Mixed Problems: Set 1 (Purification and Chromatography)

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take a liquid distilled with steam. At the boiling point, its vapour pressure is only thirty millimetres of mercury, while water's is seven hundred and thirty. So water's vapour pressure is about twenty four times bigger. Then the distillate should be almost pure water, with just a trace of the liquid, right? But that is not what happens! Today we will solve it, in these mixed problems on purification and chromatography. Questions like these come up often in JEE Main. So, let's begin.

0:38Idea 1: pick the method by the property

Idea number one: pick the method by the property it uses. The drawing on the right is a distillation set-up, and we will come to it soon. So tell me, some camphor has a little sand mixed in. Which method would you use to purify it? Yes, exactly, we use sublimation! On heating, camphor passes straight into vapour without melting, and the vapour turns back into a solid on a cold surface. Sand does not sublime, so it stays behind. The plus sign on screen just means mixed with. Okay? Now two liquids: diethyl ether, boiling at thirty five degrees Celsius, and toluene, at one hundred and eleven. The small circle before the C means degrees. These boiling points are far apart, by much more than twenty or thirty degrees, so simple distillation is enough. Right? Next, two miscible liquids, boiling at eighty and eighty six degrees. Two liquids again, so simple distillation again, right? No, no, careful, that is the trap here! Six degrees apart is close, so we need fractional distillation. In the drawing, vapours rising through the fractionating column condense and re-vaporise again and again, so they become richer in the lower-boiling liquid, which collects first. With me? Last, an essential oil has to be taken out of leaves without decomposing it. It is steam-volatile and immiscible with water, so we use steam distillation. Clear?

2:08Idea 2: steam distillation in numbers

Idea number two: steam distillation, in numbers. Here p is a vapour pressure, with a small w below it for water, org for the organic liquid, and atm for atmospheric. The mixture boils when p w plus p org equals atmospheric pressure, so it boils below one hundred degrees Celsius. Okay? In the distillate, the mole ratio equals the pressure ratio. So n org over n w, where n means moles, equals p org over p w. With me? But questions usually ask for mass, not moles. So how do we turn this mole ratio into a mass ratio? Yes, multiply each number of moles by its own molar mass, capital M! So m org over m w, with a small m for mass, equals p org times M org, over p w times M w. Got it? So a high molar mass makes up for a low vapour pressure. Keep this in mind, because it is the key to our opening puzzle. Right? Now a famous pair: ortho-nitrophenol and para-nitrophenol, written with a small o and a small p on screen. They have the same formula and the same molar mass. So steam distillation cannot tell them apart, right? No, no, careful, they behave very differently! In ortho-nitrophenol, the hydrogen bond is intramolecular, inside the same molecule, so there is no free O H group left to bond with water. H-bond on screen means hydrogen bond. So the ortho isomer is steam-volatile. Okay? Para-nitrophenol forms intermolecular hydrogen bonds, between different molecules, so it has a higher boiling point, and it is not steam-volatile. So steam distillation separates the two. Clear?

3:52Idea 3: chromatography and polarity

Idea number three: chromatography. In adsorption chromatography, the stationary phase is a solid adsorbent, silica gel or alumina. Column chromatography, like the column in the drawing, and TLC, which means thin-layer chromatography, are of this kind. Okay? Now, paper chromatography. Paper is a solid, so this must be adsorption chromatography too, right? No, and this is a common trap! Paper chromatography is partition chromatography. Water held in the pores of the paper is the stationary phase, and each component is shared between that water and the moving solvent. Clear? Now, silica and alumina hold polar compounds strongly. In the drawing, the eluent, which is the mobile phase, flows down through the adsorbent, and the mixture splits into bands. With me? So here is a statement from a JEE-style question: in column chromatography on silica, the most polar component is eluted first. True or false? That statement is false, and the reason is simple. Held strongly means it moves slowly. In the drawing, the most polar band is held near the top, and the least polar band moves fastest, so it is collected first. Washing the components out of the column one after another, with the eluent, is called elution. Got it? On a TLC plate too, a polar compound is held back, so it has a low R f value, the capital R with a small f below it. And R f is always less than or equal to one, because no compound overtakes the solvent front. Right?

5:25Example 1: spot distance for Rf 0.65 and an 8 cm front

Now, example one, a quick one from this set. Here, a compound has an R f value of zero point six five. How far does it move when the solvent front moves eight centimetres? Pause the video and try it first. Done? Okay, let's check. R f is the spot's distance over the solvent front's distance. So, to find the spot's distance, do we multiply eight by zero point six five, or divide? Yes, we multiply, because the spot's distance equals R f times the solvent front's distance. If you divided, you would get more than eight centimetres, beyond the solvent front, and that is impossible. Okay? So zero point six five times eight gives five point two centimetres. That is less than eight, just as it must be. Clear?

6:17Example 2: steam distillation, mass of liquid per g water

Now example two, and this is our opening puzzle. Here, a liquid of molar mass one hundred and fifty, capital M on screen, is steam-distilled. At the boiling point, its vapour pressure is thirty m m, millimetres of mercury, and water's is seven hundred and thirty. Find the mass of the liquid per gram of water in the distillate. Pause the video and try it first. Done? Okay, let's check. Many students write thirty over seven hundred and thirty, about zero point zero four grams, and stop. The wavy equals sign means about. That matches the puzzle, almost pure water. But is it right? Oh no, careful, this is exactly the trap! The pressure ratio gives the mole ratio, n org over n w. But the question asks for mass. Okay? So multiply by the molar masses: one hundred and fifty for the liquid, and eighteen for water. That gives thirty times one hundred and fifty, over seven hundred and thirty times eighteen. With me? Now, four thousand five hundred over thirteen thousand one hundred and forty gives about zero point three four. Right? So zero point three four grams of the liquid come over per gram of water, about eight times the mole ratio, because the molar mass, one hundred and fifty, is about eight times water's eighteen. So a high molar mass makes up for a low vapour pressure, and that solves our puzzle. Got it?

7:45Quick revision: method choice, steam distillation, Rf

So, quick revision. Pick the method by the property: does it sublime, are the boiling points far apart, is it steam-volatile? In steam distillation, the mass ratio is the pressure ratio times the molar mass ratio, so a high molar mass makes up for a low vapour pressure. And silica holds polar compounds strongly: they come out last and have low R f values, and R f is never more than one. Now go and try the three practice problems on this page. All the best!