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  • - [Instructor] In a previous video,

  • we introduced ourselves to the idea of average atomic mass,

  • which we began to realize could be a very useful way

  • of thinking about a mass at an atomic level,

  • or at a molecular level.

  • But, what we're gonna do in this video

  • is connect it to the masses that we might actually see

  • in a chemistry lab.

  • You're very unlikely to just be dealing with one atom,

  • or just a few atoms, or just a few molecules.

  • You're more likely to deal with several grams

  • of an actual substance.

  • So, how do we go from the masses at an atomic scale

  • to the masses, masses of samples

  • that you see in an actual chemistry lab,

  • or in, I guess you could say, r-scale of the world.

  • Well, the chemistry community has come up with

  • a useful tool.

  • They said, all right, let's think about a given element.

  • So, say, lithium.

  • We know its average atomic mass is 6.94,

  • 6.94 unified atomic mass units per atom, atom of lithium.

  • What if there were a certain number of atoms of lithium

  • such that if I have that number,

  • so times certain, certain number of atoms,

  • then I will actually end up with 6.94 grams,

  • grams of lithium.

  • And, this number of atoms is 6.02214076

  • times 10 to the 23rd power.

  • So, if you have a sample with this number of lithium atoms,

  • that sample is going to have a mass of 6.94 grams.

  • Whatever its average atomic mass is

  • in terms of unified atomic mass units,

  • if you have that number of the atom,

  • you will have a mass of that same number in terms of grams.

  • Now, you might be saying, is there a name for this number,

  • and there is indeed a name,

  • and it is called Avogadro's number,

  • named in honor of the early 19th century Italian chemist,

  • Amedeo Avogadro.

  • And, in most contexts, because you're not normally

  • dealing with data with this many significant digits,

  • we will usually approximate it as

  • 6.022 times 10 to the 23rd power.

  • Now, there's another word that it's very useful

  • to familiarize yourself with in chemistry,

  • and that's the idea of a mole.

  • Now, what is a mole?

  • It is not a little mark on your cheek.

  • It is not a burrowing animal.

  • Actually, it is both of those things,

  • but, in a chemistry context,

  • a mole is just saying you have this much of something.

  • The word mole was first used by the German chemist

  • Wilhelm Ostwald at the end of the 19th century,

  • and he came up with the word

  • because of its relation to molecule.

  • Now, what does that mean?

  • Well, think about the word dozen.

  • If I say I've got a dozen of eggs,

  • how many eggs do I have?

  • Well, if I have a dozen of eggs,

  • that means I have 12 eggs.

  • So, if I say I have a mole of lithium atoms,

  • how many lithium atoms do I have?

  • That means that I have 6.02214076

  • times 10 the to 23rd lithium atoms.

  • Exact same idea, it's just that Avogadro's number

  • is much hairier of a number than a dozen.

  • So, let's use our new found powers of the mole

  • and Avogadro's number to start doing some useful things.

  • Let's say that someone were to walk up to you and say,

  • hey, you, I have a 15.4 milligram sample of germanium.

  • How many atoms of germanium am I dealing with?

  • Pause this video and try to think about that.

  • So, let me clear out some space

  • the periodic table of elements was taking up.

  • All right, so we started off

  • with 15.4 milligrams of germanium.

  • The first step might be hey, let's convert this

  • to grams of germanium.

  • And so, we can do a little bit of dimensional analysis.

  • We can just multiply this,

  • for every one gram of germanium

  • that is equivalent to 1,000 milligrams,

  • milligrams of germanium.

  • And so, if you essentially multiply by one thousandth

  • or divide by 1,000, we're gonna get the grams of germanium.

  • And, you can see that in the dimensional analysis

  • by seeing that that is going to cancel out with that

  • leaving us with just the grams of germanium.

  • And, now that we have an expression for grams of germanium,

  • we can think about moles of germanium.

  • So, how do we do that?

  • Well, we're going to multiply by some quantity,

  • and in the denominator we're going to want

  • grams of germanium for the dimensional analysis to work out,

  • grams of germanium, and in the numerator we want

  • the new expression to be in terms of moles of germanium.

  • So, one mole of germanium is equal to

  • how many grams of germanium?

  • Well, we see it right over here.

  • Germanium's molar mass is 72.63 grams per mole.

  • So, for every mole, we have 72.63 grams of germanium.

  • And, you can see that the units work out.

  • These grams of germanium are going to cancel

  • with the grams of germanium

  • just leaving us with moles of germanium.

  • In an actual chemistry practice,

  • finding out the moles of a substance

  • might actually be the most useful thing,

  • but if you wanted to find out the actual atoms of germanium

  • that we're dealing with, we will just multiply by

  • the number of atoms you have per mole.

  • And, this is going to be true for any element.

  • For every mole, you have Avogadro's number of atoms.

  • And, we're going to approximate that as

  • 6.022 times 10 to the 23rd atoms, atoms of germanium,

  • for every one mole, mole of germanium.

  • And so, just to review what we just did,

  • we had milligrams of germanium.

  • You multiply these two together,

  • you'll have grams of germanium, which makes sense,

  • you're essentially just dividing by 1,000.

  • If you were to multiply your grams of germanium

  • times the moles per gram,

  • which is really just the reciprocal

  • of this molar mass we got here,

  • and just to make sure where it makes sense,

  • the units work out nice with the dimensional analysis,

  • this right over here tells you your moles,

  • moles of germanium.

  • And then, if you take your moles

  • and then you multiply it by Avogadro's number,

  • it tells you how many atoms of germanium we have,

  • and that makes sense.

  • If I told you I had a certain number of dozen of eggs,

  • if I wanted to know how many eggs that is

  • I would multiply by 12.

  • So, this whole expression is the number of atoms,

  • atoms of germanium.

  • So, we have 15.4 milligrams.

  • If we wanna figure out how many grams we have,

  • we then divide by 1,000,

  • that's what our dimensional analysis tells us,

  • and it also makes logical sense,

  • divided by 1,000.

  • So, this is how many grams we have.

  • And then, if we wanna figure out how many moles,

  • and it's going to be a small fraction of a mole

  • because a mole is 72.63 grams per mole,

  • we have a small fraction of a gram,

  • much less 72.63 grams.

  • And so, we saw from our analysis

  • to figure out the number of moles,

  • we're now going to essentially divide by 72.63,

  • so divided by 72.63 is equal to,

  • this is the number of moles of germanium we have.

  • And, if we wanna figure out the number of atoms

  • of germanium, we'll then multiply that

  • times Avogadro's number.

  • So, times 6.022 times 10 to the 23rd,

  • and this EE button means times-10-to-the,

  • EE 23rd power, so that's how you do it on a calculator.

  • And then, that gives us this many atoms.

  • And, let's see, just to get our significant digits here,

  • our significant figures, out of all of the things

  • we multiplied, see we had four significant digits here,

  • four significant digits here,

  • but we only had three over here,

  • so I'm going to round to three significant digits.

  • So, I'll go to 1.28 times 10 to the 20th atoms.