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礦物的化學成分 Mineral Chemistry

礦物的化學成分 Mineral Chemistry. 礦物化學成分的研究意義. 礦物 的 化學成分 是 區別不同礦物 的 重要依據 ; 礦物化學成分 的 變化特點 常作為 反映礦物形成條件 的 標誌 ; 礦物化學成分 是人類 利用礦物資源 的一個重要方面。. Chemical Analysis. Stability Field (T, P, fO 2 , pH 2 O, pCO 2 , etc.). Microstructure

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礦物的化學成分 Mineral Chemistry

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  1. 礦物的化學成分Mineral Chemistry

  2. 礦物化學成分的研究意義 • 礦物的化學成分是區別不同礦物的重要依據; • 礦物化學成分的變化特點常作為反映礦物形成條件的標誌; • 礦物化學成分是人類利用礦物資源的一個重要方面。

  3. Chemical Analysis

  4. Stability Field (T, P, fO2, pH2O, pCO2, etc.) Microstructure (records history of phase transformations - use to work out what conditions rock has experienced) Change in physical properties (e.g. change in elastic properties changes speed of seismic waves) Principles of mineral behaviour A Happy Stable Mineral! Chemical composition (what elements are present) Crystal structure (how they are arranged) Phase transformation! (change in structure and/or change in way elements are distributed in structure)

  5. Mineral Chemistry Mineral properties = f(structure + chemistry) But not independent: structure = f(chem, T, P) Compositions are conventionally given as wt% oxides (unless sulfides, halides, etc.) I'd prefer mole % actually, but inherited this system Difference between Fo = Mg2SiO4 and Fo= 51.5% SiO2 and 48.5% MgO

  6. Composition of the Earth’s Crust Most common silicates are from these O alone = 94 vol. % of crust Perhaps good to think of crust as a packed O array with interspersed metal cations in the interstices! Analogy works for minerals too (they make up the crust)

  7. Mineral Groups Element Abundances Silica (SiO4)4- SILICATES Common cations that bond with silica anions All others: 1.5%

  8. Chemistry of igneous rocks • The chemical composition of rocks is determined by analyzing a powder of the rock. Routine geochemical analysis of geologic materials can be carried out using either or a combination of the following two techiques: • X-ray Fluoresence Spectroscopy (XRF) to determine both major and trace elements • Atomic Absorbtion Spectrometry (AAS) to determine both major and trace elements

  9. Reading chemical data • Reported as oxides • Al as Al2O3 • Fe2+ and Fe3+ often reported together as FeO (i.e., all Fe2+) • conventional order: • SiO2 Al2O3 (Fe2O3) TiO2 Cr2O3 FeO MnO MgO CaO BaO Na2O K2O F Cl

  10. What are the units of these numbers? Grams of oxide per gram of material Example chemical analysis

  11. Methods of analysis (major elements) • “Wet chemical” - chemistry class stuff • Electron probe - fairly common, moderately easy, microscopic, highly precise & accurate • SEM EDS - here in building, easy, fairly precise & accurate, microscopic • XRF (X-ray fluorescence) - whole-rock chemistry, very common, very easy • others - both whole-rock, and microanalysis

  12. SEM / Electron Probe • because beam is charged (e–), can focus with magnets • beam as small as 2-3 µm • analyzes volume ~10µm diameter

  13. SEM / Electron Probe • electrons in beam knock out orbital electrons, others cascade down • each “fall” emits a photon with characteristic energy, depending on element • elements produce characteristic x-ray spectra

  14. Limitations of SEM EDS • Gives chemistry, not structure • contrasts with XRD - structure, not chemistry • Used for minerals, not whole rocks • contrasts with XRF • Sensitive to concentrations down to ~1 wt % • Contrast with EPMA (~100 ppm) • Contrast with Mass Spectrometry (~1-10 ppb?)

  15. Secondary Ion Mass Spectroscopy (SIMS) • Detection limit: ppm-ppb • Chemical element: H-U • Lateral resolution: 5 μm • Source of bombarding primary ions: O- or Cs+ • It is commonly applied to determine the abundance and distribution of rare earth elements.

  16. Analysis recalculation • converts oxides into mineral formulas • need to know what the mineral is to begin with, usually • really an idealization - must either assume an ideal number of cations or oxygens

  17. Step 1 - Moles of oxide

  18. Step 1 - Moles of oxide

  19. Step 1 - Moles of oxide

  20. Step 2 - Normalize Oxygens

  21. Step 2 - Normalize Oxygens

  22. Step 2 - Normalize Oxygens

  23. 2.931 = Moles of oxygens per gram mineral We know one mole of mineral contains 8 moles of oxygens (for a feldspar) So we have: 2.931 (mol O / g mineral) 8 (mol O / mol mineral) And the quotient: 8 / 2.931 (g mineral / mol mineral) Multiply each value by this quotient... Step 2 - Normalize Oxygens

  24. Step 3 - Calculate Cations

  25. Step 3 - Calculate Cations

  26. Step 3 - Calculate Cations

  27. Step 3 - Calculate Cations

  28. K0.030Na0.582Ca0.394Al1.365Si2.626O8 Step 4 - Allocate cations to sites • How would you plot this on a ternary feldspar plot?

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