Empirical and Molecular Formulae

Transcript

0:00Introduction

Hello everyone! So, one small puzzle before we start. Analysis of a compound gives forty percent carbon, six point six seven percent hydrogen, and the rest oxygen. One student writes its formula as C H two O, okay? Another student writes C two H four O two. Now here is the funny part: both formulas give exactly the same percentages! Forty percent carbon in both. So, who is right? Hmm, the percentages alone cannot decide, right? In this video we will find the one extra clue that settles it. And look, getting a formula from percentage composition comes in JEE again and again. So, let's begin.

0:46Empirical formula vs molecular formula

So, there are two kinds of formula, and this whole video is about the difference between them. First, the empirical formula. It gives the simplest whole number ratio of atoms in a compound, okay? Only the ratio, nothing about how big the molecule is. Second, the molecular formula. It gives the actual number of atoms in one molecule. The real count, not just the ratio. Clear? Now think. If the empirical formula is only the simplest ratio, how is it connected to the molecular formula? Yes, exactly! The molecular formula is just n times the empirical formula, okay? Every subscript in the empirical formula gets multiplied by the same number n. And how do we find n? n equals the molar mass, divided by the empirical formula mass. So, to go from the empirical formula to the molecular formula, you need one extra piece of information: the molar mass. Got it? Now look at our puzzle again: C H two O, and C two H four O two. Carbon to hydrogen to oxygen is one is to two is to one in both, so the percentages are the same. Percentages give you only the ratio, that is, only the empirical formula. To choose between the two, we need the molar mass. Hold that thought; we will come back to it.

2:04Four steps: from percentages to the formula

Alright, now the method. Given the percentage composition, how do we get the empirical formula? Step one: we take one hundred grams of the compound. Why exactly one hundred? Because then the percentages simply become grams! Forty percent carbon becomes forty grams of carbon. No calculation at all. Really easy. Step two: divide each mass by its atomic mass. That gives the number of moles of each element, okay? The formula counts atoms, so we compare moles, not grams. Step three: divide all the mole numbers by the smallest one. Now the smallest becomes one, and the others show the ratio directly. And step four is the topper's trick. Learn this little table by heart. If a ratio ends in point five, multiply everything by two. Point three three or point six seven, multiply by three. Point two five or point seven five, multiply by four. Now all the numbers are whole numbers. And there you go, that is your empirical formula.

3:08JEE trap: do not round too early

Now, the commonest mistake in this topic. Suppose after step three your ratio comes out as one is to one point five, okay? One point five is close to two, so just round it and write one is to two, right? No, no, careful! This is the JEE trap. Look, the rule is: round only when the ratio is within about zero point one of a whole number. One point five is a full half away from two. Round it, and you get a completely different formula! So, it ends in point five? Multiply both numbers by two, that is the times two written on the arrow. One is to one point five becomes two is to three. One point five is not two, okay? Same story with one point three three. Some students round it down to one. Wrong again! Point three three means multiply by three, the times three on the arrow, so one is to one point three three becomes three is to four. Rounding too early is the commonest mistake here. Don't do it! With me?

4:12Vapour density gives the molar mass

Now, the molar mass. Sometimes the question gives it to you directly. But often it comes hidden as the vapour density. Vapour density is the density of the gas, divided by the density of hydrogen gas, H two, at the same temperature and pressure, the T and P you see on screen, okay? And this ratio works out to M by two, where M is the molar mass of the gas. But why divide by two? Yes, because two is the molar mass of H two! So we are just comparing M with two. Clear? So, remember it in one line: M equals two times V D, the short form of vapour density. Not equal to the vapour density. Two times! Write this down; we will need it very soon.

4:58Example 1: CH2Cl empirical formula to C2H4Cl2

Now, first example. Read it with me: a compound contains four point zero seven percent hydrogen, twenty four point two seven percent carbon, and seventy one point six five percent chlorine. Its molar mass is ninety eight point nine six grams per mole. Find its empirical formula and its molecular formula. Use hydrogen one, carbon twelve, and chlorine thirty five point five. Pause the video, and try it first. Done? Okay, let's check. Take one hundred grams, so the percentages become grams. Hydrogen: four point zero seven divided by one, that is four point zero seven moles. Carbon: twenty four point two seven divided by twelve, about two point zero two. Chlorine: seventy one point six five divided by thirty five point five, also about two point zero two. Now divide everything by the smallest, two point zero two. Hydrogen gives about two, carbon one, and chlorine one. All three are within zero point one of whole numbers, so here rounding is safe. Two is to one is to one, and the empirical formula is C H two C L, okay? Its mass: carbon twelve, plus two hydrogens at one each, plus chlorine thirty five point five. Forty nine point five. So, n equals ninety eight point nine six divided by forty nine point five. That is almost exactly two. Multiply every subscript by two: C two H four C L two. Nice and clean.

6:37Example 2: vapour density turns CH2O into C2H4O2

Now the JEE pattern question, and it is our puzzle from the start. An organic compound contains forty percent carbon, six point six seven percent hydrogen, and the rest oxygen. Its vapour density is thirty. Find its molecular formula. Pause the video, and try it. Done? Okay, let's check. First, the oxygen. The rest means one hundred minus forty minus six point six seven, that is fifty three point three three percent. Now the moles. Carbon: forty by twelve, three point three three. Hydrogen: six point six seven by one, six point six seven. Oxygen: fifty three point three three by sixteen, three point three three. Divide by the smallest, three point three three: one, two, one. So the empirical formula is C H two O, and its mass is twelve plus two plus sixteen, thirty. Now the molar mass. The vapour density is thirty, so the molar mass is thirty, and n is one. Final answer C H two O, right? No, no, careful! The trap again. M equals two times the vapour density. Two times thirty, that is sixty. So, n equals sixty by thirty, that is two. Molecular formula: C two H four O two, for example, acetic acid. And there, our puzzle is solved! The percentages fixed only the ratio, C H two O, and the vapour density decided the real molecule. Got it?

8:18Quick revision: four steps, rounding trap, M = 2 x VD

So, quick revision. Percent to grams, grams to moles, then divide by the smallest. Never round one point five or one point three three: multiply by two or by three. And n is the molar mass divided by the empirical formula mass, where the molar mass is two times the vapour density. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!