What is the empirical formula of the compound with a mass percent composition of 70.0% Fe and 30.0% O? (Type your answer using the format CH4 for CH4
2 answers
Fe2O3
Take a 100 g sample which gives you
70.0 g Fe and 30.0 g O.
mols Fe = 70.0/55.85 = about 1.25
mols O = 30.0/16 = 1.875
Find the ratio; the easy way to do that is to divide the smaller number by itself and do that to the other number too.
1.25/1.25 = 1
1.875/1.25 = 1.5
This obviously 2:3 so the formula is
Fe2O3.
70.0 g Fe and 30.0 g O.
mols Fe = 70.0/55.85 = about 1.25
mols O = 30.0/16 = 1.875
Find the ratio; the easy way to do that is to divide the smaller number by itself and do that to the other number too.
1.25/1.25 = 1
1.875/1.25 = 1.5
This obviously 2:3 so the formula is
Fe2O3.
1 answer