Halogens, Sulphur, Phosphorus and Oxygen

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. Suppose a chemist takes zero point four zero gram of an organic compound and analyses it for chlorine. The precipitate at the end weighs zero point five seven four gram, heavier than the whole sample! So where did the extra mass come from? We answer it in Example one. Today we find the percentage of halogens, sulphur, phosphorus and oxygen in a compound, and these calculations come up often in JEE Main. So, let's begin.

0:33Carius' method: the sealed tube

So, for a halogen we use Carius' method. The compound goes into a hard-glass tube, the Carius tube, with fuming nitric acid, HNO three, and silver nitrate, A g N O three; the small three is a subscript, and the plus signs mean we add them together. Then the tube is sealed. Now tell me, why seal it? Because fuming nitric acid must act at a high temperature, yes! In the sealed tube, heated in a furnace, it works under pressure without escaping. So carbon and hydrogen are oxidised completely, and the halogen comes out as halide ions. Right? Then why did we add silver nitrate along with the acid? To precipitate the halogen at once as silver halide, exactly! On screen it is A g X, where that letter X stands for any halogen. Clear? The precipitate is then washed, dried and weighed, and that mass is all we need. With me?

1:33One pattern for every element

Now, one pattern works for every element. Per cent element equals a fraction, times m p p t over w, times one hundred. Here p p t means precipitate, so m with the small p p t below is the mass of the precipitate, and w is the mass of compound taken. The fraction is the mass of the element in one mole of precipitate, over the molar mass of the precipitate. Okay? So, for silver chloride, what part of the mass is chlorine? Thirty five point five over one hundred and forty three point five, correct! That is chlorine's atomic mass over the molar mass of silver chloride. In the same way, silver bromide gives eighty over one hundred and eighty eight, and silver iodide gives one hundred and twenty seven over two hundred and thirty five. Got it?

2:22Sulphur and phosphorus: new precipitates

Next, let us do sulphur. The compound is heated with fuming nitric acid, or, as the brackets say, with sodium peroxide, N a two O two. The sulphur becomes sulphuric acid, H two S O four, which is precipitated as barium sulphate, B a S O four. Okay? One sulphur, thirty two, in a molar mass of two hundred and thirty three, so the fraction is thirty two over two hundred and thirty three. Right? Now let us do phosphorus. It becomes phosphoric acid, H three P O four, which is precipitated as M g N H four P O four, and then ignited, that is, strongly heated, to give M g two P two O seven. Clear? Phosphorus is thirty one, and the molar mass is two hundred and twenty two. So the fraction is thirty one over two hundred and twenty two, right? No, no, wait, that is the trap! Look at the formula: P two, so two phosphorus atoms. The top is two times thirty one, which is sixty two, so the fraction is sixty two over two hundred and twenty two. Always count the atoms. Got it? Or it is precipitated as ammonium phosphomolybdate: N H four in brackets, three times, then P O four, a dot, and twelve M o O three. The dot just joins two parts of one formula. With one phosphorus, the fraction is thirty one over one thousand eight hundred and seventy seven. With me?

3:47Oxygen: usually by difference

Finally, let us look at oxygen. In most problems we do not measure oxygen at all. So how do we get its percentage? By difference, yes, exactly! Per cent oxygen is one hundred, minus the sum of the other percentages. So every other element must be estimated first. Okay? Oxygen can also be estimated directly, by converting it to carbon monoxide, C O, which liberates iodine from I two O five. Five C O plus I two O five gives I two plus five C O two; the arrow means gives. Clear?

4:25Example 1: % chlorine from 0.574 g of AgCl in Carius method

Now Example one, our puzzle from the start. Zero point four zero gram of a compound gives zero point five seven four gram of silver chloride in Carius' method; the small g means gram. Find the percentage of chlorine. Pause the video and try it first. Done? Okay, let's check. Use the pattern: thirty five point five over one hundred and forty three point five, times zero point five seven four over zero point four zero, times one hundred. Okay? Here is a quick way to do the arithmetic. Zero point five seven four over one hundred and forty three point five gives zero point zero zero four mole of silver chloride. One chlorine in each, so chlorine is zero point zero zero four times thirty five point five, which gives zero point one four two gram. Right? So per cent chlorine is zero point one four two over zero point four zero, times one hundred, which gives thirty five point five per cent. And that solves our puzzle: most of the precipitate's mass is silver, which came from the silver nitrate, see?

5:28Example 2: % H by difference and empirical formula C3H7Br

Now Example two, a harder one. The compound has only carbon, hydrogen and bromine. Zero point two five gram gives zero point three eight two gram of silver bromide, and zero point five zero gram gives zero point five three seven gram of carbon dioxide. Find the percentage of hydrogen, and the empirical formula. Pause the video and try it first. Done? Let me test you first. For bromine, which mass is w: zero point two five, or zero point five zero? It is zero point two five, because the silver bromide came from that sample; using one mass for both is a common trap. So per cent bromine is eighty over one hundred and eighty eight, times zero point three eight two over zero point two five, times one hundred, which gives sixty five point zero per cent. Okay? The carbon dioxide came from the zero point five zero gram sample, and twelve over forty four is carbon's part of carbon dioxide, from Liebig's method. So per cent carbon is twelve over forty four, times zero point five three seven over zero point five zero, times one hundred, which gives twenty nine point three per cent. Clear? But look, there is no hydrogen data at all. So how do we get the hydrogen? By difference, just like oxygen, yes! The compound has only carbon, hydrogen and bromine, so per cent hydrogen is one hundred, minus the sum of sixty five point zero and twenty nine point three, which is ninety four point three. That leaves five point seven per cent. Right? Now let us find the empirical formula. Divide each percentage by its atomic mass to get moles in one hundred gram. Carbon gives two point four four, hydrogen gives five point seven, and bromine, sixty five over eighty, gives zero point eight one. With me? Divide each by the smallest, zero point eight one, and we get about three, about seven, and one; the colons mean a ratio. The double arrow means this gives the empirical formula, C three H seven B r, okay?

7:35Quick revision: AgX, BaSO4, Mg2P2O7, oxygen by difference

So, quick revision. For halogens, weigh the silver halide from the Carius tube, and use the fractions on screen. Sulphur is weighed as barium sulphate, thirty two over two hundred and thirty three, and phosphorus as M g two P two O seven, sixty two over two hundred and twenty two; always count the atoms. And oxygen is usually found by difference, as one hundred minus the sum of the others. Now go and try the practice problems on this page. You will do well. All the best!