Empirical Formula from Percentages

Transcript

0:00Introduction

Hello everyone! Let us begin with a strange result. Suppose a lab analyses a compound and reports only percentages of carbon, hydrogen and oxygen. You do the working, and the atoms come out in the ratio one, to one and a half, to one. One and a half hydrogen atoms? Can a molecule really have half an atom? Of course not, right? So what has gone wrong, and how do we fix it? That is today's topic: finding the empirical formula from percentages. It comes up often in JEE, usually inside a bigger question on organic analysis. So, let's begin.

0:40From percentages to a formula

So, we are given percentages, and we want a formula, which counts atoms. Step one: take one hundred grams of the compound. The small g on screen means grams. Now, why exactly one hundred grams? Why not ten, or fifty? Because then each percentage is simply a mass in grams, that's why! The little sign on screen is the per cent sign. So every percentage, read just as it is, becomes grams of that element, with no calculation at all. Right? Now, very often the question gives you carbon and hydrogen, and simply says the rest is oxygen. Then we find oxygen by difference: one hundred, minus the sum of the others. Never skip this, or oxygen disappears from your formula completely. Okay? Step two: change grams to moles. Moles equals mass over atomic mass; the line in the middle is a fraction bar. On screen, C is carbon, H is hydrogen and O is oxygen; their atomic masses are twelve, one and sixteen. Why moles? Because moles count atoms, and a formula counts atoms, not grams. Got it? But these mole values usually come out as messy decimals. So tell me, how do we turn them into a simple ratio of atoms? Step three: we divide every mole value by the smallest one, exactly! The smallest becomes one, and the others tell us how many times bigger they are. If all of them land near whole numbers, those are the subscripts, the small numbers written low after each symbol in the empirical formula. Clear?

2:21Step 4: not a whole number? Multiply, never round

Now step four, the step that decides your marks. After dividing, you may round a value only when it is within about zero point zero five of a whole number. Just a hair away from a whole number, then you may round. With me? But suppose one ratio comes out as one point five. It is far from one and far from two, so you cannot round it. So what do you do? You multiply every ratio by two, yes, exactly! One point five times two gives three, a whole number. On screen, the letter x stands for any whole number, the arrow means leads to, and the cross means multiply. So for anything ending in point five, multiply all by two. Okay? Ending in point three three, or point six seven? Those are thirds, so multiply all by three. One point three three times three gives about four, a whole number again. Right? And point two five, or point seven five? Those are quarters, so multiply all by four. Clear? Now the trap, and it is a big one. Never round one point three three, or one point five, down to one. That changes the formula, and the answer is simply wrong. And remember, when you multiply, multiply every ratio, not just the awkward one. Got it?

3:43Example 1: C 52.17%, H 13.04%, oxygen by difference to C2H6O

Now, Example one. Carbon is fifty two point one seven per cent, hydrogen is thirteen point zero four per cent, and the rest is oxygen. Find the empirical formula. Pause the video and try it first. Done? Okay, let's check. Oxygen is not given, so we find it by difference. One hundred, minus fifty two point one seven, minus thirteen point zero four, gives thirty four point seven nine per cent. In one hundred grams, these are just grams. Right? Now moles. For carbon, fifty two point one seven over twelve gives four point three five. For hydrogen, thirteen point zero four over one is still thirteen point zero four. Okay? And for oxygen, thirty four point seven nine over sixteen gives two point one seven. That is the smallest of the three, so it is our divisor. Clear? Now you make the next move. Divide each mole value by two point one seven. What ratio do you get? Carbon, four point three five over two point one seven, is very close to two. Hydrogen, thirteen point zero four over two point one seven, is very close to six. And oxygen is exactly one. Both are well within zero point zero five of a whole number, so here rounding is allowed. The colons in the ratio are read as to, so this is two to six to one. Right? So the empirical formula is C two H six O, with no subscript written when there is just one atom. And notice what saved us: finding oxygen first. Skip that step, and there is no oxygen in your formula at all. That is the trap in this question, okay?

5:30Example 2: ratio 1 to 1.5 to 1, multiply by 2 to get C2H3O2

Now, Example two, a little harder. Carbon is forty point six eight per cent, hydrogen is five point zero eight per cent, and oxygen is fifty four point two four per cent. This time all three are given. Find the empirical formula. Pause the video and try it first. Done? Okay, let's check. For carbon, forty point six eight over twelve gives three point three nine. For hydrogen, five point zero eight over one is five point zero eight. For oxygen, fifty four point two four over sixteen gives three point three nine again. Okay? The smallest is three point three nine, so we divide by it. Carbon gives one, oxygen gives one, and hydrogen, five point zero eight over three point three nine, gives one point five. There it is, our one and a half hydrogen atoms from the start, do you see? So, easy, we just round one point five down to one, and the formula is CHO, right? No, no, careful, that is exactly the trap! One point five is not near a whole number, so we never round it. It ends in point five, so we multiply all the ratios by two, not just hydrogen. Got it? Times two, carbon becomes two, hydrogen becomes three, and oxygen becomes two. So the ratio is two to three to two, and the double arrow on screen leads us to the empirical formula, C two H three O two. Half an atom was never the answer; it was just a signal to multiply by two, and that solves our mystery. Clear? And look at succinic acid, C four H six O four. Halve every subscript, and you get C two H three O two, exactly our empirical formula. So the empirical formula is the simplest ratio, and the real molecule can be a whole number multiple of it. With me?

7:28Quick revision: 100 g, moles, divide by smallest, multiply

So, quick revision. Take one hundred grams, find oxygen by difference if needed, and divide each mass by its atomic mass. Divide by the smallest, and round only when you are within about zero point zero five of a whole number. Otherwise multiply everything: by two for point five, by three for point three three or point six seven, and by four for point two five or point seven five. Now go and try the practice problems on this page. You will do well. All the best!