Question #cf1f9

1 Answer
Dec 18, 2015

Answer:

Here's what's going on here.

Explanation:

The idea here is that all the phosphorus that was initially part of the washing powder will now be a part of the precipitate.

This means that in order to figure out how much phosphorus was present initially, all you have to do is figure out how much phosphorus you have in that #"0.085-g"# sample of magnesium pyrophospate, #"Mg"_2"P"_2"O"_7#.

Now, to do that, you need to figure out the percent composition of phosphorus in magnesium pyrophosphate, i.e. how many grams of phosphorus you get per #"100 g"# of compound.

You will need to know two things

  • the molar mass of magnesium pyrophosphate, which is equal to #"222.553 g/mol"#

  • the molar mass of phosphorus, which is equal to #"30.973761 g/mol"#

The idea here is that every mole of magnesium pyrosulfate will contain two moles of phosphorus, since you have two atoms of phosphorus per formula unit of the compound.

So, one mole of magnesium pyrosulfate has a mass of #"222.553 g"#. Since this mole contains two moles of phosphorus, and knowing that one mole of phosphorus has a mass of #"30.97376 g"#, you can say that

#(2 xx 30.973761 color(red)(cancel(color(black)("g"))))/(222.553 color(red)(cancel(color(black)("g")))) xx 100 = "27.835% P"#

This means that every #"100 g"# of magnesium pyrosulfate will contain #"27.835 g"# of phosphorus.

The precipitate you obtained will thus contain

#0.085 color(red)(cancel(color(black)("g Mg"_2"P"_2"O"_7))) * "27.835 g P"/(100color(red)(cancel(color(black)("g Mg"_2"P"_2"O"_7)))) = "0.02366 g P"#

This is how much phosphorus was present in the washing powder. Since the total mass of the washing powder was equal to #"2 g"#, the percentage by mass of phosphorus was

#(0.02366 color(red)(cancel(color(black)("g"))))/(2 color(red)(cancel(color(black)("g")))) xx 100 = "1.183%"#

Now, I assume that the total mass of the sample was #"2.0 g"#, since #"2 g"# would only justify one sig fig for the answer. Rounded to two sig figs, the answer will indeed be

#"% P " = color(green)(" 1.2%")#