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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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工作實例。根據百分比構成數據確定經驗公式|可汗學院 (Worked example: Determining an empirical formula from percent composition data | Khan Academy)

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    林宜悉 發佈於 2021 年 01 月 14 日
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