# How many atoms of phosphorus are in 3.40 mol of copper(ll) phosphate?

Jun 15, 2017

$4.09 \cdot {10}^{24}$

#### Explanation:

You know that $1$ mole of copper(II) phosphate, "Cu"_ 3("PO"_4)_2, contains

• three moles of copper(II) cations, $3 \times {\text{Cu}}^{2 +}$
• two moles of phosphate anions, $2 \times {\text{PO}}_{4}^{3 -}$

Since $1$ mole of phospahte anions contains

• one mole of phosphorus atoms, $1 \times \text{P}$
• four moles of oxygen, $4 \times \text{O}$

you can say that $1$ mole of copper(II) phosphate contains

2 xx ( 1 xx "P") = "2 moles of P"

Consequently, you can say that your sample contains

3.40 color(red)(cancel(color(black)("moles Cu"_3("PO"_4)_2))) * "2 moles P"/(1color(red)(cancel(color(black)("mole Cu"_3("PO"_4)_2))))

$\text{ = 6.80 moles P}$

To convert this to atoms of phosphorus, use Avogadro's constant, which tells you the number of atoms present in $1$ mole of any given element.

6.80 color(red)(cancel(color(black)("moles P"))) * overbrace((6.022 * 10^(23)color(white)(.)"atoms of P")/(1color(red)(cancel(color(black)("mole P")))))^(color(blue)("Avogadro's constant"))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{4.09 \cdot {10}^{24} \textcolor{w h i t e}{.} \text{atoms of P}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the number of moles of copper(II) phosphate.

Jun 15, 2017

There are $4.09 \times {10}^{24}$ atoms of phosphorus in $\text{3.40 mol}$ copper(II) phosphate.

#### Explanation:

The formula unit for copper(II) phosphate is $\text{Cu"_3"(PO"_4)_2}$. If the formula unit represents 1 mole of copper(II) phosphate , then there are 2 moles of phosphorus per 1 mole copper(II) phosphate formula units.

One mole of atoms = $6.022 \times {10}^{23} \text{ atoms}$.

During the process of answering this question, we will go from:

color(red)("mol Cu"_3"(PO"_4")"_2"$\rightarrow$color(blue)("mol P"$\rightarrow$color(purple)("atoms P"

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color(red)("Mol Cu"_3"(PO"_4")"_2" to color(blue)("Mol P"

Multiply mol $\text{Cu"_3"(PO"_4)_2}$ by mol ratio between $\text{P}$ and $\text{Cu"_3"(PO"_4)_2}$, 2:1.

3.40 color(red)cancel(color(black)("mol Cu"_3"(PO"_4")"_2))xx(2"mol P")/(1color(red)cancel(color(black)("mol Cu"_3"(PO"_4")"_2)))="6.80 mol P"

color(blue)("Mol P"$\rightarrow$color(purple)("Atoms P"

Multiply mol $\text{P}$ by $6.022 \times {10}^{23} \text{atoms/mol}$.

6.8color(red)cancel(color(black)("mol P"))xx(6.022xx10^23"atoms P")/(1color(red)cancel(color(black)("mol P")))=4.09xx10^24" atoms P" rounded to three sig figs

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You can put all the steps into one equation.

color(red)(3.40 color(black)cancel(color(red)("mol Cu"_3"(PO"_4")"_2)))xx(color(blue)(2"mol P"))/(color(red)1color(black)cancel(color(red)("mol Cu"_3"(PO"_4")"_2)))xx(color(purple)(6.022xx10^23"atoms P"))/(1color(black)cancel(color(blue)("mol P")))=color(purple)(4.09xx10^24" atoms P"