Question

Choose the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively and its molecular mass is 160.
(A) $FeO$
(B) $F{e_3}{O_4}$
(C) $F{e_2}{O_3}$
(D) $Fe{O_2}$

Answer 1

As the ratio of the number of iron and oxygen atoms are different for all the given compounds, by calculating the molecular mass of each of the compound can is a quick approach.

Complete step by step solution:
The first step in calculating molecular formula is by converting the given mass percentage into mass in grams. Let us consider the amount of compound in the starting is 100g. So, Iron is 69.9g, oxygen is 30.1g.
Now convert the masses into moles. By using the formula of number of moles (n), which is the ratio of weight of the substance to its gram molecular mass.
Number of moles of iron $\dfrac{{69.9}}{{55.84}} = 1.25$
Number of moles of oxygen $\dfrac{{30.1}}{{16}} = 1.88$
Next we have to divide each moles by the smallest value obtained, 1.25
$\dfrac{{1.25}}{{1.25}} = 1,\dfrac{{1.88}}{{1.25}} = 1.5$. This means 1 mole of Fe and 1.5 mole of O respectively.
Multiply the number of moles of Oxygen and Iron with 2 to get the whole numbers we get,
(1 mole of Fe) 2 = 2 moles of Iron,(1.5 moles of O) 2 = 3 moles of Oxygen.
Using the number of moles obtained the empirical formula, is $F{e_2}{O_3}$ .
Calculate the molecular mass of the compound obtained and compare it with the given molecular mass. The molecular mass of $F{e_2}{O_3}$ is 2(55.84) + 3(16) = 159.68 $\simeq$ 160g.
The calculated molecular molar mass is equal to the given molecular mass i.e.160g.

So, the Molecular formula of the given Iron oxide is Option (C) $F{e_2}{O_3}$.

Note: To convert the number of moles of oxygen which is in fraction into the whole number, it must be remembered that all the moles of the constituent atoms have to be multiplied by the specific number. If only that particular atom with fractional moles is converted then the law of constant proportions does not obey.