Question

Determine the empirical formula of an oxide of iron which has 69.9 % iron and 30.1% dioxygen by mass

Answer 1

hello friend!!

%of o2=30.1
%of fe=69.9

=> relative moles of iron in iron oxide_:-
= %of o2 by mass/atomic mass of iron

69.9/55.85= 1.25(approx)

now,
relative moles of o2 in iron oxide:-

=> % of o2 by mass of fe/ atomic mass of o2.

=> 30.1/16.oo
= 1.88

simplest molar ration to o2 is here:-

1.25 : 1.88
=>1:1.5

=> emperical formulae is 2:3 which is :- fe2 o3.

answer......

thank you!