When 1.5173g of an a organic iron compound contains Fe,C,HH and O was burned in oxygen 2.838g CO2 and 0.8122g H2O wereproduce.in a separate experiment to determine mass % of iron, 0.3355g of the compound yield 0.0758g Fe2O3.What are the empirical formula of the compound

Answer 1

This is a long problem but not complicated.
First, determine %Fe.
am = atomic mass
mm = molar mass
0,0758 g Fe2O3 x (2 am Fe./mm Fe2O3) = approx 0.053 and
%Fe in that 0.3355 g sample is (0.053/0.3355)*100 = approx 15.81%,
Now g Fe in the initial sample of 1.5173 = 0.1581*1.5173 = about 0.24 but you need a more accurate answer than that. Remember that is g Fe in the initial sample.
g C in the sample is mass CO2 x (am C/mmCO2) = ?
g H in the sample is mass H2O x (2*am H/mmH2O) = ?
g O in the sample is 1.5173 - gC - g H = ?

Now convert all of the grams to mols.
mols Fe = g Fe/am Fe = ?
mols C = g C/am C = ?
mols H = g H/am H = ?
mols O = g O/am O = ?

Now you want to find the ratio of the elements to one another an the formula will be FewCxHyOz and the job is find the values of w,x,y,z. The easy way to do that is to divide all of the numbers by the smallest number of the 4. I ran through the calculations and that number is for Fe and for mols Fe I obtained 0.00429. Values for w(= 1 of course), x,y,z follow. Post your work if you get stuck.