Question

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

Answer 1

`4.09 * 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 xx "Cu"^(2+)`
• two moles of phosphate anions, `2 xx "PO"_4^(3-)`

Since `1` mole of phospahte anions contains

• one mole of phosphorus atoms, `1 xx "P"`
• four moles of oxygen, `4 xx "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))))`

`" = 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"))`

`= color(darkgreen)(ul(color(black)(4.09 * 10^(24)color(white)(.)"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.

Answer 2

There are `4.09xx10^24` atoms of phosphorus in `"3.40 mol"` copper(II) phosphate.

Explanation:

The formula unit for copper(II) phosphate is `"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 = `6.022xx10^23" atoms"`.

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

`color(red)("mol Cu"_3"(PO"_4")"_2"` `rarr` `color(blue)("mol P"` `rarr` `color(purple)("atoms P"`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

Multiply mol `"Cu"_3"(PO"_4)_2"` by mol ratio between `"P"` and `"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"` `rarr` `color(purple)("Atoms P"`

Multiply mol `"P"` by `6.022xx10^23"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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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"`