Question

The nonstoichiometric oxide of iron called wustite has empirical formula $F{e}_{x}O$. It has a density of $5.75g/c{m}^3,a$ cubic unit cell with cell constant of $431pm$ and an $FCC$ arrangement of oxygen atoms. Find the fraction $Fe(II)$ in total iron.

Answer 1

Answer: 0.895

$5.75=\frac{(56x+16)4}{{\left(4.31\times 10^{-8}\right)}^3\times 6.02\times 10^{23}}$
$\Rightarrow x=0.95$
Using charge balance we write for $z$ moles of $+2$ and rest $+3$ iron.
$2(z)+3(0.95-z)=2$
$\Rightarrow 2z+2.85-3z=2$
$\Rightarrow z=0.85$
$\therefore$ fraction is $\frac{0.85}{0.95}=0.895$