# Question #e3e98

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The trick here is to use the given mole ratio to find a relationship between the

You know that the compound contains a mixture of

This is where the *charge* of the compound comes into play. The compound must be *neutral*, so the

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) ions100 ions of oxygen

Let's say that you have

#x + y = 93#

Now focus on the charge. You have **100** moles of

#underbrace(x * (2+))_(color(blue)("total charge of iron(II) ions")) + overbrace(y * (3+))^(color(red)("total charge of iron(III) ions")) = |100 * (2-)|#

Use the first equation to get

#2 * (93-y) + 3y = 200#

#186 - 2y + 3y = 200#

#y = 200 - 186 implies y = 14#

Therefore, you have

#93 - 14 = 79 -># 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