Question

Determine the empirical formula of an oxide of iron which has $69.9\%$ iron and $30.1\%$ oxygen. $(Fe = 55.85amu;O = 16.00amu)$ :
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

In this question firstly we have to sort out the given quantities, ( the percentage of elements and the atomic masses in this case ). Then we have to use the unitary method in order to apply the mole analysis in the question as the elements react and form the compounds in the specific mole ratio. So we will divide the given mass per $100$ g to get the moles. Then we have to find the ratio and we will get the answer.

Complete step by step answer:
Firstly we have to sort out whatever we know in the question. They are :
The amount of the iron in the iron oxide : $69.9\%$
The amount of the oxygen in the iron oxide : $30.1\%$
The atomic weight of the iron : $Fe = 55.85amu$
The atomic weight of oxygen : $O = 16.00amu$
we get that the amount of both the elements in the 100 g of iron oxide is $69.9g$ iron and $30.1g$oxygen.
In order-a to get the number of moles in per $100g$ of compound we should divide the given content by the atomic weights of the respective ones.
Step 1:
So,
The number of the moles of iron present in of $100g$ iron oxide are :
$\dfrac{{69.9}}{{55.85}} = 1.25$
The number of the moles of oxygen present in of $100g$ iron oxide are :
$\dfrac{{30.1}}{{16}} = 1.88$
Step 2:
Now we have to obtain the ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide. It is :
$\dfrac{{1.88}}{{1.25}} = 1.5:1 = 3:2$
Hence, the formulated chemical formula of the iron oxide will be : $F{e_2}{O_3}$

So, the correct answer is Option A .

Note:
For the cases in which we don't get the $100g$ automatically, we will first have to make it to it and then in order for that we have to obtain mass accordingly too. After that we will get the answer by acquiring the rest of the process.