I just finished an experiment where I precipitated 1.44 grams of MgNH4PO4 x 6H2O
ID: 572357 • Letter: I
Question
I just finished an experiment where I precipitated 1.44 grams of MgNH4PO4 x 6H2O.
The gravimetric analysis of phosphorous in this experiment is based on the precipitation of phosphorous as MgNH4PO4 . 6H2O from a solution that contains the monohydrogen phosphate ion (HPO42- ), ammonium ions, and magnesium ions. The balanced reaction is: 5 H2O + HPO42- + NH4+ + Mg2+ + OH1- =====> MgNH4PO4 . 6H2O
What is the theoretical weight of MgNH4PO4 x 6H2O?
Values obtained from experiment:
1.44 grams= Air-dry weight of MgNH4PO46 H2O
245.74 g/mol = Molecular weight of MgNH4PO46H2O
0.1815 grams = Grams of phosphorus in plant food
15.1% = Phosphorus in plant food
34.60% = %P2O5 equivalent to % P above
30% P2O5 in 15-30-15 plant food
Explanation / Answer
If the plant food contains 52% P2O5,
The P content would be 13.09%,
Calculated as
In P2O5 43.6421% P contains So,
100% P2O5 = 43.6421% P
30% P2O5 = ?
P = (30 * 43.6421)/100 = 13.09%
The amount of P in sample would be
0.1815 grams = Grams of phosphorus in plant food
15.1% = Phosphorus in plant food
15.1% P in plant food = 100% plant food contain
0.1815 gm P in plant = How many gm of sample
Sample weight = (0.1815 * 100)/15.1 = 1.201 gm of sample
So, P in sample is 1.201gm 13.09% = 0.15708 gm 0.16 gm 2 = 0.32 gm
Since,
1.44 gm of MgNH4PO4 contain 0.3248 gm P
So, You got 100% yield that why theoretical weight is 1.44 gm