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Problem 88 Hard Difficulty

When a 2.118 -g sample of copper is heated in an atmosphere in which the amount of oxygen present is
restricted, the sample gains 0.2666 g of oxygen in forming a reddish-brown oxide. However, when 2.118 g of copper is heated in a stream of pure oxygen, the sample gains 0.5332 $\mathrm{g}$ of oxygen. Calculate the empirical formulas of the two oxides of copper.

Answer

Hence limit will be $\frac{-1}{6}$

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Lizabeth T.

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Video Transcript

problem. We have to chemical equations involving copper and oxygen. So copper metal reacts with oxygen gas to produce on unknown copper oxide X Just call it copper X oxygen. Why is, you know, unknown. But here, the amount of oxygen reacting is restricted. But in the other chemical, that's number this one in two. So in the second chemical equation, we have copper again, reacting with oxygen to make a copper oxide. But in this case, we have an excess amount of copper of oxygen's, so to denote excess, I'm just gonna put X s, which sounds like access, but yeah, So there's corpora, except, um, I'll call the at this time copper a oxygen be, since it's a completely different, you know, compound. All right, so what we've been given is the mass of each reacted. So we started off with 2.118 grams of copper on both you know, equations, 2.18 grams. And here we have to your 0.266 grams of oxygen, gas, throw. And on the excess oxygen side, we have your 0.5332 And just by looking at this, we can see that this figure is about double the amount of oxygen than this. That sort of gives us a hint. Maybe Teoh what the chemical equation will be formula. All right, So the chemical or the mass of the copper oxides here are negligible. And don't matter in as to, you know, finding out It's Moeller. It's, you know, molar ratios X and Y A B because, like way assume that these occasions are going fully forward and our that will get 100% percent 100% yield. So we're going to react all of this mess. So this compound is made up of all of this amount of copper and olives. My oxygen, Yeah, I hope that makes sense. So right now, in order for us to find the mole ratios of thes copper oxides, we need to work in malls. Yeah, that's it. Suggest so to find malls. We know that moles is equivalent to right over here mass over molar mass, something We can just look up on our periodic table atomic mass of copper and oxygen, like separately. And then we can just divide, which will give us basically 0.33 Gramps or moles all right of copper and zero point zero 16 six moles off oxygen over here. You just take this mass divided by the time it mass equivalent to molar mass of copper, which is, like 62.5. And that will also give us, you know, the same amount of moles. 0.33 and then for oxygen. We take this mass and divide it by, um, oxygen's mullah mats, which is, like, 16 on. And that will give us also 0.33 moles. Okay, so, no, these are basically X. This column right here, the most columns Basically what? This is basically A and B, but we need entered your values we can't work with, you know, thes that's the most. So in order for us to do that, we will take the larger, more quantity and divided by the smaller one to see, like what? Reach you the urn into each other and, like, what's the whole number ratio? So this is basically double if you should tell you how he basically double this number and these air recently equivalents like 1 to 1. And so this is to toe because it's double. Oh, yeah, That's basically your ex. Otherwise, and so this is copper, too. Let's take another color. And just first a race that's yeah, yeah. So this one over here is copper to I wanted Red, not what stopped. Oh, it's happening over here. Green. Well, OK, it's OK. Yes, Copper to oxygen. I m in just copper oxygen. 1212 to 1. So, yeah, we can say Tell that this is this copper over here is copper Roman numeral one That's its charge and this year's Carper to this year's copper one.

McMaster University
Top Chemistry 101 Educators
Lizabeth T.

Numerade Educator

Allea C.

University of Maryland - University College

JH
Jacquelin H.

Brown University

Jake R.

University of Toronto