A mixture of FeO and Fe3O4 (i.e. FeO.Fe2O3 ) actually consists of 2 units of FeO and 1 unit of Fe2O3. In case of FeO, The percentage of Fe in FeO =
Atomic weight of Fe
Formula weight of FeO
×100 =
56
72
×100=77.78% The percentage of Fe in Fe2O3 =
(Atomic weight of Fe)×2
Formula weight of Fe2O3
×100 =
112
160
×100=70.00% Suppose, 'x'g of FeO (in total) and 'y' g of Fe2O3 form the mixture that contains 75% Fe. Hence mathematically, 77.78%of x+70.00%y=75%of (x+y) or 77.78% of x+70.00%y=75.00 of x+75.00% of y or (77.78−75.00%) of x=(75.00−70.00)% of y or 2.77% of x=5.00% of y or
x
y
=
5.00%
2.78
=
500
278
=
250
139
or x:y=250:139 Now, percent amount of FeO in mixture (x) =
250
(250+139)
×100=
25000
389
%=64.3% Hence, percent amount of Fe2O3 in mixture (y) =(100−64.3)%=35.7%