Question

Question #4ed88

Answer 1

`"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

`color(red)(2)"M"^color(blue)(3+) + color(blue)(3)"O"^(color(red)(2-)) -> "M"_ color(red)(2) "O"_color(blue)(3)`

This tells you that one mole of this metal oxide contains `color(red)(2)` moles of the unknown metal for every `color(blue)(3)` moles of oxygen.

Now, the oxide is `68.4%` metal by mass. This means that if you take a `"100-g"` sample of the oxide, it will contain

• `"68.4 g M"`

• `"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 `"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 `"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, `"Cr"`, which has a molar mass of `"51.9981 g mol"^(-1)`. The oxide is chromium(II) oxide, `"Cr"_2"O"_3`.