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Determine the empirical formula of an oxide of iron which has `69.9%` iron and `30.1%` dioxygen by mass.

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% of ion by mass =69.9 % [Given]
% of oygen by mass =30.1 % [Given]
Ralative moles of iron oxide :
`= (% "of iron by mass")/("Atom mass of iron ")`
`=(69.9)/(55.85)`
=1.25
Ralative moles of oxygen iniron oxide :
`=(% "of oxygen by mass ") /("Atomic mass or oxygen")`
`=(30.1)/(16.00)`
=1.88
Simplest molar ratio of iron to oxygen :
=1.25: 1.88
=1: 1.5
`~- 2:3`
`therefore ` THe empirical formula of the oxide is `Fe_(2) O_(3)`

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