitroaniline; the small italic p means para. On screen, an N H two group and an N O two group sit on a benzene ring, written C six H four, and the dashes are bonds. Its molar mass, capital M, is indeed one hundred and thirty eight. Mystery solved, and neither method was wrong. With me?

7:58Quick revision: percent N, meq vs mmol, what Kjeldahl misses

So, quick revision. For Kjeldahl's method, per cent N is one point four times the milliequivalents, over w. For H C l, milliequivalents equal millimoles, and for sulphuric acid, they are two times the millimoles. Dumas' method counts all the nitrogen, while Kjeldahl's method misses nitro, azo and ring nitrogen. And use this chapter well: Part A as a thirty minute test, Part B to join the tools, and Part C to check every statement with its reason. Now open Part A, set your timer for thirty minutes, and start. All the best!