Question

You heat 3.970 g of a mixture of Fe3O4 and FeO to form 4.195 g Fe2O3. The mass percent of FeO originally in the mixture was:

Answer 1

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

(remember , we only have molar mass of FeO as well as molar mass for Fe3O4 , we do not have their mass)

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 %