# How many moles of aluminium cations are present in "5.1 g" of aluminium oxide?

Sep 9, 2016

#### Answer:

${\text{0.10 moles Al}}^{3 +}$

#### Explanation:

The idea here is that each formula unit of aluminium oxide, ${\text{Al"_color(red)(2)"O}}_{\textcolor{b l u e}{3}}$, consists of

• two aluminium cations, $\textcolor{red}{2} \times {\text{Al}}^{3 +}$
• three oxygen anions, $\textcolor{b l u e}{3} \times {\text{O}}^{2 -}$

This means that one mole of aluminium oxide, which is simply a very, very large collection of aluminium oxide formula units, will contain

• two moles of aluminium cations, $\textcolor{red}{2} \times {\text{Al}}^{3 +}$
• three moles of oxygen anions, $\textcolor{b l u e}{3} \times {\text{O}}^{2 -}$

So all you have to do in order to find the number of moles of aluminium cations present in your sample is figure out how many moles of aluminium oxide you have in the sample.

To do that, use the compound's molar mass

5.1 color(red)(cancel(color(black)("g"))) * ("1 mole Al"_2"O"_3)/(101.96color(red)(cancel(color(black)("g")))) = "0.05002 moles Al"_2"O"_3

Therefore, you can say that your sample contains

0.05002 color(red)(cancel(color(black)("moles Al"_2"O"_3))) * (color(red)(2)color(white)(a)"moles Al"^(3+))/(1color(red)(cancel(color(black)("mole Al"_2"O"_3)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.10 moles Al"^(3+))color(white)(a/a)|)))

The answer is rounded to two sig figs.