Question

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

Answer 1

Hint- Here, we will consider that there are 100 molecules of the metal oxide ${{\text{M}}_{0.98}}{\text{O}}$ present. Then, amounts of ${{\text{M}}^{3 + }}$ present in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ is determined and then, we will apply the formula i.e., Fraction of the metal which exists as ${{\text{M}}^{3 + }}$ = $\dfrac{{{\text{Amount of }}{{\text{M}}^{3 + }}{\text{ in metal M }}}}{{{\text{Amount of M in }}{{\text{M}}_{0.98}}{\text{O}}}} \times 100$.

Complete answer:
Let us suppose there are 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ metal oxide present. According to the composition of ${{\text{M}}_{0.98}}{\text{O}}$, we can say that for 1 molecule of ${{\text{M}}_{0.98}}{\text{O}}$, 0.98 atoms of metal M are there (present as ${{\text{M}}^{2 + }}$ and ${{\text{M}}^{3 + }}$) and 1 atom of oxygen (O) is present.
For 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$, $0.98 \times 100 = 98$ atoms of metal M are there (present as ${{\text{M}}^{2 + }}$ and ${{\text{M}}^{3 + }}$) and $1 \times 100 = 100$ atoms of oxygen (${{\text{O}}^{ - 2}}$) is present.
As we know that oxygen ion occurs as ${{\text{O}}^{ - 2}}$ i.e., usually have a charge of -2 when written in ion form
Now, let us assume that x amount of ${{\text{M}}^{3 + }}$ exists in the compound ${{\text{M}}_{0.98}}{\text{O}}$ when 100 molecules of this compound are taken.
Amount of ${{\text{M}}^{3 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ = x
Since, total amount of metal M in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ = 98
Also, Total amount of metal M in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ = Amount of ${{\text{M}}^{3 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ + Amount of ${{\text{M}}^{2 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$
By substituting the values in the above equation, we get
$\Rightarrow$ 98 = x + Amount of ${{\text{M}}^{2 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$
$\Rightarrow$ Amount of ${{\text{M}}^{2 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ = 98 – x
As we know that the overall charge on any compound should be zero (i.e., the compounds are neutral)
For the metal oxide (compound) ${{\text{M}}_{0.98}}{\text{O}}$, we can write
+ 3(Amount of ${{\text{M}}^{3 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$) + 2(Amount of ${{\text{M}}^{2 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$) - 2(Amount of ${{\text{O}}^{ - 2}}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$) = 0
$\Rightarrow + 3\left( x \right) + 2\left( {98 - x} \right) - 2\left( {100} \right) = 0 \\\ \Rightarrow 3x + 196 - 2x - 200 = 0 \\\ \Rightarrow x - 4 = 0 \\\ \Rightarrow x = 4 \\\$
Therefore, the amount of ${{\text{M}}^{3 + }}$ in 100 molecules of ${{\text{M}}_{0.98}}{\text{O}}$ (or the amount of ${{\text{M}}^{3 + }}$ in 98 atoms in metal M) is equal to 4
Fraction of the metal which exists as ${{\text{M}}^{3 + }}$ in terms of percentage = $\dfrac{{{\text{Amount of }}{{\text{M}}^{3 + }}{\text{ in metal M }}}}{{{\text{Amount of M in }}{{\text{M}}_{0.98}}{\text{O}}}} \times 100 = \dfrac{4}{{98}} \times 100 = 4.08$%
Therefore, the required fraction in terms of percentage is equal to 4.08%

Hence, option A is correct.

Note- In this particular problem, if we were asked for the fraction of the metal which exists as ${{\text{M}}^{2 + }}$ instead of ${{\text{M}}^{3 + }}$ then the formula used would be given by Fraction of the metal which exists as ${{\text{M}}^{2 + }}$ in terms of percentage = $\dfrac{{{\text{Amount of }}{{\text{M}}^{2 + }}{\text{ in metal M }}}}{{{\text{Amount of M in }}{{\text{M}}_{0.98}}{\text{O}}}} \times 100$ where the amount of ${{\text{M}}^{2 + }}$ in metal M is equal to (98-x) = (98-4) = 94.