Question

Experimentally it was found that a metal oxide has formula ${M_{0.98}}O$ . Metal M is present as ${M^{2 + }}$ and ${M^{3 + }}$ in its oxide. Fraction of the metal which exists as ${M^{3 + }}$ would be:
A. $5.08\% $
B. \[7.01\% \]
C. \[4.08\% \]
D. \[6.05\% \]

Answer 1

Hint: The given oxide is not a stoichiometric compound. In order to determine the fraction of charge of ${M^{3 + }}$ first we will balance the charge of both cation and anion i.e. metal and oxygen. After that we will find the amount of ${M^{3 + }}$ present in the compound.

Complete step by step answer:
According to question first we will
Consider one mole of the oxide then, we have 1 mole of ${M_{0.98}}O$ and ${O_2}^ - $

${M_{0.98}}$​ have both ${M^{3 + }}$ and ${M^{2 + }}$ present in the 1 mole.

Let moles of ${M^{3 + }}$ be $x$ and moles of ${M^{2 + }}$ be $0.98 - x$

Since the net charge in the compound is zero therefore first we will balance the charge

Now, by balancing the charge, we get

Metal have +3 and -2 while oxygen have -2 charge therefore we have
$3x + 2(0.98 - x) - 2 = 0$
On further solving we get
$x = 0.04$

Therefore the percentage of ${M^{3 + }}$ is given by
$\% {\text{ }}of{\text{ }}{M^{( + 3)}} = \dfrac{x}{{0.98}} \times 100$
On substituting the values we have
$\% {\text{ }}of{\text{ }}{M^{( + 3)}} = \dfrac{{0.04}}{{0.98}} \times 100 = 4.08$
Hence finally we conclude that the fraction of ${M^{3 + }}$ present in the given compound is 4.08 percent of the metal.

So, the correct option is C.

Note:
Metal oxides are crystalline solids that contain a metal cation and an oxide anion. They typically react with water to form bases or with acids to form salts. Thus, these compounds are often called basic oxides.