Procedure Summary
A fertiliser sample was dissolved such that another ionic compound* could be added that would collect the sulfate ions present in solution and precipitate them out into a solid that could be filtered, dried, and then weighed. Now having obtained the mass of product, the molar mass of the compound can be used to determine the quantity of sulfur contained within it, then compared against the initial mass of the fertiliser sample.
* The ionic compound used was Barium Chloride, this was chosen using the solubility table of ions for its ability to form BaSO4, which has low solubility.
* The ionic compound used was Barium Chloride, this was chosen using the solubility table of ions for its ability to form BaSO4, which has low solubility.
Experimental Data
| Group | Fertiliser Sample Mass (g) | Filter Paper Mass (g) | Filter Paper + BaSO4 Mass (g) | BaSO4 Mass (g) |
|---|---|---|---|---|
| Group 1 | 0.95 ±0.005 | 2.08 ±0.005 | 3.26 ±0.005 | 1.18 ±0.01 |
| Group 2 | 1.04 ±0.005 | 2.15 ±0.005 | 3.14 ±0.005 | 1.26 ±0.01 |
Created with the HTML Table Generator
Results
The sulfur content in the fertiliser is calculated using the mass % composition formula;
% S = m(S) / m(fertilizer) x 100%
where S = Sulfur
m(S) = Mass of Sulfur
So the mass of sulfur must be calculated - this can be determined via analysing how much sulfur is contained in the Barium Sulfate precipitate.
Mass of Barium Sulfate:
By subtracting the mass of the filter paper from the mass of the dried product on the filter paper, the mass of the Barium Sulfate precipitate is found. Using group 1's data;
m(BaSO4) = m(paper + BaSO4) - m(paper)
= 1.18g ±0.01g
= 1.18g ±0.8% (error is represented in relative uncertainty)
Moles of Sulfur:
Since every unit of BaSO4 contains exactly one sulfur atom, then the number of moles of BaSO4 is equal to the number of Sulfur atoms.
n(BaSO4) = n(S)
n(BaSO4) = m/M since n=m/M, but we need to calculate the molar mass to proceed
Molar Mass:
Mr(BaSO4) = 137.34amu + 32.06amu + 4(16.00amu)
= 233.40amu
M(BaSO4) = 233.40 g/mol
Moles of Barium Sulfate:
n(BaSO4) = (1.18g ±0.8%) / 233.40g/mol
= 5.06E-3 mol ±0.8% (note: E-3 is shorthand for x 10 ^ -3, since I cannot use superscripts on my webpage here)
Moles of Sulfur:
As stated from earlier above, n(BaSO4) = n(S), therefore
n(S) = 5.06E-3 mol ±0.8%
Mass of Sulfur:
m(S) = n(S) x M(S)
= (5.06E-3 mol ±0.8%) x 32.06g/mol
= 0.162g ±0.8%
Percent composition of Sulfur:
% Sulfur = m(S) / m(fertiliser) x 100% note: m(fertiliser) = 0.95g ±0.005g = 0.95g ±0.5% from data table
= (0.162g ±0.8%) / (0.95g ±0.5%) x 100%
= 17% ±1.3%
% S = m(S) / m(fertilizer) x 100%
where S = Sulfur
m(S) = Mass of Sulfur
So the mass of sulfur must be calculated - this can be determined via analysing how much sulfur is contained in the Barium Sulfate precipitate.
Mass of Barium Sulfate:
By subtracting the mass of the filter paper from the mass of the dried product on the filter paper, the mass of the Barium Sulfate precipitate is found. Using group 1's data;
m(BaSO4) = m(paper + BaSO4) - m(paper)
= 1.18g ±0.01g
= 1.18g ±0.8% (error is represented in relative uncertainty)
Moles of Sulfur:
Since every unit of BaSO4 contains exactly one sulfur atom, then the number of moles of BaSO4 is equal to the number of Sulfur atoms.
n(BaSO4) = n(S)
n(BaSO4) = m/M since n=m/M, but we need to calculate the molar mass to proceed
Molar Mass:
Mr(BaSO4) = 137.34amu + 32.06amu + 4(16.00amu)
= 233.40amu
M(BaSO4) = 233.40 g/mol
Moles of Barium Sulfate:
n(BaSO4) = (1.18g ±0.8%) / 233.40g/mol
= 5.06E-3 mol ±0.8% (note: E-3 is shorthand for x 10 ^ -3, since I cannot use superscripts on my webpage here)
Moles of Sulfur:
As stated from earlier above, n(BaSO4) = n(S), therefore
n(S) = 5.06E-3 mol ±0.8%
Mass of Sulfur:
m(S) = n(S) x M(S)
= (5.06E-3 mol ±0.8%) x 32.06g/mol
= 0.162g ±0.8%
Percent composition of Sulfur:
% Sulfur = m(S) / m(fertiliser) x 100% note: m(fertiliser) = 0.95g ±0.005g = 0.95g ±0.5% from data table
= (0.162g ±0.8%) / (0.95g ±0.5%) x 100%
= 17% ±1.3%
Discussion
(This is a student exercise to write from this activity)
Conclusion
The aim of the investigation was to verify the authenticity of the manufacturer's claim of 17% sulfur content, and it was expected that experimental results would differ greatly from the manufacturer's label. The data obtained through this experimental analysis closely agrees with the claim with a result of 17% ±1.3% sulfur. Consequently, this provides supportive evidence on the accuracy of the manufacturer's quantities claimed on the labels and manufacturing processes.
