What Is the Mass Percent of Iron in the Compound?

[Complexity]

Homework Statement

0.3106 g of an iron-containing compound yield 0.07017 g Fe2O3 upon oxidation. What is the mass percent of the iron in the compound??

Homework Equations

stoichiometry?

The Attempt at a Solution

(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)(2 moles Fe/1 mole Fe2O3)(0.3106g Fe)(1 mole Fe) = 2.729468003x10^-4 grams

Then I took 2.72x10^-4 grams / 0.07017 compound grams of Fe2O3 = .00388979 x 100 = .3889 percent .

What did I do wrong?? Was I off on sig figs?

 

[carguies]

If an oxidation happens, you have to begin in a lower oxidation state, you only have two options:
Feº , or FeO

In both options it are a 2:1 proportion front a Fe2O3

 

[Borek]

(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)

OK. Number of moles of Fe₂O₃.

(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)(2 moles Fe/1 mole Fe2O3)

OK. Number of moles of Fe in Fe₂O₃.

(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)(2 moles Fe/1 mole Fe2O3)(0.3106g Fe)(1 mole Fe) = 2.729468003x10^-4 grams

No idea. It looks like ostrich, but has a head of giraffe.

 

[carguies]

Ok, I find the mistake; ¿Why you multiply by the mass of Fe 0.3106?

Compound with Fe 0.3106

Fe in the compound "x"

¿How much Fe produces Fe2O3?

0.07017 g Fe2O3 (1 mol Fe2O3 / 159.7 g Fe2O3)(2 mol Fe / 1 mol Fe2O3)---> mol of Fe

mol Fe x (55.845 g Fe / 1 mol Fe) = 0.04907 g Fe = x

Then

Rock = 0.3106 g
Fe in rock = 0.04907 g

Percentage (0.04907/0.3106) x 100 = 15.8%

 

[Borek]

Now it has a beak, as usual.