Sulphur and Phosphorus

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. We take zero point zero eight four grams of an organic compound, and heat it with fuming nitric acid in a sealed tube. At the end, we weigh zero point two three three grams of barium sulphate. Almost three times heavier than the compound we started with! Wait, how can the product weigh more than the whole sample? And here is the bigger surprise: from these two weighings and the molar mass, we will find how many sulphur atoms sit in one molecule of that compound. Look, JEE loves these numericals. They look scary, but each one is just one line. So, let's begin.

0:42Sulphur: the Carius tube

So, first, sulphur. We take a weighed sample of the compound, its mass is w, and heat it with fuming nitric acid, HNO three, inside a Carius tube. Yes, the same sealed tube we used for the halogens, okay? Now, what happens to the sulphur? Follow the arrows on screen. The sulphur becomes sulphuric acid, H two S O four. Then it is precipitated as barium sulphate, B a S O four, and we weigh that. Its formula mass is two hundred and thirty three, the number in brackets. Clear? But wait. We weighed barium sulphate, not sulphur. So tell me, in two hundred and thirty three grams of barium sulphate, how many grams are sulphur? Thirty two, yes! One sulphur atom in the formula, and sulphur is thirty two. The rest is barium and oxygen, just extra weight. That is why the precipitate in our puzzle was so heavy, right? So, the percentage of sulphur equals thirty two over two hundred and thirty three, times m of barium sulphate over w, times one hundred. Thirty two over two thirty three changes the mass of the precipitate into the mass of sulphur. Then divide by the sample mass w, and multiply by a hundred. And that's it.

1:57Phosphorus: two ways to weigh it

Alright, now phosphorus. First, the phosphorus in the compound is oxidised to phosphoric acid, H three P O four. After that, there are two ways to precipitate it and weigh it. Both come in JEE, so let's see both. Way one. It is precipitated as M g N H four P O four. But we do not weigh this one, okay? It is ignited, and it turns into M g two P two O seven, the formula after the arrow, with formula mass two hundred and twenty two. That is the mass we weigh. Now, the factor. Phosphorus is thirty one, and the precipitate is two hundred and twenty two. So the factor is thirty one over two hundred and twenty two, right? No, no, careful! This is the JEE trap. Look at the formula again. How many phosphorus atoms do you see? Two! M g two P two O seven has two phosphorus atoms. So the phosphorus in it weighs two times thirty one, that is sixty two. The percentage of phosphorus equals sixty two over two hundred and twenty two, times m of the precipitate over w, times one hundred. With thirty one on top, your answer comes out exactly half. Got it? Way two. It is precipitated as ammonium phosphomolybdate. The formula is long: N H four in brackets, three times, then P O four, a dot, and twelve M o O three. And its formula mass is huge: one thousand eight hundred and seventy seven. How many phosphorus atoms here? Just one. So the factor is thirty one over one thousand eight hundred and seventy seven. The percentage of phosphorus equals thirty one over one thousand eight hundred and seventy seven, times m, the mass of this precipitate, over w, times one hundred. See, phosphorus is a tiny part of this precipitate. So even a small amount of phosphorus gives a heavy precipitate to weigh. That is why this method is sensitive for small amounts of phosphorus. Smart, right?

4:00Topper's trick: read the factor from the formula

Now let me share a trick that toppers use. Do not mug up three different factors. Read each one from the formula of the precipitate: the number of atoms of the element, times its atomic mass, over the formula mass of the precipitate. Okay? Barium sulphate has one sulphur atom. So, one times thirty two, over two hundred and thirty three. That is our sulphur factor. M g two P two O seven has two phosphorus atoms. So, two times thirty one, over two hundred and twenty two: sixty two over two twenty two, and the trap is gone! And phosphomolybdate? How many phosphorus atoms in its formula? Just one, yes! So, one times thirty one, over one thousand eight hundred and seventy seven. Exactly the same logic. And in moles, the same idea is even faster. Moles of the element equal moles of the precipitate, times the number of atoms per formula. We will use exactly this in both examples. With me?

4:59Example 1: percentage of phosphorus from Mg2P2O7

Now, Example one. Zero point two zero grams of an organic phosphorus compound gives zero point two two two grams of M g two P two O seven. Find the percentage of phosphorus. Pause the video, and try it first. Done? Okay, let's check. First, moles of the precipitate. Zero point two two two divided by two hundred and twenty two: zero point zero zero one mole. Nice clean number, isn't it? So, zero point zero zero one mole of phosphorus too, right? No, no, careful! If you do that, you get fifteen point five percent, exactly half, and that is wrong. Remember the trap? Each formula unit has two phosphorus atoms. So, moles of phosphorus are two times zero point zero zero one, that is zero point zero zero two mole. Times thirty one: zero point zero six two grams of phosphorus. Now the percentage. Zero point zero six two divided by zero point two zero, times one hundred. Thirty one percent! And with the factor, sixty two over two twenty two, you get the same thirty one percent. Exactly the same.

6:09Example 2: sulphur atoms per molecule from BaSO4

Now the harder one, and it is our puzzle from the start. Zero point zero eight four grams of a compound, with molar mass eighty four, gives zero point two three three grams of barium sulphate. How many sulphur atoms are there in one molecule? Pause the video and try. Hint: think in moles. Alright, first a quick guess. The precipitate is almost three times heavier than the compound. So three sulphur atoms per molecule, right? No, no, careful! This is the trap. Most of that weight is barium and oxygen, remember? Atoms are counted in moles, not in grams. So, moles of the compound: zero point zero eight four divided by eighty four. Zero point zero zero one mole. Moles of barium sulphate: zero point two three three divided by two hundred and thirty three. Also zero point zero zero one mole. And each barium sulphate has one sulphur atom, so that is zero point zero zero one mole of sulphur. Now divide. Zero point zero zero one mole of sulphur, from zero point zero zero one mole of the compound. One! Just one sulphur atom per molecule. And see, thiophene, C four H four S, has one sulphur atom and molar mass eighty four, so it fits. Mystery solved! The extra weight was barium and oxygen, not extra sulphur. All clear?

7:37Quick revision: factors 32/233, 62/222 and 31/1877

So, quick revision. Sulphur: weigh barium sulphate, two hundred and thirty three, and the factor is thirty two over two thirty three. Phosphorus: weigh M g two P two O seven, two hundred and twenty two, and the factor is sixty two over two twenty two, because of the two phosphorus atoms. Or weigh ammonium phosphomolybdate, one thousand eight hundred and seventy seven, and the factor is thirty one over that. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!