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Ch.3 - Molecules, Compounds & Chemical Equations

Chapter 3, Problem 88a

A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound. a. 1.245 g Ni, 5.381 g I

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Hi everyone. Today, we have a question telling us that a decomposition of a certain compound yields the following mass is 16.77 g of magnesium, 22.13 g of sulfur, 44.16 g of oxygen. And we need to identify the empirical formula of this compound. So first we need to change all of these two moles. So we have 16 .77 g magnesium. We're going to multiply that by one mole Over magnesium is molar mass, which is 24 0.31. So our grams are going to cancel out here And that is going to equal zero .69 moles. And for our software, we have 44 .16 g of oxygen. And we're going to multiply by one mole over 16 g of oxygen. And our grams here are going to cancel out. And that is going to equal 2.76 moles. And we have our sulfur, which is 22.13 g Times one Mole over Sulfur, smaller mass, which is 32.07. And our grams are going to cancel out. And that's going to leave us with 0. moles. And now we need to divide by our smallest, which is 0.69. So 0.69 divided by 0.69 equals, one, 2.76, divide by 0.69 equals four, 0.69 divided by 0.69 equals one. So that means our empirical formula will be magnesium sulfur and four oxygen's. And that is our final answer. Thank you for watching Bye.