Worked example: Determining an empirical formula from combustion data | AP Chemistry | Khan Academy

We are told that a sample of a compound containing only carbon and hydrogen atoms is completely combusted, producing 5.65 grams of carbon dioxide and 3.47 grams of H2O, or water. What is the empirical formula of the compound? So pause this video and see if you can work through that all right.

Now let's just try to make sure we understand what's going on. They say that I have some mystery compound; it only contains carbon and hydrogen. So it's going to have some number of carbon, I'll call that x, and some number of hydrogens, I'll call that y. We're going to put it in the presence of molecular oxygen, and it's going to combust. After it's combusted, I'm going to end up with some carbon dioxide and some water. What I just drew here, this is a chemical reaction that I'm describing. I haven't balanced it; I could try to even with the x's and y's, but that's not the point of this video.

The point of this video is they tell us how many grams of the carbon dioxide we have and how many grams of the water we have. They tell us that right over there. And so what we need to do is say, “Alright, from that we can figure out how many moles of carbon dioxide we have, how many moles of water we have, and from that, we can figure out how many moles of carbon did we start with and how many moles of hydrogen did we start with.” If we look at those ratios, then we can come up with the empirical formula of the compound.

So just to start, because I'm going to be thinking about moles and molar masses, let’s just get the average atomic mass for carbon, hydrogen, and oxygen for us to work with. I’ll get out our handy periodic table. We can see hydrogen has an average atomic mass of 1.008. Let me write that down: hydrogen is at 1.008. Then we have carbon, and carbon is at 12.01, so carbon is at 12.01. We could also think about them in terms of molar masses; we could say this is grams per mole.

Last but not least, we have oxygen, and oxygen is at 16.00 grams per mole. It's the average atomic mass, but then we can think of that as molar mass; that number is molar mass. So oxygen is at 16.00 grams per mole. Now we can try to figure out how many moles of C in the product do we have. We can see that all the carbon in the product is in the carbon dioxide that's in the product. So we have 5.65 grams of CO2.

We can think about how many moles of CO2 that is. So times one mole of CO2 for every how many grams of CO2? Well, we just have to think about, actually, let me just put it right over here: CO2. You’re going to have, let’s see, you have one carbon and two oxygens, so it’s going to be 12.01 plus 2 times 16. 2 times 16.00 grams per mole. So let’s see, this would get us to... this is 32 plus 12.01, so that is 44.01, and that’s grams per mole.

But now we’re thinking about moles per gram, so it’s going to be 1 over 44.01. If we did just this, the grams of CO2 would cancel with the grams of CO2, and this would give us moles of CO2. But I care about moles of carbon in the product. So how many moles of carbon are there for every mole of CO2? Well, we know that we have one mole of carbon for every one mole of CO2. Every carbon dioxide molecule has one carbon in it.

What is this going to get us? So we have 5.65 divided by 44.01 and then times one. So I don’t have to do anything there. That is equal to... I'll round it to three digits here, so 0.128. So this is 0.128, and my units here are... let’s see, the grams of carbon dioxide cancel with the grams of carbon dioxide, the moles of carbon dioxide cancel with the moles of carbon dioxide, so I am exactly where I want to be. This is how many moles of carbon that I have.

You can do the dimensional analysis, but it also makes intuitive sense. Hopefully, if this is how many grams of carbon dioxide we have, and a mole of carbon dioxide is going to have a mass of 44.01 grams, well then 5.65 over this is going to tell us what fraction of a mole we have of carbon dioxide. Whatever that number of moles we have of carbon dioxide is going to be the same as the moles of carbon because we have one atom of carbon for every carbon dioxide molecule.

So that all makes sense, and now let's do the same for hydrogen. Let's think about moles of hydrogen in the product, and it’s going to be the same exercise. If you’re so inspired— and if you didn’t calculate in the beginning— I encourage you to try to do this part on your own.

Alright, so all of the hydrogen is in the water. We know that we have in our product 3.47 grams of water, 3.47 grams of H2O. Now let’s think about how many moles of H2O that is. So that’s going to be... let’s see, every one mole of H2O is going to have a mass of how many grams of H2O? We could do that up here: H2O. It’s going to be... we have two hydrogens, so it’s going to be 2 times 1.008 plus the mass; the average atomic mass of the oxygen is going to be plus 16, but we can also view that as what would be the mass in grams if you had a mole of it.

This is going to be in grams per mole. Let’s see, 2 times 1.008, this part over here is 2.016, and then you add 16 to it; it’s going to be 18.016 grams per mole. If I just calculated this, this would give me how many moles of water I have in my product, but I care about moles of hydrogen. So how many moles of hydrogen do I have for every mole of water? For how many moles of hydrogen for every mole of water? Well, I’m going to have two moles of hydrogen for every mole of water because, in each water molecule, I have two hydrogens.

That’s going to cancel out with that, and we’re just going to be left with moles of hydrogen. This is going to be equal to... I’ll take my 3.47 grams of water, divide it by how many grams a mole of water, what its mass would be, so divided by 18.016. This is how many moles of water I have, and now for every molecule of water I have two hydrogens. Then I will multiply by 2, times 2, which is equal to... I’ll just round three digits right over here: 0.385. So now we know the number of hydrogen atoms; we know the number of carbon atoms.

To figure out the empirical formula of the compound, we can think about the ratio between the two. I’m going to find the ratio of hydrogens to carbons, and that is going to be equal to... I have 0.385 moles of hydrogen over 0.128 moles of carbon. What is this equal to? It looks like it's going to be roughly 3, but let me verify that in my head. That seems so... I already have the hydrogen there, and so if I divide it by 0.128, I get... yep, pretty close to 3.

Now actually, I think if I... yep, pretty close to 3. So there you go; this is approximately three. I can say with pretty good confidence this was very close to three. I can say for every carbon I have three hydrogens in my original compound, and this thing right over here—the empirical formula of our original compound—for every one carbon, I have three hydrogens. So CH3, and we are done.