Question

Two metallic oxides contain 27.6% and 30% oxygen respectively. If the formula of the first oxide is\[{{\text{X}}_{\text{3}}}{{\text{O}}_{\text{4}}}\], that of the second will be:
$
  A.XO \\
  B.X{O_2} \\
  C.{X_2}{O_5} \\
  D.{X_2}{O_3} \\
 $

Answer 1

Hint: We do not know the mass of metal but we know the mass of oxygen. This will help us to solve the given problem. Also the percentage of oxygen is given in both the metal oxide.
Formula used:
We must also know the concept of mass percentage. It is a method of representing the concentration of an element in a compound or a component in a mixture. It is calculated as:
\[{\text{Mass percentage = }}\dfrac{{{\text{Mass of a component }}}}{{{\text{Total mass of the mixture}}}} \times 100\]

Complete step by step solution:
Given that, the formula of first oxide =\[{{\text{X}}_{\text{3}}}{{\text{O}}_{\text{4}}}\]
Let us consider the mass of the metal (X) = x
Then the percentage of metal in \[{{\text{X}}_{\text{3}}}{{\text{O}}_{\text{4}}}\]​= $\dfrac{{3x}}{{3x + 64}} \times 100$
But, we can get the percentage of metal by subtracting the percentage of oxygen from 100.
Therefore, \[\left( {100 - 27.6} \right){\text{ }} = {\text{ }}72.4{\text{ }}\% \]
Now, we can find the mass of metal by substituting the percentage of metal in mass percentage equation
So, $\dfrac{{3x}}{{3x + 64}} \times 100 = 72.4$
From the calculation we got the value of \[x{\text{ }} = {\text{ }}56\]
Therefore, the mass of metal ‘M’ is\[56\].
In another unknown oxide, given oxygen is\[30\% \].
Therefore, we can get the percentage of unknown metal by subtracting the percentage of oxygen from 100. the metal has to be=$100 - 30$=\[70\% \].
So, the ratio of mass of metal (M) and oxygen (O) can be done by dividing the mass percentage by corresponding molecular weight.
The mass of oxygen is \[16\] and that of metal is \[56\] (‘x’ already found)
We must know that
=$\dfrac{{{\text{ratio of mass of metal}}}}{{{\text{mass percentage of metal}}}}$
We can write as $M:O$
Therefore, substituting the mass of metal and oxygen and percentage of metal and oxygen
$\dfrac{{70}}{{56}}:\dfrac{{30}}{{16}}$
We get this by simple multiplication
$1.25:1.875$
Therefore, we can round off the value as
$2:3$
Hence, we can conclude that the molecular formula of second metal oxide is\[{{\text{M}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]. And the correct option is D.

Note: We can find that which metal forms an oxide having general chemical formula as \[{{\text{M}}_{\text{2}}}{{\text{O}}_{\text{3}}}\].
We can see that the metal has 3 valencies. We know group III A metal has 3 valency. Therefore, it must have the 3rd shell with three valence electrons. And which gives the electronic configuration as 2, 8, and 3. That means metal is Aluminium (Al). Hence, the metal forms an oxide as\[A{l_2}{O_3}\].