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

I got about `"15.6%"` `"P"_2"O"_5` `"w/w"`.

An overview of the scenario is as follows:

1. `"0.5000 g"` fertilizer (of some unknown `%"P"_2"O"_5`) is dissolved in water to form a `"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 `"1.00-mL"` aliquot of A (presumably via volumetric pipette) is diluted to `"100.00 mL"` (also presumably in a `100.00` `[pm "0.02 mL"` `]` volumetric flask), together with a developing agent that allows specifically the `"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 `"L/mg"`).
2. Use it to calculate the concentration in `"mg/L"` from the absorbance graph of solution B.
3. From the `"P"_2"O"_5` concentration of B, get the `"P"_2"O"_5` concentration of A.
4. From the `"P"_2"O"_5` concentration of A, get the mass of A that is actually `"P"_2"O"_5`.
5. From the experimental `"P"_2"O"_5` mass, get the mass percent of `"P"_2"O"_5` in the original fertilizer.

OBTAINING THE SLOPE

From Beer's Law, we have

`A = epsilonbc`,

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

If we plot the first three absorbances as `y` and the first three concentrations as `x` (the concentration in `"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"`

`= (0.214 + 0.001_(789))/(0.276_(05) "L/mg")`

`=` `0.781_(7)` `"mg/L"` `=>` `color(green)("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 `"1-mL"` aliquot of `A` by a factor of `100`:

`c_AV_A = c_BV_B`

`color(green)(c_A) = (c_BV_B)/(V_A)`

`= ("0.781 mg/L")("100.00 mL"/"1.00 mL")`

`=` `color(green)("78.1 mg/L")`

But, recall that the `"1.00 mL"` was not the original volume - it was merely an aliquot of the original `"1.00000-L"` solution made.

MASS OF `bb("P"_2"O"_5)` IN ORIGINAL SOLUTION

Thus, to find the mass of `"P"_2"O"_5` in the original solution (the one with the `"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`

`= color(green)("0.0781 g P"_2"O"_5)`

MASS PERCENT OF `bb("P"_2"O"_5)` IN ORIGINAL SOLUTION

Therefore, the `%"P"_2"O"_5` `"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)%)`