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

  1. #"0.5000 g"# fertilizer (of some unknown #%"P"_2"O"_5#) is dissolved in water to form #"1-L"# of solution A.

  2. A #"1.00-mL"# aliquot of A is diluted to #"100 mL"#, together with a developing agent that allows the #"P"_2"O"_5# in the solution to be detected to form solution B.

  3. Solution B is analyzed to find its absorbance at #"600 nm"# to be #0.000, 0.325, 0.553# at concentrations of #0.0, 1.2, 2.0# #"mg/L"#.

1 Answer
Dec 22, 2016

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)%)#