Question

determine the empirical formula of an oxide of iron which has 69.9 % iron and 30.1% dioxygen by mass

Answer 1

Answer : The empirical formula of the compound is, $Fe_2O_3$

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Fe = 69.9 g

Mass of O = 30.1 g

Molar mass of Fe = 55.85 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of Fe = $\frac{\text{ given mass of Fe}}{\text{ molar mass of Fe}}= \frac{69.9g}{55.85g/mole}=1.25moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30.1g}{16g/mole}=1.88moles$

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

For Fe = $\frac{1.25}{1.25}=1$

For O = $\frac{1.88}{1.25}=1.5$

The ratio of Fe : O = 1 : 1.5 = 2 : 3

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

Thus, the Empirical formula = $Fe_2O_3$

Answer 2

Emperical Formula Of Iron Oxide is Fe2O3

Explanation:

Percent of Fe by mass = 69.9 % [As Given Above]

Percent of O2 by mass = 30.1 % [As Given Above]

Relative moles of Fe in Iron Oxide:

= $\frac{Percent Of Iron By Mass}{Atomic Mass Of Iron}$

= $\frac{69.9}{55.85}$

= 1.25

Relative moles of O in Iron Oxide:

= $\frac{Percent Of Oxygen By Mass}{Atomic Mass Of Oxygen}$

= $\frac{30.1}{16.00}$

= 1.88

Simplest molar ratio of Fe to O:

= 1.25: 1.88

= 1: 1.5

≈ 2:3

Therefore, empirical formula of iron oxide is Fe2O3.