# Find the percent "P"_2"O"_5 by mass in the fertilizer?

## $\text{0.5000 g}$ fertilizer (of some unknown %"P"_2"O"_5) is dissolved in water to form $\text{1-L}$ of solution A. A $\text{1.00-mL}$ aliquot of A is diluted to $\text{100 mL}$, together with a developing agent that allows the ${\text{P"_2"O}}_{5}$ in the solution to be detected to form solution B. Solution B is analyzed to find its absorbance at $\text{600 nm}$ to be $0.000 , 0.325 , 0.553$ at concentrations of $0.0 , 1.2 , 2.0$ $\text{mg/L}$.

Dec 22, 2016

I got about $\text{15.6%}$ ${\text{P"_2"O}}_{5}$ $\text{w/w}$.

An overview of the scenario is as follows:

1. $\text{0.5000 g}$ fertilizer (of some unknown %"P"_2"O"_5) is dissolved in water to form a $\text{1.00000 L}$ (pm "0.02 mL" ) solution (presumably in a volumetric flask, which is why I assume 5 decimal places).
We call this solution A.

2. A $\text{1.00-mL}$ aliquot of A (presumably via volumetric pipette) is diluted to $\text{100.00 mL}$ (also presumably in a $100.00$ [pm "0.02 mL" ] volumetric flask), together with a developing agent that allows specifically the ${\text{P"_2"O}}_{5}$ in the solution to be detected via UV-VIS Spectroscopy.
We call this solution B.

3. Only solution B is analyzed, and it has a known absorbance.

Thus, our calculations shall follow these steps:

1. Obtain the slope (the extinction coefficient, multiplied by the path length, the result in units of $\text{L/mg}$).
2. Use it to calculate the concentration in $\text{mg/L}$ from the absorbance graph of solution B.
3. From the ${\text{P"_2"O}}_{5}$ concentration of B, get the ${\text{P"_2"O}}_{5}$ concentration of A.
4. From the ${\text{P"_2"O}}_{5}$ concentration of A, get the mass of A that is actually ${\text{P"_2"O}}_{5}$.
5. From the experimental ${\text{P"_2"O}}_{5}$ mass, get the mass percent of ${\text{P"_2"O}}_{5}$ in the original fertilizer.

OBTAINING THE SLOPE

From Beer's Law, we have

$A = \epsilon b c$,

where $\epsilon$ is the molar absorptivity (extinction coefficient), $b$ is the path length of the cuvette (usually $\text{1 cm}$), and $c$ is the concentration of the analyte in $\text{mg/L}$.

If we plot the first three absorbances as $y$ and the first three concentrations as $x$ (the concentration in $\text{mg/L}$), this is the graph we'd get, say, in Excel:

If you aren't familiar with Excel, here is a resource for you to look at.

GETTING THE CONCENTRATION OF SOLN B

The slope was acquired (from the LINEST function), so we can get the concentration for solution B:

color(green)(c_B) = A/(epsilonb) = (A_B + "y-int")/"slope"

$= \frac{0.214 + {0.001}_{789}}{{0.276}_{05} \text{L/mg}}$

$=$ ${0.781}_{7}$ $\text{mg/L}$ $\implies$ $\textcolor{g r e e n}{\text{0.781 mg/L}}$,

where subscripts indicate the digits past the last significant digit.

GETTING THE CONCENTRATION OF SOLN A

Next, we can get the concentration of solution $A$, ${c}_{A}$ by realizing that we had diluted a $\text{1-mL}$ aliquot of $A$ by a factor of $100$:

${c}_{A} {V}_{A} = {c}_{B} {V}_{B}$

$\textcolor{g r e e n}{{c}_{A}} = \frac{{c}_{B} {V}_{B}}{{V}_{A}}$

$= \left(\text{0.781 mg/L")("100.00 mL"/"1.00 mL}\right)$

$=$ $\textcolor{g r e e n}{\text{78.1 mg/L}}$

But, recall that the $\text{1.00 mL}$ was not the original volume - it was merely an aliquot of the original $\text{1.00000-L}$ solution made.

MASS OF $\boldsymbol{{\text{P"_2"O}}_{5}}$ IN ORIGINAL SOLUTION

Thus, to find the mass of ${\text{P"_2"O}}_{5}$ in the original solution (the one with the $\text{0.5000 g}$ of fertilizer dissolved in it):

color(green)(m_("P"_2"O"_5)) = "78.1 mg/L" xx "1.00000 L" = "78.1 mg P"_2"O"_5

$= \textcolor{g r e e n}{{\text{0.0781 g P"_2"O}}_{5}}$

MASS PERCENT OF $\boldsymbol{{\text{P"_2"O}}_{5}}$ IN ORIGINAL SOLUTION

Therefore, the %"P"_2"O"_5 $\text{w/w}$ is:

color(blue)(%"P"_2"O"_5) color(blue)("w/w") = m_("P"_2"O"_5)/(m_"fertilizer")xx100%

= ("0.0781 g P"_2"O"_5)/"0.5000 g fertilizer used"xx100%

~~ color(blue)(15.6_(3)%)