Nitrogen: Dumas' and Kjeldahl's Methods

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You want to know how much nitrogen is in an organic compound. Easy, isn't it? Burn it, collect the nitrogen gas, and measure it. So let us just burn it in air, right? Hmm, wait. There is a hidden problem with that plan. Keep it in mind; we will come back to it in a minute. So how do chemists really do it? Today we learn Dumas' method and Kjeldahl's method, with the arithmetic of both. These numericals come up often in JEE Main, so let's begin.

0:35Dumas' method: catch the nitrogen as a gas

So, method one is Dumas' method. We heat the compound with copper two oxide, C u O, in a stream of carbon dioxide, C O two. The small two written low is a subscript. Okay? Now, back to our puzzle. Why do we heat it in carbon dioxide, and not simply in air? Yes, exactly! Air is mostly nitrogen, so it would add its own nitrogen to ours. Carbon dioxide has none, and it sweeps our nitrogen forward. That solves our puzzle. Clear? Here is the reaction. C x H y N z is a general compound; the subscripts x, y and z count its atoms. It needs two x plus y over two units of copper two oxide; that is the number in round brackets. The arrow means gives: x molecules of carbon dioxide, y over two of water, and two x plus y over two of copper, while the nitrogen comes out free as nitrogen gas, N two, z over two molecules of it. With me? Some nitrogen may come out as oxides of nitrogen. So the gases pass over hot copper gauze, which reduces those oxides to nitrogen gas. Simple! Finally, the gases are collected over KOH solution, potassium hydroxide, in a nitrometer. It absorbs all the carbon dioxide, so only nitrogen is left, along with a little water vapour, and we measure its volume. Got it?

2:02Dumas: from volume to percentage

Now, the arithmetic. Our nitrogen was collected over a solution in water. So is its pressure just the atmospheric pressure? No, and this is the first trap. The gas also holds water vapour. So p, the pressure of the nitrogen, equals p atm, the atmospheric pressure, minus the aqueous tension, the pressure of the water vapour. The small atm written low is a subscript. Okay? Next, we reduce the volume to standard temperature and pressure, STP on screen. V naught, V with a small zero, equals V times p over seven hundred and sixty, times two hundred and seventy three over T, the temperature in kelvin. The dots mean times, and the pressure is in millimetres of mercury. Right? Now the percentage. Per cent N equals twenty eight over twenty two thousand four hundred, times V naught in millilitres over w, times one hundred. Why this fraction? Because one mole of nitrogen gas, twenty eight grams, fills twenty two thousand four hundred millilitres at standard conditions. And w is the mass of the compound in grams. Clear?

3:12Kjeldahl's method: catch the nitrogen as ammonia

Now, method two is Kjeldahl's method. We heat the compound with concentrated sulphuric acid, H two S O four. Potassium sulphate, K two S O four, raises the boiling point, and copper sulphate, C u S O four, is the catalyst. Okay? The nitrogen of the compound becomes ammonium sulphate: N H four in round brackets, taken twice, then S O four. So all the nitrogen is now trapped in this salt. Next, we add sodium hydroxide, N a O H, and the ammonia, N H three, is set free. We distil it into a known excess of standard acid, that is, more acid than it needs. Right? So some acid is left over, and we titrate it with standard alkali. This is called a back titration. Got it?

3:59Kjeldahl: the arithmetic, and the catch

Now the arithmetic. Millimoles of ammonia equal the milliequivalents of acid taken, minus the milliequivalents of alkali used in the back titration. The short forms on screen stand for these words. Clear? Now a quick question for you: ten millimoles of sulphuric acid is how many milliequivalents? It is twenty, not ten, so be careful here. Sulphuric acid has two acidic hydrogens, so its milliequivalents are two times its millimoles. Forgetting this two is a very common mistake. Got it? And the percentage: per cent N equals one point four, times the milliequivalents used by the ammonia, over w in grams. Why one point four? Each milliequivalent is one millimole of ammonia, with fourteen milligrams of nitrogen; change to grams and multiply by one hundred. Right? Kjeldahl's method is quick and simple. So can we use it for every compound, say, nitrobenzene? No, we cannot, and this point comes up often. Nitrogen in nitro and azo groups, or in a ring like pyridine, does not become ammonium sulphate. For these, use Dumas' method, which works for all organic nitrogen. With me?

5:15Example 1: Dumas numerical with aqueous tension correction

Now, Example one. Zero point two five grams of a compound gives forty millilitres of nitrogen over KOH, at three hundred kelvin and seven hundred and fifty millimetres. The aqueous tension is twenty five millimetres. Find the percentage of nitrogen. Pause the video and try it first. Done? Okay, let's check. First, subtract the aqueous tension: seven hundred and fifty minus twenty five gives seven hundred and twenty five millimetres. Okay? Now to standard conditions. V naught equals forty, times seven hundred and twenty five over seven hundred and sixty, times two hundred and seventy three over three hundred. That gives thirty four point seven millilitres. Very good! The mass of nitrogen is twenty eight over twenty two thousand four hundred, times thirty four point seven, which gives zero point zero four three four grams. With me? Finally, zero point zero four three four over zero point two five, times one hundred, gives seventeen point four per cent. So the compound has seventeen point four per cent nitrogen. Clear?

6:24Example 2: Kjeldahl back titration to identify the compound

Now a harder one, Example two. Zero point four zero grams of a compound is Kjeldahlised, and the ammonia goes into twenty five millilitres of zero point five molar sulphuric acid; capital M means molar. The excess acid needs fifteen millilitres of one molar sodium hydroxide. Is it acetamide, benzamide, urea or thiourea? Pause the video and try it. Done? Okay, acid first. Twenty five times zero point five is twelve point five, so twelve point five milliequivalents, right? Oh no, careful, that is our trap again! Twelve point five is the millimoles, and sulphuric acid counts two times. So the acid taken is twenty five milliequivalents. Okay? Sodium hydroxide used is fifteen times one, which gives fifteen milliequivalents. So the ammonia took twenty five minus fifteen, which is ten millimoles. Right? Now our formula: one point four times ten, over zero point four zero, gives thirty five per cent. With me? From their formulas, acetamide has twenty three point seven per cent nitrogen, and benzamide eleven point six. Urea has forty six point seven, and thiourea thirty six point eight. So which one is nearest to thirty five? Yes, it is thiourea, at thirty six point eight per cent. Urea is far too high, and acetamide and benzamide are far too low. Well done if you got it!

8:00Quick revision: Dumas steps, Kjeldahl formula and limits

So, quick revision. Dumas: heat with copper two oxide in carbon dioxide, collect the nitrogen over potassium hydroxide, subtract the aqueous tension, and reduce to standard conditions. Kjeldahl: ammonia into excess acid, per cent N equals one point four times milliequivalents over w, and for sulphuric acid, milliequivalents are two times millimoles. And Kjeldahl fails for nitro, azo and ring nitrogen, while Dumas works for all. Now try the three practice problems on this page. All the best!