Consider a 5.313 g mixture of FeO and Fe3O4..!!! HELP?

Question

Consider a 5.313 g mixture of FeO and Fe3O4. You react this mixture with an excess of oxygen to form 5.749 g Fe2O3. Calculate the percent by mass of FeO in the original mixture.

Answer 1

First of all Fe3O4 is a mixed oxide: it is FeO.Fe2O3 or if you prefer a 1:1 mixture of FeO and Fe2O3.

The molecular weights are
FeO : 71.8
Fe2O3 : 159.6
Fe3O4 : 231.4

Let's assume that in the original mixture you have x g FeO and y g Fe3O4. Then
x+y= 5.313 (1)

When you react the mixture with oxygen FeO (both the "free" and that from Fe3O4) will be oxidized to Fe2O3, while the Fe2O3 from Fe3O4 will remain unchanged.

In 231.4g Fe3O4 you have 159.6 g Fe2O3
in y g Fe3O4 you have ?
So ?= 159.6y/231.4= 0.690y g Fe2O3
and similarly you have (71.8/231.4)y= 0.310y g FeO from Fe3O4

The reaction with oxygen is

2FeO+ 1/2 O2 -> Fe2O3

so 2*71.8g FeO give 159.6g Fe2O3
x+0.310y g FeO give ? g Fe2O3
?= 159.6*(x+0.310y) /(2*71.8)= 1.111x+0.344y g Fe2O3 come from the reaction.

BUT remember that you will have some Fe2O3 which pre-existed in Fe3O4, so the total amount of Fe2O3 formed is that from the reaction+the amount that pre-existed, so
1.111x+0.344y +0.690y =5.749 =>
1.111x+1.034y =5.749 (2)

You have a system of two equations (1) and (2). Solving that will give you

x=3.316, y=1.997

The percent mass of FeO is x/5.313 *100%= 3.316/5.313 *100%= 62.4%

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