Complete the following table. Air-dry weight of MgNH4PO4 . 6H20 = 0.44g Grams of
ID: 994518 • Letter: C
Question
Complete the following table.
Air-dry weight of MgNH4PO4 . 6H20 = 0.44g
Grams of phosphorus in plant food = 0.056g
% P in plant food = 4.7 %
P2O5 equivalent to % P above = 21.5
% P2O5 in 10-52-10 plant food = 52%
Molecular weight of MgNH4PO4 . 6H2O = 245.42g
Theoretical weight of MgNH4PO4 . 6H2O = ?
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
yield = actual yield/ theoretical yield
Theoretical yield = actual yield / yield
= 245.2/ 21.5
= 11.40 gm
%yield = actual yield/ theoretical yield X 100
Theoretical yield = actual yield / %yield X 100
= 245.2 /21.5 X 100
= 1140.46 gm
% error = theoretical – experimental/ theoretical X 100
= 1140.46-245.2/1140.46X 100
= 78.5%
% yield = 245.2/ 52 X 100
= 471.5 gm
Sodium hydroxide will ppt the Mg+2 as Mg (OH)2. It has less M.wt than MgNH4PO4.6H2O. Ammonia acts as a ion NH4+ in MgNH4PO4.6H2O. Ammonia exits as a fertilizer (NH4+ion)
Gravimetric determination of phosphate exits as a magnesium ammonium phosphate hexa hydrate (MgNH4PO4.6H2O). The ppt neutralizes ammonia in acidic conditions.