What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

1 Answer
Apr 24, 2016

Answer:

#Fe_2O_3#

Explanation:

We assume there are #100*g# of compound, and thus there are:

#(i)# #(69.9*g)/(55.85*g*mol^-1)# #=# #1.25*mol*Fe#, and

#(ii)# #(30.0*g)/(15.999*g*mol^-1)# #=# #1.88*mol*O#.

If we divide thru by the lowest molar quantity we get a formula of #FeO_(1.5)#. Because, by definition the empirical formula is the simplest WHOLE number ratio, we double this result to give #Fe_2O_3#.