# Calculate the percent phosphorus?

Jun 9, 2018

(a) $\text{P = 7.112 %}$; (b) $\text{P"_2"O"_5 = "16.30 %}$

#### Explanation:

(a) $\text{% P}$

Step 1. Calculate the mass of $\text{P}$ in the precipitate

$\text{Mass of P" = 1.1682 color(red)(cancel(color(black)("g precipitate"))) × "30.97 g P"/(1876.5 color(red)(cancel(color(black)("g precipitate")))) = "0.019 280 g P}$

Step 2. Calculate the percent of $\text{P}$ in the sample

$\text{% P" = "Mass of P"/"Mass of sample" × 100 % = ("0.019 280" color(red)(cancel(color(black)("g"))))/(0.2711 color(red)(cancel(color(black)("g")))) × 100 % = "7.112 %}$

(b) ${\text{% P"_2"O}}_{5}$

Step 3. Calculate the mass of ${\text{P"_2"O}}_{5}$ equivalent to $\text{P}$ in the precipitate

The skeleton chemical equation is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m m} 1876.5 \textcolor{w h i t e}{m m m m m m m m l l} 141.94$
color(white)(mmm)2("NH"_4)_3"PO"_4"·12 MoO"_3 + … → "P"_2"O"_5 + …

${\text{Mass of P"_2"O"_5 = 1.1682 color(red)(cancel(color(black)("g precipitate"))) × ("141.94 g P"_2"O"_5)/(3753.0 color(red)(cancel(color(black)("g precipitate")))) = "0.044 182 g P"_2"O}}_{5}$

Step 4. Calculate the ${\text{% P"_2"O}}_{5}$

$\text{% P"_2"O"_5 = ("Mass of P"_2"O"_5)/"Mass of sample" × 100 % = ("0.044 182" color(red)(cancel(color(black)("g"))))/(0.2711 color(red)(cancel(color(black)("g")))) × 100 % = "16.30 %}$