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Gravimetric determination of phosphorus in plant food lab answers

08/01/2021 Client: saad24vbs Deadline: 10 Days

GRAVIMETRIC DETERMINATION OF PHOSPHORUS


IN PLANT FOOD


DISCUSSION


Gravimetric analyses may be difficult and time consuming but are inherently quite


accurate. The accuracy of an analysis is often directly proportional to the time expended


in carrying it out. The ultimate use of the analytical result governs how much time and


effort the analytical chemist should expend in obtaining it. For example, before building a


mill to process gold ore, an accurate analysis of the ore is required. Mills are very


expensive to build and operate, and economic factors determine whether of not


construction of the mill is worthwhile. Because of the value of gold, the difference of


only a few hundredths of a percent of gold in an ore may be the governing factor as to


whether of not to construct a mill. While, on the other hand, the analysis of an


inexpensive commodity chemical such as a plant food requires much less accuracy; the


economic consequences of giving the consumer an extra 0.2% of an active ingredient are


usually small even for a large volume of product. Time is too valuable, whether it be the


students’ or scientists’, to be wasted in the pursuit of the ultimate in accuracy when such


is not needed.


Consumer chemicals are subject to quality control by the manufacturer and by various


consumer protection agencies. Consumer chemicals are usually analyzed both


qualitatively to determine what substances they contain and quantitatively to determine


how much of these substances are present. For example, plant foods are analyzed this


way.


Plant foods contain three essential nutrients that are likely to be lacking in soils. These


are soluble compounds of nitrogen, phosphorus and potassium. The labels on the plant


food usually have a set of numbers such as 15-30-15. These numbers mean that the plant


food is guaranteed to contain at least 15% nitrogen, 30% phosphorus (expressed as P2O5)


and 15% potassium (expressed as K2O). The rest of the product is other anions or cations


necessary to balance charge in the chemical compounds, dyes to provide a pleasing color,


and fillers.


In this experiment we will illustrate one of the quality control analyses for plant food


by gravimetric determination of its phosphorus content. Phosphorus will be determined


by precipitation of the sparingly soluble salt magnesium ammonium phosphate


hexahydrate according to the reaction:


5H2O(l) + HPO4 2-


(aq) + NH4 + (aq) + Mg


2+ (aq) + OH


- (aq)  MgNH4PO4 6H2O (s)


EXAMPLE 1:


What is the minimum phosphorus percentage of a plant food whose P2O5 percentage is


guaranteed to be 15%?


Solution: Assuming 100 g of plant food, we would have 15 g of P2O5. Using this


quantity we can calculate the amount of P in the sample:


(15 g P2O5)(1 mol P2O5/141.9 g P2O5)(2 mol P/1 mol P2O5)(30.97 g P/1 mol P) = 6.5 g P


%P = 6.5 g P/100 g sample x 100 = 6.5%


EXAMPLE 2:


If a 10.00g sample of soluble plant food yields 10.22g of MgNH4PO4  6H2O, what are


the percentages of P and P2O5 in this sample?


Solution: First, solve for the grams of P in the sample.


(10.22g MgNH4PO4  6H2O)(1 mol MgNH4PO4  6H2O/245.3g MgNH4PO4  6H2O) X


(1 mol P/1 mol MgNH4PO4  6H2O)(30.97g P/1 mol P = 1.290g P


%P = (1.290g P/10.00g Sample) X 100 = 12.90% P


Similarly, solve for grams of P2O5.


(10.22g MgNH4PO4  6H2O)(1 MOL MgNH4PO4  H2O/245.3g MgNH4PO4  6H2O) X


(1 mol P/1 mol MgNH4PO4  6H2O)(1mol P2O5/2 mol P)(141.9g P2O5/ 1 mol P2O5) =


2.956g P2O5


%P2O5 = (2.956g / 10.00g) = 29.56% P2O5


PROCEDURE


Weigh to the nearest 0.01g 1.5 to 2.0grams of your assigned brand of fertilizer onto


weighing paper. Transfer the sample quantitatively to a 250 mL beaker and record the


sample weight. Add 35 to 40 mL of distilled water and stir the mixture with a glass


stirring rod to dissolve the sample. Although plant foods are all advertised to be water-


soluble, they may contain a small amount of insoluble residue. 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 form. Allow the


mixture to sit at room temperature for 15 minutes to complete the precipitation. Collect


the precipitate on a preweighed piece of filter paper.


Transfer the precipitate to the filter paper with the aid of a rubber policeman. Wash the


precipitate with two or three 5mL portions of distilled water. Do this by adding each


portion to the beaker in which you did the precipitation to transfer any remaining


precipitate; then pour over the solid on the funnel. Finally, pour two 10 mL portions of


75% isopropyl alcohol through the filter paper. Remove the filter paper, place it on a


numbered watch glass, and store it in your drawer until the next lab period to dry.


Repeat the above procedure with two more samples. In the next period, when the


MgNH4PO4  6H2O is dry, weigh it and calculate the percentages of P and P2O5 in your


original samples.


Date ______________ Section _________ Name ____________________


Data and Calculations Sheet: GRAVIMETRIC DETERMINATION OF PHOSPHORUS IN PLANT FOOD


Data:


Trial 1 Trial 2 Trial3


Weight of sample __________ __________ __________


Weight of filter paper


And MgNH4PO4  6H2O __________ __________ __________


Weight of filter paper __________ __________ __________


Calculations: (You MUST show your work on a separate sheet)


Weight of MgNH4PO4  6H2O__________ __________ __________


Weight of P in original


Sample


__________ __________ __________


Percentage P in


original sample __________ __________ __________


Average percent phosphorus __________


Average percent P2O5 __________


Guaranteed minimum percent P2O5


from label __________


Post Lab Questions:


1.MgNH4PO4  6H2O has a solubility of 0.023g/100mL in water. Suppose a 5.02g


sample were washed with 20 mL of water. What fraction of the MgNH4PO4  6H2O


would dissolve?


2.MgNH4PO4  6H2O loses H2O stepwise as it is heated. Between 40 and 60C the


monohydrate is formed and above 100C the anhydrous material is formed. What are


the phosphorus percentages of the monohydrate and of the anhydrous material?


3. Ignition of MgNH4PO4  6H2O produces NH3, H2O and magnesium pyrophosphate,


Mg2P2O7. Complete and balance the equation for this reaction. If 5.00g of


MgNH4PO4  6H2O are ignited, how many grams of Mg2P2O7 would be formed?


4. How closely did your result for the % P (as P2O5) agree with the minimum percentage claimed on the label? If it did not agree propose an explanation for


this result.


Date ___________ Section _________ Name________________________


ADVANCED STUDY ASSIGNMENT: Gravimetric Determination of Phosphorus


in Plant Food.


1. Why would only three significant figures be required for the analysis of a


consumer chemical such as P2O5 in plant food?


2. The label on a plant food reads 23-19-17. What does this mean?


3. What is the minimum percentage of phosphorus in the plant food in question 2?


4. What is the minimum percentage of potassium in the plant food in question 3?


5. If an experiment’s results are 12.1g, 12.4g and 12.6g, find the mean, the average


deviation from the mean and the standard deviation from the mean.

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