 Okay, so this one says a substance C-X-H-Y has the composition by mass 85.63% carbon and 14.37% hydrogen. First it wants you to calculate the empirical formula of the substance and then it asks what is the molecular formula of the substance if its molar mass is 70.1 grams per mole. So remember looking at this problem percent mass doesn't really help us with this thing. Okay, so we want to convert these to molar numbers. Okay, so percent mass of the total, the whole substance is going to be 100% like that. I'll say 100.00% since there's two after the last minute. So what we can do is convert this to actual masses. Okay, so the mass of carbon, if we've got 100 grams, so the mass total is going to be 100 grams, if we've got that of the substance C-X-H-Y then the mass of carbon is going to be 85.63 grams. So the mass of hydrogen is going to be what then? 14.7 grams. 14.37, yeah. 14.37 grams of hydrogen like that. Okay, because that's 85.63% of 100 grams. That's 14.37% of 100 grams. Okay, so remember what did I say? We need to convert these to moles. Okay, because we need the molar ratio of X to Y. So carbon, one mole of carbon is 12.01 grams of carbon. Okay, that's going to cancel out our grams of carbon like that. One mole of hydrogen is 1.008 grams of hydrogen. Remember I'm getting these numbers from the periodic table. That's actually in my head. I've done these so many times, but you can look at the periodic table to convert these to moles. So what do we got? 85.63 divided by 12.01. And that equals? 7.13 moles. 7.13 moles of carbon. And then 14.37 divided by 1.08 is 14.26 moles of hydrogen like that. So now what we can do is take... Well, we've got moles and moles, okay? So we can say C7.130 moles H14.26. So hopefully that looks weird to you, right? Because you don't see normally molecular formulas having decimal points. They're usually integers, right? So what we're going to have to do is divide both of these by the smallest number, okay? Which is smaller, 7.130 or 14.26? So 7.130, 7.130 like that, okay? So this is going to give us the empirical formula provided we get integers here, right? 7.130 divided by 7.130 is 1. And then H, if you can't see that, that's going to be almost 2. It's going to be like 1.9999. So we're going to say 2, like that, okay? But we don't write C1. So the empirical formula is going to be CH2, like that. How are you fine with doing that? I've got to switch it off, Ben, because they're all dying at the same time. Okay, so the thing we want to do now is figure out, well, what is the empirical mass? We've got the molar mass over here. So we want to figure out what is the empirical mass? How do we do that? Well, we look at CH2, and we add it up. So 12.01 plus 2 times 1.008. So 2, 6, that's per mole. That's the empirical mass, okay? So how do I figure out, well, I've got the molar mass here, the empirical mass. How do I figure out what the molecular formula is? I've got to figure out, well, what's the ratio between the molar mass and the empirical mass? So the ratio is going to be the molar mass divided by the empirical mass. So that's 70.1 grams per mole divided by 14.026 grams per mole. And if I cut, this should be 5, but let's just make sure. So 70.1 divided by 14.026. Yeah, 4.9978, that's essentially 5. Are we okay with that? Anybody not okay with what we've done so far? Everybody is okay with that. Okay, wonderful. So what do we do? We take this empirical formula, and we multiply each of those things by 5, okay? So we're going to erase down here. So we got CH2 multiplied by 5. But when we do that, we put the 5 there, multiply it down there. That's going to equal C5H10, because 1 times 5 is 5, 2 times 5 is 10. So the molecular formula for this is C5H10. Any questions? No. Pretty straightforward, right? Good.