Empirical formula from mass composition | Chemistry | Khan Academy
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Empirical formula from mass composition | Chemistry | Khan Academy

October 13, 2019

Sal: What I want to do
in this video is start with mass composition and
see if we can figure out the empirical formula of
the molecule that we’re dealing with based on
the mass composition. Let’s say that we have a bag
and we’re able to measure that this bag is 73 percent, it’s
73 percent mercury and it is, the remainder of the
bag, 27 percent chlorine. Based just on this can we
figure out the likely empirical formula for the molecule
that we have in that bag? I encourage you to pause the video and try to see if you can figure
it out on your own. Well one way to think about it is let’s just assume a number. This is all the information we have, let’s just assume we have 100 grams of it. We could assume a thousand
grams or 10,000 grams or 57 grams, but I’ll pick
100 grams because it will make the numbers easy to
work with in our head. Let’s just assume, let me make it clear that I’m assuming this. I’m going to assume that I
have 100 grams of this molecule that is 73 percent mercury
and 27 percent chlorine. If I assume that that means
that the 73 percent that is mercury is going to be
73 grams and the 27 percent that is chlorine is going
to be 27 grams of chlorine. Let me make it clear this is
mercury and this is chlorine. Now I just need to think about, well how many moles of
mercury is 73 grams and how many moles of chlorine is 27 grams. To do that I’ll look up the
periodic table right here. I have the atomic weight
which is the weighted average of the atomic masses
that’s found in nature. The atomic weight here
for mercury is 200.59. That means, let me write
this right over here. One mole of, one mole of mercury is, we could say is 200.59
grams, 200.59 grams. Similarly we could look up the
atomic weight for chlorine. Chlorine right over here, 35.453. We could say one mole of chlorine, and once again this is a
weighted average of all of the isotopes of chlorine
as found in nature. I guess we’ll just go with that number. One mole of chlorine is going to be 35.453, 35.453, 35.453 grams. Given this information right over here how many moles of mercury is this, roughly, and how many moles of
chlorine is this, roughly? I say roughly because getting
an empirical formula from measurements of mass composition
is going to be necessarily a messy affair, it’s not
going to come out completely, the numbers aren’t going to
work out completely exact, so that’s why I said roughly. So how many moles is this? This is going to be 73
over 200.59 of a mole. If a mole is 200.59 we have
73, this is the fraction of a mole that we have, moles of mercury. Remember, moles are just a
number, Avogadro’s number of something, but let’s
just figure out what it is. If we take 73 divided by 200.59
we get .36, I’ll just say 0.364, and once again this
is, so approximately 0.364. That’s how many moles
of mercury that we have. We can do the same thing for chlorine. This is going to be 27 over
35.453 moles of chlorine, which is approximately equal
to, get our calculator out. 27 divided by 35.453 is equal
to .76, I’ll just say two. So 0.762 moles of chlorine. What’s going to be the ratio
of mercury to chlorine? Or I guess we could say since chlorine, there’s more of that, chlorine to mercury. Remember, this is just a number. When I say 0.762 moles, this is just 0.762 times Avogadro’s number of chlorine atoms. This is 0.364 times Avogadro’s
number of mercury atoms. We can literally think
of this as the ratio. This is a certain number of moles, this is another number of moles. Well what’s the ratio, let’s see. What’s the ratio of chlorine to mercury? Well you can eyeball
it, it looks like it’s roughly two to one, and
you can verify that if you take that number and
you divide it by .3639, and once again I’m just going
to get the rough approximate. You can see it’s pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be
some measuring error right over here, but you can say
that it looks like roughly, this is what I’m talking about when you’re trying to find the empirical formula for mass composition it tends
to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines. Based on these measurements
right over here it’s very likely that you have
mercury two chloride. The reason why it’s called
mercury two chloride is because, well I won’t go into too
much detail right over here but chlorine is
highly electronegative, it’s an oxidizing agent,
it likes to take other people’s electrons or hog
other people electrons. In this case it’s hogging, since each of the chlorine likes to hog one electron, this case two chlorines are
going to hog two electrons, so it’s hogging two
electrons from the mercury. When you lose electrons
or when your electrons are being hogged you’re being oxidized, so the oxidation state on
mercury right over here is two. Two of its electrons are being hogged, one by each of the two chlorines. This is mercury two chloride, where the two is the oxidation state of the mercury. This is what we likely have. The ratio, we have two chlorines for every one mercury, roughly.

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  2. Wait, so is Cl2 an oxidizing agent because it's on the right side of the periodic table? How do you know which one's the oxidizing agent?

  3. If youre trying to find the empirical formula, with 3 different elements, how do you find the ratio? Do you divide by the smallest number?

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