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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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Answer» % of iron by mass = 69.9 % [Given] % of oxygen by mass = 30.1 % [Given] Atomic mass of iron = 55.85 amu Atomic mass of oxygen = 16.00 amu Relative moles of iron in iron oxide =%mass of iron/Atomic mass of iron = 69.9/55.85 = 1.25 Relative moles of oxygen in iron oxide = %mass of oxygen /Atomic mass of oxygen = 30.01/16=1.88 Simplest molar ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3 ∴ The empirical formula of the iron oxide is Fe2O3. |
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