 OK, wonderful. So let's try this next problem. So it says a quantity of 7.480 grams of an organic compound is dissolved into water to make 300 milliliters of solution. The solution has an osmotic pressure of 1.43 ATM at 27 degrees Celsius. The analysis of this compound shows that. So that's the elemental analysis. It shows that it contains 41.8% carbon, 4.7% hydrogen, 37.3% oxygen, and 16.3% nitrogen. So calculate the molecular formula for this compound. So if you don't remember how to do empirical formula then this problem will be very difficult for you. OK, it probably won't be able to get away. Anyways, so this is that table of elemental analysis that it just gave us. So the best way to do this is to remember this is mass percent. So we can just pretend like we have 100 grams of that particular compound to make it much easier. So if we have 100 grams of the compound all together, we can say that instead of being the mass percent, we can say that that's the actual mass in grams. So if we have 100, that's total is going to be 100 point over there. So if we say that's 100.0 grams, 41.8% of 100 is 41.8%. Is everybody OK with that? So we can change the units of these to grams, grams, grams, grams instead of those percentages. Is everybody cool with doing that? Yes. OK, wonderful. So we've got, let's make these closer to the thing. So we can, so CHO in. Now we're going to have to change this to moles. So how do we do that? We need to remember the mole or mass. OK, remember the periodic table. So and I remember them again in my head. So look at the periodic table if you don't remember. So carbon, so when I cancel out grams, we want to get moles. So grams on the bottom, moles on the top, remember a dimensional one else. 12.01 grams of carbon is one mole of carbon. So here, one mole hydrogen, 1.008 grams hydrogen, one mole oxygen, 16.00 grams oxygen, one mole nitrogen, 14.01. OK, so hopefully you can see, cancel, cancel, cancel. In each one of those, now we get moles. So let's go through these together, 41.04. Now we can get the empirical formula of this thing. So how do we do that? I'm sure you guys remember, right? You guys remember? Yeah. Yeah. OK, good. So you can count me out. So 3.48, so C 3.48, H 4.66, O 2.33, and 1.16. Is that what you guys would have done? Yes. Yeah, so that's right, that's what you do. So what do we do now? Remember, we divide by the smallest number, right? Smallest number. Yeah, the smallest number, OK. So which one is the smallest one? 1.16. Yeah, the 1.16. So that thing is called the empirical formula. Empirical formula. So if you don't recall how to do that, you've got to remind yourself that's how to do it. OK, so that's how to calculate the empirical formula. Now let's go back and solve what this problem says. So it wants us to figure out what the molar mass of the thing is, OK? So that's the empirical formula, not the molecular formula, OK? So that's the smallest whole number ratio of the atoms within this molecule. So it doesn't give us the actual molecular formula. We could calculate the empirical formula, which actually we should do. Let's do that right now. So how do we do that? 3 times 12, so let's say the empirical mass is the next thing we're going to have to calculate. So 3 times 12.01 grams per mole plus 4 times 1.008 grams per mole plus 2 times 16.00 grams per mole plus 1 times 14.01, so I get 86.02, right? 07 grams per mole as the empirical mass, OK? That's probably not the molar mass, OK? It may be, but it's probably not, OK? So we've got to figure out this osmotic pressure problem in order to do that, OK? So I'm going to erase most of this stuff, except for what we just figured out. Is everybody OK with that? Yes. OK, I wonder. Going to, just so I can use more of the board here. OK, so if you guys remember, tell me how. There are more too. So pi, which is osmotic pressure, equals m. What is m in this case? Malarity. Malarity, OK? So you've got to remember that's the molarity. And what was the rest? RT. What is R in this case? A gas constant. So is that one that we just did? 0.0821 liter ATM per mole kelp, OK? So remember, what does that do for us? It helps us to remind us what units we want our stuff in. OK, what units we want these guys in here, OK? So do we have the proper units? Degree Celsius, is that the proper units? No, so let's change that. So 273, 6300 kelp, OK? So molarity, what do we need for molarity? That's going to be, so let's go over here. So molarity equals the number of moles of the solute, right? Divided by the what? Volume of the solution. So that's the solution, not the solvent, like we did in the last one with molality, OK? So do we have volume of solution? Yes, but is it in the right unit? No. So we're going to have to convert it. Is everybody OK with that? So milliliters, we're going to want it in what? Liters. So how do we convert it? 1,000, 1 liter, cancel, cancel, and maybe 0, 3, 0. So that's how much solution we got. Do we have the number of moles of the solute? The number of moles of it, can we? What do we need to know to figure out the number of moles of it, not the empirical mass, the what? The molar mass, right? Do we have the molar mass? No, we don't. This is the empirical mass. We've got to figure out what the molar mass is before we can figure out the, OK? So switching this over, we've got to figure out the molarity, sorry, before we can figure out the number of moles, OK? And from there, we're going to figure out the molar mass. So we've got to rearrange this equation effectively is what we're doing, OK? So we're going to solve for molarity, right? So hopefully, you guys remember how to rearrange algebraically. So it's all for molarity, right? So it's going to be pi divided by RT, like that. To expand my r, 0.0821 liter, 8 m, or 1 mol kelvin, 300 kelvin. Up at the top here, 1.43, 8 m, OK? So what's that going to give us? 8 m cancels with 8 m, kelvin cancels with kelvin. 1 divided by liters divided by moles gives us what? Moles per liter, OK? So that's going to be our unit here. So moles per liter, and that's units of what? Molarity, yes. So, and that's what we want, right? So 1.43 divided by, and my molarity is 0.0581, OK? So remember, molarity is what? The number of moles of solute divided by the volume of the solution, OK? The other thing we know is what the molar mass is, right? So what's the equation for molar mass, right? So it's the mass in grams divided by the number of moles, like that, OK? So let's just draw it, this number of moles, OK? So we've got this, right? The molarity, we've got the volume of the solution, right? So we could figure out the number of moles, right? Is everybody OK with that? Yes. Let's rearrange this. So the number of moles of solute is going to equal the molarity times the volume of the solution. So 0.0581 moles per 1 liter times 0.300 liters, like that. So that gives us the number of moles. Is everybody OK with that? 0174 moles of the stuff, OK? So the mass divided by the number of moles, here, right? That's going to give us the molar mass. Is everybody OK with that? Does that make sense? Does that make sense, OK? So do we have the mass of it in grams? Yes. 430 grams divided by the number of moles, 0.0174 moles per mole, that's going to give us the molar mass, OK? So if this number is 86.7, then this is the molecular form, OK? Does everybody understand what I'm saying? Right. Why don't you move the decimal place with calculus? Do what now? You move the decimal place with calculus? Well, they're right there. That's right. I thought it was calculus. Yeah, I kind of got it. OK, so let's just continue the problem, OK? 7.48 divided by, so what did I get? 429 grams per mole, right? That's the 366, 366, OK? Everybody remembers their 366 moles, OK? So is 429 the same number as 86.07? Is that the same number, guys? No, it's not the same number. So is this the molecular formula here? No. So we're going to have to figure out what the molecular formula is by taking the ratio of the molar molecular mass, the molar mass, divided by the empirical mass, OK? So you guys remember how to do that, I'm sure. I'm going to erase this bottom part down here, OK? So how do I get that mass ratio? So that's going to be the big one, the molar mass, divided by the empirical mass, right? So 429 grams per mole divided by 86.07 grams per mole. Grams per mole, cancels, of course. 4.989, so what is that? So that equals 4.99, so that's approximately what? 5, OK? So 5 there, so what are we going to do to the empirical formula? We're going to multiply by 5, OK? So just take 5C3H4O2M, and so what's the molecular formula? So that's going to be C3 times 5 is 15, H4 times 5 is 20, O5 times 2, 10, and M1 times 5 is 5. That's how you do that problem, OK? So the crux of this problem, of course, is to know that of my approach. All the other stuff is stuff that you learned in general chemistry, OK? Any other questions on this one? No. OK, wonderful, thank you.