Molar Mass: Vapour Density, Silver Salts and Chloroplatinates

Transcript

0:00Introduction

Hello everyone! Let us start with a strange idea. You have an unknown organic acid, and you want its molar mass. So chemists turn it into its silver salt, heat the salt strongly, and weigh the silver metal left behind. Just a pinch of silver, and from it, the molar mass of the acid! How can weighing silver tell us anything about the acid? Keep that question in mind; Example one will answer it. Today we learn three ways to find a molar mass: vapour density, silver salts, and chloroplatinates. These can appear in JEE, usually inside bigger problems on organic analysis. So, let's begin.

0:41Why we need the molar mass

So, why do we need the molar mass at all? The empirical formula gives only the simplest ratio of atoms. The molecular formula is the empirical formula, held in brackets, taken n times; the small n written low is a subscript. Okay? Now let me ask you this. To find n, what must we measure for the compound? The molar mass, yes, exactly! n equals capital M, the molar mass, over the empirical formula mass; the long line is a fraction bar, read as over. Right? For example, CHO two has empirical formula mass forty five. If the molar mass is ninety, then n is two; the double arrow means so. Double every subscript, and we get C two H two O four, oxalic acid. With me?

1:31Vapour density and Victor Meyer's method

Now, the first method, for compounds that vaporise easily. Vapour density, written V D, compares the vapour with hydrogen gas, H two, whose molar mass is two, okay? So if the vapour density is thirty, is the molar mass also thirty? No, no, careful, that is the trap! M equals two times V D, so thirty gives sixty grams per mole. The g mol with a small minus one written up high is read as grams per mole. Clear? In Victor Meyer's method, a weighed volatile liquid is vaporised, and its vapour displaces its own volume of air, which we measure. Then M equals w times R times T, over p times V, where w is the mass taken, R the gas constant, T the temperature, p the pressure, and V that volume of air. Got it? Now, one small detail. If the air is collected over water, first subtract the aqueous tension, the pressure of the water vapour, from p, okay?

2:30Silver salt method: for organic acids

Now, organic acids, the method from our puzzle. Here n is the basicity of the acid, and its silver salt is RCOOA g, in brackets, taken n times. Capital R is the rest of the molecule, and A g is silver. Ignite the salt, that is, heat it strongly, and only silver is left. Right? One mole of salt gives n moles of silver, of atomic mass one hundred and eight. So M salt equals w salt times one hundred and eight times n, over w A g, the mass of silver; the small words written low are subscripts. With me? But we want the acid. The salt has n silver atoms of mass one hundred and eight, so the acid is M salt minus one hundred and eight times n, right? Not quite, and this is the trap. Each COOA g was a COOH group, so silver, one hundred and eight, took the place of hydrogen, one; the brackets hold these atomic masses. So each group changes by one hundred and eight minus one, which is one hundred and seven. Clear? So M acid equals M salt, minus one hundred and seven times n. One hundred and seven, not one hundred and eight. Okay?

3:50Chloroplatinate method: for organic bases

Now, organic bases. Their chloroplatinate salt is B two, H two P t C l six, in brackets, taken n times. Capital B is the base, n its acidity, and P t is platinum, atomic mass one hundred and ninety five. Ignite it, and only platinum is left. Got it? It is the same pattern as silver. M salt equals w salt times one hundred and ninety five times n, over w P t, the mass of platinum left. Right? To reach the base, we remove the H two P t C l six part. Two for hydrogen, plus one hundred and ninety five, plus six times thirty five point five for chlorine, gives four hundred and ten. With me? So we subtract four hundred and ten times n from M salt. Is what remains the molar mass of the base? No, careful, the salt has B two, two molecules of the base. So we must divide by two: M base equals M salt minus four hundred and ten times n, all over two. Clear?

4:59Example 1: silver salt of monobasic acid gives benzoic acid

Now, Example one, our puzzle from the start. Zero point four five eight grams of the silver salt of a monobasic acid leaves zero point two one six grams of silver on ignition; the small g means grams. Find the molar mass of the acid. Pause the video and try it first. Done? Okay, let's check. Monobasic means basicity one, so n equals one. Right? For the salt, zero point four five eight times one hundred and eight, over zero point two one six. Here is a quick trick. One hundred and eight over zero point two one six is exactly five hundred. So the salt is zero point four five eight times five hundred, which gives two hundred and twenty nine. Okay? Now, to get the acid, do we subtract one hundred and eight, or one hundred and seven? One hundred and seven, yes, exactly! Two hundred and twenty nine minus one hundred and seven gives one hundred and twenty two, which is benzoic acid. So that is our puzzle solved: a pinch of silver gave us the molar mass. See?

6:13Example 2: chloroplatinate of monoacidic base gives aniline

Now a harder one, Example two. Zero point five nine six grams of the chloroplatinate of a monoacidic base leaves zero point one nine five grams of platinum. Find the molar mass of the base. Pause the video and try it first. Done? Okay, let's check. Monoacidic means acidity one, so n equals one. Clear? For the salt, zero point five nine six times one hundred and ninety five, over zero point one nine five. One hundred and ninety five over zero point one nine five is one thousand, so the salt is zero point five nine six times one thousand, which is five hundred and ninety six. With me? Subtract four hundred and ten for the H two P t C l six, and five hundred and ninety six minus four hundred and ten gives one hundred and eighty six. So the base is one hundred and eighty six, right? No, no, wait, that is the trap! The salt has B two, so one hundred and eighty six is two base molecules. One hundred and eighty six over two gives ninety three, and that is aniline. Got it?

7:29Quick revision: VD, silver salt and chloroplatinate formulas

So, quick revision. For a vapour, M equals two times V D, or use Victor Meyer's method. For acids, subtract one hundred and seven times n from the silver salt. For bases, subtract four hundred and ten times n from the chloroplatinate, then divide by two. Then n equals M over the empirical formula mass, and that gives the molecular formula. Now go and try the practice problems on this page. You will do well. All the best!