Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Here is a small challenge. Zero point one mole of a hydrocarbon is burnt, and out come seventeen point six grams of carbon dioxide and nine grams of water. From just these few numbers, can you write its formula? Hmm, it looks like too little information, right? By the end, you will do it in a minute. This is Part A of the unit problem bank: twelve quick JEE Main level questions, which the course suggests trying as a thirty minute test. We will build four quick tools, then solve two problems together. So, let's begin.

0:38Tool 1: choose the method by the property

So, tool one: choose the method by the property. Some solids turn straight into vapour on heating, without melting, and the arrow on screen means turns into. That is sublimation. Now, question one of this set: iodine is mixed with sand. Which method separates them? Sublimation, yes, exactly, because iodine sublimes and the sand stays behind. The double arrow on screen means therefore. Okay? Next, some liquids decompose at their normal boiling point. Lowering the outside pressure lowers the boiling point, so we use distillation under reduced pressure. Question two: glycerol, acetone or ethanol, which one is purified this way? Glycerol is the right answer, because it decomposes at its normal boiling point. Right?

1:31Tool 2: Rf values and polarity

Now tool two, chromatography. On screen is a developed TLC plate, with the base line at the bottom and the dashed solvent front near the top. R f, a capital R with a small f below it, is the spot's distance over the solvent front's distance, both from the base line. Question three: the front moves six centimetres, and a spot moves four point two. What is R f? Four point two over six gives zero point seven. It has no units, and it never exceeds one, because no spot overtakes the solvent front. Clear? Silica is polar, so it holds a polar compound strongly, and it moves less, with a low R f, like spot B on the plate. Question twelve: on a silica column run with hexane, which comes out last: benzene, chlorobenzene or phenol? Phenol is right, because its O H group makes it the most polar of the three, so silica holds it longest and it is eluted last. Benzene comes out first, then chlorobenzene. With me?

2:37Tool 3: nitrogen, from Lassaigne to Kjeldahl

Now tool three is about nitrogen. The chart shows the tests on the sodium fusion extract; we follow the nitrogen branch on the left. Fused with sodium, the compound's carbon and nitrogen give N a C N, sodium cyanide. Question four: which ion in this extract shows nitrogen? It is the cyanide ion, C N minus, where the small raised minus sign is its charge. Sodium cyanide needs carbon as well as nitrogen, so the test works only for compounds that contain carbon. And if sulphur is also present, too little sodium gives thiocyanate instead, which gives a blood red colour rather than Prussian blue. Okay? Boiled with iron two sulphate, six cyanide ions join one iron two plus ion. The square brackets hold the complex, F e, then C N in round brackets six times, with a raised four minus charge: hexacyanoferrate two. Iron three and acid then give Prussian blue, at the bottom of the branch. Got it? Now Kjeldahl's method measures the nitrogen. Concentrated sulphuric acid turns the nitrogen into ammonium sulphate: N H four in round brackets, twice, then S O four. Sodium hydroxide then sets free ammonia, N H three. Question eleven: why is this method not used for nitrobenzene? Because nitro nitrogen is not converted into ammonium sulphate, so the method misses it. The same holds for azo and ring nitrogen, as in pyridine, while the Dumas method works for all.

4:13Tool 4: percentage of an element

Right? Tool four: the percentage of an element. In the Carius method, a halogen ends up as a silver halide precipitate, written p p t for short. Then the per cent of the element is its mass in one mole of precipitate, over the molar mass of the precipitate, times m p p t, the mass of precipitate, over w, the mass of the sample, times one hundred. Okay? Question five says that zero point five grams of a compound gives zero point seven one seven five grams of silver chloride. Its molar mass is one hundred and forty three point five, and thirty five point five of that is chlorine. Can you finish it? Zero point seven one seven five over one hundred and forty three point five gives zero point zero zero five moles of silver chloride, A g C l. So chlorine is zero point zero zero five times thirty five point five, which is zero point one seven seven five grams. Over zero point five, times one hundred, that is thirty five point five per cent. Right?

5:21Example 1: Kjeldahl percent N from 25 mL of 1 M HCl

Alright, example one, an easy one, question six of this set. The ammonia from zero point seven grams of a compound neutralises twenty five millilitres of one molar hydrochloric acid, H C l, where the capital M means molar. Find the percentage of nitrogen. Pause the video and try it first. Done? Okay, let's check. Kjeldahl's formula says per cent nitrogen equals one point four, times the milliequivalents, m e q, of acid used by the ammonia, over w, the mass of the sample in grams. The one point four comes from fourteen milligrams of nitrogen per milliequivalent. Okay? Now, twenty five millilitres of one molar acid is twenty five millimoles. So the milliequivalents are double that, fifty, right? No, no, careful, that doubling is only for sulphuric acid, which gives two H plus ions. Hydrochloric acid gives one H plus, the raised plus being its charge, so the milliequivalents equal the millimoles, which is twenty five. Clear? So, one point four times twenty five gives thirty five, and thirty five over zero point seven gives fifty per cent. With the doubling you would get one hundred per cent, which is impossible, right?

6:39Example 2: hydrocarbon formula from CO2 and water masses

Now, remember our challenge from the start? Let's solve it, with question ten of this set. Zero point one mole of a hydrocarbon gives seventeen point six grams of C O two, carbon dioxide, and nine grams of water. Find its formula. Pause the video and try it first. Done? Okay, let's check. Call the hydrocarbon C x H y, where the small x and y below the letters count its carbon and hydrogen atoms. Each carbon ends up in one C O two, hence the x in front, and every two hydrogens make one water, H two O, which gives y over two. The oxygen used, x plus y over four, is not needed here. With me? For the moles, seventeen point six over forty four gives zero point four moles of carbon dioxide, and nine over eighteen gives zero point five moles of water. Okay? Let us do carbon first. The moles of carbon dioxide are zero point one times x, and that equals zero point four, so x equals four. Got it? Now for hydrogen, the water, zero point five moles, is five times the hydrocarbon. So y equals five, right? Oh no, careful, each water molecule carries two hydrogen atoms! So zero point one times y over two equals zero point five, which makes y over two equal to five, and y equals ten. Clear? So the formula is C four H ten. Challenge solved with just two masses, and the moles did the rest. Really simple, isn't it?

8:16Quick revision: method choice, Rf, nitrogen tests, formulas

So, quick revision. First, choose the method by the property; polar compounds have low R f and come out last. Second, nitrogen shows up as cyanide, and Kjeldahl's method misses nitro, azo and ring nitrogen. Third, Kjeldahl's per cent nitrogen is one point four times m e q over w, and the moles of carbon dioxide and water give x and y. Now go and try the practice problems on this page, and time yourself. All the best!