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Making a Liquid Fertilizer
Transcript of Making a Liquid Fertilizer
(%composition by mass/100)* 100g= (.87/100)*100=.87g Phosphorus needed. This means we need (.87g /30.974 g/mol)=.0281 mol P
Because there is 1 mol of P for each mol Na3PO4, we need .0281 mol Na3PO4.
To find grams of Na3PO4 needed, take .0281 mol*molar mass=mass needed= .0281mol*147.964g/mol= 4.16 g Na3PO4 needed in a 100g solution. Divide 4.16g by 5 to get .832 g Na3PO4 needed in a 20g solution. Repeat for K, N, and Fe:
K: (2.49/100)*100g= 2.49K which means we need 2.49/molar mass K=2.49g/39.098 g/mol= .064 mol needed. Since there is 1 mol of K in 1 mol of KCl, we need .064 mol KCl. To find # grams of KCl needed, take .064*molar mass KCl=.064*(39.098+35.453)=4.77g
Divide by 5 to get the mass of KCl needed for a 20 g solution: 4.77g/5=.954g KCl By: Kaelyn Holgerson, Bret Andersen, Laura Hayes, Jacob Garay Making a Liquid Fertilizer! Purpose Why Is this Important? Procedure References Data, Observations, and Calculations: >Flinn Scientific, Inc. Discussion Data - Creating liquid fertilizer is difficult. High concentration of solids needed.
Error - Measuring errors, but ineffectual to results.
Take 2 - Try different compounds. Expand list of available compounds.
Outcome - Unsuccessful. Not aqueous. Too much Solid. To find % by mass of Phosphorus in Scotts Turf Builder from 2.0% by total mass of P2O5 in the fertilizer: Take ((molar mass of P)*2/(molar mass of P2O5))*.02 to get (30.974*2 g/Mol)/(141.948 g/Mol)= .87% Phosphorus by mass.
Repeat this step with Potassium in K20, Nitrogen, and Iron as FeSO4:
K: (2*39.098 g/Mol)/(2*39.098+16)g/Mol*.03=2.49% K by mass
N: 16.0% because it was given in elemental form.
Fe: (55.847 g/Mol)/(151.913 g/Mol)*.175=6.43% Fe by mass. www.flinnsci.com/Documents/MSDS/IJ/Iron(III)Nitrate.pdf (accessed 10/3/12)
sciencelab.www.sciencelab.com/msds.php?msdsId=9927076 (accessed 9/19/12)
sciencelab.www.sciencelab.com/msds.php?msdsId=9926684 (accessed 9/19/12)
sciencelab.www.sciencelab.com/msds.php?msdsId=9925028 (accessed 9/19/12)
sciencelab.www.sciencelab.com/msds.php?msdsId=9927321 (accessed 9/19/12) •Create a 10.0 mL of an aqueous fertilizer whose concentration is...
16 % Nitrogen
2.0% of the Phosphorus in P2O5
3.0% of the Potassium in K2O
17.5% of the Iron in FeSO4
•pH needs to between 6.0 and 7.0.
•Ion Concentration in Molarity
•Data with why compounds were used or excluded >science lab >CHEM1075H Lab Moodle Page Now we need to find the # grams needed for our arbitrary 20 g solution: Data, Observations, and Calculations Part 2 To calculate the mass of H2O needed for the solution:
20g total solution = 10.95g (NH4)2CO3 + .832g Na3PO4 + .954g KCl + 5.57g Fe(NO3)3 + Xg H2O
X = 1.69g H2O
for part 1 of the lab, equation only included the mass of (NH4)2CO3, Na3PO4, and KCl
equation was changed for part 2 to include Fe(NO3)3 Data, Observations, and Calculations Part 3 1.Calculate the amount of pure Nitrogen, Phosphorus, Potassium and Iron needed out of the 16.0% Nitrogen, 2.0% of the Phosphorus in P2O5, 3.0% of the Potassium in K2O and 17.5% of the Iron in FeSO4.
2.Choose the compounds with N, P, K, and Fe.
3.Calculate the amount of solid compounded needed for each compound
4.Create a chart with this information
5.Find which compounds if combined form a precipitate. Exclude those compound combinations
6.Once certain combinations are narrowed down, the amount of solid added should be considered. The greater the amount of a compound the more likely this substance will form a paste.
7.Form this data compounds used should be chooses, these were (NH4)2CO3, NaPO4, KCl, FeNO39H2O, and deionized H2O. N: (16/100)*100=16g N which means we need 16g/14.007g/mol=1.14 mol N.
There are 2 moles of N for each mol (NH4)2CO3, so we need 1.14/2=.57 moles of (NH4)2CO3.
To find grams: .57mol*(1.008g/mol*8+2*14.007g/mol+16*3+12.011g/mol)=54.77g (NH4)2CO3
To find grams for a 20 g solution take 54.77/5=10.95g (NH4)2CO3 Procedure Continued 8.Look up the MSDS sheet for (NH4)2CO3, NaPO4, KCl, FeNO39H2O, and deionized H2O.
9.Twenty grams of solution where chosen which at the end of the experiment be decreased to 10g. This gives a little extra solution to work with.
10.The amount of solid substance for 20g should be calculated.
11.An electronic balance should be used to measure out .66g of NaPO4, .92g of KCL, 10.9g of (NH4)2CO3, and 5.57 gFeNO39H2O using weigh paper.
12.Pour the solid compounds into a 40mL beaker.
13.Add 1.9 grams of deionized water to 40 mL beaker using 10 mL graduated cylinder.
14.Heat the solution using a heating plate on the 3rd setting stir while heating.
15.Solution then got very cold and started bubbling so it should be taken off the heat and placed under the fume hood. Fe:(6.43/100)*100g=6.43g Fe needed in 100g solution. # moles= 6.43g/molar mass Fe=6.43g/55.847g/mol= .115 mol Fe needed. There is 1 mol of Fe for every mol of Fe(NO3)3, so we need .115 mol Fe(NO3)3. To find grams of Fe(NO3)3 needed, take molar mass Fe(NO3)3*#moles=
.115*241.868 g/mol=27.8148g Fe(NO3)3 for a 100g solution. Meaning we need 27.8148/5=5.56g Fe(NO3)3 for our 20 g solution. KCl: slightly toxic if ingested, body tissue irritant, avoid contact with strong oxidizers and acids, health and exposure hazards: 1, dispose of safely in waste container
Na3PO4: avoid contact with skin and eyes, reacts violently with water + strong acid, avoid heat and flame, dispose of safely in waste container
(NH4)2CO3: avoid contact with skin and eyes, avoid ingestion, nonflammable, avoid acids, dispose of safely in waste container
Fe(NO3)3: slightly toxic by ingestion and inhalation, irritant to body tissues, avoid all body contact, strong oxidizer, dangerous fire risk, nonflammable solid; health, exposure, storage hazard:1 , reactivity hazard: 3 MSDS information for compounds: Here ^ & Here ^ Fertilizers help plants develop healthy roots, and healthier plants overall.
Liquid fertilizers are easier to use. Especially useful for smaller gardens. Conclusion After mixture of the (NH4)2CO3, Na3PO4, KCL, and deionized H20 in Part 1, our solutes did not entirely dissolve. To fix this, we heated our solution on setting 3 on the hot plate, at which point we observed that there were no visible solids left in the solution, leading us to believe that all solutes had dissolved. in Part 1. + = After the addition of Fe(NO3)3 to the solid mixture that we had previously created in part 1, we added the new mixture to the deionized water. However, the iron compound began to react with one of our other compounds when we added them to water. Furthermore, our mixture did not have nearly enough water to dissolve the "solutes" leaving us with a part-liquid/part-solid heterogeneous mixture. https://moodle2.umn.edu/course/view.php?id+1076 (accessed 10/3/12)