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Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are `69.9` and `30.1`, respectively.A. FeOB. `Fe_3O_4`C. `Fe_2O_3`D. `FeO_2`

Answer» For element Fe, mole of atoms `=(69.9)/(56)=1.25`
For element O, mole of atoms `=(30.1)/(16)=1.88`
Mole ratio of Fe `=(1.25)/(1.25)=1`,
Mole ratio of O `=(1.88)/(1.25)=1.5`
Simplest whole number ratio of Fe and O= 2, 3
Empirical formula of compound `=Fe_2O_3`
Molecular mass of `Fe_2O_3=160`
`n=("Molecular mass")/("Empirical formula mass")=160/160=1`
Molecular formula `=Fe_2O_3`


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