Question

Please help with this problem

Answer 1

Answer : The empirical formula of the iron oxide is $Fe_2O_3$.

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 = $\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}= \frac{0.315g}{56g/mole}=0.005625moles$

Moles of O = $\frac{\text{ Mass of O}}{\text{ Molar mass of O}}= \frac{0.135g}{16g/mole}=0.0084375moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Fe = $\frac{0.005625}{0.005625}=1$

For O = $\frac{0.0084375}{0.005625}=1.5$

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 = $\frac{1\times 2}{1.5\times 2}=2:3$

The mole ratio of the element is represented by subscripts in empirical formula.

Therefore, the empirical formula = $Fe_{2}O_{3}$.