Question 9cc47

Jun 26, 2017

The mass percentage of $\text{Fe"^"2+}$ in ${\text{Fe"_0.93"O}}_{1.00}$ is 64.93 %.

Explanation:

Assume that we have 100 formula units of ${\text{Fe"_0.93"O}}_{1.00}$.

Then we have 93 $\text{Fe}$ atoms and 100 $\text{O}$ atoms.

The $\text{Fe}$ atoms are a mix of $\text{Fe"^"2+}$ and $\text{Fe"^"3+}$ ions, and the $\text{O}$ atoms are present as $\text{O"^"2-}$ ions.

The total charge of the $\text{O"^"2-}$ ions is 200-, so the total charge of the $\text{Fe"^"2+}$ and $\text{Fe"^"3+}$ ions must be 200+.

Let $x = \text{number of Fe"^"2+"color(white)(l) "ions}$.
Then $\text{93 – x" = "number of Fe"^"3+"color(white)(l) "ions}$.

The total + charge of these ions is

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

$2 x + 279 - 3 x = 200$

$x = 279 - 200 = 79$

So, 100 formula units of ${\text{Fe"_0.93"O}}_{1.00}$ contain

$\text{79 Fe"^"2+}$
$\text{14 Fe"^"3+}$
$\text{100 O"^"2-}$

"% Fe"^"2+" = (79 × 55.84 color(red)(cancel(color(black)("u"))))/((79 × 55.84 + 14 × 55.84 + 100 ×16.00) color(red)(cancel(color(black)("u")))) × 100 % = 4411/6793 × 100 % = 64.93 %#