Question #4b22d

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Answer 1 · 2018-01-15T18:12:00

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

The molar mass of $"AlPO"_4$ is $26.98 + 30.97 + 4\cdot 16.00 = 121.95 \frac{\text{g}}{\text{mol}}$ .

Now, we convert $"25 g"$ $"AlPO"_4$ to moles $"AlPO"_4$ :

$25\text{ g }\cdot\frac{\text{mol}}{121.95\text{ g}} = "0.205 mol AlPO"_4$

We can convert this to formula units of $"AlPO"_4$ using Avogadro's constant, $6.02\times 10^{23}$ .

$"0.205 moles" \cdot 6.02\times 10^{23}\text{ mol}^{-1} = 1.23\times 10^{23}$ $"formula units AlPO"_4$

Since there are 4 oxygen atoms per formula unit of $"AlPO"_4$ , we multiply this by 4 to get our answer:

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