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Gravimetric determination of phosphorus in plant food lab answers

24/11/2021 Client: muhammad11 Deadline: 2 Day

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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