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A 0.100 g sample of a compound containing $\mathrm{C}, \mathrm{H},$ and $\mathrm{O}$ is burned in oxygen, producing $0.1783 \mathrm{g} \mathrm{CO}_{2}$ and $0.0734 \mathrm{g}$ $\mathrm{H}_{2} \mathrm{O} .$ What is the empirical formula of the compound?

$$\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}$$

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let's go through how to find the empirical formula in a combustion reaction. So we're given 0.1 g of some molecule, and we're told it makes this much carbon dioxide and this much water vapor. So this carbon here, it's going to be converted into carbon dioxide. And it's the carbon in the carbon dioxide, and this hydrogen here is gonna be converted to this hydrogen in water vapor. So we need to find the massive carbon and the mass of hydrogen. It will go here and, well, find out how much carbon we have. So we have 0.1783 g of carbon dioxide. We'll divide that by the atomic mass of the entire carbon dioxide molecule. That's gonna be 44. And then we multiply that number. Bye. The mass of just the carbon because that's all we care about is the carbon. We'll do that. And when we do that, we see that 0.1783 divided by 44 times 12 is 0.486 So we have 0.4 86 g of carbon. All right, we've already got carbon over here, so I don't want to run it twice. All right, So for this hydrogen, we're going to take the 0.734 and we divided by the mass of water. And then we need to multiply it by how maney parts we have of hydrogen. So hydrogen is one atomic mass. But we have two molecules, so we're gonna add it to their We do. 0.70734 divided by 18. Arms to and it gives you point 00 815 0.0 a one. Bob Grant. All right, so this is how much carbon we have. This is how much hydrogen have so we can find out how much oxygen we have since we're given the mass of the sample. So the oxygen, it's just gonna be present Missile on it up here. This is gonna be present in the remaining mountain over here, So we'll do 0.1, and we'll take away 0.486 and then we'll take away another 0.815 And that gives you 0.0 43 25 Graham. All right, so this is how much we have of each element in this 0.1 sample, but we need the moles of each. How many moles? We have to find the Mueller ratio for the empirical formula. So we're gonna divide each one of these Bother Mueller. Matt. So carbons is 12. Cartagines is one oxygen, and then we'll see what we get. So 0.4 86 divide about 12. It's gonna be 0.4 or five. All right. And this is the number of moles, and then we have 0.815 divided by one. That's the massive hydrogen. All right, And then we have point 043 to 5, divided by 16. And that gives you one second. It gives you 0.27 old. So this is how many moles we have of each. So next we just need to find the molar ratio, and we divide each one of these numbers. Ah, the smallest number here, which is 0.27 So we'll divide this number by 0.27 and we get 1.5. This number, divided by 0.0 27 is going to give us three and then this to about about itself is one. If we write this out, we'll have C 1.5 h three. Oh, and then understood one. All right, so we need to find the lowest number for all of the's three to behold. And we could just multiply this 1.5 by two. And you see that we'll have C three h six, 02 and that will be our empirical formula.