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

2 Answers
Jun 15, 2017

Answer:

#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.

Jun 15, 2017

Answer:

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"#