FeO + Fe3O3 + 1/2O2 → 2Fe203
→ Molar mass of Fe2o3 = 159.69g/mol
→ Molar mass of FeO = 71.84g/mol
→Molar mass Fe3o4 = 231.53g/mol
1. first of all , let consider the amount of Fe in Fe2O3
Atomic Mass of Fe = 58.845 g/mol
Percentage of Fe in Fe2)3 = ( 2 x mass of Fe / Molar Mass Fe203) = 2x(55.845) /159.69g/mol
= 0.6994 * 100 = 69.94 %
2. Mass of Fe in 4.195g of Fe2O3 will then be :
0.06994 * 4.195g = 2.934 g
3.(i) percentage of Fe in FeO = (55.845 / 71.84g/mol ) = 0.77735* 100 =77.735
(ii)percentage of Fe in Fe3O4 = (3x 55.845)/231.53g/mol= 0.72359*100 =72.359
<em>(remember , we only have molar mass of FeO as well as molar mass for Fe3O4 , we do not have their ma</em>ss)
4. so we will say; let x be the mass for Fe3O4
and ; let y be the mass of FeO
However , the mass of FeO and FeO3O4 must equal 3.970 as given in the question.
meaning x +y = 3.970; therefore x = (3.970-y)
But, we also know from the above percentage calculations; that :
• Mass of Fe in FeO = 0.77735 = 0.77735 ,y
,
• Mass of Fe in Fe3O4 = 0.72359 = 0.72359,x
5. now lets solve for x and y by equating the above:
0.77735 y + 0.72359x =2.934 g
0.77735 y + 0.72359(3.970-y) =2.934 g
solving for y; you will get :
0.07735y + 2.8726 -0.72359y = 2.934
0.05385y = 0.061347
therefore y = 1.142 g = mass of FeO
Finally :
Mass percentage= (Mass FeO /given mass of the mixture ) * 100
= 1.142g/3.970 *100
= 28. 7 %
