Paul, Meg, Dominique, Sunil, Cassi, and Emily
October 13-16, 2009

The Goal
The goal of this experiment is to illustrate an application of gravimetric analysis to a consumer product. The purpose of this experiment is to illustrate one of the quality-control analyses for plant food by gravimetric determination of its phosphorus content.

The Materials
balance, six 250mL or larger beakers, filter paper, three funnels, funnel support, ring stand, three stirring rods with rubber policeman


The Chemicals
75% aqueous isopropyl alcohol, 10% aqueous MgSO4*7H2O, 2 M NH3 (aq), and plant food

The Procedure
Reference: Nelson, John H., and Kenneth C. Kemp. Laboratory Experiments, Chemistry: The Central Science. 10th ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2006. pg 87

I. Weigh 2.5g of the unknown sample to the nearest hundredth gram, using weighing paper.
II. Transfer sample to 250mL beaker & record the sample mass.
III. Add 40 mL of distilled water & stir mixture w/ glass stirring rod to dissolve sample.
IV. If sample does not completely dissolve, remove insoluble material by filtration.
V. Add 45mL of 10% MgSO4*7H2O solution to the filtrate.
VI. Add 150mL of 2 M NH3 (aq) slowly while stirring.
VII. A white precipitate MgNH4PO4*6H2O will form.
VII. Allow mixture to sit at room temperature for 15 minutes to complete precipitation.
VIII. Obtain a filter paper.
IX. Fold the paper as illustrated, weigh it, & then fit it into a funnel.
X. Wet the paper w/ distilled water to hold it in place in the funnel.
VIII. Transfer the precipitate & the solution from the beaker onto the filter using a rubber policeman.
XI. Wash the precipitate w/ two 5mL portions of distilled water.
XII. Pour two 10mL portions of 75% isopropyl alcohol through the filter paper.
XIII. Remove filter paper & store it to dry.
XIV. Repeat above procedure with 2 more samples.
XV. When MgNH4PO4*6H2O is dry, weigh filter paper plus MgNH4PO4*6H2O.
XVI. Record the mass & calculate percentages of phosphorous & P2O3 in orignal samples.

The Data and Observations
When the miracle gro was added to the distilled water, the water turned blue. The NH3 that was added to Sunil and Paul's group was cloudy because the original NH3 had run out.
After the solution was filtered, the cloudy solution was clear.A white precipitate formed in all of the trials. This precipitate however, clogged the filter paper. This caused the experiment to take much longer than expected

	Mass of Sample (g)	Mass of Filter Paper (g)	Mass of Filter Paper and MgNH4PO4+6H20 (g)
Sunil/ Paul	Trial 1- 2.50 Trial 2. 2.50	Trial 1- .78 g Trial 2- .78 g	Trial 1- 1.91 Trial 2- 1.59
Meg/ Dominique	Trial 1 - 2.50 g Trial 3 - 2.50 g	Trial 1 - 2.16 g Trial 3 - 2.16 g	Trial 1- 3.25 g Trial 3- 3.59 g
Cassi/ Emily	Trial 1- 2.50g	Trial 1- 1.12g	Trial 1- 2.18g

The Calculations
Molar Masses needed in calculations:
Molar Mass of P2O5 = 2(30.97) + 5(16) = 141.94 g
Molar Mass of MgNH4PO4 + 6H20 = 24.31 + 14.01 + 4(1.01) + 30.97 + 4(16) + 12(1.01) + 6(16) =245.4 g

All use same formula so it only needs the work of one sample.

Paul and Sunil's results were used in these sample calculations.
Find Mass of P2O5 = .620g MgNH4PO4 + 6H2O (1 mole / 245.4 g ) (1 mole P / 1 mole) (1 mole P2O5 / 2 mole P) (141.9 g P2O5 / 1 mole P2O5) = .179g P2O5

Find mass of P = .620g MgNH4PO4 + 6H2O (1 mole / 245.4 g) (1 mole P / 1 mole) (30.97 g P / 1 mole P) = .0782g P

%P = (.0782g P / 2.50g sample) X 100= 3.13% P
and
%P2O5 = (.179g P2O5 / 2.50g sample) X 100= 7.16% P2O5

Mean = .0782g + .0379g/ 2= .0581

Deviation= l .0782- .0581l= .0201

All other formulas are the same and were used from other groups.


Formulas: using Dominique's and Meg's Results
Mass of Filter Paper
(small plus large)
1.24 + .92 = 2.16
Mean = .180 + .138 / 2 = .159
Deviations from mean = l .159 - .180 I = .021
I .159 - .138 I = .021
Average Deviation from the mean = .021 + .021 / 2 = .021
Relative Deviation = average deviation / mean = .021 / .159 = .132
Standard Deviation = square root of the sum of squares of the deviations from the mean / number of obervations - 1
square root of (.021)(.021) + (.021)(.021) / 2-1 = .0297 / 1
Relative Standard Deviation = standard deviation / number of experiments = .0297 / 2 = .0148

Cassi and Emily's Results

Mass of P2O5= (0.08)(2.50)=0.2g P2O5 in original
Mass of P= (0.2g)(1 mol P2O5/141.9g P2O5)X(2 mol P/1 mol P2O5)X(30.97g P/1 mol)= 0.087 g P

The Results

Results from Calculations	Mass of P (g) in original sample (rounded to 3 sig figs)	Mass of P2O5 (g) (rounded to 3 sig figs)	Mass (g) of MgNH4PO4 + 6H2O (rounded to 3 sig figs)	% of P (rounded to 3 sig figs)	% of P2O5 (rounded to 3 sig figs)	Mean	Deviations from Mean	Standard Deviation (rounded to 3 sig figs)	Relative Standard Deviation (rounded to 3 sig figs)
Dominique & Meg	.0873 g	.200 g	trial 1- 1.09 g trail 3- 1.43g	trial 1- 5.50% trial 3- 7.22%	trial 1-12.6% trial 3- 16.5%	.159	.021 & .021	.0297	.0148 or 1.48%
Paul & Sunil	Trial 1- .0782 Trial 2- .0379	Trial 1- .179 g Trial 2- ..0867	Trial 1- .620 Trial 2- .300	Trial 1-3.13 Trial 2-1.52	Trial 1-7.16 Trial 2-3.47	.0581	.0201 & .0201	.0284	.0142 or 1.42%
Emily & Cassi	0.087g	0.20g	1.06g	3.55%	8.00%	0.087 *see calculations*	*see calculations*	*see calculations*	*see calculations*

	Trail 1	Trial 2	Trial 3
Paul and Sunil	1.91 g	1.59 g	N/A
Emily and Cassi	2.18 g	N/A	N/A
Meg and Dominique	3.25 g	N/A	3.59 g

The Discussion
This experiment has proven to be a difficult method of accurately obtaining correct figures to precisely determine the levels of phosphorus in plant food. Many trials were necessary to ensure that at least one trial was effectively completed. Obstacles that affected the results mainly came form the methods of filtration and insufficient time to properly drain and filter the solution. When the original filtration method (filter paper in a regular funnel) proved to be slow, it was immediately realized that this process needed to be altered in order to be completed in a timely manner. Thye constant moving and switching of the solutions provided a larger margin for error. Relocation of the solution gave a greater chance for some of it to be lost, and in one case, they were confused to such an extent that some solutions were completely mixed up. This experiment could have been executed differently to guarantee more specific results; however, given the circumstances and a thorough evaluation, this experiment was conducted successfully. The results yielded were as exact as could be derived. The time allotted and the materials supplied were taken advantage of, and the experiment served its purpose.