Finding formula of reduced scandium oxide using its mass and mass of obtained metal

Question

An oxide of scandium, with a mass of $\pu{1.423 g},$ is chemically reduced with $\ce{H2}$ to give $\ce{H2O}$ and $\pu{0.929 g}$ of $\ce{Sc}$ metal. What is the formula of scandium oxide? How much water is formed?

I know that the reaction is of the form:

$$\ce{Sc_xO_y + H2 -> H2O + Sc},$$

where $x$ and $y$ are the units to define scandium oxide molecular formula (not empirical).

I don’t know how to get the constants using the info given, and what the info given tells me. I typically use a linear system of equations to solve for the reactants and products. In this case, we’d have:

$$ \begin{align} \ce{Sc}&: &\quad Ax - D &= 0;\\ \ce{O}&: &\quad Ay - C &= 0;\\ \ce{H}&: &\quad 2B - 2C &= 0. \end{align} $$

Usually, we would have an equation where $x$ and $y$ aren’t present and you can solve the system of equations. Here, it prevents you from doing the balancing. Instead, you’re just guessing. Any suggestions when linear algebra doesn’t work?

Answer 1

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.