posted by Julia .
need to calculate percent yield of MgNH4PO4 . 6H2O obtained in my experiment using 10-52-10 plant food label as a guide to determine the theoretical yield of MgNH4PO4 . 6H2O.
My calculations so far:
air dry weight of MgNH4PO4 . 6H2O = 2.30 g
Grams of phosphorusin MgNH4PO4 6H20 = 0.290g
%P in plant food = 24.16
% P2O5 in equivalent to %P above = 55.45
% p2O5 in 10-52-10 plant food = 52
How do I proceed to calculate percent yield?
I forgot to add that in my experiment I used 1.2 grams of 10-52-10 plant food ( + 4 grams MgSO4 7H2O) and obtained 2.30 g of MgNH4PO4 6H20
1.2 x 0.52 = ??g P2O5 if that's what the 52 means; i.e., 52% P2O5.
Then ??g P2O5 x (2*molar mass MgNH4PO4.6H2O/molar mass P2O5) = xx = grams MgNH4PO4.6H2O you should have obtained which would be the theoretical yield. I ran that on my calculator and obtained something like 2.16 g which means your air dried sample may not have been dry (over 100% yield).