Question

Certain element X forms two oxides. The MAss percentage of oxygen in the two oxides is 43.7 % 56. 3 % respectively. Find the formula of the second is the formula of the first oxide is X4O6

Answer 1

In the first oxide, oxygen = 43.7 parts by mass
thus element X = 100 - 43.7 = 56.3 parts by mass

As the formula of the oxide is X₄O₆ , this means

56.3 parts by mass of element X = 4 atoms of element X
and 43.7 parts of oxygen = 6 atoms of oxygen

In the second oxide,
oxygen = 56.3 parts by mass
element X = 100 - 56.3 = 43.7 parts
But 56.3 parts by mass of element= 4 atoms of element
thus 43.7 parts of element = $\frac{4}{56.3}\times 43.7$atoms of element = 3.10 atoms of element
Also 43.7 parts by mass of oxygen = 6 atoms of oxygen
thus 56.3 parts by mass of oxygen = $\frac{6}{43.7}\times 56.3$ atoms of oxygen = 7.73 atoms of oxygen

Hence ratio of X : O in the second oxide
3.10 : 7.73
= 3.10/3.10 : 7.73/3.10
= 1 : 2.5
= 2 : 5

Therefore formula of second oxide = X₂O₅