Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 3 - Exercises - Page 150b: 48

Answer

a. The compound would be called aluminum chloride, and its empirical formula would be AlCl$_{3}$. b. The compound would be called sodium oxide, and its empirical formula would be Na$_{2}$O. c. The compound would be called strontium fluoride, and its empirical formula would be SrF$_{2}$. d. The compound would be called calcium selenide, and its empirical formula would be CaSe.

Work Step by Step

a. Al (aluminum) has an oxidation number of 3+ whereas Cl (chlorine) is a halogen and has an oxidation number of 1-. To have a neutral compound, the positive and negative charges must balance one another. We would need three ions of chlorine for every aluminum ion. The compound would be called aluminum chloride, and its empirical formula would be AlCl$_{3}$. b. Na (sodium) is an alkali metal and has an oxidation number of 1+ whereas O (oxygen) is in group 16 and has an oxidation number of 2-. To have a neutral compound, the positive and negative charges must balance one another. We would need two ions of sodium for every oxygen ion. The compound would be called sodium oxide, and its empirical formula would be Na$_{2}$O. c. Sr (strontium) is an alkaline earth metal and has an oxidation number of 2+ whereas F (fluorine) is a halogen and has an oxidation number of 1-. To have a neutral compound, the positive and negative charges must balance one another. We would need two ions of fluorine for every ion of strontium. The compound would be called strontium fluoride, and its empirical formula would be SrF$_{2}$. d. Ca (calcium) is an alkaline earth metal and has an oxidation number of 2+ whereas Se (selenium) is in group 16 and has an oxidation number of 2-. To have a neutral compound, the positive and negative charges must balance one another, and they do in this compound. The compound would be called calcium selenide, and its empirical formula would be CaSe because in an empirical formula, we find the smallest whole number ratio of atoms for elements in that compound instead of the actual number of atoms in each formula unit of that compound.
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