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# Gravimetric Determination of Phosphorus in Plant Food

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Introduction

﻿Gravimetric Determination of Phosphorus in Plant Food Abstract: Gravimetric analysis can be used to determine the percentage of phosphorus in plant food. A precipitant of know composition is produced and weighed to find percent of phosphorus in compound. From the mass and known composition of the precipitate, the amount of the original ion can be determined. In doing so, percents of phosphorus and average percent phosphorus of sample plant food were determined. Introduction: Gravimetric analysis is a quantitative method of classical analysis. The element to be determined is isolated in a solid compound of known identity and definite composition. The mass of the element that was present in the original sample can be determined from the mass of this compound. Plant foods contain three essential nutrients that are not readily available from soils. These are soluble compounds of nitrogen, phosphorus, and potassium. A typical label on a plant food will 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 remaining of the product is fillers, dyes and other anions and cations to balance the charge in the chemical compounds. In this experiment, we will illustrate aquality control analysis for the determination of phosphorus in plant food by gravimetric analysis. Phosphorus will be determined by precipitation of the insoluble salt magnesium ammonium phosphate hexahydrate according to the reaction: 5H2O(l) ...read more.

Middle

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.) ...read more.

Conclusion

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 ...read more.

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