Mixed Problems: Set 3 (Estimation and Formulae)

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Zero point three five grams of an organic compound gives zero point five seven four grams of silver chloride. Look at that: the precipitate is heavier than the compound itself! So, did the chlorine somehow grow? Hold that thought; we will settle it in Example one. This is Set three of the mixed problems: estimation and formulae. These quick numericals come up often in JEE Main, and one simple pattern solves most of them. So, let's begin.

0:36One pattern for C, H, halogens, S and P

So, here is the one pattern. Per cent of an element equals the mass of that element in one mole of the precipitate, over the molar mass of the precipitate, times m p p t over w, times one hundred. The first sign is the per cent sign, the long lines are fraction bars, and the crosses mean times. Here p p t is short for precipitate: small m, with p p t written low, is its mass, and w is the mass of the compound. Okay? Combustion works the same way. C O two, with the two written low, is carbon dioxide: twelve grams of carbon sit in forty four grams of it, so the fraction is twelve over forty four. Now hydrogen, from water, which is eighteen grams per mole; hydrogen's atomic mass is one. So hydrogen is one over eighteen of the water, right? No, no, careful, that is the trap! Each water molecule, H two O, has two hydrogen atoms, so hydrogen is two over eighteen of the water. Always count the atoms of the element in one formula. Clear? Now the Carius method, for halogens and sulphur. Chlorine is thirty five point five over one hundred and forty three point five of the silver chloride, A g C l. Bromine is eighty over one hundred and eighty eight of the silver bromide, A g B r. And sulphur is thirty two over two hundred and thirty three of the barium sulphate, B a S O four. Got it? And phosphorus is weighed as magnesium pyrophosphate, M g two P two O seven. The atomic mass of phosphorus is thirty one, but each formula has two phosphorus atoms. So the fraction is sixty two over two hundred and twenty two, the same atom counting again. With me?

2:19Nitrogen: Dumas and Kjeldahl

Now nitrogen, which has two methods. In Dumas' method, nitrogen gas, N two, is collected over an aqueous solution, so it is moist: it carries water vapour. So its own pressure, small p, is the atmospheric pressure, p with the letters a t m written low, minus the aqueous tension. Right? Then we change the volume to STP, standard temperature and pressure. V naught, the V with a small zero, equals capital V, the volume of gas we collected, times p over seven hundred and sixty, times two hundred and seventy three over capital T, the temperature in kelvin. Here the dots also mean times. Okay? And the percentage of nitrogen equals twenty eight over twenty two thousand four hundred, times V naught in millilitres over w, times one hundred. Twenty eight grams of nitrogen gas fill that many millilitres at STP, and m L means millilitres. Clear? Kjeldahl's method turns the nitrogen into ammonia, N H three, caught in acid. Per cent of nitrogen equals one point four, times the milliequivalents of acid used by the ammonia, over w in grams. Here m e q is short for milliequivalents, and g is grams. Got it? Now a quick question. The acid is often sulphuric acid, H two S O four. Are its milliequivalents the same as its millimoles? No, they are not, and many students slip here! Sulphuric acid has two acidic hydrogens, so milliequivalents are two times the millimoles, m m o l on screen. And remember, Kjeldahl's method does not work for nitrogen in nitro and azo groups, or in a ring. With me?

4:00Formulae: from percentages to the molecule

Now the formulae. For the empirical formula, take one hundred grams, so each percentage becomes a mass. Then divide each mass by its atomic mass to get the moles. Right? Next, divide all of them by the smallest. And careful here: one point three three, or one point five, must never be rounded to one, because that changes the formula. Okay? For the molecular formula, we need the molar mass. Capital M, the molar mass, is two times V D, the vapour density. Then small n equals M over the empirical formula mass. Clear? And the molecular formula is the empirical formula in brackets, with n written low outside: n units of it. Now we have all our tools. Let's go!

4:47Example 1: percentage of chlorine from 0.574 g AgCl

Now Example one, an easy one from Set three: our puzzle from the start. Zero point three five grams of a compound gives zero point five seven four grams of silver chloride. Find the percentage of chlorine. Pause the video and try it first. Done? Okay, let's check. Use the pattern: chlorine is thirty five point five over one hundred and forty three point five of the silver chloride. So the percentage of chlorine is that fraction, times zero point five seven four over zero point three five, times one hundred. Right? First, the mass of chlorine. Thirty five point five over one hundred and forty three point five, times zero point five seven four, gives zero point one four two grams. The rest of the precipitate is silver, which is why it weighs more than the compound. So the puzzle is solved, and the chlorine never grew at all. Clear? Now zero point one four two over zero point three five, times one hundred, gives forty point six per cent. And forty point six is less than one hundred, so the answer makes sense. With me?

5:54Example 2: Dumas method, percentage N from moist N2 gas

Now Example two, a step harder: a medium one from Set three. Zero point one five grams of a compound gives twenty five millilitres of moist nitrogen gas, at two hundred and ninety kelvin and seven hundred and sixty millimetres of mercury. The aqueous tension is twenty millimetres. Find the percentage of nitrogen. On screen, capital K is kelvin, and m m is millimetres. Pause the video and try it first. Done? Okay, let's check. The pressure given is seven hundred and sixty millimetres. So we put seven hundred and sixty straight into the formula, right? Oh no, careful, the gas is moist! Part of that pressure comes from water vapour. So the dry nitrogen has seven hundred and sixty minus twenty, which is seven hundred and forty millimetres. Okay? Now change to STP, standard temperature and pressure. Twenty five, times seven hundred and forty over seven hundred and sixty, times two hundred and seventy three over two hundred and ninety, gives twenty two point nine millilitres. Right? Next, the mass of nitrogen, with a topper's trick. Twenty eight over twenty two thousand four hundred is just one over eight hundred. So twenty two point nine over eight hundred gives zero point zero two eight six grams. Clear? And the percentage: zero point zero two eight six over zero point one five, times one hundred, gives nineteen point one per cent. Got it? And if you forgot the aqueous tension, you would get nineteen point six per cent, the trap answer. So in Dumas' method, always subtract the aqueous tension first. With me?

7:33Quick revision: atom counts, aqueous tension, Kjeldahl meq

So, quick revision. First, count the atoms: two over eighteen for hydrogen, and sixty two over two hundred and twenty two for phosphorus. In Dumas' method, subtract the aqueous tension before you change to STP, okay? In Kjeldahl's method with sulphuric acid, milliequivalents are two times millimoles. And for formulae, divide by the atomic masses, then by the smallest, and find n from the molar mass. Now go and try the practice problems on this page. All the best!