Eudiometry

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here is an unknown gas. No balance, no weighing, only a tube with markings. I take ten millilitres of the gas, add eighty millilitres of oxygen, and pass a spark. After cooling, seventy millilitres are left. Then I pass the gas through KOH, and only fifty are left. From just these volumes, can you tell me the formula of the gas? Sounds impossible, right? Today we learn eudiometry: gas analysis by volume. And look, in JEE this is a favourite: a few volumes are given, and you must find the formula. So, let's begin.

0:40The eudiometer: spark, cool, absorb

So, what is a eudiometer? Just a graduated tube, with volume markings, okay? We put our gas inside, add excess oxygen, more than it needs, and explode the mixture with a spark. That is the first arrow in the picture. After the spark, the tube has carbon dioxide, water, and the oxygen that was not used. Now we cool the tube. Tell me, what happens to the water? Yes, exactly, it condenses into liquid! So, on cooling, the water condenses. And remember, liquid water takes up no gas volume, only gases do. So the volume in the tube falls. Clear? Next, we pass the gas through KOH, potassium hydroxide. It absorbs carbon dioxide, so the volume falls again, and this fall is exactly the carbon dioxide. What is left now? Only the unused oxygen, see the last box. And alkaline pyrogallol absorbs oxygen. So remember the pair: KOH for carbon dioxide, pyrogallol for oxygen. Got it?

1:47Burning CxHy in oxygen

Now, the gas is usually a hydrocarbon. We write it as C x H y: x carbon atoms and y hydrogen atoms, okay? Here is its balanced equation. C x H y, plus x plus y by four O two, gives x C O two, plus y by two H two O, as a liquid: that is the small l in brackets. Don't just mug this up. Where does this x plus y by four come from? Come on, let's balance it together. Every carbon atom goes into one C O two, so x carbons give x C O two. And every two hydrogen atoms make one water molecule, so y hydrogens give y by two waters. Now the oxygen. Each C O two needs one O two, so that is x. Each water needs only half an O two, so y by two waters need y by four. Total: x plus y by four. That's it, that is the whole equation. Clear?

2:44Per volume of CxHy: what each reading gives

Now the beautiful part. For gases, the coefficients are also the volume ratios, remember? So for every one volume of hydrocarbon, the oxygen used is x plus y by four volumes. Next, cooling. Before the spark, we had one volume of hydrocarbon, plus all the oxygen. After cooling, which gases are left in the tube? Yes, the x volumes of carbon dioxide, and the unused oxygen. The water is liquid now. So what disappeared? The one volume of hydrocarbon, and the x plus y by four of oxygen that was used. And what appeared? x volumes of carbon dioxide. One plus x plus y by four, minus x: the x cancels! So the contraction on cooling is one plus y by four per volume. The unused oxygen is there before and after, so it cancels too. And the last one is easy. KOH absorbs only the carbon dioxide, and there are x volumes of it. So the further contraction with KOH is just x. See, this fall gives you the carbon count directly.

3:51Topper's trick: x first, then y, then check

Now let me share a trick that toppers use. Step one: always start with the KOH fall. Divide it by the volume of the gas, and you get x straight away. Step two: divide the fall on cooling by the volume of the gas. That equals one plus y by four, so out comes y. Step three: check with the oxygen used. Per volume of gas, it must be x plus y by four. Two readings giving the same y? Then your answer is right, no doubt about it. But careful, here is the JEE trap. The question tells you the oxygen added. Is that the oxygen used? No way! The oxygen was in excess, so some is always left over. So, oxygen used equals oxygen taken, minus oxygen left. And the oxygen left is the last volume, after the KOH step, because only unused oxygen remains. With me?

4:48Example 1: hydrocarbon formula from gas volumes

Now, remember our puzzle? Let's finish it. Ten millilitres of a gaseous hydrocarbon is burnt with eighty millilitres of oxygen. After cooling, the volume is seventy millilitres, and after passing through KOH it is fifty. Find the formula. Pause the video, and try it first. Done? Okay, let's check. Step one, the KOH fall. Before KOH the volume was seventy, and after it, fifty. So the carbon dioxide is twenty millilitres. Not fifty, right? Fifty is what is left. Twenty divided by ten: x equals two. Now the oxygen. Many students write: ten times x plus y by four equals eighty. That gives y equals twenty four. Hmm. Was all eighty millilitres of oxygen really used? No, careful! That is the trap. Fifty millilitres are left after the KOH step, and that is all unused oxygen. So the oxygen used is eighty minus fifty: thirty millilitres. Now, ten times x plus y by four equals thirty. So two plus y by four equals three, y by four is one, and y equals four. And the topper's check. Ten plus eighty, ninety millilitres went in, and seventy were left after cooling: a fall of twenty. Twenty by ten is two, and one plus y by four is also two. Perfect match! So the formula is C two H four. And there you go, puzzle solved, without weighing anything.

6:26Example 2: composition of an H2 and CH4 mixture

Now a harder one, a mixture, JEE level. Thirty millilitres of a mixture of hydrogen and methane is exploded with one hundred millilitres of oxygen. After cooling, the volume is seventy five millilitres, and after KOH it is fifty five. Find the composition. Pause the video and try it. Hint: start with KOH, as always. Done? Okay, let's check. Before any calculation, tell me: does hydrogen give any carbon dioxide? No! Hydrogen has no carbon at all. So all the carbon dioxide comes from methane: seventy five minus fifty five, twenty millilitres. Methane is C H four, one carbon, so each volume of methane gives one volume of carbon dioxide. So methane is twenty millilitres. The rest is hydrogen: thirty minus twenty, ten millilitres. Done in two lines! But a topper always checks, right? Now the topper's check: the oxygen used. For hydrogen, use our formula with x zero and y two: y by four is one half. Ten times half, five. For methane, x is one and y is four, so x plus y by four is two volumes of oxygen per volume: twenty times two, forty. Five plus forty: forty five millilitres. So the oxygen left should be one hundred minus forty five, fifty five millilitres. And the tube after KOH shows exactly fifty five! Absolutely perfect. So the mixture has ten millilitres of hydrogen and twenty of methane. See how the KOH reading cracked the whole question?

8:10Quick revision: O2 used, cooling fall and KOH fall

So, quick revision. Per volume of hydrocarbon, oxygen used is x plus y by four. The fall on cooling is one plus y by four, and the fall with KOH is x. And oxygen used is oxygen taken minus oxygen left. Always start with the KOH reading. Now go and try the practice problems on this page. All the best!