L #39

Moderators: Chem_Mod, Chem_Admin

Kayla Ikemiya 1E
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

L #39

Postby Kayla Ikemiya 1E » Thu Oct 05, 2017 12:45 am

why do you use O instead of O2 for this problem?

An Dang 3F
Posts: 20
Joined: Fri Sep 29, 2017 7:07 am

Re: L #39

Postby An Dang 3F » Thu Oct 05, 2017 1:20 am

The empirical formula will still be the same even if you use 02 by changing the molar mass to 32 grams and solving for the moles. Then you would divide both the Tin and the Oxygen by the lowest number, and the result will be Sn02.

Charles Ang 1E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: L #39

Postby Charles Ang 1E » Thu Oct 05, 2017 10:44 am

For part B of this question, how is the name of the oxide determined? To my understanding, the roman numeral refers to the subscript of the metal. Why is SnO2 called tin(IV) oxide?

Maeve Gallagher 1J
Posts: 56
Joined: Fri Sep 29, 2017 7:07 am

Re: L #39

Postby Maeve Gallagher 1J » Thu Oct 05, 2017 12:00 pm

The Roman numeral refers to the charge on tin. I believe that some metals can have a few different charges, so the the IV is clarifying that this is tin with a charge of +4, creating the need for two oxygen atoms each with a charge of -2. Hope that helps !

Vincent Grospe 3C
Posts: 20
Joined: Fri Sep 29, 2017 7:07 am

Re: L #39

Postby Vincent Grospe 3C » Thu Oct 05, 2017 12:37 pm

When oxygen is alone (as well as hydrogen, nitrogen, chlorine, bromine, iodine, and fluorine -- HONClBrIF), it exists as a diatomic molecule -- hence, the subscript of 2. However, oxygen is apart of the compound -- it does not stand alone. So you would use the molar mass of one oxygen atom.

However, you can use two moles of oxygen atom (32 g/mol) as part of your stoichometry calculations, but you must add an additional stoichiometry factor (or mole ratio) of 2 moles of O/1 mole of O2.


Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 4 guests