Answer:
About 2.24 × 10²⁵ oxygen atoms
Explanation:
We want to determine the amount of oxygen atoms in 1665 grams of iron (II) phosphate (Fe₃(PO₄)₂).
To do so, we can convert from grams to moles of iron (II) phosphate, moles of iron (II) phosphate to moles of oxygen, and finally to atoms of oxygen.
Find the molar mass of iron (II) phosphate:
[tex]\displaystyle \begin{aligned} \text{Molecular Weight} & = 3(55.85 + 2(30.97) + 8(16.00))\text{ g/mol} \\ \\ & = 357.49\text{ g/mol}\end{aligned}[/tex]
From the chemical formula, we can see that for each molecule of iron (II) phosphate, there are eight oxygen atoms. In other words, each mole of iron (II) phosphate has eight moles of oxygen atoms.
Finally, each mole of a substance contains 6.02 × 10²³ amount of that substances.
With the initial amount, multiply:
[tex]\displaystyle \begin{aligned} 1665 \text{ g Fe$_3$(PO$_4$)$_2$} \cdot \frac{1 \text{ mol Fe$_3$(PO$_4$)$_2$}}{357.49 \text{ g Fe$_3$(PO$_4$)$_2$}}& \cdot \frac{8\text{ mol O}}{1 \text{ mol Fe$_3$(PO$_4$)$_2$}} \\ \\ &\cdot \frac{6.02\times 10^{23} \text{ atom O}}{1 \text{ mol O}} \\ \\ &=2.24\times 10^{25} \text{ atom O}\end{aligned}[/tex]
In conclusion, there are about 2.24 × 10²⁵ oxygen atoms in 1665 grams of iron (II) phosphate.
