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A couple of videos ago we figured out how to go from the

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empirical formula to the mass composition.

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And all I mean by that, mass composition, is that you could

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start with the empirical formula H2O, which is also its

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molecular formula.

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And then we were able to figure out what percentage of

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this was water and what

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percentage of this was hydrogen.

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The way we did it is we said, OK, oxygen's mass is 16 atomic

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mass units.

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Hydrogen is one.

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The mass of the entire molecule is 2 times hydrogen.

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2 times 1.

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Plus 16 for that one oxygen, so it's 18.

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And then we said, the composition of

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oxygen is just 16/18.

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So oxygen is 16/18.

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I don't have my calculator anymore.

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I think the number was something on the order of 88%

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or 88.9% oxygen.

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So now let's see if we can go the other way.

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Let's see if we could start with mass composition, the

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percent composition of the different elements, and then

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go to empirical formula.

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And I think the first video I made, I

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spelled empirical wrong.

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Which is good reason for me not to do spelling videos.

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I think I spelled this with an e right here, because I

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pronounce it empricial.

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But it's spelled with an i.

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Empirical formula.

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So back to the empirical formula.

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So let's do some exercises here to see

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if we can get someplace.

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So let's say that I start off with-- I need space, my

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periodic table is right down there.

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Let's say I am told that I have a bag of stuff.

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And 75% of that bag, let's say it is mercury.

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Hg.

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And say that the other 25% of the bag is chlorine.

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So the question that I'm asked is, what is the empirical

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formula of the stuff that I have here, assuming that it's

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all one molecule or one type of molecule?

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So what is the empirical formula?

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So the way to think about this is, let's assume that you have

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100 grams of this stuff.

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Right?

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And the reason why I'm picking 100 is because 100 is a useful

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number to deal with.

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So if I have 100 grams of the stuff, all right.

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So let's say I assume that you have 100 grams. If you have

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100 grams, how many grams of mercury do you have?

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Well that means that 75% of that, that means you have 75

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grams of mercury, Hg.

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And that means you have 25 grams of chlorine.

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You might be saying, where did get these numbers from?

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Well, I got the numbers because I assumed 100 grams. I

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could have assumed 112 grams or 7 grams, but those would

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have been harder numbers to work with.

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So I'm just assuming 100 grams. So the question is, in

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75 grams of mercury, how many moles of mercury do I have?

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So I need to convert from grams to moles.

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Let's go the other way.

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One mole of mercury-- this is at least how my brain thinks--

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let me pick a suitable color-- so one mole of mercury so 6.02

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times 10 to the 23 mercury atoms.

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What's the mass of that?

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Well it's equal to whatever mercury's mass number is in

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grams. So let's go look up mercury's mass number.

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So mercury is a transition metal.

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It's one of the few metals that are liquid at room

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temperature.

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Right there we have mercury.

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Its mass number, let's pick the 200.

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It has a mass number of 200.

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Obviously, there's some mercury out there that has a

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mass of 201.

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But for simplicity, let's say it's 200.

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So 1 mole of mercury is 200 grams. How did I get that?

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Mercury's mass number.

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One atom of mercury has a mass of 200 atomic mass units.

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So one mole of mercury-- 6.02 times 10 to the 23 mercury

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atoms-- has that many grams of mass.

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So instead of 200 atomic mass units per atom, we have 200

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grams per mole of that atom.

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It's an easy calculation.

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You just look up it atomic mass number and say it's that

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many grams. You have a mole of the substance.

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Fair enough.

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And what about one mole of chlorine?

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We do the same exercise.

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We go down to our periodic table.

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Chlorine has a mass number of, let's say, 35.

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It has various isotopes on this planet.

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But let's say the mass number is 35.

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So one chlorine atom weighs-- well now, we shouldn't say

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weighs, we have to be very careful.

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One chlorine atom has a mass of 35 atomic mass units.

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So 6.02 times 10 to the 23 chlorine atoms, or one mole of

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chlorine atoms will have a mass of 35.

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35 grams.

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So how many moles of mercury do I have?

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Let me get my calculator going.

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So I have 75 grams of mercury.

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One mole of mercury would have a mass of 200 grams. So I

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could just take 75 divided by 200 and it tells me how many

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moles I have. 75 divided by 200 is equal to

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0.375 moles of mercury.

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How did I figure that?

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One mole of mercury would be 200 grams. I only have 75

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grams of mercury, so I have a small fraction of it, roughly

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0.375 moles of mercury.

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I've just expressed the number of atoms of mercury I have.

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Moles tells me it's 6.02 times 10 to the 23 atoms of mercury.

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So I have 0.375 times Avogadro's number of mercury

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atoms. Fair enough.

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Now let's do the chlorine.

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I have 25 grams of chlorine.

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One mole of chlorine is 35 grams. So I

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have 25 over 35 moles.

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So what's that?

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So 25 divided by 35 is equal to 0.714 moles of chlorine.

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Now these numbers, they're not exact.

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Because if you take twice this number up here, you don't get

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twice this number right here.

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If you get twice this, you don't get this.

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But this is roughly twice as many moles of chlorine as I do

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of mercury.

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So that tells me that for every mercury atom-- remember,

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moles is just a number-- for every mercury atom I have, I

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have two chlorine atoms. So the empirical formula here is

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for every mercury I have two chlorines.

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I have two chlorines, right there.

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And the numbers didn't work out almost exactly right.

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Especially in the real world, when you're actually trying to

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figure out things empirically, your numbers seldom will.

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And maybe there's some random other things running out there

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in terms of other things that are contributing to the mass.

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But this is close enough to know that the ratio of mercury

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to chlorine is roughly 1:2.

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Right?

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This is a number.

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For every one chlorine atom, you have roughly two.

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You have twice as many.

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I guess the other way of thinking about it is you have

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twice as many chlorine atoms in the

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bag as you have mercury.

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Roughly.

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Although this is a little bit more.

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This is close to 0.75.

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But it's close enough for you to know that you're dealing

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with mercury chloride.

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Right there.

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Let's do another one of these.

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Let's see how the numbers turn out for this one.

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Let's say you have another bag that is 9% magnesium.

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And let's say the remainder of the bag, 91%, is iodine.

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91% is iodine.

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So the way to do all of these is, you do the same thing.

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Assume you have 100 grams. So if you have 100 grams, of

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which you have 9 grams of magnesium.

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And you have 91 grams of iodine.

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And then figure out how much a mole of magnesium-- what would

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be the mass of a mole of magnesium and

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then a mole of iodine.

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So let me write here.

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One mole of magnesium.

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And we want to figure out one mole of iodine.

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Let's figure those out.

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So magnesium's mass number is, let's just go with 24.

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Let's say we have the isotope that's 24.

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Magnesium is 24 and since we're already down here,

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what's iodine.

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127.

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It's 127.

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Iodine is one of the halogens.

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So let's see.

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You have 127 and you have 24.

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Let me write those down.

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So you have 127 and you have 24.

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So one atom of iodine has a mass of 127 atomic mass units.

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So a whole mole all of it, 6.02 times 10 to the 23 iodine

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atoms will have a mass of 127 grams. You just take the

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atomic weight, or the atomic mass, and the mass will be

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that many grams when you have this many of the atom.

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So then one mole of magnesium will be 24 grams.

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So now we just figure out how many moles of each of these we

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have. We have 9 grams of magnesium.

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So what fraction of a mole is that?

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A mole is 24 grams. So this is equal to

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9/24 moles of magnesium.

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And what is 9 over 24?

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9 divided by 24 is equal to 0.375.

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So this is equal to 0.375, which is similar to what we

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had in the last problem when that 0.375 showed up.

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And 91 grams of iodine is what?

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91 grams of iodine.

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Well, in a whole mole of iodine you're going to have

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127 grams. So we have 91/127 moles of iodine.

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And what's 91 over 127?

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I have a feeling we're going to have very similar numbers.

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121 divided by 127 is equal to 0.716.

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I should do another problem with better ratios than this.

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These have the same ratios as the last problem, 0.72,

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roughly, moles of iodine.

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And this is moles of magnesium.

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So we have roughly twice the number of iodine atoms as we

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do of magnesium atoms. Right?

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For every one magnesium atom we have roughly two iodine

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atoms. I know this isn't exactly 1:2,

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but it's pretty close.

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So the formula here is magnesium iodide.

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Right there.

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And that's the empirical formula.

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We don't know, maybe in every atom maybe you had two

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magnesiums and four iodines.

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We don't know.

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All we know is that the ratio of magnesium to

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iodine here is 1:2.

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In the next video, I'm going to try to look for ones that

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have more interesting ratios, because I don't want to do two

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problems that both have the same ratio.

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But hopefully you found this a little bit helpful.

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