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Gravimetric Analysis

Key Concepts

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

  • An analyte is the substance to be analysed.

  • A precipitating reagent is the reactant used to precipitate the analyte.2

  • The precipitate must be a pure substance of definite chemical composition.

  • One advantage of gravimetric analysis compared to volumetric analysis (titrimetric analysis) is that there is greater likelihood of any impurities being seen, and therefore a correction can be applied.

  • One disadvantage of gravimetric analysis is that it is generally more time-consuming.

Experimental Procedure:

  1. Weigh the sample to be analysed

  2. Dissolve the sample in a suitable solvent, eg, water

  3. Add an excess of the precipitating reagent to precipitate the analyte

  4. Filter the mixture to separate the precipitate from the solution3

  5. Wash the precipitate to remove any impurities4

  6. Dry the precipitate by heating to remove water

  7. Cool the precipitate in a dessicator to prevent the precipitate absorbing moisture from the air

  8. Weigh the cooled precipitate

  9. Repeat the drying and weighing process until a constant mass for the precipitate is achieved

  10. Calculate the percent by mass of analyte in the sample

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General calculation of the percent by mass of analyte in a sample:

  1. Write the balanced chemical equation for the precipitation reaction

  2. Calculate the moles of precipitate: moles = mass ÷ molar mass

  3. Calculate moles of analyte from the balanced chemical equation using the mole ratio of analyte : precipitate
    (also known as the stoichiometric ratio of analyte to precipitate)

  4. Calculate mass of analyte: mass = moles × molar mass

  5. Calculate percent by mass of analyte in sample: (mass analyte ÷ mass sample) × 100

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Sources of Error:

  • Incomplete precipitation results in a value for the percentage of analyte in the sample that is too low.
  • Incomplete drying of the sample results in a value for the percentage of analyte in the sample that is too high
  • Other ions in the sample may also be precipitated resulting in a value for the percentage of analyte in the sample that is too high

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Worked Example

Question:

A 2.00 g sample of limestone was dissolved in hydrochloric acid and all the calcium present in the sample was converted to Ca2+(aq).

Excess ammonium oxalate solution, (NH4)2C2O4(aq), was added to the solution to precipitate the calcium ions as calcium oxalate, CaC2O4(s).

The precipitate was filtered, dried and weighed to a constant mass of 2.43 g.

Determine the percentage by mass of calcium in the limestone sample.

Answer:

  1. Wite the balanced chemical equation for the precipitation reaction:

    Ca2+(aq) + C2O42-(aq) → CaC2O4(s)

  2. Calculate the moles of calcium oxalate precipitated.

    moles(CaC2O4(s)) = mass ÷ molar mass

    moles(CaC2O4(s)) = 2.43 ÷ (40.08 + 2 x 12.01 + 4 x 16.00)

    moles(CaC2O4(s)) = 2.43 ÷ 128.10

    moles(CaC2O4(s)) = 0.019 mol

  3. Find the moles of Ca2+(aq).

    From the balanced chemical equation, the mole ratio of Ca2+ : CaC2O4(s) is 1 : 1

    So, moles(Ca2+(aq)) = moles(CaC2O4(s)) = 0.019 mol

  4. Calculate the mass of calcium in grams

    mass (Ca) = moles × molar mass

    mass (Ca) = 0.019 × 40.08 = 0.76 g

  5. Calculate the percentage by mass of calcium in the original sample:

    %Ca = (mass Ca ÷ mass sample) × 100

    %Ca = (0.76 ÷ 2.00) x 100 = 38%

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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:

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

  2. 100.00 mL of this solution was transferred to clean beaker and acidified with a few milliltres of dilute hydrochloric acid.

  3. The solution was warmed gently.

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

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

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

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

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

  9. 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.

STOP STOP! State the Question.
  What is the question asking you to do?

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

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?

  1. Precipitation in 100.00 mL sample: Ba2+(aq) + SO42-(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.

  2. moles BaSO4(s) = mass BaSO4 ÷ molar mass BaSO4
    Assumes the barium sulfate does not contain any impurities.

  3. moles SO42-(aq) in 100.00 mL solution = moles BaSO4(s) in precipitate

  4. mass SO42- in 100.00 mL sample = moles SO42- × molar mass SO42-

  5. mass SO42- in original 500.00 mL solution = 500/100 × mass SO42- in 100.00 mL sample
    = 5 × mass SO42- in 100.00 mL sample made with 1.150 g of fertiliser

  6. % by mass SO42- = (mass SO42-/mass fertiliser) × 100
GO GO with the Game Plan
 

  1. Precipitation in 100.00 mL sample: Ba2+(aq) + SO42-(aq) → BaSO4(s)

  2. moles BaSO4(s) = mass BaSO4 ÷ molar mass BaSO4

    n(BaSO4(100 mL)) = 0.436 ÷ (137.3 + 32.06 + 4 × 16.00) = 0.436 ÷ 233.36 = 1.868 x 10-3 mol

  3. moles SO42-(aq) in 100.00 mL solution = moles BaSO4(s) in precipitate

    n(SO42-(in 100 mL)) = 1.868 x 10-3 mol

  4. mass SO42- in 100.00 mL sample = moles SO42- × molar mass SO42-

    m(SO42-(in 100 mL)) = 1.868 x 10-3 × (32.06 + 4 × 16.00) = 1.868 x 10-3 × 96.06 = 0.1794 g

  5. mass SO42- in original 500.00 mL solution = 500/100 × mass SO42- in 100.00 mL sample
    = 5 × mass SO42- in 100.00 mL sample made with 1.150 g of fertiliser

    m(SO42-(in 500 mL)) = 5 × 0.1794 = 0.897 g

  6. % by mass SO42- = (mass SO42-/mass fertiliser) × 100

    % by mass SO42- = (0.897/1.150) × 100 = 78.0%

PAUSE PAUSE to Ponder Plausibility
  Is your answer plausible?

The percentage of SO42- 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(SO42-(sample) = 78.0/100 × 1.150 = 0.897 g
Convert this mass to moles of (SO42-(sample):
moles(SO42-(sample) = mass ÷ molar mass = 0.897 ÷ (32.06 + 4 × 16.00) = 9.338 × 10-3 mol in 500 mL
Calculate moles(SO42-) in 100 mL of this solution:
moles(SO42-(100 mL) = 100/500 × 9.338 × 10-3 = 1.8676 × 10-3 mol in 100 mL
Calculate moles(BaSO4) that will precipitate:
moles(SO42-(100 mL) = moles(BaSO4(s)) = 1.8676 × 10-3 mol
Convert moles(BaSO4) to mass(BaSO4):
mass(BaSO4(s)) = moles(BaSO4(s)) × molar mass(BaSO4(s)) = 1.8676 × 10-3 × (137.3 + 32.06 + 4 × 16.00) = 0.436 g
Compare mass(BaSO4) calculate 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 %


What would you like to do now?

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. The precipitate formed must be stable and must be so slightly soluble that no appreciable loss occurs when the precipitate is separated by filtration and weighed.

3. The particles making up the precipitate must be large enough so that they do not pass through the filtering medium, and, the 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.

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

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