Finding formula of reduced scandium oxide using its mass and mass of obtained metal
Since scandium oxide is an ionic compound, the method is exactly the same as obtaining the empirical formula, from the ratio of the amounts of elements $n(\ce{Sc})$ and $n(\ce{O}).$
$$n(\ce{Sc})=\frac{m(\ce{Sc})}{M(\ce{Sc})}=\frac{\pu{0.929 g}}{\pu{44.96 g mol-1}}=\pu{2.07E-2 mol}$$
The mass difference comes from the loss of oxygen atoms:
$$n(\ce{O})=\frac{m(\ce{Sc_xO_y})-m(\ce{Sc})}{M(\ce{O})}=\frac{\pu{1.423 g}-\pu{0.929 g}}{\pu{16.00 g mol-1}}=\pu{3.09E-2 mol}$$
The ratio between the amounts of elements is close to $2:3$, therefore the formula is $\ce{Sc2O3}$. You can check that using oxidation states.