Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Question
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Solution
Sure, here are the steps to determine the empirical formula:
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Assume that you have 100g of the substance. This means you have 69.9g of iron (Fe) and 30.1g of oxygen (O).
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Convert the mass of each element to moles. The atomic mass of Fe is about 55.85g/mol and the atomic mass of O is about 16.00g/mol. So, you have (69.9g ÷ 55.85g/mol) = 1.25 mol of Fe and (30.1g ÷ 16.00g/mol) = 1.88 mol of O.
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Write the ratio of the moles. This is Fe1.25O1.88.
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To find the empirical formula, we need to divide by the smallest number of moles, which is 1.25. So, the ratio becomes Fe1O1.5.
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Since we can't have half of an oxygen atom in the formula, we multiply through by 2 to get the empirical formula Fe2O3.
So, the empirical formula of the oxide of iron in the problem is Fe2O3.
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