Question

Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are $$69.9$$ and $30.1$ respectively .

Answer 1

Hint : This kind of problem is approached by finding the empirical formula of a compound. Then we calculate n which is the ratio of molar mass of the compound to the empirical mass of the compound. Then we multiply n with the empirical formula and molecular formula of compound.

Complete answer:
> It is given in the problem that the iron oxide contains the mass percent of iron and oxygen are $$69.9$$ and 30.1 respectively .It means that $100$ g of iron oxide contains $69.9$ g of iron and $30.1$ g of oxygen. Now we will calculate number of moles of iron present in $100$ g of iron oxide .So number of moles of iron = weight in gram of iron /atomic weight of iron $={\displaystyle \frac{69.9}{55.8}}=1.25$ mole
> Now we will determine the number of moles of oxygen present in $100$ g of iron oxide .
Number of moles of oxygen =weight in gram of oxygen / atomic weight of oxygen$={\displaystyle \frac{30.1}{16.0}}$$=1.88$mole.
> Now we will calculate the ratio of number of moles of iron to number of moles of oxygen thus the ratio of iron to oxygen is ; $1.25:1.88=1:1.5$ ,on multiplying with $2$we get the ratio $2:3$ .
- Therefore the empirical formula of iron oxide is $F{e}_2{O}_3$ .Hence empirical formula mass of $F{e}_2{O}_3$will be equals to $2\times 55.85+3\times 16$ gram. We know that the molar mass of $F{e}_2{O}_3$ is $159.69$.
n= molar mass /Empirical formula mass $={\displaystyle \frac{159.69}{159.7}}=1$
We know that we can obtain molecular formula of $F{e}_2{O}_3$ by multiplying the empirical formula of with n .Thus the molecular formula of iron oxide is $F{e}_2{O}_3$.

Note : We have calculated molecular formula finding the number of moles of iron and oxygen .This gives us empirical formula then we calculated n by dividing molar mass to empirical mass of iron oxide .