# Question 3f2eb

##### 1 Answer
Dec 23, 2016

Here's what I got.

#### Explanation:

For starters, I don't think that you were given correct information here. More specifically, I think that the problem should read

A trivalent metal oxide contains 68.4% metal by mass. What is the atomic mass of the metal?

If that's the case, you can find the answer to that question here.

That said, I'll solve the problem with the information given to you to show you that you're dealing with inaccurate data.

So, your strategy here will be to convert the percent composition of oxygen to moles of oxygen, then use the fact that the metal, $\text{M}$, has a valency of $3$ to find the number of moles of metal in a set sample.

To make the calculations easier, let's pick $\text{100 g}$ of this metal oxide as our sample. You know that this sample will contain

• $\text{68.4% oxygen " = " 68.4 g O}$
• $\text{31.6% metal " = " 31.6 g M}$

Use the molar mass of oxygen to calculate how many moles of oxygen you have in this sample

68.4 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0 color(red)(cancel(color(black)("g")))) = "4.275 moles O"

Now, oxygen forms $2 -$ anions; this should tell you that the chemical formula of the metal oxide will be

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

Here you need $3$ oxygen anions to balance the positive charge coming from $2$ metal cations.

Consequently, you can say that this sample will contain

4.275 color(red)(cancel(color(black)("moles O"))) * (color(blue)(2)color(white)(.)"moles M")/(color(purple)(3)color(red)(cancel(color(black)("moles O")))) = "2.85 moles M"#

Since you know that you have $\text{31.6 g}$ of metal in this sample, you can say that the molar mass of the metal will be

${M}_{\text{M metal" = "31.6 g"/"2.85 moles" = "11.1 g mol}}^{- 1}$

This is equivalent to an atomic mass of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{m}_{\text{a metal" = "11.1 u}}}}}$

The closest match to this value is actually boron, $\text{B}$, which has an atomic mass of $\approx \text{10.8 u}$. Despite the fact that boron does form oxides, it can't really be classified as a metal.

Boron is actually a metalloid, an element that shares properties of both metals and nonmetals.

If boron is indeed the answer, then the unknown oxide is

${\text{B"_2"O}}_{3} \to$ boron trioxide