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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Vsauce on February 17, 2019, 01:25:04 PM

Title: Metallic Oxide with 52,9411% metal
Post by: Vsauce 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
M₂O_n = the oxide
A_M = the atomic mass of the metal
n = the valence of the metal
(2 x A_M x 100)/(2 x A_M+n x 16 )=52,9411 % Metal
200 x A_M = 105,8822 x A_M + 847,0576 x n
(16 x n x 100)/(2 x A_M+n x 16 )=47,0589 % Oxygen
1600 x n = 94,1178 x AM + 847,0576 x n
94,1178 x A_M = 847,0576 x n
A_M = 9 x n

Title: Re: Metallic Oxide with 52,9411% metal
Post by: AWK 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.