 So let's do this one. It's one of these empirical formula ones. So it says, what is the empirical formula of an ingredient in buffering tablets that has the percent composition of carbon at 14.25 percent, oxygen at 56.93 percent, and magnesium at 28.83 percent of my max. Okay? So notice those are percentages by max. Okay? So let's write down what we've got. So it says percent mass of carbon is 14.25 percent. Percent mass of oxygen is 56.93 percent. And then the percent mass of magnesium is 28.83 percent. Okay, so this is percentage of what, you know, who knows how much mass you might have started out. So because you don't know that, the way I do it is just to set it all equal to 100 grams of unknown. Okay? So if we do that, if we say the mass of the unknown equals 100 grams like that, is everybody okay with what I'm doing so far? What is the mass of carbon going to be? 14.25. Okay? Is everybody okay with that? So if you do that, it really helps out a lot. Okay? So let's just do that first step. When we do that, can I just do that and not have to go over the working out of it and say grams of carbon, grams of oxygen, grams of magnesium like that? Is everybody okay with that? Okay, so the next thing we want to do is to, well, we've got to figure out the formula. So we got to put it in those whole number ratios. Okay? But we can't do it by ratios of mass because the thing's way different. Right? So we got to do a ratio of moles. Okay? It's like, you know, like a broken record every time you convert things to moles. Okay? So grams to moles, we know that that's the atomic mass of these atoms. Right? So more of the molar mass, I should say. So in carbon's case, 12.01 grams of carbon per one mole of carbon, oxygen, so it's 12.01 grams of oxygen per one mole of oxygen. And for magnesium, what is it? 34, you know, 34.44. 12, 24.31. It's cool with that. Okay? And then you just cancel, cancel our number of moles from 100 grams of total. Right? So let's just do it together. 14.25 divided by 12.01. And you guys can help me out if you want. 1.1865. And I'm just going to go out to a bunch of digits. Not that it matters because we're just going to cut them all off when we show the molecular form. Okay, so 56.93 divided by 16. 3.558. 3 divided by 24.31. What did you get? Okay, so already, hopefully you guys can see that something's going to happen, right? What do you guys see? And magnesium are going to be what? The same number of carbon and magnesium in this conga. How did I figure that out? Because they got the same number of moles. Okay? So what are we going to do now? We're going to just take our molecular formula, p, y, o, z, like that. We're going to figure out what x, y, and z are. Okay? How do we do that? Well, we take, we look here and look for the smallest ratio. Okay? Or the smallest number, right? Which one's the smallest? Well, this one, really, right? But respectively, these two are the same. So it's going to be magnesium, 6.1859 divided by 1.1k59. Carbon, 1.1859 at the bottom, 3.558. So carbon is also 1.1. See, that's what I'm saying. You guys, you need to check me before I get back. And then oxygen's going to be 3.558 divided by 1.1. So I got those right. So carbon, 6.5 magnesium, 5.9 oxygen. Okay? So now what do we do? We just divide. So magnesium, 1 carbon. So it's a, well, carbon's going to be effectively 1, right? And then oxygen is 3.558 divided into 1.118. What do you guys think? 3.003. Yeah, something, 3. Okay. So do we leave these ones in there? No. Nope. So the formula, MgCO, magnesium carbonate. What did we learn about magnesium carbonate earlier today? It reacts with an acid, carbon dioxide. It's a gas forming reaction, right? It's also, it reacts with an acid, so it must be a base. What kind of base? You guys cool with that? Okay. Do you guys have any questions more on this type of problem? Yeah. Yeah, then you're going to multiply through. Multiply through. All of them, yes. Okay, we'll try an example with that. Okay, any questions on this particular problem? Everybody cool with this type of thing? You will have some on the exam, so make sure you're okay with it.