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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 sampled material
Chapters 1–2. Separate crystal framework, trapped melt and whole rock.
Build the mass balance
Chapters 3–5. Calculate inventories, ratios and freezing effects.
Test the inverse claim
Chapters 6–8. Estimate a fraction and challenge its uniqueness.
Read the evidence
Chapters 9–10. Check phase amounts, assumptions and misconceptions.
Practise and continue
Chapters 11–12. Work changed mixtures and choose connected reading.
Full chapter index · Practice and worked checks · Petrology subject root
A whole-rock analysis measures the material placed in the sample, not automatically the liquid from which its crystals grew. A crystal-rich rock can combine an accumulated crystal framework with liquid retained between the crystals. Treating that mixture as a frozen sample of the parent melt can therefore change the geological story before any sophisticated calculation begins.
This guide uses fictional concentrations and anonymous elements X and Y to make the distinction inspectable. Start with the Petrology root for the broader evidence framework. No composition here is a reported natural specimen, and the model does not identify a unique magma source.
Chapter index
Define the sampled material · Chapters 1–2
1. Ask what the analysed rock contains
2. Define a closed two-reservoir model
Build the mass balance · Chapters 3–5
3. Work two elements through one mixture
4. Inspect the ratio before interpreting its cause
5. Follow what freezing does and does not preserve
Test the inverse claim · Chapters 6–8
6. Invert the calculation only with specified endmembers
7. Test whether the inverse problem has one answer
8. Separate mass fraction from volume fraction
Read the evidence · Chapters 9–10
9. Build an evidence check for a real sample
10. Repair common misconceptions
Practise and continue · Chapters 11–12
11. Practise with changed compositions
12. Choose a connected reading route
CHAPTER 1 OF 12 · Define the sampled material
1. Ask what the analysed rock contains
A cumulate is associated with the concentration of crystals relative to liquid during an igneous history. Its present whole-rock composition can include both the crystal framework and material crystallised from interstitial liquid. The resulting rock need not have the composition of that liquid alone.
Do not reduce this to a universal picture of crystals simply falling onto a floor. Crystal growth, segregation, compaction and liquid movement can contribute in different settings. The question here is narrower: what does a specified combination of framework and retained liquid do to the bulk chemical measurement?
Separate three quantities in the notebook: framework composition, trapped-liquid composition at the chosen stage, and final whole-rock composition. A real investigation must also establish how later exchange or alteration affected them. A familiar rock name does not settle those relationships.
CHAPTER 2 OF 12 · Define the sampled material
2. Define a closed two-reservoir model
Let f be the mass fraction of trapped liquid in the crystal-rich mixture at a stated stage. Let C_l be an element’s concentration in that liquid and C_c its concentration in the entire crystal framework. C_c is already a framework average, not the concentration in whichever individual mineral is most conspicuous.
The framework mass fraction is 1 − f. For a closed mixture on compatible mass-concentration bases, whole-rock concentration is C_w = (1 − f)C_c + fC_l. This is conservation of the element, not a rule selecting how much liquid the rock actually retained.
We assume representative sampling, known endmembers and no net gain or loss of the model element. Later crystallisation can redistribute the element among minerals without changing its whole-rock total if that boundary remains closed. Later transport across the boundary requires a different ledger.
Take a fictional 1,000 kg mixture containing 800 kg crystal framework and 200 kg trapped liquid, so f = 0.20. Framework concentrations are 2 mg/kg for X and 80 mg/kg for Y. Liquid concentrations are 20 mg/kg for X and 40 mg/kg for Y.
| Reservoir | Mass | X concentration | Y concentration |
|---|---|---|---|
| Crystal framework | 800 kg | 2 mg/kg | 80 mg/kg |
| Trapped liquid | 200 kg | 20 mg/kg | 40 mg/kg |
| Whole mixture | 1,000 kg | 5.6 mg/kg | 72 mg/kg |
For X, the framework holds 1,600 mg and liquid holds 4,000 mg. Their 5,600 mg total divided by 1,000 kg gives 5.6 mg/kg. For Y, the corresponding inventories are 64,000 and 8,000 mg, giving 72 mg/kg overall.
The whole rock is depleted in X relative to the liquid but enriched in Y. Neither observation alone proves that the liquid had those whole-rock concentrations. The reservoir weights explain the differences directly under this constructed model.
CHAPTER 4 OF 12 · Build the mass balance
4. Inspect the ratio before interpreting its cause
The liquid has Y/X = 40/20 = 2 on these compatible concentration units. The whole mixture has Y/X = 72/5.6 ≈ 12.86. Crystal accumulation and retained-liquid proportion can therefore change a chemical ratio without requiring the liquid endmember itself to change.
Ratios of mixtures generally are not simple weighted averages of endmember ratios. Here the framework ratio is 80/2 = 40. Weighting the two ratios by 0.80 and 0.20 would give 32.4, which is not 12.86. Mix the element amounts or concentrations first, then divide.
The mistaken move is treating a ratio like an independently conserved substance. X and Y inventories are conserved in this model; their ratio is calculated from those inventories. A ratio plot can be useful, but its interpretation needs the underlying mixing relationships attached.
CHAPTER 5 OF 12 · Build the mass balance
5. Follow what freezing does and does not preserve
Suppose the trapped liquid crystallises completely within the same closed 1,000 kg sample. X may move into newly formed phases or crystal rims. Its original liquid concentration is no longer directly measurable as a present liquid, but the whole-rock X inventory remains 5,600 mg under our closure assumption.
The fraction of late-grown minerals in the final rock need not equal the former trapped-liquid mass fraction. Existing crystals can grow from the liquid, reactions can modify phases, and densities can differ. A present textural area fraction cannot simply be relabelled an earlier liquid mass fraction.
Read texture and chemistry together. Interstitial relationships, zoning and phase compositions can help test the history. A closed whole-rock chemical total can survive internal redistribution while particular mineral compositions change substantially. These are different scales of conservation.
CHAPTER 6 OF 12 · Test the inverse claim
6. Invert the calculation only with specified endmembers
If C_c and C_l are known and different, rearrange the mixing relation: f = (C_w − C_c)/(C_l − C_c). With X, f = (5.6 − 2)/(20 − 2) = 0.20. With Y, f = (72 − 80)/(40 − 80) = 0.20.
| Element and endmembers | Whole-rock value | Inferred trapped-liquid mass fraction |
|---|---|---|
| X: framework 2, liquid 20 mg/kg | 5.6 mg/kg | 0.20 |
| Y: framework 80, liquid 40 mg/kg | 72 mg/kg | 0.20 |
| Y with the same assumed endmembers | 64 mg/kg | 0.40 |
The third row is a changed observation. If it belonged to the same specimen as the first row, the two estimates would disagree under these endmembers. Do not quietly average 0.20 and 0.40 and call the model confirmed. Investigate measurement uncertainty, sampling, endmember assumptions or processes outside the closed two-reservoir description.
CHAPTER 7 OF 12 · Test the inverse claim
7. Test whether the inverse problem has one answer
Knowing C_w alone does not determine f when endmember compositions are unknown. For X = 5.6 mg/kg and framework X = 2 mg/kg, f = 0.20 implies liquid X = 20 mg/kg. But f = 0.40 implies liquid X = 11 mg/kg. Both combinations reproduce the same whole-rock X.
Additional elements can constrain the account if their endmembers and relationships are independently supported. Adding measurements while allowing every endmember to vary freely may leave the problem underdetermined. More decimal places do not create the missing constraints.
If C_l and C_c are nearly equal, the denominator in the inversion is small. That element then provides little leverage on f, and small compositional errors can change the estimate greatly. If they are identical, that element cannot determine f through this equation at all.
CHAPTER 8 OF 12 · Test the inverse claim
8. Separate mass fraction from volume fraction
Our f is a mass fraction because concentrations are in mg/kg. A petrographic description may instead estimate volume fractions. Converting needs appropriate densities: liquid mass fraction is ρ_lV_l/[ρ_lV_l + ρ_cV_c] in a representative two-component description.
For an invented 20% liquid volume fraction, ρ_l = 2.5 and ρ_c = 3.0 in common density units, the liquid mass fraction is 0.5/(0.5 + 2.4) ≈ 0.1724. Substituting 0.20 into the mass-based chemical equation would answer a different model.
Real solidification also changes volume and phase assemblage. A modern measured pore fraction is not automatically the original melt fraction. Preserve the stage, basis and density assumptions of every fraction rather than using the word proportion as if it erased those distinctions.
Begin with provenance, field relationships, textures and a representative sample. Record whether the analysis concerns whole rock, selected mineral separates, mineral cores or rims. Examine whether alteration, veins, weathering or sample preparation could have changed the measured population.
Then state how endmembers were estimated and which observations are genuinely independent tests. A liquid composition calculated from the same whole-rock value cannot be offered as independent confirmation of the assumed trapped-liquid fraction. Preserve the assumptions connecting mineral measurements with any inferred liquid.
The Geochemistry guide develops chemical reservoirs and interpretation. The Mineralogy guide helps connect phase identity with structure and evidence. Neither a chemical fit nor a photograph alone supplies the complete history.
Does an enriched whole-rock element prove an enriched parent liquid?
No. A framework rich in that element can raise the bulk concentration. Test the phase proportions and compositions before attributing the whole-rock value to liquid evolution.
Does a low incompatible-element concentration prove greater partial melting?
It can have several explanations, including crystal accumulation, differing sources and later processes. The batch and fractional melting guide describes a different reservoir history. Choose the physical model before reading its concentration trend.
Is every estimated fraction between zero and one acceptable?
That is a necessary range check, not validation. Several elements, texture, sampling and endmember evidence still need to agree adequately.
What does an estimate outside zero to one mean?
The stated endmembers and closed mixing model cannot reproduce that central value as a physical fraction. Investigate uncertainty and assumptions; do not clip the result silently.
Use a new fictional framework containing X = 4 and Y = 60 mg/kg, and liquid containing X = 24 and Y = 20 mg/kg. Let trapped-liquid mass fraction be 0.30. Predict whether each whole-rock concentration rises or falls relative to the framework.
X becomes 0.70 × 4 + 0.30 × 24 = 10 mg/kg. Y becomes 0.70 × 60 + 0.30 × 20 = 48 mg/kg. The inversions give (10 − 4)/(24 − 4) = 0.30 and (48 − 60)/(20 − 60) = 0.30. Whole-rock Y/X is 4.8; liquid Y/X is about 0.833.
Now change the observed whole-rock X to 30 mg/kg while retaining the same endmembers. The inferred fraction is 1.30, outside the physical interval. Explain why this challenges the model before proposing the next measurement. Repeating the calculation with a capped fraction would conceal the disagreement.
Rock formation and reservoir history
Return to Petrology and the longer rock-formation guide. Compare batch and fractional melting with non-modal melting and mineral exhaustion. Those models explain different histories from the crystal-rich mixture examined here.
Primary research and chemical evidence
Namur, Humphreys and Holness’s Skaergaard study, published in 2014, examines interstitial-liquid crystallisation and mineral zoning in a real intrusion. It supplies a specialist research context; our anonymous-element ledgers are original teaching examples, not Skaergaard observations. Continue through Geochemistry and Mineralogy for complementary evidence.
Learning transfer
Connect to The Gold Standard of Education and How X Works. The transferable habit is identifying which material a measurement represents before using it as evidence about an earlier reservoir.
Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.