Asked by Anonymous
A compound contains 69.9% iron and 30.1% oxygen. What is its empirical formula?
Answers
Answered by
DrBob222
Take a 100 g sample which gives you
69.9 g Fe
30.1 g O
mols Fe = grams/atomic mass = ?
mols O = grams/atomic mass = ?
Now determine the ratio of one element to the other with the smallest being no less than 1.00. The easy way to do that is to divide both numbers by the smallest number, then determine the whole number ratio from that.
69.9 g Fe
30.1 g O
mols Fe = grams/atomic mass = ?
mols O = grams/atomic mass = ?
Now determine the ratio of one element to the other with the smallest being no less than 1.00. The easy way to do that is to divide both numbers by the smallest number, then determine the whole number ratio from that.
Answered by
Anonymous
69.9 g Fe /55.845 g Fe = 1.252 mol Fe
30.1 g O /15.9994 g O = 1.881 mol O
1.252 mol Fe/1.252 = 1 mol Fe
1.881 mol O/1.252 = 1.502 mol O
FeO2?
30.1 g O /15.9994 g O = 1.881 mol O
1.252 mol Fe/1.252 = 1 mol Fe
1.881 mol O/1.252 = 1.502 mol O
FeO2?
Answered by
Amit
Fe2oe
Answered by
Amit
Fe2o3
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