- Empirical, molecular, and structural formulas
- Worked example: Calculating mass percent
- Worked example: Determining an empirical formula from percent composition data
- Worked example: Determining an empirical formula from combustion data
- Elemental composition of pure substances
We can use percent composition data to determine a compound's empirical formula, which is the simplest whole-number ratio of elements in the compound. Created by Sal Khan.
- [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.