[claudia]

Hello there! i am struggling over this problem here...hopefully someone could take a look at it and help  me out

1. What percent of iron is there an iron ore containing Fe3O4 is 0.2824g of ore gave 0.0917g of Fe2O3 by gravimetric analysis?

2. What percent of the ore was Fe3O4?



I came up with this answer: 
{ 55.85/[3(55.85)+4(16.00)] }* 100 =24.12% of Fe

{ 16.00/[3(55.85)+4(16.00)] }*100=6.91% O

--> 24.12+6.91=31.03% so 100%-31.03%=68.97% of ore?
does this make sense? can anyone help me out i am extremely lost and confused...??
i would really greatly appreciate it....
Claudia

[enahs]

M.W. Fe2O3 ~159g/mol

~55.8 g/mol Fe * 2 = ~111.6 g/mol Fe in Fe₂O₃
(~111.6/~159)*100 = ~70.2% Fe in Fe₂O₃.
0.702*0.0917g Fe₂O₃ = ~0.065 g Fe in the collected Fe₂O₃

There is no more then ~0.065g total, so for question 1.
(~0.065g/0.2842)*100 = ~22.9%


~0.065g Fe * 1mol Fe/ ~55.8 g = ~1.2 x 10^-3 mol Fe

~1.2 x 10^-3 mol Fe * 1 mol Fe₃O₄  = ~3.9 x 10^-4 mol Fe₃O₄
                             3 mol Fe

~3.9 x 10^-4 mol Fe₃O₄ * ~231.4g/mol = ~0.09 g Fe₃O₄

(~0.09 g Fe₃O₄ / 0.2842 g) * 100 =~31.8%



There are many ways to work it, and many I would consider a lot easier (i.e. a lot less teps). But that is not to hard and hopefully you can follow it.

I of course rounded a lot and was not very precise, so you have to redo the work.