Question #cf1f9
1 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
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
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
#(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
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
#(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
#"% P " = color(green)(" 1.2%")#