Question

Determine the empirical formula of an oxide of iron which has $69.9\%$iron and $30.1\%$ oxygen by mass:
A.$F{e_2}{O_3}$
B.$F{e_4}{O_6}$
C.$F{e_8}{O_{12}}$
D.None of the above

Answer 1

Empirical formula or simplest formula provides the lowest whole number ratio of atoms in a compound. Basically, the relative number of atoms of every element in the compound is provided by this formula. The percentage composition of a compound directly leads to its empirical formula.

Formula used:
Number of moles $= \dfrac{{given\,mass}}{{molar\,mass}}$

Complete step by step answer:
Basically, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in the compound. Now, let’s find out the empirical formula of an oxide of iron which has $69.9\%$ iron and $30.1\%$ oxygen by mass.
Now, 100g of iron oxide contains $69.9\%$ iron and $30.1\%$ oxygen. So, the number of moles of iron present in 100g of iron oxide according to the formula will be:
$\dfrac{{69.9}}{{55.8}} = 1.25$
Similarly, number of moles of oxygen present in 100g iron oxide will be:
$\dfrac{{30.1}}{{32}} = 0.94$
Now, the ratio of number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide will be:
$= \dfrac{{2 \times 0.94}}{{1.25}}$
$= 1.5:1$
$= 3:2$
Therefore, the formula of the iron oxide is $F{e_2}{O_3}$

Hence, the correct option is A.

Note: Don’t get confused between molecular formula and empirical formula. Basically, the empirical formula of a compound gives the simplest ratio of the number of atoms present whereas the molecular formula gives the actual number of each different atom present in a molecule. Moreover, the molecular formula is commonly used and is a multiple of empirical formula. The general statement that relates the molecular and empirical formula is:

Molecular formula $= n \times$ empirical formula