In general, the key to successful gravimetric analyses is having components in two or more solutions, which when mixed together will react to form an insoluble product that will be able to be separated from the solution by filtration. Today we will take one of the phosphorous-containing ions present when plant foods dissolve and combine it with magnesium and ammonia to make a substance that is insoluble, MgNH4PO4•6H2O. [The “•6H2O” refers to the “waters of hydration”, i.e. there are 6 water molecules which are part of the crystal structure – they aren’t covalently bound to the rest of the species, but they are held rigidly in place.]
Mg2+(aq) + NH3(aq) + HPO42- (aq) + 6 H2O(l) → MgNH4PO4•6H2O(s)
The calculation goes something like this:
- A mass of MgNH4PO4•6H2O is collected, and it can be related to the number of moles of MgNH4PO4•6H2O by knowing the molecular weight (you figure it out).
- The precipitation equation indicates that the number of moles of MgNH4PO4•6H2O is the same as the number of moles of HPO42-,
- The number of moles of HPO42- is twice the number of moles of P2O5 present in the plant food sample originally.
- One can find the mass of P2O5 in the plant food sample by knowing its molecular weight (you figure that one out too).
- Finally, knowing the mass of the P2O5 and the mass of the original plant food sample, one can easily determine the weight percent P2O5 in the plant food.
The difficult part of this experiment is in controlling the acidity of the solution so that the only phosphate species is the HPO42- ion
If we make the solution too basic, there will be PO43- rather than HPO42-, and we won’t form the desired precipitate. Additionally, if there are too many OH- ions (i.e., the solution is too basic) they will precipitate with Mg2+ to form Mg(OH)2, and we don’t want that to happen either. If the solution is not basic enough, H2PO4- will be formed , and it will also not precipitate. You will slowly add ammonia until it just begins to become basic. Adding too fast or too much can lead to the coprecipitation of Mg(OH)2.
Procedure:
Weigh by difference to the nearest hundredth gram of the sample plant food. Transfer the sample quantitatively to a 250-mL beaker and record the sample mass. Add 35 - 40 mL of distilled water and stir mixture with a stirring rod to dissolve the sample. Although plant foods are advertised to be water soluble, it may contain small traces of insoluble residue. If the sample does not completely dissolve, remove the insoluble material by means of filtration. To the filtrate add about 45 mL of a 10% MgSO4•7H2O solution. Then add approximately 150 mL of 2 M NH3(aq) slowly while stirring. A white precipitate of MgNH4PO4•6H2O will soon form upon allowing mixture to sit at room temperature for 15 minutes to complete the precipitation.
Next the precipitate must collect on a preweighed piece of filter paper[1]. Obtain a filter paper (three of these will be needed) and weigh it accurately. (Be certain that it is weighed after it has been folded and torn, not before.) Fold the paper as illustrated in Figure 1 and fit into a glass funnel. Be certain that the filter paper is opened in the funnel so that one side has three pieces and one side has one piece of paper against the funnel, not two pieces on each side. Wet the paper with distilled water to hold it in place in the funnel. Completely and quantitatively transfer the precipitate and all of the solution from the beaker onto the filter[2]. Wash the precipitate with two or three 5-mL portions of distilled water; pour over the solid on the funnel. Finally, pour two 10 mL portions of 75% isopropyl alcohol through the filter paper. Once properly filtrated, remove the filter paper, place it on a numbered watch glass[3]. Repeat the above procedure with two more samples. Once the three samples of MgNH4PO4•6H2O is thoroughly dry, weigh the filter papers containing MgNH4PO4•6H2O. Record the mass and calculate the percentages of phosphorus and P2O5 in the original samples.
In means of calculating the percentage of phosphorus and P2O5 in the original the following calculations must be performed:
If for example a 10.00g sample of soluble plant food yields 10.22g of MgNH4PO4•6H2O
First, solving for the grams of P in this sample:
(10.22g MgNH4PO4•6H2O) (1 mol MgNH4PO4•6H2O / 245.4g MgNH4PO4•6H2O) • (1 mol P / 1 mol MgNH4PO4•6H2O) (30.97g P / 1 mol P) = 1.290 P
Thus:
(1.290g P / 10.00g Sample) • 100 = 12.90% P
Similarly, solving for grams of P2O5:
(10.22g MgNH4PO4•6H2O) (1 mol MgNH4PO4•6H2O / 245.4g MgNH4PO4•6H2O) • (1 mol P / 1 mol MgNH4PO4•6H2O) ( 1 mol P2O5 / 2 mol P) (141.9g P2O5 / 1 mol P2O5) = 2.955g P2O5
Thus:
(2.955g / 10.00g sample) • 100 = 29.55% P2O5
As means of estimating the precision of the results, it is desirable to calculate the standard deviation. In doing so the mean must be determined then take that mean and subtract each individual result. This will give three values in which will then be added together and divided by the number of those given values, which is three. That calculated value is the standard deviation.
Here's an example:
Four given results are 1.0, 2.0, 3.0, and 4.0
Mean = 1.0 + 2.0 + 3.0 + 4.0 / 4 = 2.5
The deviations from the mean are:
|2.5 - 1.0| = 1.5
|2.5 - 2.0| = 0.5
|2.5 - 3.0| = 0.5
|2.5 - 4.0| = 1.5
The average deviation from the mean is therefore:
1.5 + 0.5 + 0.5 + 1.5 / 4 = 1.0
Tables:
Results Data Sheet:
Chemicals used Data Sheet:
Figures:
Figure 1:
Figure 2:
Figure 3:
(Figure[3] credit to classmates Kerigan Medeiros & Zulay Holland)
Post Lab Questions:
1. The following percentages of chloride were found in a sample: 22.52%, 22.14%, 22.61%, and 22.75%. (a) Find the mean, the standard deviation, and the relative standard deviation. (b) Can any result be discarded ?
(a) Mean = 22.50%, Standard Deviation = 0.26, Relative Standard Deviation = 1.16
(b) No
2. What is the percent phosphorus in MgKPO4 • 6H2O?
(1) (P) / (266.48) (100%) = 11.62%
Thus: 11.62% P
3. MgNH4PO4•6H2O has a solubility of 0.023 g/100 mL in water. Suppose a 5.02-g sample were washed with 20 mL of water. What fraction of the MgNH4PO4•6H2O would be lost?
(.023g) (.20) = .0046
(.0046 / 5.02g) (100) = .092%
Thus: 0.092%
4. MgNH4PO4•6H2O loses H2O stepwise as it is heated. Between 40o and 60oC the monohydrate is formed, and above 100o the anhydrous material is formed. What are the phosphorus percentages of the monohydrates and of the anhydrous material?
MgNH4PO4•H2O = 155.34
(30.973) (100) / 155.34 = 19.9
MgNH4PO4 = 137.32
(30.973) (100) / 137.32 = 22.6
Thus: Monohydrate = 19.9% P & Anhydrous = 22.6% P
5: Ignition of MgNH4PO4 • 6H2O produces NH3, H2O, and magnesium pyrophosphate, Mg2P2O7. Complete and balance the equation for this reaction. If 2.50g of MgNH4PO4 • 6H2O are ignited. how many grams of Mg2P2O7 would be formed?
1.13g Mg2P2O7
6: What is the percentage of P2O5 in Mg2P2O7?
(141.94) / (222.56) (100) = 63.7% P2O5
Thus: 63.7% P2O5
7: Today sodium content in food is an important health concern. How many milligrams of sodium are present in one ounce of table salt (NaCl)?
11152mg