A 2.000 g sample of an unknown metal, M, was completely burned in excess O_2 to yield .02224 mol of the medal oxide, M_2O_3. What is the metal?

Dec 30, 2015

Scandium.

Explanation:

The idea here is that you need to write a balanced chemical equation for this reaction. If you take $\text{M}$ to be the metal, and ${\text{M"_2"O}}_{3}$ to be the metal oxide, you can say that

$\textcolor{red}{4} {\text{M"_text((s]) + 3"O"_text(2(g]) -> color(purple)(2)"M"_2"O}}_{\textrm{3 \left(s\right]}}$

Notice that you have a $\textcolor{red}{4} : \textcolor{p u r p \le}{2}$ mole ratio between the metal and the oxide. This tells you that the reaction will always consume twice as many moles of metal than the number of moles of oxide produced.

So, you know that the reaction produced $0.02224$ moles of metal oxide. This means that it must have consumed

0.02224 color(red)(cancel(color(black)("moles M"_2"O"_3))) * (color(red)(4)" moles M")/(color(purple)(2)color(red)(cancel(color(black)("moles M"_2"O"_3)))) = "0.04448 moles M"

Now, a substance's molar mass tells you the mass of one mole of that substance. In your case, you know that $0.04448$ moles have a mass of $\text{2.000 g}$, which means that one mole will have a mass of

1 color(red)(cancel(color(black)("mole M"))) * "2.000 g"/(0.04448color(red)(cancel(color(black)("moles M")))) = "44.964 g"

The molar mass of the metal will thus be

${M}_{M} = \textcolor{g r e e n}{\text{44.96 g}} \to$ rounded to four sig figs.

A quick look in the periodic table will show that the closest math to this value is the molar mass of scandium, $\text{Sc}$, which has a molar mass of $\text{44.9559 g/mol}$.

The compound is called scandium(III) oxide, ${\text{Sc"_2"O}}_{3}$, and it features scandium in its $3 +$ oxidation state.