Question

determine the empirical formula of an oxide of iron which has 69.9 % iron and 30.1% dioxygenby mass

Answer 1

% of iron by mass is 69.9
% of oxygen by mass is 30.1
now,
relative mole of iron in this oxide = (% of iron by mass)/(atomic mass of iron)= 69.9/56=1.25

similarly relative moles of oxygen =(% of oxygen by mass)/(atomic mass of oxygen=)30.1/16=1.88

and we know that empirical formula is the simplest whole number ratio , therefore 1.25 : 1.88 is equals to 1 : 1.5
it can also be written as 2 : 3
therefore empirical formula will be Fe₂O₃