Question

Two metallic oxides contain 27.6% and 30% oxygen respectively. If the formula of first metal oxide is ${{M}_{3}}{{O}_{4}}$, that of second will be:
[A] $MO$
[B] $M{{O}_{2}}$
[C] ${{M}_{2}}{{O}_{5}}$
[D] ${{M}_{2}}{{O}_{3}}$

Answer 1

Hint: To find out the formula of the second metal oxide, we have to calculate the percentage of metal in the first metal oxide. After that, we have to compare the ratios of metal to oxygen in both the metal oxides and from that ratio the formula can be easily figured out.

Complete step by step answer:
Let us consider, the percentage of the metal present in the first metal oxide is x% and the total mass of the compound is 100g.
As we know, the given percentage of oxygen content of the first metal oxide = 27.6%

As the formula of the first metal oxide states that it contains 3 moles of the metal and 4 moles of oxygen,
1mole of oxygen is equivalent to 16g
Therefore, 4 moles of oxygen is $(16\times 4)$i.e. 64g
Similarly, 3 moles of metal corresponds to 3 times x.

So, % of ${{M}_{3}}{{O}_{4}}$= $[3x\div (3x+64)]\times 100$

But, as the question suggests, % of oxygen= 27.6 %
Therefore, % of metal= (100-27.6) %= 72.4%

Comparing above equations, we get,
72.4 = $[3x\div (3x+64)]\times 100$
Or, x=56

For the second oxide, %of oxygen= 30%
Therefore, %of metal= (100-30) %= 70%
$M\div O=(70\div 56)\div (30\div 16)$= 1.25:1.875 $\approx $2$\div $3, which explains that the second metal oxide contains 2 mole of the metal and 3 mole of the oxide.
Therefore, the formula of the second metal oxide is ${{M}_{2}}{{O}_{3}}$.
The correct answer is option [D] ${{M}_{2}}{{O}_{3}}$.

Note: It is important to remember here that while calculating the ratio, we are taking the percentage of the metal and the oxide as mentioned in the question and dividing it by the molar mass which gives us the number of moles of the required metal oxide.