Complete the following table. Air-dry weight of MgNH4PO4 . 6H20 = 1.24g Molecula
ID: 1022158 • Letter: C
Question
Complete the following table.
Air-dry weight of MgNH4PO4 . 6H20 = 1.24g
Molecular Weight of MgNH4Po4 6 H20 = 245.42
Grams of phosphorus in plant food = 0.156g
% P in plant food = 13%
P2O5 equivalent to % P above = 29.8
% P2O5 in 10-52-10 plant food = 52%
Theoretical weight of MgNH4PO4 . 6H2O = ? I got 2.16g
1) Compare your % P2O5 value with the theoretical value by calculating the percent error?
2)Now calculate the percent yield of MgNH4PO4 . 6H2O obtained in your experiment using the 10-52-10 label as a guide to determining the theoretical yield of MgNH4PO4 . 6H2O?
3) Why was ammonia used instead of sodium hydroxide to isolate MgNH4pO4 . 6H2O in this experiment?
Explanation / Answer
From the given data
1) Percent error = ((52 - 29.8)/52) x 100 = 42.69%
2) 2 moles of P2O5 is obtained from 1 mole of MgNH4PO4.6H2O
% P2O5 in plant food = 52%
Then theoretical yield = 0.52 x 2 x 245.42/141.94 = 1.80 g
[pl. note this value would change with change in grams of plant food]
3) NaOH is a stronger base than NH3. NaOH addition would produce hydroxide of Mg, as Mg(OH)2 which is gel like and hard to work with. On the other hand addition of NH3 in the form of aqueous solution as NH4OH provides ammonium ions which forms MgNH4PO4, a higher molecular weight, powder (formed after drying) material which is easy to work with and gives more reproducible results in the analysis.