Question

Question 15-16 is based on the following given information: A sample of ferrous oxide has actual formula $F{e}_{0.93}O$. In this sample what fraction of metal ions are $F{e}^{2+}$ ions?

• A. 0.93
• B. 0.81
• C. 0.79
• D. 0.14

Answer 1

The correct option is B 0.81

Let the formula of sample be

$\left(F{e}^{2+}{)}_x\right(F{e}^{3+}{)}_{y}O$.

On looking at the given formula of the compound

x + y = 0.93 ... (1)

Total positive charge on ferrous and ferric ions should balance the two

Units of negative charge on oxygen. Therefore,

2x + 3y = 2 ... (2)

$\Rightarrow$ $x+\left(\frac{3}{2}\right)y=1$ …(3)

On subtracting equation (1) from equation (3) we have

$\left(\frac{y}{2}\right)$ = 0.07 => y =0.14

On putting the value of y in equation (1) we get,

x + 0.14 = 0.93

$\Rightarrow$ x = 0.93 – 0.14

x = 0.79

Fraction of $F{e}^{2+}$ ions present in the sample $=\left(\frac{0.79}{0.93}\right)=0.81$