# Question 4ed88

May 24, 2016

#### Answer:

${\text{51.9 g mol}}^{- 1}$

#### Explanation:

The idea here is that you need to use the fact that you're dealing with a trivalent metal to write the general formula of the metal oxide.

A trivalent metal is simply a metal that has a valency of $3$, i.e. a metal that has three valence electrons. This metal will thus have an oxidation state of $+ 3$ on the metal oxide.

The general formula of the oxide will be

$\textcolor{red}{2} {\text{M"^color(blue)(3+) + color(blue)(3)"O"^(color(red)(2-)) -> "M"_ color(red)(2) "O}}_{\textcolor{b l u e}{3}}$

This tells you that one mole of this metal oxide contains $\textcolor{red}{2}$ moles of the unknown metal for every $\textcolor{b l u e}{3}$ moles of oxygen.

Now, the oxide is 68.4% metal by mass. This means that if you take a $\text{100-g}$ sample of the oxide, it will contain

• $\text{68.4 g M}$

• $\text{31.6 g O}$

Use oxygen's molar mass to determine how many moles of oxygen would be present

31.6 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.975 moles O"

Now, according to the mole ratio that exists between the two elements in the oxide, the sample will also contain

1.975 color(red)(cancel(color(black)("moles O"))) * (color(red)(2)color(white)(a)"moles M")/(color(blue)(3)color(red)(cancel(color(black)("moles O")))) = "1.317 moles M"

So, if $\text{68.4 g}$ of this metal are equivalent to $1.317$ moles, it follows that $1$ mole of the metal will have a mass of

1 color(red)(cancel(color(black)("mole M"))) * "68.4 g"/(1.317color(red)(cancel(color(black)("moles M")))) = "51.9 g"

So, if one mole of this metal has a mass of $\text{51.9 g}$, it follows that the molar mass of the metal is

"molar mass of metal" = color(green)(|bar(ul(color(white)(a/a)"51.9 g mol"^(-1)color(white)(a/a)|)))#

The answer is rounded to three sig figs.

As a side note, your unknown metal is most likely chromium, $\text{Cr}$, which has a molar mass of ${\text{51.9981 g mol}}^{- 1}$. The oxide is chromium(II) oxide, ${\text{Cr"_2"O}}_{3}$.