Gravimetric_Determination_of_Phosp._in_Plant_Food

Gravimetric_Determination_of_Phosp._in_Plant_Food - ‘...

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Unformatted text preview: ‘- K , . -i--l'l"'. Gravimetric ‘ Determination of Phosphorus in Plant Food To illustrate an application of gravimetric analysis to a consumer product. Apparatus balance funnel support beakers (6), any combination, ring stand 250 mL or larger stirring rods (3) with filter paper (Whatman No. 40) rubber policeman funnels (3) Chemicals :9 75% aqueous isopropyl alcohol" 2 M NH3(aq) 10% aqueous MgSO4 - 7 H20:E plant food Analytical chemistry is concerned with determining how much of one or more constituents is present in a particular sample of material. Two common quantitative methods used in analytical chemistry are gravimetric and volumetric analysis. Gravimetric analysis derives its name from the fact that the constituent being determined is isolated in some weighable form. Volu- metric analysis, on the other hand, derives its name from the fact that the amount of a constituent being determined involves measuring the volume of a reagent. Volumetric analyses are generally less time consuming and less ac- curate than gravimetric analyses. Gravimetric analyses may be difficult and time consuming, but they are in— herently quite accurate. The accuracy of an analysis is often directly propor- tional to the time expended in carrying it out. The ultimate use of the analytical result governs how much time and effOri‘ 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 or not con- struction 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 or not to construct a mill. 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. "Epsom salts, household ammonia (non-sudsy), and rubbing alcohol, respectively, may be substituted for these reagents. 88 Experiment 9 o Gravimetric Determination of Phosphorus in Plant Food Consumer chemicals are subject to quality control by the manufacturer and by various consumer protection agencies. Consumer chemicals are usu- ally 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 P205), and 15% potassium (ex— pressed as K20). 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. EXAMPLE 9.1 What is the minimum percentage of phosphorus in a plant food whose P205 per- centage is guaranteed to be 15%? SOLUTION: Assuming 100 g of plant food, we would have 15 g of P205. Using this quantity, we can calculate the amount of P in the sample: 1molP205)( 21110112 )(30.97gP)_65 P 1:11.9ng05 11110119205 1molP ' g (15 s P205)( Thus, g P 6.5 g %P = X 100 — X 100 — 6.5% g sample 100 g In this experiment we will illustrate one of the quality—control analyses for plant food by gravimetric determination of its phosphorus content. Phos- phorus will be determined by precipitation of the sparingly soluble salt mag- nesium ammonium phosphate hexahydrate according to the reaction 5HZO(I) + HPOf’Qaq) + NH4+(aq) + Mg2+(aq) + OH—(aq) —> MgNH4PO4 - 6H20(s) EXAMPLE 9.2 If a 10.00-g sample of soluble plant food yields 10.22 g of MgNH4PO4 - 6H20, what are the percentages of P and P205 in this sample? SOLUTION: First, solving for the grams of P in the sample 1 mol MgNH4PO4 - 6H20 ) 245.4 g MgNH4PO4 - 6HZO 30.97 P X 2 1.290 g P 1 mol MgNH4PO4 - 6H20 1 mol P (10.22 g MgNH4PO4- 6HZO)( Thus, g P 1.290 g P %P = w X 100 I —~-- X 100 = 12.90%P g sample 10.00 g sample C C E IL‘ 5' E 6' E E e.- c,- E C C E e- u- I— o— u- u- u- h- .- UUUUHUUUH I I u I“ I‘ t . Laboratory Experiments Similarly, solving for grams of P205: 1 mol ' (10.22 g MgNH4PO4 - 6H20)( ) .4 g ‘ 1 l 1 1P 0 141.9 P O x( m P m0 2 5X g 2 5) — 2955ng05 1 mol ' 2 11101 P 1 mol P205 2.955 g and 0/0 P205 = 10.00 g x 100 e 29.55 513205 Weigh by difference to the nearest hundredth gram 1.5 to 2.5 g of your un- l PROCEDURE known sample, using weighing paper. Transfer the sample quantitatively ' to a 250-mL beaker and record the sample mass. Add 35 to 40 mL of dis- tilled 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. If your sample does not completely dissolve, remove the insoluble material by filtration. To the fil- trate add about 45 mL of a 10% MgSO4 - 7HZO solution. Then add approx- imately 150 mL of 2 M NH3(aa) slowly while stirring. A white precipitate of MgNH4PO4 - 61-120 will form. Allow the mixture to sit at room tempera- ture for 15 minutes to complete the precipitation. Collect the precipitate on a preweighed piece of filter paper (Figure 9.1). Fold and crease lightly. Seal the moistened edge of the filter Tear off comer paper against the unequally. funnel, making sure Open out to form ' that the paper over a cone with one piece ‘ the bottom portion of paper against one side and three pieces of paper against the other side of the funnel. is set firmiy against the tunnel to prevent air from being sucked down the side of the paper. Pour down a glass rod to aid in transfer. The filtrate should run down the walls of the beaker. The weight of the water column hastens filtration. ‘ Use a rubber policeman to transfer the last traces of precipitate from the beaker. A FIGURE 9.] Filter paper use. (Filtration of MgNH4PO4 ' 6H20 is slow. Time may be saved byfilteriag by suction with a Biichnerfimnel.) 89 _. 90 Experiment 9 0 Gravim'etric Determination of Phosphorus in Plant Food Instructor demonstration Obtain a filter paper (three of these will be needed) and weigh it accurate- ly. (Be certain that you weigh the paper after it has been folded and torn, not before.) Fold the paper as illustrated in Figure 9.1 and fit it into a glass fun- nel. Be certain that you open the filter paper 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. Why? Your instructor will also demonstrate this for you. Wet the paper with distilled water to hold it in place in the funnel. Com— pletely and quantitatively transfer the precipitate and all of the solution from the beaker onto the filter, using a rubber policeman (your laboratory instruc- tor will show you how to use a rubber policeman). Wash the precipitate with two or three 5—mL 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 to dry in your locker until the next period. Repeat the above procedure with two more samples. In the next period, when the MgNH4PO4 - 6H20 is thoroughly dry, weigh the filter papers plus MgNH4PO4 ' 6HZO. Record the mass and calculate the percentages of phos— phorus and P205 in your original samples. Standard Deviation As a means of estimating the precision of your results, it is desirable to cal- culate the standard deviation. Before we illustrate how to do this, however, we will define some of the terms above as well as some additional ones that are necessary. Accuracy: measure of how closely individual measurements agree with the correct (true) value. Precision: the closeness of agreement among several measurements of the same quantity; the reproducibility of a measurement. Error: difference between the true result and the determined result. Determinate errors: errors in method or performance that can be discov- ered and eliminated. Indeterminate errors: random errors that are incapable of discovery but which can be treated by statistics. Mean: arithmetic mean or average (,u), where sum of results number of results For example, if an experirnent's results are 1, 3, and 5, then _1+3+5_ 3 3 )1. Median: the midpoint of the results for an odd number of results and the average of the two middle results for an even number of results (In). For example, if an experiment’s results are 1, 3, and 5, then m : 3. If results are 1.0, 3.0, 4.0, and 5.0, then 3.0+. =_40=35 m 2 e. g e E e' e E e.- e.- c.- e- e- E II - uni UUUUUUUU ll‘llllllflllll The scatter about the mean or median—that is, the deviations from the mean or median—are measures of precision. Thus the smaller the devia- tions, the more reproducible or precise the measurements. EXAMPLE 9.3 If an experiment’s results are 1.0, 2.0, 3.0, and 4.0, calculate the mean, the devia- tions from the mean, the average deviation from the mean, and the relative aver- age deviation from the mean. SOLUTION: The mean is calculated as follows: 1.0 + 2.0 + 3.0 + 4.0 _ 10.0 ,u, — 4 , 4 2.5 The deviations from the mean are ‘25 s 1.0] 2 1.5 12.5 — 2.0! = 0.5 I25 - 3.0I = 0.5 [2.5 — 4.0; = 1.5 The symbol I I means absolute value, so all differences are positive. The aver- - age deviation from the mean is therefore 1. + . +0. +1. 5 054 5 5:10 The relative average deviation from the mean is calculated by dividing the aver- age deviation from the mean by the mean. Thus, Relative deviation 2 = 0.40 This can be expreSSed as 40%, 400 parts/ thousand (ppt), or 40,000 parts/million (ppm). Note, if the mean were larger, say 100 instead of 2.5, and the average deviation still 1.0, the relative deviation becomes 1.0/100 or 1.0% (10ppt). In this case the precision is better because the relative deviation is smaller. Standard deviation (3) is related to statistics and is a better measure of pre— cision and is calculated using the formula \/ sum of the squares of the deviations from the mean S = number of observations - 1 Eva — #2 N—l where s = standard deviation from the mean, X,- = members of the set, p, = mean, and N = number of members in the set of data. The symbol 2,- means to sum over the members. EXAMPLE 9.4 An experiment's results are 1, 3, and 5. Calculate the mean, the deviations from the mean, the standard deviation, and the relative standard deviation for the data. SOLUTION: The mean is as follows: 1+3+5 =—=3 ”‘ 3 Laboratory Experiments 91 92 Experiment 9 0 Gravimetric Determination of Phosphorus in Plant Food The deviations from the mean are I X1- — id = deviation I1—M=2 13-3I=0 5—m22 f+02+fi 3r1 M+0+4 2 f' _ E = 4:2 s=———- The results of this experiment would be reported as 3 :t 2. The relative standard deviation is g 2 0.7, or 70% Calculate the standard deviation of your data and report the results on your report sheet. The standard deviation may be used to determine whether a result should be retained or discarded. As a rule of thumb, you should discard any result that is more than two standard deviations from the mean. For example, if you had a result of 49.65% and you had determined that your percentage of phosphorus was 49.25 :l: 0.09%, this result (49.65%) should be discarded. This is because 5 = 0.09 and [49.25 - 49.65I = 0.40, which is greater than 2 X 0.09. This result is more than two standard deviations from the mean. PRE LAB Before beginning this experiment in the laboratory, you should be able to an— QUESTIONS swer the following questions; 1. Which method of analysis generally is the faster method, gravimetric or volumetric? ' 2. Why would only three significant figures be required for the analysis of a consumer chemical such as P205 in plant food? The label on a plant food reads 23-19-17. What does this mean? 4. What is the minimum percentage of phosphorus in the plant food in question 3? 5. What is the minimum percentage of potassium in the plant food in question 3? What is the percent sulfate in CuSO4 - 5H20? Does standard deviation give a measure of accuracy or precision? If an experiment’s results are 12.1, 12.4, and 12.6, find the mean, the av- erage deviation from the mean, the standard deviation from the mean, and the relative average deviation from the mean. 9. What is meant by the term indeterminate errors? 10. Differentiate between qualitative and quantitative analysis. 11. What is meant by the term accuracy? “any:cannnnnnnnnnnnnnnn. ...
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This note was uploaded on 02/06/2012 for the course BIO 210 taught by Professor Tricca during the Fall '11 term at Canada College.

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Gravimetric_Determination_of_Phosp._in_Plant_Food - ‘...

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