Your goal this week is to determine if a given plant food sample has as much Phosphorus as the label indicates. You are writing a short form memo to the VP for quality control detail in the results of your gravimetric analysis of the supplied sample.
In your summary/results you are reporting an average (of all your runs) of %mass P₂O₅ in the sample as well as a standard devation. It is the average that you will compare to other groups for your commentary section.

Gravimetric analysis is a quantitative (i.e. how much?) 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 a quality 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:

5 H₂O(l) + HPO₄^2-(aq) + NH₄⁺(aq) + Mg²⁺(aq) + OH^-(aq) → MgNH₄PO₄·6H₂O(s)
The ·6H₂O at the end of the product shows that it is a ‘hydrate’. This means that there are 6 water molecules (garden variety) bonded to each MgNH₄PO₄ molecule. This means that when you calculate the molar mass of MgNH₄PO₄·6H₂O(s) you need to include the mass of the 6 water molecules.

The % P₂O₅ (which is what the label says) can be calculated using a psuedo balanced chemical equation :
? NH₃ + ? MgSO₄·7H₂O + 1 P₂O₅ (in sample)  → 2 MgNH₄PO₄·6H₂O(s) + other crap 
If you know the mass of product, you can easily calculate the mass of P₂O₅  that reacted, which is how much was in your sample. Divide that amount by the original sample amount and you are done.

Procedure in brief:
You will prepare 3 samples by adding ~0.10 g of plant food sample into a 150 mL beaker. Add ~10 mL water and stir to dissolve the sample. When it is completely dissolved (you might have to filter it) add 5 mL of 10% MgSO₄·7H₂O solution to the beaker. Slowly add 20 mL of 2M NH₃ to the solution while stirring. Let the solution sit for 15 min to complete the precipitation. Filter the solution with a Buchner funnel through a pre-weighed filter paper. Rinse the filter paper with isopropyl alcohol before placing in drying oven.
    Note about your lab report: It is a short form memo, what is above is NOT what you put in for a procedure. You need to use your own words, speak in past tense (you ‘did’ it) and remove things like references to the beaker, because that is not important. Also, you need to talk about the calculations as well in the procedure.

Pre lab questions:
1. What is the molecular weight of MgNH₄PO₄·6H₂O?
2.  A 0.100 g sample of plant food was reacted (via the given procedure) and 0.0963 g of  MgNH₄PO₄·6H₂O was recovered. What is the mass % of P₂O₅ in the original sample?
-- On a separate piece of paper make some semblance of a data sheet. What will you need to know/measure for the experiment? Show the data sheet to your instructor upon entering lab.

~MEO 11.2.10