Question

Two molecules of metal M have 27.6% and 30% oxygen by weight. If the formula of the first oxide is ${{M}_{3}}{{O}_{4}}$. What is the formula of second oxide?

Answer 1

Hint An oxide is a chemical compound which contains at least one oxygen atom with any other element. Oxide is di anion of oxygen represented as ${{O}^{2-}}$ in this state the oxygen is present in -2 oxidation state and acts as an anion which carries negative charge.

Complete Step by step solution:
According to the question in the first oxide oxygen = 27.6, which represents that metal will be given by $100-27.6=72.4$parts by mass.
Formula of the oxide is given as ${{M}_{3}}{{O}_{4}}$ which describe that
72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen which is also equal to the 27.6 parts of mass of oxygen. Now we can say that 72.4 parts by mass of metal contains 3 atoms of metal. Now on the other case 30 parts by mass of oxygen are given from which metal can be calculated i.e. $100-30=70$ parts by mass of metal.
Now we calculated that 72.4 parts by mass of metal contains 3 atoms of metal then 70 parts by mass of metal contains:
$\dfrac{3}{72.4}\times 70=2.9$ atoms of metal
Similarly 27.6 parts by mass of oxygen contains 4 atoms of oxygen then 30 parts by mass of oxygen contains:
$\dfrac{4}{27.6}\times 30=4.35$ atoms of oxygen
Hence ration of metal to oxygen in second oxide will be = $2.90:4.35$ i.e. $1:1.5$ which can be equal to $2:3$ i.e. metal to oxide ratio for second oxide can be written as ${{M}_{2}}{{O}_{3}}$.

Note: Oxides can be formed due to electronegative nature of oxygen which forms stable chemical bonds with almost all metals and they give corresponding oxides of that metal. Ratio of metal oxide tells us about how many metals and oxygen atoms are connected with each other.