Earlier AIEEE and JEE Main Questions (2004–2016): Detection and Estimation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Imagine a bottle with no label. You know the compound inside is one of four: acetamide, benzamide, urea or thiourea. And you are allowed just one experiment: estimate its nitrogen, with one titration. Can that single titration tell you which compound is in the bottle? It was asked in the two thousand four paper, and we solve it in Example two. Today we go through past paper questions, from two thousand four to two thousand sixteen, on the detection and estimation of elements. Questions like these come up often, and they test just a few ideas. So, let's begin.

0:41Idea 1: Lassaigne's test for nitrogen

So, idea number one is detection. In Lassaigne's test, we fuse the compound with a small piece of sodium. On screen, N a is sodium, and the arrow means gives: sodium plus carbon plus nitrogen gives N a C N, sodium cyanide. This sodium cyanide ends up in the sodium fusion extract. Okay? Now the extract is boiled with freshly made iron two sulphate, F e S O four; the small four written low is a subscript. Air oxidises some of the iron to iron three, written with the Roman numeral three in round brackets, or ferric chloride, F e C l three, is added for it. Then the mixture is acidified with concentrated sulphuric acid, H two S O four, right? Here is a question from the two thousand four paper. If nitrogen is present, which compound is formed in this test? It is Prussian blue, exactly! Its formula is F e four, then square brackets around F e and C N taken six times, and that whole bracket taken three times. The square brackets hold the complex ion, and the round brackets hold the cyanide group. Its name is iron three hexacyanoferrate two, the Roman numerals again giving the oxidation states. Clear? Now, another option in that paper was N a four, then F e, C N taken five times, and NOS, in square brackets. Oh no, careful, do not pick that one! That violet complex comes from the sulphur test, not from nitrogen. Got it?

2:16Idea 2: purifying glycerol

Idea number two, from the two thousand sixteen paper, is about purification. In the soap industry, glycerol is separated from spent lye. Which kind of distillation suits it best? Here is the clue: glycerol decomposes at its normal boiling point. So if we simply boil it, we break it down. With me? So, we lower the external pressure. The double arrow means this leads to a lower boiling point, so the glycerol distils at a lower temperature, without decomposing. Okay? And that is the answer: distillation under reduced pressure, the method for liquids that decompose at or below their normal boiling point. Right?

2:58Idea 3: Kjeldahl's back titration

Idea number three is Kjeldahl's method for nitrogen. The nitrogen of the compound finally comes out as ammonia, N H three, and it is passed into a known excess of standard acid. The acid left over is titrated with alkali: a back titration. Clear? So, millimoles of ammonia equal the milliequivalents of acid taken, minus the milliequivalents of alkali used. The short forms on screen, m mol and m e q, stand for millimoles and milliequivalents. Right? And for sulphuric acid, milliequivalents are two times millimoles, because it has two acidic hydrogens. Remember this two; we will need it soon. Okay? Then, per cent N equals one point four, times the milliequivalents used by the ammonia, over w, the mass of the compound in grams. The one point four comes from this: each millimole of ammonia carries fourteen milligrams of nitrogen; change to grams and multiply by one hundred. With me? And one condition, which comes up often. Kjeldahl's method does not work for nitrogen in nitro or azo groups, or in a ring like pyridine, because that nitrogen is not converted under these conditions. Got it?

4:11Idea 4: Carius' method for halogens

Idea number four is Carius' method. The compound is heated in a sealed tube with fuming nitric acid, HNO three, and silver nitrate, A g N O three. The halogen comes out as silver halide, A g X, where X stands for chlorine, bromine or iodine. Okay? Then one pattern gives the percentage. Per cent of the element equals the mass of the element in one mole of the precipitate, over the molar mass of the precipitate, times m p p t over w, times one hundred. Here p p t is short for precipitate, m p p t is its mass, and w is the mass of the compound. Right? Now tell me, for silver bromide, A g B r, what is that first fraction? Silver is one hundred and eight and bromine is eighty, so silver bromide is one hundred and eighty eight. So the fraction is eighty over one hundred and eighty eight, exactly! Clear?

5:05Example 1: percent bromine from 141 mg of AgBr (2015)

Now, Example one, from the two thousand fifteen paper. In Carius' method, two hundred and fifty milligrams of an organic compound gives one hundred and forty one milligrams of silver bromide; m g means milligrams. Find the percentage of bromine. Pause the video and try it first. Done? Okay, let's check. Per cent bromine equals eighty over one hundred and eighty eight, times one hundred and forty one over two hundred and fifty, times one hundred. Both masses are in milligrams, so we need not change any units. Okay? Now a small trick to save time. One hundred and forty one over one hundred and eighty eight is exactly three quarters. So the bromine is three quarters of eighty, which gives sixty milligrams. Right? Finally, sixty over two hundred and fifty, times one hundred, gives twenty four per cent. So the compound has twenty four per cent bromine. With me?

6:04Example 2: one Kjeldahl titration identifies urea (2004)

Now, Example two, from the two thousand four paper, and this is our puzzle from the start. The ammonia from zero point three zero gram of an organic compound is passed into one hundred millilitres of zero point one molar sulphuric acid; capital M means molar. The excess acid needs twenty millilitres of zero point five molar sodium hydroxide, N a O H, okay? Is the compound acetamide, benzamide, urea or thiourea? Pause the video and try it. Done? Okay, acid first. One hundred times zero point one is ten, so ten milliequivalents of acid, right? Oh no, careful, that is the trap from idea three! Ten is the millimoles, and sulphuric acid counts two times. So the acid taken is twenty milliequivalents. Okay? Sodium hydroxide used is twenty times zero point five, which gives ten milliequivalents. So the ammonia took twenty minus ten, which is ten millimoles. Right? Each millimole of ammonia carries fourteen milligrams of nitrogen, so the nitrogen is ten times fourteen, one hundred and forty milligrams, which is zero point one four gram. Zero point one four over zero point three zero, times one hundred, gives forty six point seven 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 in our bottle? It is urea, exactly! Urea has two nitrogen atoms, twenty eight grams in a molar mass of sixty, which is forty six point seven per cent, an exact match. So yes, one titration really did name the compound in the bottle. Got it?

7:54Quick revision: Prussian blue, Kjeldahl meq, Carius 80/188

So, quick revision. In Lassaigne's test, nitrogen shows up as Prussian blue. For a liquid that decomposes at its boiling point, distil under reduced pressure. In Kjeldahl's method, for sulphuric acid, milliequivalents are two times millimoles, and per cent N is one point four times the milliequivalents used by the ammonia, over w. And in Carius' method, per cent bromine is eighty over one hundred and eighty eight, times the mass of silver bromide over w, times one hundred. Now go and try the two practice problems on this page. All the best!