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# Worked example: Determining an empirical formula from combustion data

AP.Chem:
SPQ‑2 (EU)
,
SPQ‑2.A (LO)
,
SPQ‑2.A.3 (EK)

## Video transcript

- [Instructor] 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. 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 and after it's combusted, I'm going to end up with some carbon dioxide and some water. And 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 tells us that right over there. And so what we need to do is say, all right, 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? And 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 molar moles and molar masses and the mass of a mole of a molecule or an atom, let's just get the average atomic mass for carbon, hydrogen, and oxygen for us to work with. So 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. So we have hydrogen is at 1.008, and then we have carbon, and carbon is at 12.01. So carbon is at 12.01, and we could also think about them in terms of molar masses. We could say this is gram per mole, grams per mole, and then last, but not least, we have oxygen and then oxygen is at 16.00 grams per mole. It's the average atomic mass, but whether we can think of that as molar mass, that number is molar mass. So oxygen is at 16.00 grams per mole. And so now we can try to figure out how many moles of C in the product do we have? So we can see that all of the carbon in the product is in the carbon dioxide that's in the product, and 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 is going to be 12.01 plus two times 16. Two times 16.00 grams per mole, and 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 one over 44.01 and so if we did just this, the grams of CO2 would cancel 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, and so what is this going to get us? So we have 5.65 divided by, 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 the grams of carbon dioxide. The moles of carbon dioxide cancel the moles of carbon dioxide, so I am exactly where I want to be. This is how many moles of carbon that I have. And 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 how, what fraction of a mole we have of carbon dioxide and then whatever that number of moles we have of carbon dioxide is gonna be the same as the moles of carbon, 'cause 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. So let's think about moles of hydrogen in the product, and it's going to be the same exercise. And if you're so inspired and you didn't calculate it in the beginning, I encourage you to try to do this part on your own. All right, so all of the hydrogen is in the water, so and we know that we have, in our product, 3.47 grams of water. 3.47 grams of H2O, and 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? And we could do that up here. H2O, it's going to be, we have two hydrogens, so it's going to be two 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? And so this is going to be in grams per mole, and so this is going to be, let's see, two times 1.008. This part over here is 2.016, and then you add 16 to it. It's going to be 18.016. 18.016, and then 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. And so how many moles of hydrogen do I have for every mole of water? So 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. And so that's going to cancel out with that and we're just going to be left with moles of hydrogen. And so this is going to be equal to, I'll take my 3.47 grams of water, divide it by the, how much, 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. Now for every, for every molecule of water, I have two hydrogens, so then I will multiply by two. Times two is equal to zero point, I'll just round three digits right over here. 0.385. 0.385 moles of hydrogen. So now we know the number of hydrogen atoms. We know the number of carbon atoms, and to figure out the empirical formula of the compound, we can think about the ratio between the two, so I'm gonna find the ratio of hydrogens to carbons. And that is going to be equal to, I have 0.385 moles of hydrogen over, over 0.128 moles of carbon. 0.128 moles of carbon, and what is this equal to? And it looks like it's going to be roughly three, 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 three. Now actually, I think if I, yep, pretty close to three, so there you go. This is approximately three, and so I can, with pretty good confidence, this was very close to three. I could say for every carbon, I have three hydrogens in my original compound, in this thing right over here. So the empirical formula of our original compound for every one carbon, I have three hydrogens. So CH3, and we are done.
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