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Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Define the reservoirs
Chapters 1–2. Separate melt mass, concentration and crystal removal.
Follow the changing melt
Chapters 3–5. Calculate fractionation, recharge and the next stage.
Test the explanation
Chapters 6–8. Close inventories and challenge a single-stage fit.
Check the geological claim
Chapters 9–10. Compare evidence, assumptions and common questions.
Practise and continue
Chapters 11–12. Change the recharge and follow connected guides.
Full chapter index · Practice and worked checks · Petrology subject root
A melt becomes richer in an element. Was that caused by removing crystals, adding new magma, or both? A concentration alone cannot settle the history, because it is a ratio whose numerator and denominator can change separately.
This guide follows an original two-element example through fractional crystallisation, a recharge pulse and renewed crystallisation. You will keep a complete mass ledger, reset the reference reservoir at the right moment and test whether a simpler explanation survives. The quantities are fictional; the calculation is a deliberately restricted chemical model, not a reconstruction of a particular volcano.
Chapter index
Define the reservoirs · Chapters 1–2
1. Start with the reservoir you are describing
2. Give crystal removal a precise rule
Follow the changing melt · Chapters 3–5
3. Derive the liquid rule before using it
4. Work the first crystallisation stage
5. Add recharge, then begin a new stage
Test the explanation · Chapters 6–8
6. Audit both elements independently
7. Challenge a single closed-system explanation
8. Keep a forward calculation separate from an inferred history
Check the geological claim · Chapters 9–10
9. Check whether the chemical model is physically credible
10. Questions readers ask
Practise and continue · Chapters 11–12
11. Transfer checks with worked answers
12. Continue by the missing part of your explanation
CHAPTER 1 OF 12 · Define the reservoirs
1. Start with the reservoir you are describing
Magma can contain liquid, crystals and gas. Here we model only a homogeneous silicate liquid and the crystals removed from chemical exchange with it. The reported final composition belongs to that liquid. An analysed crystal-bearing rock would require a further account of its component proportions.
Write element inventory as X = MC, with melt mass M and concentration C in compatible units. A 1,000 kg liquid containing 10 mg/kg of an element holds 10,000 mg, or 10 g. Increasing concentration does not necessarily mean that more of the element remains in the reservoir.
Recharge means incoming magma adds material to a resident system. In this model it arrives as crystal-free liquid and mixes completely. Assimilation of surrounding rock, crystal entrainment, volatile loss and incomplete mixing are separate processes that we have not included.
Let D be the concentration in an infinitesimal newly separated crystal assemblage divided by the concentration in its coexisting liquid. We use a bulk, mass-based partition coefficient for that assemblage. A mineral-specific coefficient and the bulk coefficient need not be equal.
For D below one, the assemblage takes up relatively little of the element compared with the liquid. For D above one, it preferentially removes the element. These descriptions apply to the specified assemblage and conditions; they are not permanent labels independent of mineralogy.
Assume fixed D, a well-mixed liquid, equilibrium at the crystallising interface and no subsequent exchange with separated crystals. The MIT fractional-crystallisation notes explain the corresponding Rayleigh model. Our recharge sequence below is an original application of that rule and conservation of mass.
CHAPTER 3 OF 12 · Follow the changing melt
3. Derive the liquid rule before using it
Let dM be a negative change in liquid mass. The accompanying change in its element inventory is d(MC) = DC dM: the removed mass has the instantaneous crystal composition. Expanding the left side gives M dC + C dM = DC dM.
Thus dC/C = (D − 1)dM/M. Integrating with constant D gives C/C₀ = F^(D − 1), where F = M/M₀ is the fraction of the stage’s starting liquid mass still present. The remaining inventory fraction is F^D.
Those are different quantities. With D = 0.50 and F = 0.25, concentration doubles while inventory halves. Three quarters of the liquid mass has crystallised out, carrying half the element away. This direction check protects the calculation from confusing enrichment with creation of material.
Begin with 1,000 kg of liquid. Anonymous element X has concentration 10 mg/kg and Dₓ = 0.50. Element Y has concentration 50 mg/kg and Dᵧ = 2.0. Treat those coefficients as exact model inputs throughout both stages.
Remove crystals until 250 kg of liquid remains, so F₁ = 0.25. X reaches 10 × 0.25^(−0.50) = 20 mg/kg. Y reaches 50 × 0.25¹ = 12.5 mg/kg. The remaining liquid holds 5,000 mg X and 3,125 mg Y.
The initial inventories were 10,000 mg X and 50,000 mg Y. The first 750 kg of removed crystals therefore contains 5,000 mg X and 46,875 mg Y. These are integrated inventories, not the composition of the last crystals to form.
Add 250 kg of liquid containing 2 mg/kg X and 80 mg/kg Y. Recharge brings 500 mg X and 20,000 mg Y. After complete mixing, the liquid mass is 500 kg, holding 5,500 mg X and 23,125 mg Y.
Divide inventories by the new mass: Cₓ = 11 mg/kg and Cᵧ = 46.25 mg/kg. Averaging the two concentrations works here only because the resident and incoming liquid masses happen to be equal. Unequal masses require weighted inventories.
Now crystallise this mixed liquid down to 250 kg. The new fraction is F₂ = 250/500 = 0.50. X finishes at 11 × √2 ≈ 15.556 mg/kg, while Y finishes at 46.25 × 0.50 = 23.125 mg/kg.
| Liquid state | Mass, kg | X, mg/kg | Y, mg/kg |
|---|---|---|---|
| Initial liquid | 1,000 | 10.000 | 50.000 |
| After first crystal removal | 250 | 20.000 | 12.500 |
| After recharge and mixing | 500 | 11.000 | 46.250 |
| After second crystal removal | 250 | 15.556 | 23.125 |
F₂ belongs to the mixed reservoir. Multiplying F₁F₂ gives 0.125, but the final mass is 0.25 of the original 1,000 kg because recharge added material between stages. There is no single unchanged starting reservoir across the whole sequence.
The second stage leaves approximately 3,889.09 mg X and 5,781.25 mg Y in the final liquid. Subtract those from the mixed inventories to obtain the second removed-crystal inventories.
| Element | Initial plus recharge | First crystals | Second crystals | Final liquid |
|---|---|---|---|---|
| X, mg | 10,500.00 | 5,000.00 | 1,610.91 | 3,889.09 |
| Y, mg | 70,000.00 | 46,875.00 | 17,343.75 | 5,781.25 |
Each output row sums to its input. Total material also closes: 1,000 + 250 = 750 + 250 + 250 kg. Passing one check would not excuse failing the other. Concentrations can look plausible while an incorrect denominator breaks conservation.
The mean X concentration in the second removed crystals is about 6.444 mg/kg. The instantaneous concentration of the last crystals is DₓCₓ ≈ 7.778 mg/kg. Their difference is expected: the accumulated crystals formed from liquid whose composition changed continuously.
CHAPTER 7 OF 12 · Test the explanation
7. Challenge a single closed-system explanation
Suppose someone ignores recharge and treats the final 250 kg as the residue of one 1,000 kg crystallisation episode. With F = 0.25, the closed model predicts X = 20 mg/kg and Y = 12.5 mg/kg. Neither equals the final pair in our ledger.
Allowing F to float does not solve both constraints. X alone implies F = (15.556/10)^(−2) ≈ 0.4132. Y alone implies F = 23.125/50 = 0.4625. Under the stipulated exact coefficients and inputs, one common fraction cannot reproduce both values.
This rejects that particular closed model, not every possible history without recharge. Different starting compositions, variable partitioning, alteration or analytical bias could change the comparison. With real uncertainties, test whether the discrepancy is meaningfully resolved before declaring the model inadequate.
CHAPTER 8 OF 12 · Test the explanation
8. Keep a forward calculation separate from an inferred history
Our calculation predicts an outcome from a specified sequence. Reversing the problem is harder. Two final concentrations do not uniquely determine recharge mass, recharge composition, both crystallised fractions and changing mineralogy.
Even a perfect two-element match can leave many candidate histories. Add constraints that ask different questions: mineral compositions, zoning, textures, field relationships and suitable isotopic or chronological evidence. Check whether they genuinely distinguish the alternatives rather than share the same assumption.
Order matters too. Incoming liquid mixed before crystal removal experiences that removal; liquid added afterwards does not. A model must state the sequence and track every reservoir. A final blend cannot generally be substituted for a complete sequence of fractionation and mixing.
CHAPTER 9 OF 12 · Check the geological claim
9. Check whether the chemical model is physically credible
Fixed D is a useful teaching boundary. In a real system, phase proportions and partition coefficients can change with composition, temperature and pressure. A large crystallisation interval particularly needs scrutiny before one constant bulk coefficient is carried through it.
The recharge calculation also omits an energy balance. Hot incoming liquid can heat, dissolve or crystallise material; complete mixing may take time. Chemical conservation alone does not establish that the assumed phase assemblage or sequence is thermally attainable.
Before fitting data, identify the sampled material, evaluate alteration, state endmembers and uncertainties, justify the phase model and compare independent observations. The volcanology guide connects these chemical questions with the wider magma system. A trace-element ledger by itself is not an eruption forecast.
Does recharge always dilute an incompatible element?
No. The mixed concentration moves toward the incoming concentration. Recharge richer than the resident liquid raises it; recharge of equal concentration leaves it unchanged while increasing mass and inventory. The partition coefficient does not determine the incoming composition.
Is fractional crystallisation just partial melting backwards?
They share partitioning and conservation principles, but their reservoir histories and exchange assumptions differ. Use the equation belonging to the stated process. Do not reverse the name and retain a formula without rechecking its derivation.
Can a rock analysis be inserted as the final liquid concentration?
Only with evidence that it represents the liquid relevant to the model. Accumulated crystals or trapped interstitial liquid change a whole-rock analysis. Identify the material before choosing its mathematical role.
First change only the recharge X concentration from 2 to 20 mg/kg. Predict the direction before calculating. The resident 250 kg already contains 20 mg/kg, so adding an equal mass at the same concentration leaves the mixed concentration at 20 mg/kg. Renewed crystallisation gives 20√2 ≈ 28.284 mg/kg.
Next add only 100 kg of the original 2 mg/kg recharge to the first-stage residue. The mixed X inventory is 5,000 + 200 = 5,200 mg in 350 kg, giving 14.857 mg/kg. Crystallising to 250 kg now requires F₂ = 250/350, giving approximately 17.579 mg/kg. Keeping F₂ at 0.50 would describe a different final mass.
Finally ask why a final X value alone cannot prove recharge. An adjusted initial concentration or a different closed fraction can reproduce a single scalar value. A convincing explanation needs several compatible constraints and an explicit comparison of alternatives.
CHAPTER 12 OF 12 · Practise and continue
12. Continue by the missing part of your explanation
Reservoirs and geological context
Return to the Petrology root for the wider rock-evidence sequence. Use Petrology: magma, sediment and metamorphism for formation processes and Geochemistry: reservoirs and reactions for inventories and interpretation.
Compare neighbouring quantitative models
Read batch and fractional melting for source-to-melt transfer, non-modal melting and mineral exhaustion for changing assemblages, and trapped melt and whole-rock composition for the material actually analysed.
Sources and learning transfer
The MIT lecture on fractional crystallisation provides the partitioning framework; its course lecture directory separates crystallisation from other melting models. The recharge ledger, invented elements and exercises here are our own constructed cases.
Continue through How X Works and the Gold Standard of Education. The transferable skill is to keep quantities, reservoirs and assumptions visible when a system changes between steps.
Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.