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secondary 4 | Chemistry
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Date Posted: 3 years ago
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Mixture of oxides of calcium, iron means there's Fe, Ca and O atoms/ions present.
Percentage of O by mass = 100% - the other two percentages = 100% - 18.5% - 51.9% = 29.6%
Then, divide all the % by the relative atomic mass of the respective elements.
For Fe, 51.9% ÷ 56 = 0.9268% (4s.f)
For Ca,18.5% ÷ 40 = 0.4625% (exact)
For O, 29.6% ÷ 16 = 1.85% (exact)
Compare 0.9268, 1.85 and 0.4625. Find the simplest ratio via dividing by the smallest value.
1.85 ÷ 0.4625 = 4
0.9268 ÷ 0.4625 ≈ 2.004 = 2 (nearest whole number)
The ratio of Fe to Ca to O = 2 : 1 : 4
Empirical formula : CaFe2O4
Percentage of O by mass = 100% - the other two percentages = 100% - 18.5% - 51.9% = 29.6%
Then, divide all the % by the relative atomic mass of the respective elements.
For Fe, 51.9% ÷ 56 = 0.9268% (4s.f)
For Ca,18.5% ÷ 40 = 0.4625% (exact)
For O, 29.6% ÷ 16 = 1.85% (exact)
Compare 0.9268, 1.85 and 0.4625. Find the simplest ratio via dividing by the smallest value.
1.85 ÷ 0.4625 = 4
0.9268 ÷ 0.4625 ≈ 2.004 = 2 (nearest whole number)
The ratio of Fe to Ca to O = 2 : 1 : 4
Empirical formula : CaFe2O4
Note that the ratio at the end is mole ratio.
In every 1 mole of ferrite there is 1 mole of Ca ions, 2 moles of Fe ions and 4 moles of O ions
Now 1 mole just means 6.022 × 1⅔
So for every single ferrite formula unit / compound, there is 1 Ca, 2 Fe and 4 O ions.
In every 1 mole of ferrite there is 1 mole of Ca ions, 2 moles of Fe ions and 4 moles of O ions
Now 1 mole just means 6.022 × 1⅔
So for every single ferrite formula unit / compound, there is 1 Ca, 2 Fe and 4 O ions.
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