Solution:
Solution
In a cubic packed structure,
$$O^{2-}$$ present in ccp=4 $$O^{2-}$$ ions
Tetrahedral voids=8
Octahedral voids=4
Thus, $$X^{-}$$ ions=$$\cfrac{8}{5}$$
$$Y^{+3}$$ ions=$$\cfrac{4}{2}=2$$
So, the molecular formula is $$X_{\cfrac{8}{5}}Y_{2}O_{4}$$.
In simplest form, $$X_{4}Y_{5}O_{10}$$