Question

empirical formula of an oxide of iron 72.4%iron by mass is

Answer 1

HELLO MACHII ✋....

1. Assume you have 100 grams of this stuff.
You then have:
72.4 g of Fe
27.6 g of O

2. Convert each to moles.
72.4 g of Fe/55.85 = 1.296 mol Fe
27.6 g of O/16 = 1.725 mol O

3. Divide by the least moles number.
1.296 mol Fe/1.296=1
1.725 mol O/1.296=1.33

4. Write out the formula.
FeO1.33
The 1.33 = 4/3

✨So, just multiply by the lowest common number to get the 4/3 to be a whole number,
which happens to be a 3, so
the empirical formula is Fe3O4.




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