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  • - [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.

- [Instructor] We are told that a sample of a compound

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実践例。燃焼データから経験式を決定する|AP化学|カーンアカデミー (Worked example: Determining an empirical formula from combustion data | AP Chemistry | Khan Academy)

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    林宜悉 に公開 2021 年 01 月 14 日
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