Determine the emperical formula of an iron oxide with 69.9% iron and 39.01% oxygen by mass
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Explanation:
Empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound.
Form the available data Percentage of iron = 69.9
Percentage of oxygen= 30.1 Total percentage of iron & oxygen= 69.9+39.1= 100%
The empirical formula of the compound = Fe²O³
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Answer:
The iron oxide has 69.9% iron and 30.1% dioxygen by mass.
Thus, 100 g of iron oxide contains 69.9 g iron and 30.1 g dioxygen.
The number of moles of iron present in 100 g of iron oxide are = 69.9/55.8=1.25
The number of moles of dioxygen present in 100 g of iron oxide are = 30.1/32=0.94
The ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide is= 2*0.94/1.25=1.5:1=3:2
Hence, the formula of the iron oxide is Fe2O3
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