Chemistry · Lesson
Precipitation and gravimetric analysis
Connect a characterized precipitate mass to analyte amount and retain finite solution equilibria. Distinguish formula identity, incomplete recovery and other retained matter.
Back to the Chemistry mapBy the end, you can
- Calculate a gravimetric factor for a stipulated pure solid.
- Explain finite dissolved ions at a precipitation equivalence inventory.
- Combine opposing mass effects in a known-error toy.
- Quantify an amount error from a wrongly assigned hydrate formula.
A mass is quantitative only when the weighed entity is known
Gravimetric analysis relates a measured mass to a defined chemical entity, often after a separation or conversion. The amount inference requires the weighed form, stoichiometry and relevant retained material to be characterized. A large, stable mass alone does not prove a pure analyte-bearing phase.
For selected Ba²⁺+SO₄²⁻→BaSO₄(s), one pure BaSO₄ formula unit contains one sulfate unit. Assign rounded molar masses 233.39 and 96.06 g mol⁻¹. The sulfate mass factor is 96.06/233.39≈0.411585758. An invented pure-solid mass 0.3500 g then corresponds to approximately 1.49963580 mmol BaSO₄ and 0.144055015 g sulfate.
The factor is formula bookkeeping, not proof that an unknown precipitate is pure, that every sulfate unit was recovered or that other constituents were excluded. Sample matrix, adsorption, coprecipitation and the characterized final form can affect the result.
Stoichiometric equivalence still leaves finite dissolved species
Consider an original ideal AgCl(s) model with initial chloride volume 40.0 mL at 0.00600 mol L⁻¹, silver titrant 0.0120 mol L⁻¹ and assigned dimensionless K_sp=10⁻¹⁰. Use fixed model temperature, c°=1 mol L⁻¹, unit ionic activity coefficients, pure-solid activity one and additive model volumes. No measured constant or side complex is assigned.
Initial chloride amount is 0.240 mmol, giving stoichiometric equivalence at 20.0 mL silver. At that volume, total liquid volume is 60.0 mL. With AgCl solid present, equal free Ag⁺ and Cl⁻ concentrations are √K_sp c°=10.0 μmol L⁻¹. Each dissolved amount is 0.600 μmol.
Solid amount is 240−0.600=239.4 μmol, so 99.75% of the initial chloride inventory is in solid, rather than exactly 100%. The mass balance and solubility relation coexist with the correct equivalence inventory.
At other selected volumes with remaining solid, the free-ion concentration difference follows the inventory difference while the activity product stays K_sp. Before equivalence, chloride excess dominates; after equivalence, silver excess dominates. The plotted branch is restricted to 12–28 mL, where this model retains positive solid amount. It must not be extended to an absent-solid regime without checking the phase condition.
Ideal-model pCl
Toy silver volume / mL
- Selected saturated-model points
- — Finite-solubility model in the stated interval
Underlying data
| Toy silver volume / mL | Ideal-model pCl |
|---|---|
| 12 | 2.733719368835253 · Selected saturated-model points |
| 19 | 3.6906247054698156 · Selected saturated-model points |
| 20 | 4.99999999999999 · Selected saturated-model points |
| 21 | 6.294969315045097 · Selected saturated-model points |
| 28 | 7.149784108816024 · Selected saturated-model points |
Original saturated ideal model on 12–28 mL added silver, with remaining AgCl solid throughout this plotted interval. Initial chloride is 0.240 mmol; K_sp=10⁻¹⁰, c°=1 mol L⁻¹, unit activity coefficients and no side complexes. Markers show selected pCl=−log₁₀([Cl⁻]/c°); the dashed curve solves inventory difference and the solubility product. At 20.0 mL, pCl=5 and 99.75% of chloride is in solid. No absent-solid extrapolation or measured analytical recovery is supplied.
Loss and retained extraneous matter can partly cancel
In a controlled original mass puzzle, the true pure-solid mass is supplied as 200 mg and the assigned analyte mass factor is 0.400. Suppose 2 mg of that solid is lost, while 5 mg other matter and 3 mg residual liquid are included in the measured material. Observed mass is 200−2+5+3=206 mg.
True analyte mass in the supplied original solid is 80.0 mg. Applying the factor blindly to 206 mg gives 82.4 mg, a +3.00% error relative to that supplied truth. Opposing contributions have partly canceled, but they have not become absent.
In a real sample the true amount is usually not supplied. Appropriate controls and characterization are needed to assess loss and retained material. Reproducible weighing by itself can coexist with a systematic compositional error.
Water in a formula changes the molar-mass conversion
A selected identity puzzle assigns observed mass 0.17217 g and actual form CaSO₄·2H₂O, with rounded molar mass 172.17 g mol⁻¹. It contains 1.000 mmol formula units. If the same mass were incorrectly treated as CaSO₄ with assigned molar mass 136.14 g mol⁻¹, the inferred amount would be approximately 1.26465403 mmol.
The amount error is about +26.4654%. The chosen molar-mass difference is 36.03 g mol⁻¹, associated with two waters per formula unit under these rounded inputs. The net formula ledger CaSO₄·2H₂O→CaSO₄+2H₂O accounts for the atoms.
That ledger does not establish a validated thermal conversion, a drying temperature or the phase identity of an actual weighed specimen. Drying, hydration, decomposition and final-form characterization are method-specific questions rather than interchangeable labels.
Retain the separation and phase boundary in the result
Report the intended analyte, weighed form, conversion factor, recovered amount and relevant uncertainty or unresolved effects. A theoretical solubility product does not alone establish selective analytical recovery, and a nominal phase formula does not prove the sample has that composition.
Classical gravimetry joins formula identity, equilibrium and material accounting. Those same distinctions are useful whenever an analytical result is derived from a separated fraction rather than directly from the entire target.
Worked examples
Convert the stipulated pure form
Use invented 0.3500 g BaSO₄ with the assigned rounded molar masses.
- Factor is 96.06/233.39≈0.411585758.
- Formula amount is approximately 1.49963580 mmol.
- Sulfate mass is approximately 0.144055015 g.
A pure-form inventory, without proof of actual selectivity or complete recovery.
Keep dissolved ions at the equivalence inventory
Use the selected AgCl model at 20.0 mL silver.
- Free Ag⁺ and Cl⁻ concentrations are each 10.0 μM.
- At 60.0 mL, dissolved chloride amount is 0.600 μmol.
- Solid AgCl amount is 239.4 μmol.
- Solid chloride fraction is 239.4/240=0.9975.
Finite solubility leaves 0.25% of the chosen chloride inventory dissolved.
Combine loss and extraneous retained mass
Use supplied true mass 200 mg, loss 2 mg, other matter 5 mg and residual liquid 3 mg.
- Observed mass is 206 mg.
- True analyte mass at factor 0.400 is 80.0 mg.
- Blindly inferred mass is 82.4 mg.
- Relative error is 3.00%.
Partly opposing effects do not disappear merely because their net is smaller.
Use the actual stipulated formula
Use observed 0.17217 g of CaSO₄·2H₂O.
- Correct amount is 1.000 mmol.
- Using the assigned anhydrous molar mass instead gives 1.26465403 mmol.
- Relative amount error is approximately +26.4654%.
Formula identity controls the mass-to-amount conversion.
Try it yourself
Under the same stipulated pure BaSO₄ model, double its mass to 0.7000 g. Find sulfate mass.
Show the answer and reasoning
Approximately 0.288110030 g.
The assigned gravimetric factor is unchanged, so the inferred sulfate mass doubles.
In a new controlled toy, true solid mass is 100 mg, loss 1 mg and other retained matter 4 mg. Use factor 0.400. Find inferred analyte mass and relative error.
Show the answer and reasoning
41.2 mg; +3.00%.
Measured mass is 103 mg; compare 41.2 with the supplied true analyte mass 40.0 mg.
Are both free ions exactly absent at the chosen AgCl equivalence inventory?
Show the answer and reasoning
No.
The saturated model gives 10.0 μM of each, while most inventory remains in solid.
Does the hydrate decomposition ledger establish a valid drying procedure for an actual specimen?
Show the answer and reasoning
No.
A formula balance does not specify thermal conditions, transformation completeness or final phase identity.
What to carry forward
- Identify the actual weighed form.
- A gravimetric factor is formula bookkeeping with assumptions.
- Finite solubility can coexist with equivalence.
- Opposing mass effects need characterization.
- A hydrate formula and a thermal procedure are different kinds of evidence.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.
- Glossary of terms related to solubility (IUPAC Recommendations 2008)Heinz Gamsjäger; John W.Lorimer; Pirketta Scharlin; David G.ShawSource · PDF
- IUPAC Gold Book: equivalence-pointIUPAC Gold Book Joint Subcommittee; original source identified in entrySource
- OpenStax Chemistry 2e: 4.5 Quantitative Chemical AnalysisPaul Flowers; Klaus Theopold; Richard Langley; William R.RobinsonSource
- Fitness for Purpose of Analytical Methods, third editionHelen Cantwell (editor); Eurachem Method Validation Working GroupSource · PDF
- EURONORM CRM 880-1 certificate distributed by BAMEuropean Coal and Steel Community/ECISS participating institutions; IRSID France; German Iron and Steel CRM Working Group; Bureau of Analysed Samples,UK; BAM distributorSource · PDF
- JRC: How to read our certificatesEuropean Commission Joint Research CentreSource