Question #4ed88

1 Answer
May 24, 2016

Answer:

#"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#.