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Problem 86 Hard Difficulty

Consider a hypothetical compound composed of elements $\mathrm{X}, \mathrm{Y},$ and $\mathrm{Z}$ with the empirical formula $\mathrm{X}_{2} \mathrm{YZ}_{3}$ . Given that the atomic masses of $\mathrm{X}, \mathrm{Y},$ and $\mathrm{Z}$ are 41.2 , 57.7, and 63.9, respectively, calculate the percentage composition by mass of the compound. If the molecular formula of the compound is found by molar mass determination to be actually $\mathrm{X}_{4} \mathrm{Y}_{2} \mathrm{Z}_{6},$ what is the percentage of each element present? Explain your results.

Answer

Hence limit will be $\infty$

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Top Chemistry 101 Educators
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Theodore D.

Carleton College

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Rice University

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Jacquelin H.

Brown University

Video Transcript

all right. So here we have two unknown compounds on were to find the mass percent each element in these compound and in this one. So these are two separate compounds and their unknown because they don't even have, like, proper chemical symbols. They're just using X y said to replace the proper ones. All right, so if we look closely at this compound in this compound to weaken steam, that each element But this element is a multiple of this element in terms of, you know, these compounds multiple of this cop out in terms of number off each Adam. So it was as though we just took this one and multiplied by two to get to terms to for one times 2 to 3 times juice. Six. So you can say that they have the same empirical formula describing the empirical formula. You know, the simplest of molecular formula simple is formula chemical formula using integer values. And not only would they do, they have the same empirical formula, but they also would share the same mass percent of each element. So here they've given us the more massive each of these unknown elements. So to find their mass percent sweet. Remember what mass percent is so it is very like the name, much like the name says, the mass percent of each element in the compound, but for one wall. So it's like mass of this own and in divided by the mass total but for one more of compound. So that's always the key here, one more so it's like a mass calculation, but for one more so and we need to mentor with Mass. And they've already given us the molar masses of each element, and we know that we're always working with one more. So then we know that if you have one mole of compound, this is the one that we then have two moles of X. You know the elemental sub scripts. Over here, the number of X Adams we have does tell us it's moles. So and why Over here it's one. We have one. Why Adam, per you know, compound and then for Z. We have three C Adams for come for one compound, so that will then tell us most so them. Everything here is in grams per mole, and there's column of the table. Just put that in brackets and these air own moles. This this is a table. So to them. So for mass, we just need to multiply mass My mom s moles by molar. And by, um, e once we make that division, we yet mass. So x would be you can see, like 80 82.4. Then why would just be 57.7? These are own grams. And so you would be, um, 1 91 0.7. Gramps, is it? So now if we total up this mass divine Tino mass of a compound, we would get 300 in 31 0.8 Gramps. So then to find mass percent, we just take this, um, individual mass and divided by the total mass. You get your calculator out. The one the master sent of X will be 24.83 for why would be 17 points. 39 three. There's three now we're out in the one for why one? Anyone were three. Jordan 31 57 points of seventh of it. Person. Yeah, way have it. And if we were Teoh, add this all up it with some up to 100 just to make it easier to see one for X. What said and it would be the same for both of these compounds. X two Wise at three and x four White Tuesday six Because there was empirical formula, same mass percent.

McMaster University
Top Chemistry 101 Educators
TD
Theodore D.

Carleton College

Nadia L.

Rice University

KS
Karli S.

Numerade Educator

JH
Jacquelin H.

Brown University