go to the AUS-e-TUTE homepage
home test Join AUS-e-TUTE game contact
 

 

Gravimetric Analysis

Key Concepts

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

  • An analyte is the substance to be analysed.

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

  • 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 solution

  5. Wash the precipitate to remove any impurities

  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

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

  1. Wite 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

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

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

  • Sources of Error:

  1. Incomplete precipitation results in a value for the percentage of analyte in the sample that is too low.
  2. Incomplete drying of the sample results in a value for the percentage of analyte in the sample that is too high
  3. 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

Example

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.

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

    Ca2+(aq) + C2O42-(aq) → CaC2O4(s)

  2. Calculate the moles (n) of calcium oxalate precipitated.

    n(CaC2O4(s)) = mass ÷ molar mass

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

    n(CaC2O4(s)) = 2.43 ÷ 128.10

    n(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, n(Ca2+(aq)) = n(CaC2O4(s)) = 0.019mol

  4. Calculate the mass of calcium in grams

    mass (Ca) = moles x molar mass

    mass (Ca) = 0.019 x 40.08 = 0.76 g

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

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

    %Ca = (0.76 ÷ 2.00) x 100 = 38%


What would you like to do now?
advertise on the AUS-e-TUTE website and newsletters
 
 

Search this Site

You can search this site using a key term or a concept to find tutorials, tests, exams and learning activities (games).
 

Become an AUS-e-TUTE Member

 

AUS-e-TUTE's Blog

 

Subscribe to our Free Newsletter

Email email us to
subscribe to AUS-e-TUTE's free quarterly newsletter, AUS-e-NEWS.

AUS-e-NEWS quarterly newsletter

AUS-e-NEWS is emailed out in
December, March, June, and September.

 

Ask Chris, the Chemist, a Question

The quickest way to find the definition of a term is to ask Chris, the AUS-e-TUTE Chemist.

Chris can also send you to the relevant
AUS-e-TUTE tutorial topic page.

 
 
 

Share this Page

Bookmark and Share
 
 

© AUS-e-TUTE