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.

You didn't use sig figs. You definitely should to help get your viewers in the habit.

Hey so since it's pretty hard to get noticed on youtube I was hoping that you guys would give my channel a chance. I make educational videos similar to these in order to help people do better in their classes. You don't even have to subscribe but please take a look! Thanks in advance!

I was also wondering about the sig figs. But wonderful tutorial I watched it twice! 🙂 thank you.

really good

can anyone tell me how to get that ti85 onto my pc?

I LUV IT

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?

how long will be with subtitles in spanish ?

Good video thnx

why some compounds have same empirical and molecular formula?

plz give me answer

He makes it difficult to follow along

I just realized what i was doing wrong. Thank you

Can somebody explain why he divides by .762 and not .364 first?

Learn how to use sig figs bud

Where did he get the 0.3639

IMMA FINNA BUSS ON THIS CHEMISTRY

PLEASE RESPONDWHERE DID U GET THE NUMBER 0.3639 FROM AND WHY WAS THAT NUMBER IN SPECIFIC USED?5:15If youre trying to find the empirical formula, with 3 different elements, how do you find the ratio? Do you divide by the smallest number?

i dont want to watch this

i would rather read the book