Complete the following table. Air-dry weight of MgNH4PO4 . 6H20 = 0.44g Grams of
ID: 994448 • 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
To calcualte the theoretical weight of MgNH4PO4.6H2O you should calculate the amount of the phosphorus in the sample from its mass . If you calculate it from the amount of the precipitated salt and use it to calclulate yield, you will just prove that 1=1.
The moles of ppt. are given by : ( 0.44g MgNH4PO4.6H2O)( 1 mol MgNH4PO4.H2O / 245.45 g MgNH4PO4.H2O)
= 0.00179 moles
theoretical wt. of ppt. = 0.00179 x 245.45 = 0.439 g
1) mass of P2O5 = ( 0.44 g MgNH4PO4.6H2O) ( 1 mol MgNH4PO4.6H2O/ 245.45 g MgNH4PO4.6H2O) ( 1 mol P / 1 mol MgNH4PO4.6H2o) ( 1 mol P2O5/ 2 mol P) ( 141.9 g P2O5 / 1 mol P2O5)
= 0.127 g P2O5
% P2O5 = (0.127 g / 0.44) x 100 = 28.86%