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# To find the empirical formula of tin oxide, you first react tin metal with nitric acid in a porcelain crucible. The metal is converted to tin nitrate, but, on heating the nitrate strongly, brown nitrogen dioxide gas is evolved and tin oxide is formed. In the laboratory you collect the following data:Mass of crucibleMass of crucible plus tin $14.710 \mathrm{g}$Mass of crucible after heating $15.048 \mathrm{g}$What is the empirical formula of tin oxide?

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##### Stephanie C.

University of Central Florida

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University of Kentucky

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University of Toronto

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So this question wants us to determine the empirical formula of tin oxide, which would have formed outside of this crucible and this question, trying kind of trying to confuse you by including the weight of the crucible where your calculations. But if you remove that wait, you can get just the weight of the 10 by itself and the weight of the 10 oxide by itself. So since we know that we have 1.591 grams of tin oxide and we know that it was made using 1.253 grams of 10 we can determine that the remaining 0.3 grams 0.338 grams comes from oxygen, so we can now determine how many moles of 10 and how many moles of oxygen we have. Our number of mole of 10. It's found by dividing our weight of 10 Ingram's by the molecular weight of two, or rather, the atomic weight of 10. Ingram's Permal. We should have 0.106 Mol of Tim, And if you do the same for oxygen, we should find you have 0.211 more of oxygen And if we divide the number of oxygen by a number of mole of Tim, we should find we get approximately to move oxygen for everyone moved in. And this gives us our empirical formula of s and 02

McMaster University
##### Stephanie C.

University of Central Florida

##### Morgan S.

University of Kentucky

##### Jake R.

University of Toronto

Lectures

Join Bootcamp