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• - [Instructor] Let's say that we have

• some type of a container that has

• some type of mystery molecule in it.

• So that's my mystery molecule there,

• and we're able to measure the composition

• of the mystery molecule by mass.

• We're able to see that it is 73% by mass mercury,

• and by mass it is 27% chlorine,

• so the remainder is chlorine by mass.

• So pause this video and see if you can come up with

• what is likely the empirical formula

• for our mystery molecule in here,

• and as a little bit of a hint,

• a periodic table of elements might be useful.

• All right, now let's work through this together,

• and to help us make things a little bit more tangible,

• I'm just going to assume a mass for this entire bag.

• Let's just assume it is,

• or this entire container is 100 grams.

• I could have assumed 1,000 grams or 5 grams,

• but 100 grams will make the math easy

• because our whole goal is to say, hey,

• what's the ratio between the number of moles we have

• of mercury and the number of the moles we have of chlorine

• and then that will inform the likely empirical formula.

• So if we assume 100 grams, well then we are dealing with

• a situation that our mercury, we have 73 grams of mercury,

• and we can figure out how many moles this is

• by looking at the average atomic mass of mercury.

• That's why that periodic table of elements is useful.

• We see that one mole of mercury is 200.59 grams on average,

• so we could multiply this times

• one over 200.59 moles per gram.

• So when we multiply this out, the grams will cancel out

• and we're just going to be left with

• a certain number of moles.

• So I'll take 73 and we're just going to divide it by 200.59,

• divided by 200.59

• is going to be equal to 0.36, and I'll just say 0.36

• because this is going to be a little bit

• of an estimation game, and significant digits,

• I only have two significant digits

• on the original mass of mercury, so 0.36 moles, roughly.

• I'll even say roughly right over there,

• and I can do the same thing with chlorine.

• Chlorine, if I have 27% by mass,

• 27% of 100, which I'm assuming, is 27 grams.

• And then how many grams per mole?

• If I have one mole for chlorine,

• on average on earth the average atomic mass is 35.45 grams.

• And so this is going to approximate how many moles

• because the grams are going to cancel out,

• and it makes sense that this is going to be

• a fraction of a mole because 27 grams is less than 35.45.

• We take 27 divided by 35.45.

• It gets us to 0.76, roughly, 0.76.

• And remember, we're talking about moles.

• This is how many moles of chlorine we have,

• or this is how many moles of mercury, that's a number.

• You can view that as the number of atoms of mercury

• or the number of atoms of chlorine.

• Moles are just the quantity specified by Avogadro's number,

• so this is 0.76 times Avogadro's number of chlorine atoms.

• So what's the ratio here?

• Well, it looks like for every one mercury atom,

• there is roughly two chlorine atoms.

• If I take two times 0.36, it is 0.72,

• which is roughly close, it's not exact,

• but when you're doing this type of empirical analysis,

• you're not going to get exact results,

• and it's best to assume the simplest ratio

• that gets you pretty close.

• So if we assume a ratio of two chlorine atoms

• for every one mercury atom,

• the likely empirical formula is

• for every mercury atom we will have two chlorines.

• And so this could be the likely empirical formula.

• The name of this molecule happens to be

• mercury two chloride, and I won't go in depth

• why it's called mercury two chloride,

• but that's actually what we likely had in our container.

- [Instructor] Let's say that we have

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# Worked example: Determining an empirical formula from percent composition data | Khan Academy

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林宜悉 posted on 2020/03/27
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