September 22, 2020, 07:25:52 AM
Forum Rules: Read This Before Posting

### Topic: Metallic Oxide with 52,9411% metal  (Read 596 times)

0 Members and 1 Guest are viewing this topic.

#### Vsauce

• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Metallic Oxide with 52,9411% metal
« on: February 17, 2019, 01:25:04 PM »
All the info I have is the % of metal (52,9411%). I need to identify the oxide and find a thermal decomposition reaction with oxide as a product. This is all I could do:
%oxygen= 100 – 52,9411 = 47,0589
M2On = the oxide
AM = the atomic mass of the metal
n = the valence of the metal
(2 x AM x 100)/(2 x AM+n x 16 )=52,9411 % Metal
200 x AM = 105,8822 x AM + 847,0576 x n
(16 x n x 100)/(2 x AM+n x 16 )=47,0589 % Oxygen
1600 x n = 94,1178 x AM + 847,0576 x n
94,1178 x AM = 847,0576 x n
AM = 9 x n

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7522
• Mole Snacks: +525/-88
• Gender:
##### Re: Metallic Oxide with 52,9411% metal
« Reply #1 on: February 17, 2019, 04:21:59 PM »
Now you should find a mass of metal which is a multiplicity of 9 where n (multiplicity or valence of an atom) is the number of oxygen atoms in the oxide.
AWK