# Question 3db95

Feb 4, 2016

$1.023 \cdot {10}^{- 7} {\text{moles AlPO}}_{4}$

#### Explanation:

The first thing to do here is figure out how many moles of oxygen you need per mole of aluminium phosphate, ${\text{AlPO}}_{4}$.

Aluminium phosphate is made up of aluminium cations, ${\text{Al}}^{3 +}$, and phosphate anions, ${\text{PO}}_{4}^{3 -}$, in a $1 : 1$ mole ratio. This tells you that every mole of aluminium phosphate contains One mole of aluminium cations and one mole of phosphate anions.

Now, the phosphate anion is made up of one phosphorus atom, $\text{P}$, and four oxygen atoms, $\text{O}$.

This means that every mole of phosphorus anions contains four moles of oxygen atoms.

Therefore, you can say that

$\text{1 mole AlPO"_4 -> "4 moles O}$

The next thing to do here is figure out how many moles of oxygen would contain that many atoms of oxygen. As you know, one mole of any element contains exactly $6.022 \cdot {10}^{23}$ atoms of that element - this is known as Avogadro's number.

In your case, the sample is said to contain $2.463 \cdot {10}^{17}$ atoms of oxygen, which means that it contains

2.463 * 10^(17) color(red)(cancel(color(black)("atoms O"))) * overbrace( "1 mole O"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms O")))))^(color(purple)("Avogadro's number")) = 4.090 * 10^(-7)"moles O"

Since we've already established that you need four moles of oxygen to get one mole of aluminium phosphate, you can say that

4.090 * 10^(-7)color(red)(cancel(color(black)("moles O"))) * "1 mole AlPO"_4/(4color(red)(cancel(color(black)("moles O")))) = color(green)(1.023 * 10^(-7)"moles AlPO"_4)#

The answer is rounded to four sig figs.