# Question #4b22d

Jan 15, 2018

$4.9 \setminus \times {10}^{23}$ oxygen atoms

#### Explanation:

The molar mass of ${\text{AlPO}}_{4}$ is $26.98 + 30.97 + 4 \setminus \cdot 16.00 = 121.95 \setminus \frac{\setminus \textrm{g}}{\setminus \textrm{m o l}}$.

This comes from the periodic table.

Now, we convert $\text{25 g}$ ${\text{AlPO}}_{4}$ to moles ${\text{AlPO}}_{4}$:

$25 \setminus \textrm{g} \setminus \cdot \setminus \frac{\setminus \textrm{m o l}}{121.95 \setminus \textrm{g}} = {\text{0.205 mol AlPO}}_{4}$

We can convert this to formula units of ${\text{AlPO}}_{4}$ using Avogadro's constant, $6.02 \setminus \times {10}^{23}$.

$\text{0.205 moles} \setminus \cdot 6.02 \setminus \times {10}^{23} \setminus {\textrm{m o l}}^{- 1} = 1.23 \setminus \times {10}^{23}$ ${\text{formula units AlPO}}_{4}$

Since there are 4 oxygen atoms per formula unit of ${\text{AlPO}}_{4}$, we multiply this by 4 to get our answer:

$4 \setminus \cdot 1.23 \setminus \times {10}^{23} = 4.9 \setminus \times {10}^{23}$ oxygen atoms.