Answer : The empirical formula of the iron oxide is [tex]Fe_2O_3[/tex].
Solution : Given,
Mass of iron oxide = 0.450 g
Mass of iron = 0.315 g
Molar mass of iron = 56 g/mole
Molar mass of oxygen = 16 g/mole
First we have to calculate the Mass of oxygen.
Mass of iron oxide = Mass of iron + Mass of Oxygen
0.450 g = 0.315 g + Mass of oxygen
Mass of oxygen = 0.450 - 0.315 = 0.315 g
Step 1 : convert given mass into moles.
Moles of Fe = [tex]\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}= \frac{0.315g}{56g/mole}=0.005625moles[/tex]
Moles of O = [tex]\frac{\text{ Mass of O}}{\text{ Molar mass of O}}= \frac{0.135g}{16g/mole}=0.0084375moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Fe = [tex]\frac{0.005625}{0.005625}=1[/tex]
For O = [tex]\frac{0.0084375}{0.005625}=1.5[/tex]
The ratio of Fe : O = 1 : 1.5
To make the ratio as a whole number multiply numerator and denominator by 2.
The ratio of Fe : O = [tex]\frac{1\times 2}{1.5\times 2}=2:3[/tex]
The mole ratio of the element is represented by subscripts in empirical formula.
Therefore, the empirical formula = [tex]Fe_{2}O_{3}[/tex].
