Colorimetric Analysis
Title
Colorimetric determination of the phosphorus content of lawn fertiliser
Aim
To find the phosphorus content using colorimetric analysis
Materials
On the sheet attached
Procedure
On the sheet attached
Results
Standard (mg/L) | Absorbance | 10.0 | 0.354 | 7.5 | 0.308 | 5.0 | 0.230 | 2.5 | 0.193 | 0.0 | 0 | unknown | 0.161 |
Calculations 1. What is the concentration of phosphorus ions, in mg/L, in the initial 250 mL of fertiliser solution? y = 0.0399x
Absorbance of unknown = 0.161
0.161 = 0.0399x x = 4.035 4.035 x 10 = 40.35 = 40.4 mg/L
2. Use your answer to question 1 to find the mass of phosphate ions in the initial 250 mL solution. This is the same as the mass of phosphorus in the sample of fertiliser.
40.35mg = 0.04035g
0.04035g / (250/1000) = 0.010088g
3. Calculate the percentage of phosphorus, by mass, in the fertiliser sample.
Mass (P) = Mr(P) / Mr(PO4) x 0.010088 = (31 / 95) x 0.010088 = 0.003292g
Mass (P) = (0.003292/0.095) x 100 = 3.46503 % = 3.47%
Discussion
1. If you analysed a commercial fertiliser, how well does your result for the percentage of phosphorus present agree with the manufacturer’s specification? Try to account for any difference.
The manufacturer’s specification stated 2.4% phosphorous in the fertiliser however the result conceived 3.47%. This amounts to a difference of 1.07% which differs greatly from the specification. Possible sources of this error are * Using a measuring cylinder instead of a volumetric flask when diluting * Contaminated test tubes *
2. If this analysis was performed using a colorimeter, yellow light would be used to measure the absorbance of solutions. Why is blue light not used?
Blue light cannot be used to measure absorbance of the solution because it cannot be absorbed. Instead of absorbing, the solution reflects blue light
Colorimetric determination of the phosphorus content of lawn fertiliser
Aim
To find the phosphorus content using colorimetric analysis
Materials
On the sheet attached
Procedure
On the sheet attached
Results
Standard (mg/L) | Absorbance | 10.0 | 0.354 | 7.5 | 0.308 | 5.0 | 0.230 | 2.5 | 0.193 | 0.0 | 0 | unknown | 0.161 |
Calculations 1. What is the concentration of phosphorus ions, in mg/L, in the initial 250 mL of fertiliser solution? y = 0.0399x
Absorbance of unknown = 0.161
0.161 = 0.0399x x = 4.035 4.035 x 10 = 40.35 = 40.4 mg/L
2. Use your answer to question 1 to find the mass of phosphate ions in the initial 250 mL solution. This is the same as the mass of phosphorus in the sample of fertiliser.
40.35mg = 0.04035g
0.04035g / (250/1000) = 0.010088g
3. Calculate the percentage of phosphorus, by mass, in the fertiliser sample.
Mass (P) = Mr(P) / Mr(PO4) x 0.010088 = (31 / 95) x 0.010088 = 0.003292g
Mass (P) = (0.003292/0.095) x 100 = 3.46503 % = 3.47%
Discussion
1. If you analysed a commercial fertiliser, how well does your result for the percentage of phosphorus present agree with the manufacturer’s specification? Try to account for any difference.
The manufacturer’s specification stated 2.4% phosphorous in the fertiliser however the result conceived 3.47%. This amounts to a difference of 1.07% which differs greatly from the specification. Possible sources of this error are * Using a measuring cylinder instead of a volumetric flask when diluting * Contaminated test tubes *
2. If this analysis was performed using a colorimeter, yellow light would be used to measure the absorbance of solutions. Why is blue light not used?
Blue light cannot be used to measure absorbance of the solution because it cannot be absorbed. Instead of absorbing, the solution reflects blue light