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).
To calculate the percent yield of MgNH4PO4 . 6H2O, you need the actual yield and the theoretical yield.
1. Start by finding the theoretical yield of MgNH4PO4 . 6H2O using the 10-52-10 plant food label as a guide:
- The % P2O5 in the 10-52-10 plant food is given as 52%.
- Convert this percentage to grams of P2O5 by assuming you have 100g of the plant food. Thus, 52 grams of P2O5 will be present.
2. Determine the moles of P2O5 using its molar mass:
- The molar mass of P2O5 is 141.94 g/mol.
- Divide the grams of P2O5 (52g) by its molar mass to get moles of P2O5.
3. Convert moles of P2O5 to moles of MgNH4PO4 . 6H2O:
- Balanced equation: 2NH4H2PO4 + 3MgCl2 + 26H2O -> 2MgNH4PO4 . 6H2O + 6HCl
- The stoichiometric ratio shows that 2 moles of P2O5 react to form 1 mole of MgNH4PO4 . 6H2O.
- Therefore, the moles of MgNH4PO4 . 6H2O will be half the moles of P2O5.
4. Calculate the theoretical yield of MgNH4PO4 . 6H2O:
- Multiply the moles of MgNH4PO4 . 6H2O by its molar mass (245.42 g/mol) to obtain the theoretical yield in grams.
5. Now, you need the actual yield of MgNH4PO4 . 6H2O from your experiment, which you mentioned as 2.30g.
6. Finally, calculate the percent yield:
- Divide the actual yield by the theoretical yield.
- Multiply the result by 100 to obtain the percent yield.
Therefore, you can proceed with these steps to calculate the percent yield of MgNH4PO4 . 6H2O.
To calculate the percent yield of MgNH4PO4 . 6H2O in your experiment, you need to compare the actual yield (the amount you obtained in your experiment) to the theoretical yield (the amount you would expect to obtain based on stoichiometry and the plant food label).
Here's how you can proceed:
1. Determine the theoretical yield of MgNH4PO4 . 6H2O based on the plant food label:
- You have already calculated the grams of phosphorus (P) in MgNH4PO4 . 6H2O as 0.290g.
- Convert the grams of P to grams of P2O5 (phosphorus pentoxide), using the percent of P2O5 in the compound:
The percent of P2O5 in the compound (%P2O5) is equivalent to 55.45% (from the plant food label).
So, grams of P2O5 = (0.290g P) * (55.45g P2O5 / 100g P) = 0.1609g P2O5.
- Finally, calculate the theoretical yield of MgNH4PO4 . 6H2O by considering the molar mass of P2O5 and the stoichiometry of the compound:
- Molar mass of P2O5 = 141.94 g/mol (phosphorus pentoxide)
- Stoichiometry: 1 mol P2O5 produces 1 mol of MgNH4PO4 . 6H2O
- Theoretical yield = (0.1609g P2O5) / (141.94g/mol P2O5) = 0.00113 mol MgNH4PO4 . 6H2O
2. Determine the actual yield of MgNH4PO4 . 6H2O obtained in your experiment:
- The air dry weight of MgNH4PO4 . 6H2O is given as 2.30 g.
3. Calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) * 100
- Percent yield = (2.30g / 0.00113 mol) * 100
- Percent yield = 2035.40%
Therefore, the percent yield of MgNH4PO4 . 6H2O in your experiment is 2035.40%.