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- [Instructor] We are told that a sample of a compound
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containing only carbon and hydrogen atoms
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is completely combusted,
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producing 5.65 grams of carbon dioxide
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and 3.47 grams of H2O or water.
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What is the empirical formula of the compound?
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So pause this video
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and see if you can work through that.
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All right, now let's just try to make sure we understand
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what's going on.
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They say that I have some mystery compound.
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It only contains carbon and hydrogen,
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so it's going to have some number of carbon,
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I'll call that x.
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Some number of hydrogens, I'll call that y.
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We're going to put it in the presence of molecular oxygen
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and it's going to combust
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and after it's combusted,
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I'm going to end up with some carbon dioxide and some water.
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And what I just drew here,
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this is a chemical reaction that I'm describing.
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I haven't balanced it.
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I could try to, even with the x's and y's,
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but that's not the point of this video.
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The point of this video is they tell us how many grams
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of the carbon dioxide we have
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and how many grams of the water we have,
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they tells us that right over there.
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And so what we need to do is say, all right,
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from that, we can figure out how many moles
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of carbon dioxide we have,
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how many moles of water we have,
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and from that, we can figure out how many moles
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of carbon did we start with
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and how many moles of hydrogen did we start with?
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And if we look at those ratios,
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then we can come up with the empirical formula
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of the compound.
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So just to start, because I'm going to be thinking
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about molar moles and molar masses
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and the mass of a mole of a molecule or an atom,
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let's just get the average atomic mass
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for carbon, hydrogen, and oxygen for us to work with.
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So I'll get out our handy periodic table.
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We can see hydrogen has an average atomic mass of 1.008.
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Let me write that down.
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So we have hydrogen is at 1.008,
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and then we have carbon,
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and carbon is at 12.01.
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So carbon is at 12.01,
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and we could also think about them
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in terms of molar masses.
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We could say this is gram per mole,
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grams per mole,
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and then last, but not least, we have oxygen
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and then oxygen is at 16.00 grams per mole.
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It's the average atomic mass,
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but whether we can think of that as molar mass,
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that number is molar mass.
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So oxygen
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is at 16.00 grams per mole.
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And so now we can try to figure out
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how many moles of C in the product do we have?
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So we can see that all of the carbon in the product
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is in the carbon dioxide that's in the product,
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and so we have 5.65 grams of CO2.
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We can think about how many moles of CO2 that is,
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so times one mole
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of CO2 for every how many grams of CO2.
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Well, we just have to think about,
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actually, let me just put it right over here.
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CO2,
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you're going to have, let's see,
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you have one carbon
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and two oxygens.
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So it is going to be 12.01 plus two times 16.
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Two times 16.00 grams per mole,
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and so let's see.
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This would get us to, this is 32
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plus 12.01, so that is 44.01.
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And that's grams per mole,
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but now we're thinking about moles per gram,
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so it's going to be one over 44.01
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and so if we did just this,
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the grams of CO2 would cancel the grams of CO2
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and this would give us moles of CO2,
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but I care about moles of carbon in the product.
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So how many moles of carbon are there
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for every mole of CO2?
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Well we know that we have one mole of carbon
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for every one mole of CO2.
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Every carbon dioxide molecule has one carbon in it,
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and so what is this going to get us?
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So we have 5.65
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divided by, divided by 44.01,
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and then times one, so I don't have to do anything there.
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That is equal to, I'll round it to three digits here,
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so 0.128,
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so this is 0.128
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and my units here are, let's see.
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The grams of carbon dioxide
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cancel the grams of carbon dioxide.
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The moles of carbon dioxide cancel the moles
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of carbon dioxide,
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so I am exactly where I want to be.
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This is how many moles of carbon that I have.
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And you can do the dimensional analysis,
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but it also makes intuitive sense, hopefully.
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If this is how many grams of carbon dioxide we have
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and a mole of carbon dioxide is going to have a mass
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of 44.01 grams,
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well then 5.65 over this is going to tell us how,
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what fraction of a mole we have of carbon dioxide
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and then whatever that number of moles we have
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of carbon dioxide is gonna be the same
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as the moles of carbon,
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'cause we have one atom of carbon
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for every carbon dioxide molecule,
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so that all makes sense.
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And now let's do the same for hydrogen.
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So let's think about moles of hydrogen
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in the product,
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and it's going to be the same exercise.
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And if you're so inspired
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and you didn't calculate it in the beginning,
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I encourage you to try to do this part on your own.
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All right, so all of the hydrogen is in the water,
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so and we know that we have, in our product,
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3.47 grams of water.
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3.47 grams of H2O,
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and now let's think about how many moles of H2O that is.
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So that's going to be, let's see,
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every one mole of H2O
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is going to have a mass of how many grams of H2O?
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And we could do that up here.
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H2O, it's going to be,
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we have two hydrogens,
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so it's going to be two times 1.008
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plus the mass, the average atomic mass
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of the oxygen is going to be plus 16,
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but we can also view that as
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what would be the mass in grams,
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if you had a mole of it?
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And so this is going to be in grams per mole,
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and so this is going to be, let's see,
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two times 1.008.
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This part over here is 2.016,
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and then you add 16 to it.
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It's going to be 18.016.
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18.016,
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and then if I just calculated this,
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this would give me how many moles of water I have
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in my product,
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but I care about moles of hydrogen.
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And so how many moles of hydrogen do I have
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for every mole of water?
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So for how many moles of hydrogen
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for every mole of water?
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Well I'm going to have two moles of hydrogen
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for every mole of water
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because in each water molecule,
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I have two hydrogens.
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And so that's going to cancel out with that
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and we're just going to be left with moles of hydrogen.
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And so this is going to be equal to,
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I'll take my 3.47 grams of water,
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divide it by the,
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how much, how many grams a mole of water,
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what its mass would be,
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so divided by 18.016.
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This is how many moles of water I have.
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Now for every,
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for every molecule of water, I have two hydrogens,
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so then I will multiply by two.
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Times two is equal to zero point,
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I'll just round three digits right over here.
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0.385.
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0.385 moles of hydrogen.
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So now we know the number of hydrogen atoms.
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We know the number of carbon atoms,
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and to figure out the empirical formula of the compound,
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we can think about the ratio between the two,
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so I'm gonna find the ratio of hydrogens to carbons.
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And that is going to be equal to,
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I have 0.385 moles of hydrogen
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over, over
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0.128 moles of carbon.
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0.128 moles of carbon,
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and what is this equal to?
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And it looks like it's going to be roughly three,
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but let me verify that.
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In my head, that seems so,
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I already have the hydrogen there,
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and so if I divide it by 0.128,
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I get, yep, pretty close to three.
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Now actually, I think if I, yep,
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pretty close to three, so there you go.
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This is approximately three,
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and so I can, with pretty good confidence,
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this was very close to three.
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I could say for every carbon,
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I have three hydrogens in my original compound,
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in this thing right over here.
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So the empirical formula of our original compound
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for every one carbon,
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I have three hydrogens.
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So CH3,
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and we are done.