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

AP.Chem:
SPQ‑2 (EU)
,
SPQ‑2.A (LO)
,
SPQ‑2.A.3 (EK)

## Video transcript

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 percent by mass mercury and by mass it is 27 percent chlorine so the remainder is chlorine by mass so pause this video and see if you can come up with what look if you can come up with what is likely the empirical formula for our mystery molecule in here and is 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 assume it is or this entire container is 100 grams I could have assumed a thousand grams or five grams but a hundred 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 moles we have with chlorine and then that will inform the likely empirical formula so if we assume a hundred grams well then we are dealing with the situation that our mercury we are 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 point five nine grams on average so we could multiply this times one over 200 point five nine moles per gram and 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 point five nine divided by 200 point five nine 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 even puts a roughly right over there and I can do the same thing with chlorine chlorine if I have 27 percent by mass 27 percent 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 0.45 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 point four five we take 27 divided by 35 point four five guess that's two 0.76 roughly 0.76 and remember when 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 it's is 0.76 times a varuga aha God Rose number number of chlorine atoms and so what's the ratio here well it looks like for everyone mercury atom there is roughly two chlorine atoms if I take 2 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 and so if we assume a ratio of two chlorine atoms for every one mercury atom the likely empirical formula is for every mercury atoms 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 I won't go in depth why it's called mercury two chloride but that's actually what we likely had in our container