# How many oxygen atoms are in "120.23 g Fe"_3("PO"_4)_2"?

Nov 25, 2014

$\text{120.23g}$ "Fe"_3("PO"_4)_2 contains $\text{1.6203 x 10"^24}$ $\text{atoms O}$.

#### Explanation:

There are eight moles of oxygen atoms in each mole of $\text{Fe"_3("PO"_4)_2}$. To answer this question, you must convert the given mass of $\text{Fe"_3("PO"_4)_2}$ to moles of $\text{Fe"_3("PO"_4)_2}$. Then multiply the moles of $\text{Fe"_3("PO"_4)_2}$ times $\left(8 \text{mol O)/(1 mol Fe"_3("PO"_4)_2}\right)$ to determine the number of moles of oxygen. Then multiply the moles of oxygen times $6.022 \times {\text{10}}^{23}$ $\text{atoms O}$.

Given/Known:
mass $\text{Fe"_3("PO"_4)_2}$$=$$\text{120.23 g}$
molar mass $\text{Fe"_3("PO"_4)_2}$$=$$\text{357.478g/mol}$
1 mole $\text{Fe"_3("PO"_4)_2}$ contains 8 moles oxygen atoms
1 mole O atoms = $6.022 \times {\text{10}}^{23}$ $\text{atoms}$

Unknown:
number of oxygen atoms in "120.23g Fe"_3("PO"_4)_2

Solution:

1. Convert given mass of $\text{Fe"_3("PO"_4)_2}$ to moles.

120.23color(red)cancel(color(black)("g Fe"_3("PO"_4)_2))xx(1 "mol Fe"_3("PO"_4)_2)/(357.478color(red)cancel(color(black)("g Fe"_3("PO"_4)_2)))= "0.336328 mol Fe"_3("PO"_4)_2"

2. Convert moles $\text{Fe"_3("PO"_4)_2}$ to moles O.

0.336328color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2))xx(8"mol O")/(1color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2)))="2.690624 mol O"

3. Convert moles O to atoms O.

2.690624color(red)cancel(color(black)("mol O"))xx(6.022xx10^23"atoms O")/(1color(red)cancel(color(black)("mol O")))=1.6203xx10^24"atoms O" (answer rounded to five significant figures)