The Precipitation of Lead (II) Phosphate
Joe Bagadonuts, Eiffel Tower, Ima Some
April 30, 2015 First Hour
Introduction:
Double replacement reactions often produce a precipitate. The purpose/goal of this lab
is_____________. Blah blah blah blah….
The balanced equation for this reaction is:
center equation. Use sub scripts.
Materials:
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50 mL beaker
50 mL graduated cylinder
250 mL Erlenmeyer flask (2)
2 stoppers
3-4 – test tubes
Test tube rack
Centrifuge
Oven
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100 mL graduated cylinder
750 mL Distilled water
Weigh paper
Electronic scale
Glass stir rod
pipette
___ g _________
___ g _________
Procedures:
To make the solution of ___________ blah blah blah
The stoichiometry for this reactant is: (BELOW IS AN EXAMPLE!!! SUB YOUR INFO!!)
500 mL Pb(NO3)2 1L
1000mL
0.250 mol Pb(NO3)2
1L Pb(NO3)2
331.22 g Pb(NO3)2
1 mol Pb(NO3)2
= 41.403 g Pb(NO3)2
To make the solution of ___________, blah blah blah.
Here is the stoichiometry: (BELOW IS AN EXAMPLE!!! SUB YOUR INFO!!)
250 mL Na3PO4
1L
0.100 mol Na3PO4
1000mL
1L Na3PO4
380.12 g Na3PO4
1 mol Na3PO4
= 9.503 g Na3PO4
1. Label three CLEANED test tubes as trial 1, trial 2, trial 3 with initials of team members
and hour.
2. Find the mass of each test tube and record it in the lab book.
3. In a clean 10 mL graduated cylinder, measure ____ mL of the stock ______________.
Then pour it into one of the clean test tubes.
4. In a clean 10 mL graduated cylinder, measure ____ mL of the stock ______________.
Then pour it into the same test tube that contains the ______________.
5. Repeat these steps so that there are (3-4) test tubes containing the reaction.
6. Put test tubes into the centrifuge on opposite sides to balance the centrifuge. Then
centrifuge for 1-6 minutes.
7. Remove the test tubes from the centrifuge. Decant the solution into the sink.
8. Place the tubes in the appropriate container in the oven to dry overnight.
9. The next day, remove all tubes from the oven.
10. Find the mass of each tube and record them in the lab book.
11. Find the mass of the precipitate (__________) by subtracting the mass of the tube from
the mass of the tube with the solid.
12. Clean all materials and follow directions for the disposal of any remaining stock solution.
Results:
Table 1:
Substance
Lead (II) nitrate
Sodium phosphate
Desired total
Amount of solute
This table shows how much of each solute to add to water to make the stock solutions. The
dimensional analysis for how these were created are shown in the procedures section.
Table 2:
Precipitate: lead (II)
Trial 1
Trial 2
Trial 3
phosphate
Amount retrieved (g)
EXAMPLE: 2.013
Theoretical yield (g)
EXAMPLE: 2.03 g
% yield
EXAMPLE: 99.2
% error
EXAMPLE: 0.8
The amount retrieved was calculated by subtracting the mass of the empty test tube from the
amount of the test tube with the dried precipitate. The theoretical yield was calculated using
stoichiometry and that calculation follows. The percent yield is calculated by taking the
amount retrieved in the experiment and dividing that by the theoretical yield. Then
multiplying that number by 100. Finally, the percent error is calculated by taking one
hundred and subtracting the percent yield.
Table 3:
Average % yield
EXAMPLE: 91.6
Average % error
EXAMPLE: 8.2
The average percent yield is calculated by adding up all three of the percent yield numbers
from Table 2 and then dividing by 3. The average percent error is found in the same way.
Theoretical yield: EXAMPLES ONLY!!!! YOU PUR YOUR OWN INFO, DON’T COPY!!!
50.0 mL Na3PO4 1L
0.100 mol Na3PO4 1 mol Pb3(PO4)2
1000 mL
150.0 mL Pb(NO3)2 1L
1000 mL
1 L Na3PO4
2 mol Na3PO4
0.250 mol Pb(NO3)2 1 mol Pb3(PO4)2
1L Pb(NO3)2
3 mol Pb(NO3)2
811.54 g Pb3(PO4)2 = 2.03 g Pb3(PO4)2
1 mol Pb3(PO4)2
811.54 g Pb3(PO4)2 = 10.1 g Pb3(PO4)2
Pb3(PO4)2
For each of the calculations, the first number is the volume of the reactant. The second
box converts the volume from milliliters into liters. The third box uses the molarity, which is
moles per liter, to convert into moles of the reactant. The fourth box uses the molar ratio between
the reactant and the product, lead (II) phosphate, to convert from the reactant to the product. The
last box uses the molar mass of lead (II) phosphate to give the mass of the product that could be
produced if all of the reactant was used up. From these calculations, sodium phosphate is the
limiting reagent because it will run out once 2.03 g of lead (II) phosphate is produced, while the
lead (II) nitrate could make more product if enough of the sodium phosphate was present.
The overall average percent yield calculation is: EXAMPLE ONLY DON’T COPY!
(77.2 + 98.3 + 99.2)/ 3 = 91.6%
The overall average percent error calculation is: EXAMPLE ONLY DON’T COPY!
(22.8 + 1.7 + 0.8)/3 = 8.2%
Discussion/Conclusion:
The purpose of the lab was to create a single precipitate that would be measured to an
accuracy of at least 85%. The average percent yield for this experiment was….. blah blah blah.
Since 100% yield wasn’t achieved, there are some potential errors. One error that could
have occurred was that…… blah blah blah blah.
(TWO ADDITIONAL SOURCES OF ERROR SHOULD BE DISCUSSED IN YOUR PAPER
AS WELL AS REAL LIFE APPLICATIONS!!!!!)