Determine the emperical formula of an oxide of irion which has69.9%iron and30.1%dioxygen by mass
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Hi,,
● Number of moles for each element
Fe= 69.9÷56=1.25
O=30.1÷16= 1.88
● Mole ratio by divid the lowest no
Fe=1
O= 1.88÷1.25= 1.5
☆Then correct the fraction by ×2
● So, the elements ratio are
Fe:O
2:3
● Empirical formula is Fe2O3
Hope the answer is helpful
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