What is the mole ratio of Fe_3O_4 to Fe?

Jan 27, 2018

Suppose instead I had a dozen formula units of $F {e}_{3} {O}_{4}$...

Explanation:

I think that by using simple arithmetic you should very quickly be able to tell me that I had 36 iron atoms, and 48 oxygen atoms....do you agree?

Here we have a $\text{mole}$ of $F {e}_{3} {O}_{4}$ units, i.e. a mass of $231.4 \cdot g$...but a mole specifies Avogadro's number of particles....we write ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.. And so we can use the $\text{mole}$ as a collective number precisely as we do the $\text{dozen}$.

And so we gots $1 \cdot m o l$ $F {e}_{3} {O}_{4}$...we gots $3 \cdot m o l$ of iron atoms, and $4 \cdot m o l$ of oxygen atoms...

If I got one mole of ${P}_{4} {O}_{10}$, can you tell me the numbers and the masses of the constituent atoms?