# What is the percentage by mass of phosphorus in a washing powder, in which a 0.085 g precipitate of magnesium pyrophosphate forms from a 2 g sample of powder? (Gravimetric analysis).

## The phosphorus in a 2 g sample of washing powder is precipitated as $M {g}_{2}$${P}_{2}$${O}_{7}$. The precipitate weighs 0.085 g.

Sep 10, 2017

Here's what I got.

#### Explanation:

The idea here is that the mass of phosphorus present in the washing powder will be equal to the mass of phosphorus present in the $\text{2-g}$ sample of magnesium pyrophosphate.

To find the mass of phosphorus present in the precipitate, start by calculating the percent composition of the salt. To do that, use the molar mass of magnesium pyrophosphate, ${\text{Mg"_2"P"_color(red)(2)"O}}_{7}$, and the molar mass of phosphorus.

(color(red)(2) * 30.974 color(red)(cancel(color(black)("g mol"^(-1)))))/(222.5533color(red)(cancel(color(black)("g mol"^(-1))))) * 100% = "27.835% P"

This means that every $\text{100 g}$ of magnesium pyrophosphate will contain $\text{27.835 g}$ of phosphorus.

You can thus say that your sample contains

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"

Now, you know that $\text{2 g}$ of washing powder contain $\text{0.02366 g}$ of phosphorus. THis means that the mass of phosphorus present in $\text{100 g}$ of washing powder is equal to

100 color(red)(cancel(color(black)("g washing powder"))) * "0.02366 g P"/(2color(red)(cancel(color(black)("g washing powder")))) = "1.18 g P"

Therefore, the percent concentration by mass of phosphorus in the washing powder, i.e. the number of grams of phosphorus present for every $\text{100 g}$ of washing powder, will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% concentration = 1.2% P}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the mass of washing powder.