Write formulas for each of the following compounds:
(a) Sodium peroxide
(b) Aluminum bromide
(c) Chromium(III) sulfate

Question

Write formulas for each of the following compounds:
(a) Sodium peroxide
(b) Aluminum bromide 
(c) Chromium(III) sulfate

Accepted Answer

Welcome back everyone in this example, we need to provide the formula for the given compounds below. So looking at the first compound, we have vanadium three cell fight. So let's recognize that. This roman numeral here corresponds to our adam vanadium, which we recall is represented by the symbol V. On our periodic table. And because we have this roman numeral of three, this tells us that we have a three plus charge on our adam vanadium. Now we want to look at the sulfide portion where we should recognize that this is a poly atomic ion. And recall that sulfate is represented as multiple atoms where we have sulfur bonded to three atoms of oxygen. And recall that from memory, we should recognize that this has a two minus an ion charge. So to write out our compound, we want to recognize that vanadium based on its position on the periodic table is a metal. Whereas our poly atomic ion is basically considered just a poly atomic ion. So we have the combination of a medal and a poly atomic ion, meaning that this is going to be. And let's write that out, poly atomic ion plus a medal means that this compound should be an ionic compound Meaning we would have a transfer of electrons between our two regions here. So we would have vanadium adopt the charge of our cell fight, which was a two minus charge, meaning we would have a subscript of two and then our cell fight. We're going to place in parentheses S. 03 because it's going to take the charge of vanadium, which is a subscript of three here. So this is going to be our first answer as our formula for part a vanadium three cell fight. And let's not highlight that. But this is for part A. Now let's move on to part B. Which will do the work for below. We have potassium super oxide. So recognize that potassium is referring to our adam. Kay whereas super oxide is referring to another polly atomic ion where we would recall we have oxygen specifically two oxygen atoms. So we have a subscript of two and recall that from memory, super oxide has a an ion charge of two minus or sorry not to minus but one minus. We also want to recognize that potassium because it's in group one A of our periodic table will have a plus one caddy on charge. So let's write this clearly we have K with a plus one caddy in charge and recognize that potassium based on its position is considered a medal and oxygen based on its position. Or rather super oxide in this case because it's a poly atomic ion, it's basically considered a poly atomic ion yet again. So we have the combination of poly atomic ion and metal which we know yields an ionic compound meaning we would have a transfer of electron charge here. So we would form the compound que which will have a subscript of one that we don't have to write in when it adopts the charge of super oxide one minus. Whereas our super oxide is going to be in parentheses. Oh, sub to where? Outside of the parentheses. And actually we don't have parentheses since our potassium only consisted also of a charge of one. And so this is going to be our formula for our compound potassium super oxide. As our second answer for part B. Moving on to part C. We have aluminum fluoride. So looking at the first part, we have aluminum which on our periodic table we recall is a L. Recognized that aluminum is located in Group three A of our periodic table. And we recall that Adams and group three A form A plus three caddy in charge. Now we have fluoride where on our periodic table we recall is located in group seven A corresponding to a fluorine atom with a minus one charge. Since fluoride is the name of flooring as an ion. And so based on aluminum's position on the periodic table, we would recognize aluminum as a medal. And flooring we would recognize based on its position on the periodic table as a non metal but in this case it's a non metal with an ionic charge. So we still have a ionic compound here and we're going to be transferring these electron charges where aluminum is going to adopt the charge of flooring which is just a minus one charge. And our flooring adam is going to adopt the charge of our aluminum which will give it a subscript of three. And this would be our third final answer to complete this example as the chemical formula for each of our given compounds. I hope everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.

Write formulas for each of the following compounds: (a) Sodium peroxide (b) Aluminum bromide (c) Chromium(III) sulfate

Verified Solution

Key Concepts

Ionic Compounds

Polyatomic Ions

Nomenclature Rules