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Synthesis of a Liquid Fertilizer

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by

Della Delaney

on 14 March 2014

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Transcript of Synthesis of a Liquid Fertilizer

Why these compounds?
Provide necessary elements

Each has one necessary element and does not have overlap element from each other

Solubility of these compounds was tested
Synthesis of a Liquid Fertilizer
Della delaney, Yi Herng Ong

Objective
Create an aqueous fertilizer with specific mass percent

Requirements
The volume of fertilizer cannot exceed 25 mL

pH value : 6~7

The ion concentration must be tabulated in terms of molarity

Safety cannot be compromised...

Wear lab coat and goggles all the times
Handle acids and bases carefully
Soap and water
Refer to MSDS if necessary
Take the smallest scale of compounds if possible (limit waste)
Pour the collected waste into labeled waste containers
Compounds We Used
Day 1
NaNO3- Sodium Nitrate; Nitrogen
Na3PO4- Sodium Phosphate; Phosphorous
KCl- Potassium Chloride; Potassium

Day 2
Fe(NO3)3- Iron (III) Nitrate; Iron
Background
Significance of fertilizer
What plants need to supply essential nutrients
What they cannot obtain themselves
Procedure-Day 1
Calculate the mass of every compound

Measure the compounds

Mix them with DI water

Test its pH value

Neutralize it by adding acid and bases (if pH value does not fall on 6~7)
NaNO3
(0.8 g /100 g) x (1g/1 mL) x (1 mol/14.01g) x (1 mol NaNO3/1 mol N)x 84.9947g/1 mol NaNO3) x (25 mL/100 mL)= 0.0121 g

Na3PO4
(0.2 g /100 g) x (1g/1 mL) x (1 mol/30.97g) x (1 mol Na3PO4/1 mol P)x 163.94g/1 mol Na3PO4) x (25 mL/100 mL)= 0.00265 g

NaNO3
(0.3 g /100 g) x (1g/1 mL) x (1 mol/39.10g) x (1 mol KCl/1 mol K)x 74.5513g/1 mol KCl) x (25 mL/100 mL)= 0.001425 g
Calculations-Day 1
Procedure-Day 2
Calculations- Day 2
Day 2 Results
Conclusion
25 mL Fertilizer Solution
And pH test strip
Calculate the mass of Iron (III) nitrate (Fe(NO3)3)
recalculation of NaNO3 is necessary (element of Nitrogen overlap)
Measure all the compounds
Mix them with DI water
Test its pH value
Neutralize it by adding acid and base (if pH value does not fall on 6~7)

Techniques, equations, and tools used
To create 0.1 mass percent of Iron

Fe(NO3)3

(0.1 g /100 g) x (1g/1 mL) x (1 mol/55.85g) x (1 mol Fe(NO3)3/1 mol Fe)x (241.88g/1 mol Fe(NO3)3) x (25 mL/100 mL)= 0.001075 g




Percent by Mass
Solubility
Concentration/Molarity
Precipitant Reactions
Volumetric Flask
Scale
pH paper
Calculation Day-2 (continue..)
NaNO3 (Recalculation)

Fe(NO3)3 it yields
0.1737g x 0.0043 g= 0.00074 g of Nitrogen

And we still need
0.8 g - 0.00074 g = 0.799 g of N

NaNO3
(0.799 g /100 g) x (1g/1 mL) x (1 mol/14.01g) x (1 mol NaNO3/1 mol N)x(241.88g/1 mol NaNO3) x (25 mL/100 mL)= 0.012118 g of NaNO3
25 mL of fertilizer solution
A pH of 6-7 was created, after adding NaOH (base)
No precipitate or coloration formed
Since we did not have enough iron to create a precipitate, no EDTA was necessary.

What are they used for?
Primary Nutrients
1. Nitrogen
rapid growth
fruit production
increasing seed

2. Phosphorus
root growth

3. Potassium
building proteins
photosynthesis
fruit quality
reduction of diseases

http://www.ncagr.gov/cyber/kidswrld/plant/nutrient.htm
3 Non mineral nutrients
Hydrogen
Oxygen
Carbon
Day 1 Results
13 Mineral Nutrients
Macro nutrients
Micro nutrients
Made 25 mL fertilizer
no precipitates formed
pH of between 6-7
16 chemical elements are crucial for plant's growth
Micro Nutrients
1. Iron
Essential for formation of chlorophyll
http://www.ncagr.gov/cyber/kidswrld/plant/nutrient.htm
Compatible 25 mL fertilizer solution
Correct pH level
Correct color (clear, no precipitate)
What we learned
Why this is important
Useful in the future?
Errors
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