A 45.2-mg sample of phosphorus reacts with selenium to form
131.6...

Question

A 45.2-mg sample of phosphorus reacts with selenium to form
131.6 mg of the selenide. Determine the empirical formula of
phosphorus selenide.

Accepted Answer

Hi everyone here, we have a question telling us 174.7 mg of arsenic pyrite forms from a reaction between arsenic and sulfur. And it's asking what is the empirical formula of arsenal pyrite, given the following amount, 106.39 mg of arsenic. So first we need to figure out how much sulfur we have. So we're going to do that by taking our total, Which is 174. mg. And subtracting our amount of arsenic. So 106. milligrams of arsenic And that equals 68.3, 1 mg of sulfur. Now we need to get our moles of arsenic and sulfur. So for arsenic That would be 106. g times one mold Divided by Arsenic Molar Mass. So .92g And our g are going to cancel out. And that's gonna give us 1.42 moles. Next, we're going to do the same for sulfur. So we have 68. g times one mole Divided by Sulfur smaller mass, which can be found on the periodic table. And that's 32.07 g. Our g are gonna cancel out and that's going to equal 2.13 moles. And now we need to divide our lowest or lowest. Here's arsenic. So we're gonna divide both by 1.42. So that's going to equal one And that's going to equal 1.5. Since that's not a good whole number. We're going to multiply both of these by two. So one times two equals two, 1.5 times two equals 3. So we are going to have arsenic two, sulfur three. And that is our final answer. Thank you for watching. Bye.

A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. Determine the empirical formula of phosphorus selenide.

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