Eudiometry and the Degree of Unsaturation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Ten millilitres of an unknown gaseous hydrocarbon is mixed with one hundred millilitres of oxygen, and a spark makes it explode. On cooling, the volume shrinks by twenty five millilitres. Then potassium hydroxide shrinks it by forty more. No balance, no weighing, only two readings. Can these two numbers tell us the formula of the gas? They can, and you will find it yourself by the end. Then we will count the rings and double bonds a formula can hold. Questions like these do appear in the JEE exam. So, let's begin.

0:39Burning CₓHᵧ in excess oxygen

So, we burn a gaseous hydrocarbon in excess oxygen, more than it needs. At the same temperature and pressure, equal volumes hold equal numbers of molecules, so volumes behave just like moles. With me? We write the hydrocarbon as C x H y: the small x and y below the letters count the carbon and hydrogen atoms. C x H y, plus, in brackets, x plus y over four, times O two, gives x C O two, plus y over two H two O, as a liquid. The arrow means gives, and the small l in brackets means liquid. Okay? Please don't just memorise it. Let me ask you this: where does x plus y over four come from? It is really simple. Each carbon makes one C O two, which needs one O two, so that gives x. And y hydrogens make y over two waters, each needing half an O two, so that gives y over four. Right? After the spark, we cool the tube, and the water condenses. Liquid water takes almost no space, so only the gases count. Clear?

1:53What each reading tells you, for a volume V

Now the readings. Take a volume capital V of the hydrocarbon. It gives x times V of carbon dioxide, which KOH, potassium hydroxide, absorbs. So the fall with KOH, divided by V, gives x, the carbon count. Got it? The oxygen used is x plus y over four, all times V, okay? Notice the word used, not added, because we usually add extra oxygen. Now the clever reading. After the spark and cooling, the water is a liquid. So tell me, by how much has the gas volume shrunk? Let us count. We lost the hydrocarbon, that is V, and the oxygen used, x plus y over four, times V, okay? We gained x times V of carbon dioxide. The unused oxygen is there before and after, so it cancels. With me? Subtract, and the x times V cancels too! So the contraction on explosion and cooling is one plus y over four, times V, and it gives you y. The common trap is to forget the one. But the hydrocarbon itself disappears too, and that one is its own volume. Right? If the compound also has oxygen, we write C x H y O z, with z oxygen atoms. The oxygen inside does part of the work, so oxygen used is x plus y over four minus z over two, times V, okay? And the contraction also falls by z over two, as you see on screen. For a hydrocarbon z is zero, and we are back to the lines above. Clear?

3:30Degree of unsaturation: the double-bond equivalent

Now the second idea. From a formula, how many rings and double bonds can the molecule hold? That number is the degree of unsaturation, or double-bond equivalent, DBE for short. The small letters c, h, n and x count carbon, hydrogen, nitrogen and halogen atoms. Careful, here x counts halogens, not carbons. Then DBE equals c, minus h plus x over two, plus n over two, plus one; the fraction bar keeps h plus x together on top. Okay? Capital X means a halogen, like chlorine, and in the DBE formula it is counted along with the hydrogens. Oxygen and sulphur, the O and S on screen, are simply ignored. Got it? Now, each ring, or each double bond, counts one, and a triple bond counts two. With me? Let me ask you this: benzene is C six H six. What is its DBE? Did you get four? Yes, exactly! Six, minus six over two, which is three, plus one, gives four. And benzene has one ring and three double bonds, four in total. Right? Now C three H six O, with an oxygen. Many students subtract something for it. Oh no, careful, oxygen is ignored! Three, minus three, plus one, gives one, so one double bond, or one ring. Clear? And each nitrogen atom adds a half. Nicotine is C ten H fourteen N two. Ten, minus fourteen over two, which is seven, plus two over two, which is one, plus one, gives five. Okay?

5:14Example 1: 20 mL hydrocarbon, 60 mL O2 used, 40 mL CO2

Now, Example one. Twenty millilitres of a gaseous hydrocarbon needs sixty millilitres of oxygen and gives forty millilitres of carbon dioxide. Find its formula. The small m L on screen means millilitres. Pause the video and try it first. Done? Okay, let's check. Start with the carbon dioxide: twenty times x equals forty, so x equals two. The double arrow means implies. Right? Here the question says needs, so sixty is the oxygen used. Twenty times, in brackets, x plus y over four, equals sixty, so x plus y over four is three. Okay? Put x equals two. Two plus y over four is three, so y over four is one, and y is four. Got it? So the formula is C two H four. No weighing at all; two volumes did the whole job. Clear?

6:12Example 2: excess O2 trap, cooling and KOH falls give C4H6

Now, our puzzle from the start, Example two. Ten millilitres of a gaseous hydrocarbon is exploded with one hundred millilitres of oxygen. On cooling, the volume contracts by twenty five millilitres, and by a further forty millilitres with KOH, okay? Find the formula. Pause the video and try it, and start with the KOH reading. Done? Okay, let's check. The fall with KOH is the carbon dioxide, forty millilitres. Ten times x equals forty, so x equals four, and there are four carbons. Okay? Now, many students take the one hundred millilitres given, and write ten times x plus y over four, equals one hundred. So x plus y over four is ten, right? No, no, wait, that is exactly the trap! The oxygen is in excess, so one hundred millilitres is what we added, not what was used. So we need the other reading. With me? The contraction on cooling is twenty five millilitres. Ten times, in brackets, one plus y over four, equals twenty five, so one plus y over four is two point five. Clear? Take away the one, and y over four is one point five. Four times one point five gives six, so y is six. Got it? So the formula is C four H six, and our puzzle is solved from just two readings. Now use the second idea: how many rings and double bonds can this molecule have? DBE equals four, minus six over two, plus one, which gives two. So it has a triple bond, two double bonds, a ring and a double bond, or even two rings. The DBE lists every option at once. Right?

8:04Quick revision: KOH fall, cooling contraction, DBE formula

So, quick revision. The fall with KOH is x times V, and the fall on cooling is one plus y over four, times V, okay? Oxygen used is x plus y over four, times V, not the excess oxygen added. And DBE is c, minus h plus x over two, plus n over two, plus one, with oxygen and sulphur ignored. Now go and try the practice problems on this page. All the best!