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Q.
Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is $M_3O_4$, then second one is
Some Basic Concepts of Chemistry
Solution:
In the first oxide,
Oxygen = 27.6 parts,
Oxygen = 27.6 parts,
As the formula of the first oxide is $M_3O_4$, this means 72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by mass.
In the second oxide,
Oxygen = 30.0 parts by mass and metal = 100 - 30 = 70 parts by mass.
But, 72.4 parts by mass of metal = 3 atoms of metal
$\therefore 70$ parts by mass of met $=\frac{3}{72.4}\times70$ atoms of metal
$= 2.90$ atoms of metal
Also, $27.6$ parts by mass of oxygen $= 4$ atoms of oxygen
$\therefore 30$ parts by mass of oxygen
$=\frac{4}{27.6}\times 30$ atoms of oxygen $= 4.35$ atoms of oxygen
Hence, ratio of $M : O$ in the second oxide $= 2.90 : 4.35 = 1 : 1.5 = 2 : 3$
$\therefore $ Formula of the metal oxide is $M_2O_3.$