Nitrogen: Kjeldahl's Method

Transcript

0:00Introduction

Hello everyone! Before we start, one small puzzle. I give you half a gram of an organic compound. The nitrogen is locked inside its molecules. You cannot pick the nitrogen out and put it on a balance, right? And yet, with just an acid, an alkali and a burette, a chemist tells you: fifty six percent nitrogen. How? And here is the twist. Nitrobenzene clearly contains nitrogen, but this same method cannot find it! Why? Today, Kjeldahl's method answers both questions. And you know, JEE loves Kjeldahl numericals, with their little traps. So, let's begin.

0:42Kjeldahl's method in four steps

So, the nitrogen is locked inside the organic molecule. How do we get it out, in a form we can actually count? Hmm? Not with tweezers, of course! Here is the idea: we will turn the nitrogen into a simple compound that an acid can count. Four steps. Let's go. Step one: digestion. The compound is digested with concentrated sulphuric acid, along with potassium sulphate, and copper sulphate as the catalyst, okay? And what happens to the nitrogen? It becomes ammonium sulphate, the formula you see on screen. Yes, now the nitrogen is out of the organic molecule. Clear? Step two: boil it with excess sodium hydroxide. Ammonia gas is released, N H three. So all that nitrogen has now become ammonia. Step three: we pass this ammonia into a known excess of standard acid. The ammonia uses up part of the acid, and the rest stays unused, okay? Step four: the unused acid is titrated with standard alkali. Look, we measure what is left over, not what was used. This is called back titration. Acid taken, minus acid left over, gives the acid used by ammonia. That's it, that is the whole method.

2:00From acid used to percentage of nitrogen

Now the counting. One ammonia molecule uses one H plus from the acid, okay? One to one. Really simple. So tell me. If the ammonia used twenty millimoles of H plus, how many millimoles of nitrogen were there? Twenty, yes! Each ammonia carries one nitrogen. In general, millimoles of nitrogen equal millimoles of H plus used. And that is the normality of the acid, capital N, times V, the volume of acid used, in millilitres. Each millimole of nitrogen weighs fourteen milligrams. So, the mass of nitrogen is fourteen, times N, times V, in milligrams. Now the percentage. Divide by one thousand to change milligrams into grams, divide by w, the mass of the compound in grams, and multiply by one hundred. Fourteen, times one hundred, over one thousand, is one point four. So, percentage of nitrogen equals one point four, times the normality of the acid, times the volume of acid used in millilitres, divided by the mass w, okay? This is the Kjeldahl formula. Write it down. Got it? Now one question. This V: is it the total acid you took, or something else? Only the acid actually neutralised by ammonia, exactly! Not the total acid taken. Remember this; we will need it in example two. And N is the normality, not the molarity, okay? For sulphuric acid, the normality is two times the molarity, because each sulphuric acid molecule gives two H plus. With me? Keep this in mind for the very next example.

3:39JEE trap: where Kjeldahl's method fails

Alright, now remember the twist from the start. Nitrobenzene clearly contains nitrogen. So Kjeldahl's method will find it, right? No, no, careful! This is the JEE trap. The method works when the nitrogen is amino or amide nitrogen, like in aniline and urea. That nitrogen is converted into ammonium sulphate during digestion. But nitro compounds, like nitrobenzene, and azo compounds, like azobenzene? The method does not work for them. Funny, isn't it? The nitrogen is right there, and still Kjeldahl cannot count it! Same story with nitrogen sitting in a ring, like in pyridine. Kjeldahl fails here too. Why? Under the digestion conditions, such nitrogen is not converted into ammonium sulphate, so no ammonia comes out to be counted. For these compounds, use Dumas' method. So in the exam, before you use the formula, first check what kind of nitrogen it is. Clear?

4:41Example 1: percentage of nitrogen with 10 mmol H2SO4

Right, on to example one. Remember our half a gram from the puzzle? Here it is. In a Kjeldahl estimation, the ammonia from zero point five grams of a compound neutralises ten millilitres of one molar sulphuric acid. Find the percentage of nitrogen. Pause the video, and try it first. Done? Okay, let's check. Many students write directly: one point four, times one, times ten, divided by zero point five. Twenty eight percent, right? Oh no! What did I just tell you about sulphuric acid? Ten millilitres of one molar sulphuric acid is ten millimoles of acid. But each one gives two H plus, so that is twenty millimoles of H plus. The molarity is one, but the normality is two, okay? One ammonia uses one H plus, so twenty millimoles of ammonia, and twenty millimoles of nitrogen. Twenty times fourteen: two hundred and eighty milligrams. That is zero point two eight grams. Percentage of nitrogen: zero point two eight over zero point five, times one hundred. Fifty six percent! See, that is exactly the fifty six percent from our puzzle. And with the formula directly: one point four, times two, times ten, over zero point five. Fifty six again. Same answer, both ways. Absolutely correct.

6:06Example 2: back titration of the leftover acid

Now a harder one, JEE level, with a back titration. Read it with me. The ammonia from zero point four zero grams of a compound is passed into forty millilitres of zero point two five molar sulphuric acid. The remaining acid needs twenty four millilitres of zero point five molar sodium hydroxide. Find the percentage of nitrogen. Pause and try. Hint: first find how much acid was left over. Done? Okay. First, the H plus we took. Forty millilitres, times zero point two five, times two, because it is sulphuric acid. Twenty millimoles of H plus. Next, the acid left over. Twenty four millilitres of zero point five molar sodium hydroxide neutralised it. Twenty four times zero point five: twelve millimoles. With me? So far, so good. Now, the trap. Some students take all twenty millimoles as nitrogen. Twenty times fourteen is two hundred and eighty milligrams, over zero point four zero grams. Seventy percent, right? No, no, careful! The ammonia did not use all the acid, did it? Acid used by ammonia equals acid taken, minus acid left over. Twenty minus twelve: eight millimoles. Exactly what the V in our formula means, only the acid actually used. So, eight millimoles of nitrogen. Eight times fourteen is one hundred and twelve milligrams, that is zero point one one two grams. Percentage: zero point one one two over zero point four zero, times one hundred. Twenty eight percent. See? Taken, minus left over, and the back titration question is done. Nice and easy.

7:54Quick revision: four steps, %N = 1.4 N V / w, limits

So, quick revision. Digest the compound, add sodium hydroxide, trap the ammonia in excess acid, and back titrate the rest. Percentage of nitrogen is one point four, times N, times V, over w, where V is the acid actually used, and N is two times M for sulphuric acid. And no Kjeldahl for nitro, azo or ring nitrogen: use Dumas' method. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!