# Question e3e98

Oct 12, 2015

Here's what I got.

#### Explanation:

The trick here is to use the given mole ratio to find a relationship between the ${\text{Fe}}^{2 +}$ and ${\text{Fe}}^{3 +}$ ions, on one side, and the ${\text{O}}^{2 -}$ ions on the other hand.

You know that the compound contains a mixture of ${\text{Fe}}^{3 +}$ and ${\text{Fe}}^{2 +}$ ions, but you don't know how many of each you have.

This is where the charge of the compound comes into play. The compound must be neutral, so the $\left(2 -\right)$ charge of the oxygen must be balanced by the overal charge of the iron cations.

To make the calculations as simple as possible, let's say that you have 100 moles of the compound, which will contain

• 93 iron(II) and iron(III) ions
• 100 ions of oxygen

Let's say that you have $x$ iron(II) ions and $y$ iron(III) ions. You can say that

$x + y = 93$

Now focus on the charge. You have 100 moles of ${\text{O}}^{2 -}$, which means that you must have

${\underbrace{x \cdot \left(2 +\right)}}_{\textcolor{b l u e}{\text{total charge of iron(II) ions")) + overbrace(y * (3+))^(color(red)("total charge of iron(III) ions}}} = | 100 \cdot \left(2 -\right) |$

Use the first equation to get $x = 93 - y$, then replace $x$ in the second equation

$2 \cdot \left(93 - y\right) + 3 y = 200$

$186 - 2 y + 3 y = 200$

$y = 200 - 186 \implies y = 14$

Therefore, you have $14$ iron(III) ions and

$93 - 14 = 79 \to$ iron(II) ions

Now, I don't know if you want to determine the percent composition of iron(III) ions in the total ions of iron or in the compound, so I'll show you both.

("14 Fe"^(3+)color(red)(cancel(color(black)("ions"))))/(93color(red)(cancel(color(black)("iron ions")))) xx 100 = 15.1% -> among iron ions

or

("14 Fe"^(3+)color(red)(cancel(color(black)("ions"))))/(193color(red)(cancel(color(black)("ions")))) xx 100 = 7.25% -># among total number of ions