Sulfate in Lawn Fertiliser Gravimetric Analysis

Key Concepts

Gravimetric analysis is the quantitative isolation of a substance by precipitation and weighing of the precipitate.¹

Lawn fertilisers (also known as lawn food) usually contain water soluble ammonium sulfate², (NH₄)₂SO₄

Gravimetric analysis can be used to determine the amount of sulfate present in lawn fertiliser (lawn food) by precipitation of the sulfate as insoluble barium sulfate³, BaSO₄
Ba²⁺_(aq) + SO₄^2-_(aq) → BaSO_4(s)

The mass of barium sulfate precipitated is used to calculate:
(i) moles of BaSO_4(s) precipitated
moles = mass ÷ molar mass

(ii) moles of sulfate ion, SO₄^2-, present
mole ratio BaSO₄:SO₄^2- is 1:1

(iii) mass of sulfate ion, SO₄^2-, present in the lawn fertiliser.
mass = moles × molar mass

The amount of sulfate in the lawn fertiliser is expressed as a percent by mass (w/w %)

% SO₄^2- = mass SO₄^2- (g)
mass lawn fertiliser (g) × 100

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Theory

The lawn fertiliser (lawn food) to be analysed contains ammonium sulfate, (NH₄)₂SO₄.
First we take a sample of lawn fertiliser, weigh it and record its mass in grams:

m(fertiliser) = ? g

Then we dissolve the sample in water producing an aqueous solution containing ammonium ions, NH₄⁺_(aq), and sulfate ions, SO₄^2-_(aq).
(Solubility rules tell us all ammonium salts are soluble in water)

(NH₄)₂SO₄ → 2NH₄⁺_(aq) + SO₄^2-_(aq)

Next we add an excess of barium ions, Ba²⁺_(aq), to precipitate out barium sulfate, BaSO_4(s)
(Solubility rules tell us that all sulfates are soluble except those of barium, strontium, calcium, silver, mercury(I) and lead, that is, barium sulfate is insoluble in water)

Ba²⁺_(aq) + SO₄^2-_(aq) → BaSO_4(s)

We weigh the barium sulfate precipitate and calculate the moles of BaSO_4(s), n(BaSO_4(s)):

n(BaSO_4(s)) = mass(BaSO_4(s)) ÷ molar mass(BaSO_4(s))
Use the Periodic Table to find the molar mass, M, of each element:
M(Ba) = 137.3 g mol^-1
M(S) = 32.06 g mol^-1
M(O) = 16.00 g mol^-1

Calculate molar mass of BaSO₄:
M(BaSO_4(s)) = 137.3 + 32.06 + 4 × 16.00 = 233.36 g mol^-1
n(BaSO_4(s)) = mass(BaSO_4(s)) ÷ 233.36

From the balanced chemical equation above, we can see that 1 mole of BaSO_4(s) is produced from 1 mole of SO₄^2-_(aq), so,

moles(BaSO_4(s)) = moles(SO₄^2-_(aq))
n(BaSO_4(s)) = n(SO₄^2-_(aq))

Assuming the ONLY source⁴ of sulfate ions was the original lawn fertiliser sample, we now know the moles of SO₄^2-_(aq) in our sample of lawn fertiliser.
So, we can calculate the mass of sulfate ions in grams in the sample of lawn fertiliser:

mass(SO₄^2-_(aq)) = moles(SO₄^2-_(aq)) × molar mass(SO₄^2-_(aq))
m(SO₄^2-_(aq)) = n(SO₄^2-_(aq)) × M(SO₄^2-_(aq))
Use the Periodic Table to find the molar mass, M, of each element:
M(S) = 32.06 g mol^-1
M(O) = 16.00 g mol^-1

Calculate molar mass of SO₄^2-:
M(SO_4(s)^2-) = 32.06 + 4 × 16.00 = 96.06 g mol^-1
m(SO₄^2-_(aq)) = n(SO₄^2-_(aq)) × 96.06

Which then allows us to calculate the percentage by mass of sulfate in the lawn fertiliser:

% by mass sulfate = mass sulfate in grams
mass of fertiliser used × 100

% (SO₄^2-) = m(SO₄^2-_(aq))
m(fertiliser) × 100

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Procedure:

Using a clean mortar and pestle, grind up about 1.5 g of lawn fertiliser up into fine powder.
The smaller the particles, the faster and more completely they will dissolve in a solvent.

Accurately weigh a 100 mL clean, dry, glass beaker.
Record the mass of the beaker:
m(beaker) = ? g

Transfer about 1 g of the powdered lawn fertiliser to the beaker and weigh.
Record the mass of lawn fertiliser and beaker:
m(fertiliser + beaker) = ? g
Calculate the mass in grams of fertiliser in the beaker:
m(fertiliser) in grams = m(fertiliser + beaker) - m(beaker)

Add about 50 mL of deionised water to the beaker and stir to dissolve all the ammonium sulfate in the lawn fertiliser.
When all the fertiliser is dissolved, suspend the stirring rod over the beaker and squirt deionised water on it to remove any particles adhering to its surface and wash them into the beaker.
It is important to ensure that no sulfate is lost at this point, otherwise our final mass of barium sulfate precipitate will be too low.

Filter the solution into a clean 600 mL beaker.
Wash the residue in the filter paper thoroughly with water.
Filtering removes any solid, undissolved impurities from the solution before we precipitate out the sulfate as barium sulfate.
If undissolved solids were not removed at this stage, they would increase the apparent weight of the barium sulfate precipitated for the analysis.
If any sulfate ions are adhering to the residue washing the residue with water should wash these into the beaker.
This is important because we need to prevent any loss of sulfate ions during the procedure. Loss of sulfate ions will result in a lower mass for barium sulfate, and hence, our final calculation of the mass of sulfate in the fertiliser will be too low.

Add 3.00 mL of 2.0 mol L^-1 HCl_(aq) to the filtrate and make the volume of the solution up to about 200 mL with distilled water.
Acidifying the solution slightly prevents the possible precipitation of other barium salts due to chromate, carbonate or phosphate impurities in the lawn fertiliser solution.
Barium chromate, barium carbonate and barium phosphate are all insoluble in neutral solution, but solubility increases with increased acidity.
The choice of acid to use is important. Clearly you could NOT use sulfuric acid to acidify the solution because this would increase the sulfate ion concentration of the solution so the mass of barium sulfate precipitated would be greater.

Adding more water also dilutes the lawn fertiliser solution which will help increase the size of the particles produced during the precipitation of barium sulfate.
(Particle size of a precipitate decreases as reactant concentration increases!)

Heat the solution over a Bunsen burner and bring it to the boil.
The precipitation reaction will be carried out in hot solution in order to maximise the amount of precipitate formed and in order to help increase the size of the barium sulfate particles formed.
The solubility of barium sulfate increases with increasing temperature.
This means that as the hot mixture cools, the solution becomes supersaturated resulting in the barium sulfate crystallising out of solution. This helps to ensure that all the sulfate ions in solution are precipitated as barium sulfate.
Crystal size is dependent on rate of cooling, the slower the rate of cooling the larger the particles will be.

Remove the Bunsen burner.
Slowly add about 15 mL of 0.5 mol L^-1 BaCl_2(aq) to the hot solution while stirring.
Allow the precipitate to settle.
Add another drop or two of BaCl_2(aq) to the solution.
Continue adding drops of BaCl_2(aq) to the solution until no more precipitate forms.
The addition of barium ions slowly is important because this will help increase the size of the barium sulfate particles.
The first BaSO_4(s) particles formed will act as nuclei which grow as further BaSO_4(s) precipitates.

Cover the beaker with a watch glass (clock glass) and allow to stand overnight.
This allows time for complete precipitation to occur, and time for the precipitate to settle to the bottom of the beaker.
It is important to cover the beaker to prevent impurities entering.

The next day, add a couple of drops of BaCl_2(aq) to the clear, supernatant liquid above the settled precipitate in the beaker.
If no precipitate forms, the mixture is ready to be filtered.

Weigh a filter paper and record its mass:
m(filter paper) = ? g
Barium sulfate precipitate is very fine, made up of small particles.
The pore size of filter paper available in the school laboratory may not be small enough to trap all these small particles⁵.
If you have access to a sintered glass crucible, this will have a greater ability to trap the small particles, so use this instead for the filtration.

Filter the mixture.
Place the filter paper in a glass funnel positioned in the neck of a conical flask (erlenmeyer flask).
Decant the clear, supernatant liquid through the filter paper in to a clean conical flask (erlenmeyer flask) without disturbing the settled precipitate.
Test the filtrate with a few drops of BaCl_2(aq).
If a precipitate forms, the precipitation reaction has not gone to completion and it is best to start again.
If no precipitate forms, discard the filtrate. Rinse the conical flask with distilled water, and place it under the funnel with the filter paper once more.
Transfer the precipitate from the beaker to the filter paper with the aid of a jet of hot water from a wash bottle.
Use a rubber-tipped rod (a "policeman") to remove any precipitate adhering to the sides of the beaker and wash that onto the filter paper also.

Wash the precipitate with a small amount of hot water.
Test the filtrate with a drop of 0.1 mol L^-1 AgNO_3(aq), if a white precipitate of AgCl_(s) forms, then discard the filtrate and wash the precipitate again.
Keep washing and testing the filtrate until no precipitate forms when AgNO_3(aq) is added.
Washing the precipitate removes any anions such as Cl^-_(aq), and cations such as NH₄⁺ and Ba²⁺, adhering to the surface of the precipitate⁶.
Ions adhering to the surface of the barium sulfate precipitate will increase its apparent weight so they need to be removed.
Silver chloride, AgCl, is insoluble in aqueous solutions so it can be used to test for the presence of chloride ions, Cl^-, in the water being washed off the precipitate.
Ag⁺_(aq) + Cl^-_(aq) → AgCl_(s)
When no more AgCl_(s) precipitate forms in the filtrate this means that all the chloride ions adhering to the surface of the BaSO_4(s) precipitate on the filter paper have been removed.

Place the filter paper containing the BaSO_4(s) on a watch glass (clock glass) and place this in the oven to dry.
When it is dry, allow it to cool in a dessicator, then weigh the filter paper with the precipitate and record its mass:
m(filter paper + BaSO_4(s)) = ? g
Place the filter paper with the BaSO_4(s) back in the oven after weighing for some time.
Allow it to cool in a dessicator before weighing it again, and record this second mass.
Keep repeating the cycle of heating, cooling, and weighing until a constant mass is achieved.
ONLY use this final, constant mass in your calculations.
The precipitate is dried to remove water.
If the water is not completely removed it will add to the apparent mass of the precipitate so our calculation of the amount of sulfate in the fertiliser will be too high.
Weighing to constant mass ensures that all the water has been removed.

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Sample Results and Calculations

Sample results for the experiment are shown in the table below:

item	mass / g
lawn fertiliser	1.020
filter paper	0.002
(1) filter paper + precipitate	1.973
(2) filter paper + precipitate	1.758
(3) filter paper + precipitate	1.758

Calculations:

Calculate mass of BaSO_4(s) precipitated.
Use ONLY the constant mass, 1.758 g, in your calculations (the first recorded mass must be a "wet" sample, that is, not all the water has been evaporated off during the drying process)
m(BaSO_4(s)) = mass(filter paper + precipitate) - mass(filter paper)
m(BaSO_4(s)) = 1.758 - 0.002 = 1.756 g

Calculate moles of BaSO_4(s) precipitated:
n(BaSO_4(s)) = mass(BaSO_4(s)) ÷ molar mass(BaSO_4(s))
n(BaSO_4(s)) = 1.756 ÷ (137.3 + 32.06 + 4 × 16.00)
n(BaSO_4(s)) = 1.756 ÷ 233.36
n(BaSO_4(s)) = 7.525 × 10^-3 mol
Calculate moles of sulfate ion present in solution and therefore in the lawn fertiliser sample:
n(SO₄^2-) = n(BaSO_4(s)) = 7.525 × 10^-3 mol

Calculate mass of sulfate in lawn fertiliser sample:
mass(SO₄^2-) = moles(SO₄^2-) × molar mass(SO₄^2-)
m(SO₄^2-) = 7.525 × 10^-3 × (32.06 + 4 × 16.00)
m(SO₄^2-) = 7.525 × 10^-3 × 96.06
m(SO₄^2-) = 0.723 g

Calculate percent by mass sulfate in lawn fertiliser:

% by mass sulfate = mass sulfate in grams
mass of fertiliser used × 100

% (SO₄^2-) = m(SO₄^2-_(aq))
m(fertiliser) × 100

% (SO₄^2-) = 0.723
1.020 × 100

% (SO₄^2-) = 70.86 %

Sources of Error:

Incomplete precipitation results in a value for the percentage of sulfate in fertiliser that is too low.

Particles smaller than the pore size of the filtering medium can pass through it resulting in a value for the percentage of sulfate in fertiliser that is too low.
Incomplete drying of the sample results in a value for the percentage of sulfate in fertiliser that is too high.

Other ions in the sample may also be precipitated resulting in a value for the percentage of sulfate in fertiliser that is too high.

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Problem Solving using StoPGoPS method

Question:
In order to determine the amount of sulfate in a brand of lawn fertiliser, a student followed the procedure below:

About 1 g of the fertiliser was dissolved in enough distilled water to make up 500.00 mL of solution.

100.00 mL of this solution was transferred to a clean beaker and acidified with a few millilitres of dilute hydrochloric acid.

The solution was warmed gently.

A solution of barium chloride was added slowly until no more barium sulfate precipitated out.

A piece of filter was weighed and placed in a funnel, and the funnel placed in a conical flask.

The solution containing the precipitate was filtered through the filter paper and washed several times with aliquots of hot distilled water.

The filter paper was placed on a watch glass (clock glass) and placed in an oven to dry the precipitate.

The filter paper with the precipitate was placed in a dessicator to cool before weighing.

The filter paper and precipitate was weighed.

The results of the experiment are shown below:

Substance	mass / g
lawn fertiliser	1.150
barium sulfate	0.436

Calculate the percent by mass of sulfate in this brand of lawn fertiliser.

Worked Solution:

STOP	STOP! State the Question.
	What is the question asking you to do? Calculate the percent by mass of sulfate in this lawn fertiliser. % sulfate = ? %
PAUSE	PAUSE to Prepare a Game Plan
	(1) What information (data) have you been given in the question? mass fertilser used to make original solution = 1.150 g volume of original solution = 500.00 mL volume of solution used in the analysis = 100.00 mL mass barium sulfate precipitated from 100.00 mL solution = 0.436 g (2) What is the relationship between what you know and what you need to find out? Precipitation in 100.00 mL sample: Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s) Assumes there was an excess of Ba²⁺(aq) Assumes there are no competing reactions, that is, all the sulfate in solution is converted to barium sulfate and nothing else. moles(BaSO_4(s)) = mass(BaSO₄) ÷ molar mass(BaSO₄) Assumes the barium sulfate does not contain any impurities. moles(SO₄^2-_(aq)) in 100.00 mL solution = moles(BaSO_4(s)) in precipitate mass(SO₄^2-) in 100.00 mL sample = moles(SO₄^2-) × molar mass(SO₄^2-) mass(SO₄^2-) in original 500.00 mL solution = 500/100 × mass(SO₄^2-) in 100.00 mL sample = 5 × mass(SO₄^2-) in 100.00 mL sample made with 1.150 g of fertiliser % by mass(SO₄^2-) = (mass(SO₄^2-)/mass fertiliser) × 100
GO	GO with the Game Plan
	Precipitation in 100.00 mL sample: Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s) moles BaSO₄(s) = mass BaSO₄ ÷ molar mass BaSO₄ n(BaSO_{4(100 mL)}) = 0.436 ÷ (137.3 + 32.06 + 4 × 16.00) = 0.436 ÷ 233.36 = 1.868 x 10^-3 mol moles SO₄^2-(aq) in 100.00 mL solution = moles BaSO₄(s) in precipitate n(SO₄^2-_{(in 100 mL)}) = 1.868 x 10^-3 mol mass SO₄^2- in 100.00 mL sample = moles SO₄^2- × molar mass SO₄^2- m(SO₄^2-_{(in 100 mL)}) = 1.868 x 10^-3 × (32.06 + 4 × 16.00) = 1.868 x 10^-3 × 96.06 = 0.1794 g mass SO₄^2- in original 500.00 mL solution = 500/100 × mass SO₄^2- in 100.00 mL sample = 5 × mass SO₄^2- in 100.00 mL sample made with 1.150 g of fertiliser m(SO₄^2-_{(in 500 mL)}) = 5 × 0.1794 = 0.897 g % by mass SO₄^2- = (mass SO₄^2-/mass fertiliser) × 100 % by mass SO₄^2- = (0.897/1.150) × 100 = 78.0%
PAUSE	PAUSE to Ponder Plausibility
	Is your answer plausible? The percentage of SO₄^2- in the lawn fertiliser was calculated to be less than 100%, so the answer is reasonable. We could work backwards to check our answer: If 78.0% of the lawn fertiliser is sulfate and 1.150 g of fertiliser was used then the mass of sulfate in this sample is: mass(SO₄^2-_(sample)) = 78.0/100 × 1.150 = 0.897 g Convert this mass to moles of (SO₄^2-_(sample)): moles(SO₄^2-_(sample)) = mass ÷ molar mass = 0.897 ÷ (32.06 + 4 × 16.00) = 9.338 × 10^-3 mol in 500 mL Calculate moles(SO₄^2-) in 100 mL of this solution: moles(SO₄^2-_{(100 mL)}) = 100/500 × 9.338 × 10^-3 = 1.8676 × 10^-3 mol in 100 mL Calculate moles(BaSO₄) that will precipitate: moles(SO₄^2-_{(100 mL)}) = moles(BaSO_4(s)) = 1.8676 × 10^-3 mol Convert moles(BaSO₄) to mass(BaSO₄): mass(BaSO_4(s)) = moles(BaSO_4(s)) × molar mass(BaSO_4(s)) = 1.8676 × 10^-3 × (137.3 + 32.06 + 4 × 16.00) = 0.436 g Compare mass(BaSO₄) calculated to mass(BaSO₄) that precipitated in the experiment: The mass of precipitate given in the question is also 0.436 g so we are confident our answer is correct.
STOP	STOP! State the Solution
	% by mass of sulfate in this lawn fertiliser is 78.0 %

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1. The separation of the element or the compound containing it may done a number of different ways, precipitation is one of the most common. Other methods include volatilisation and electro-analytical methods.

2. Not all lawn fertilisers contain ammonium sulfate, for example, you can buy "sulfate of iron" which is used as a fertiliser and herbicide which contains ferrous sulfate, or iron(II) sulfate, FeSO₄, which has a solubility in water of 1.89 g/100 g. Some formulations labelled as "weed'n'feed" may also contain iron(II) sulfate.

3. The solubility of barium sulfate is 0.001 g/100 g at 25°C. Because so very little of the barium sulfate dissolves in water it is said to be insoluble in water at 25°C.

4. In gravimetric analysis, it is absolutely essential that all equipment is clean to prevent contamination of the sample to be analysed, in this case, we do not want any sulfate ions adhering to the glassware we use in the analysis. Tap water would be unsuitable to use for the analysis because it is likely to contain sulfate ions, deionised water would be a better solvent to use.

5. The particles making up the precipitate must be large enough so that they do not pass through the filtering medium, and, they must remain larger than the pore size of the filter medium during the washing process.
Ordinary quantitative filter paper will retain particles up to a diameter of about 10 μm.

6. Washing removes impurities from the surface of the precipitate particles but will not remove occluded foreign substances.

% by mass sulfate =	mass sulfate in grams mass of fertiliser used	× 100
% (SO₄^2-) =	m(SO₄^2-_(aq)) m(fertiliser)	× 100