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 $ = \dfrac{{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$ = \dfrac{{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 $ = \dfrac{{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 .