Formulae from Combustion Data

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You take zero point three zero grams of an unknown compound, and you burn it completely. Out come zero point four four grams of carbon dioxide, and zero point one eight grams of water. Just three readings on a balance. Now tell me: from these three numbers, can you find the formula of the compound? Sounds impossible, right? Burn something, and then name it? Look, combustion questions are a JEE favourite: mole concept and formulae, all in one question. So, let's begin.

0:39Burn it: all C goes to CO₂, all H to H₂O

So, first the big idea. When a compound of carbon and hydrogen burns completely, maybe with some oxygen or nitrogen in it too, where does all its carbon go? Yes, into carbon dioxide! And all its hydrogen goes into water. The little arrow on screen means goes to. Clear? Now think. One mole of carbon dioxide, C O two, weighs forty four grams. How many grams of that is carbon? Twelve grams, exactly! One carbon atom, twelve, out of forty four. So, the mass of carbon equals twelve by forty four, times the mass of carbon dioxide, okay? Now water, eighteen grams per mole. What fraction of it is hydrogen? One by eighteen, right? No, no, careful! This is the first trap. Water is H two O, with two hydrogen atoms in it, remember? So, two grams of hydrogen in every eighteen grams of water. The mass of hydrogen equals two by eighteen, times the mass of water. Write one by eighteen, and you get only half the hydrogen, and the whole formula goes wrong. Don't do that!

1:52Oxygen: always by difference

Now, carbon and hydrogen are done. But what if the compound also contains oxygen? Can we find it from the oxygen inside the carbon dioxide and the water? Tempting, right? But no. Look, the compound burns in oxygen gas, O two. That oxygen also ends up inside the carbon dioxide and the water. So part of it came from the gas, not from our compound, and we cannot separate the two, okay? So, we find oxygen by difference. Mass of oxygen equals the mass of the sample, minus the mass of carbon, minus the mass of hydrogen. But only when the compound has just carbon, hydrogen and oxygen, nothing else. Clear? And one quick check that toppers always do. If carbon plus hydrogen already add up to the whole sample, the difference is zero. No oxygen at all; it is a hydrocarbon. Remember this check. We will need it in example two.

2:50Hydrocarbons: read x and y from the equation

Now a special case, hydrocarbons: only carbon and hydrogen, written C x H y, okay? Here is the balanced equation: C x H y, plus x plus y by four O two, gives x C O two, plus y by two H two O, okay? Why x plus y by four? Each carbon dioxide takes one O two, and each water takes half of one. Simple balancing, perfectly logical. Now read the equation backwards. Burn one mole of the hydrocarbon. The moles of carbon dioxide you get are exactly x, the number of carbon atoms. And the moles of water per mole? That is y, right? Hold on, careful! Look at the equation again. The moles of water per mole are y by two, because each water carries two hydrogens. So, to get y, double the moles of water. The double arrow on screen means so, or implies. Same trap as before: that little two in H two O, got it?

3:56Topper's trick: go straight to moles

Now a trick that toppers use. For the formula we need moles, not grams, right? So why not go straight to moles? Every carbon dioxide has one carbon, so the moles of carbon equal the moles of carbon dioxide: its mass divided by forty four. And every water has two hydrogens, so the moles of hydrogen equal two times the moles of water: two times its mass by eighteen. But if you must find oxygen by difference, you still need the grams of carbon and hydrogen, okay? Then, the steps you already know for any empirical formula. Take the ratio of moles, turn it into the simplest whole numbers, and that is the empirical formula. And if the molar mass is given, n equals the molar mass divided by the empirical formula mass. Multiply the empirical formula by n, and that's it, you have the molecular formula.

4:50Example 1: CH2O from 0.44 g CO2 and 0.18 g water

So, remember our puzzle? Let's solve it now. Zero point three zero grams of an organic compound gives zero point four four grams of carbon dioxide, and zero point one eight grams of water, on combustion. Find its empirical formula. Pause the video, and try it first. Done? Okay, let's check. Carbon first: twelve by forty four, times zero point four four. The forty four cancels nicely, and we get zero point one two grams of carbon. Hydrogen: two by eighteen, times zero point one eight. Zero point zero two grams of hydrogen. Two by eighteen, not one by eighteen, right? Carbon plus hydrogen is only zero point one four grams. But the sample was zero point three zero grams. So what is the rest? Yes, oxygen! By difference: zero point three zero, minus zero point one two, minus zero point zero two. Zero point one six grams of oxygen. Now grams to moles. Carbon, zero point one two by twelve: zero point zero one. Hydrogen, zero point zero two by one: zero point zero two. Oxygen, zero point one six by sixteen: zero point zero one. So the ratio is one is to two is to one, and the empirical formula is C H two O, okay? Three readings from a balance, and we found the formula. Puzzle solved!

6:24Example 2: molecular formula of a hydrocarbon, C3H4

Now a harder one, JEE pattern. One point zero gram of a hydrocarbon, on complete combustion, gives three point three grams of carbon dioxide and zero point nine grams of water. Its molar mass is about forty. Find the molecular formula. Pause and try, and careful with the ratio! Done? Okay, let's check. Carbon: twelve by forty four, times three point three, is zero point nine grams. In moles, zero point nine by twelve: zero point zero seven five moles. Hydrogen: two by eighteen, times zero point nine, is zero point one gram. Divided by one, that is zero point one mole. The question says hydrocarbon, but let's do the topper's check anyway. Is there any oxygen? Zero point nine plus zero point one: one point zero gram, the whole sample! So no oxygen. Confirmed, then: a pure hydrocarbon. Now the ratio. Divide both by the smaller one, zero point zero seven five. Carbon becomes one, and hydrogen becomes one point three three. So just round it: one is to one, formula C H, right? No, no, careful! This is the JEE trap. One point three three is not one. When a ratio ends in point three three, multiply everything by three. One is to one point three three becomes three is to four. So the empirical formula is C three H four, and its mass is three twelves plus four: forty. The molar mass is about forty, so n is forty by forty, that is one. The molecular formula is C three H four itself: propyne, or propadiene. See how fast that was?

8:16Quick revision: C from CO2, H from water, O by difference

So, quick revision. Carbon is twelve by forty four of the carbon dioxide; hydrogen, two by eighteen of the water. Oxygen only by difference, and if carbon plus hydrogen make the whole sample, there is no oxygen. For a hydrocarbon, x is the moles of carbon dioxide, and y is two times the moles of water, per mole burnt. Now go and try the practice problems on this page. All the best!