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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

• 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