Emperical formula of an oxide of iron

Jan 2020
16
0
Mumbai(Bombay), Maharashtra State, India.
How to determine the emperical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

My attempt to answer this question:-

average automic mass of iron is \(\displaystyle 55.845g*mol^{-1}\) and that of dioxygen is \(\displaystyle 15.999g*mol^{-1}\) Now, I found three types of iron oxides in the periodic table having chemical formula \(\displaystyle FeO,Fe_2O_3, FeO*Fe_2O_3\). I don't understand how to proceed further?

So if any member knows the answer to this question, may reply.
 
Jan 2020
16
0
Mumbai(Bombay), Maharashtra State, India.
Hello,
My answer:- The emperical formula of iron oxide which has 69.9% iron and 30.1% dixygen is $Fe_4O_6$.

I don't know whether my answer is correct or wrong. If any member knows the correct answer may reply.
 
Jan 2020
16
0
Mumbai(Bombay), Maharashtra State, India.
My answer in the previous post can be written as $Fe_2O_3$ as an emperical formula.