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Solid Solution in Minerals

Lecture 5 Crystal Chemistry Part 4: Compositional Variation of Minerals 1. Solid Solution 2. Mineral Formula Calculations. Solid Solution in Minerals. Where atomic sites are occupied by variable proportions of two or more different ions Dependent on:

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Solid Solution in Minerals

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  1. Lecture 5 Crystal ChemistryPart 4:Compositional Variation of Minerals 1. Solid Solution2. Mineral Formula Calculations

  2. Solid Solution in Minerals Where atomic sites are occupied by variable proportions of two or more different ions Dependent on: • Similar ionic size (differ by less than 15-30%) • Must have electrostatic neutrality • Atomic sites are more accommodating at higher temperatures … BUT as temperatures cool exsolution can occur

  3. Types of Solid Solution 1)Substitutional Solid Solution Simple cationic or anionic substitution e.g. Olivine (Mg,Fe)2SiO4; Sphalerite (Fe,Zn)S Coupled substitution e.g. Plagioclase (Ca,Na)Al(1-2)Si(3-2)O8 (Ca2+ + Al3+ = Na+ + Si4+) neutrality preserved

  4. Yellow, green (SiO4)-4 Purple Be tetrahedra Blue Al+3 in voids Types of Solid Solution 2)Interstitial Solid Solution Occurrence of ions and molecules within large voids within certain minerals (e.g., Beryl) Beryl, arguably considered a ring silicate (a Cyclosilicate)

  5. Types of Solid Solution 3)Omission Solid Solution Exchange of single higher charge cation for two or more lower charged cations which creates a vacancy (e.g. Pyrrhotite Fe(1-x)S) with x = Fe++ ranging 0-0.2 within regions of the crystal Where Fe+2 absent from some octohedral sites, some Iron probably Fe+3 to restore electrical neutrality Two Ferric Fe+3 ions balance charge for each three missing Ferrous Fe+2 ion

  6. Mineral Formula Calculations • Chemical analyses are usually reported in weight percent of elements or elemental oxides • To calculate mineral formula requires transforming weight percent into atomic percent or molecular percent

  7. Ion Complexes of Important Cations (with cation valence in parentheses) • SiO2 TiO2 (+4) • Al2O3 Cr2O3 Fe2O3 (+3) • MgO MnO FeO CaO(+2) • Na2O K2O H2O (+1)

  8. Problem 1Calculate a formula for these Weight Percents • Oxide Wt% MolWt Moles Moles Moles Oxide Oxide Cation Oxygen • SiO2 59.85 60.086 .996* .996 1.992 • MgO 40.15 40.312 .996 .996 .996 • total 100% 2.998 • Mole ratios Mg : Si : O = 1 : 1 : 3 • Formula is: MgSiO3 Enstatite Checked 9 Sept 2011 CLS *59.85/60.086

  9. Problem 2Formula to weight percents • Kyanite is Al2SiO5 • Calculate the weight percents of the oxides: – SiO2 – Al2O3

  10. Problem 2 p2 Kyanite: Al2SiO5 • Oxide Moles MolWt Grams Wt% PFU Oxide Oxide • SiO2 1 60.086 60.086 60/162 37.08 • Al2O3 1 101.963 101.963 102/162 62.92 • Formula weight 162.049 100% Checked 9 Sept 2011 CLS

  11. Problem 3: Solid SolutionsWeight percents to formula • Alkali Feldspars may exist with any composition between NaAlSi3O8 (Albite) and KAlSi3O8(Sanidine, Orthoclase and Microcline) • Formula has 8 oxygens: (Na,K)AlSi3O8 • The alkalis may substitute in any ratio, but total alkalis (Na + K) to Al is 1 to 1.

  12. Problem 3 (cont’) Solid SolutionsWeight percents to Formula • Oxide Wt% MolWt Moles Moles Moles Oxide Oxide Cation Oxygen • SiO2 68.20 60.086 1.1350 1.1350 2.2701 • Al2O3 19.29 101.963 0.1892 0.3784 .5676 • Na2O 10.20 61.9796 0.1646 0.3291 .1646 • K2O 2.32 94.204 0.0246 0.0493 .0246 • 100.00 3.0269 • Units: Wt% [g/FU] / MolWt [g/mole]  moles\FU • 3.0269 oxygens is wrong for this mineral. Multiply cations by 8.000/ 3.0269 oxygen correction • Mole ratios Na 0.87, K 0.13, Al 1.000, Si ~3.0, calculated as cations per 8 oxygens • Notice, now Na + K = 1.00, as required Checked 9 Sept 2011 CLS Answer (Na.87,K.13)AlSi3O8

  13. Various Simple Solid Solutions • Alkali Feldspars • NaAlSi3O8 - KAlSi3O8 • Orthopyroxenes: • MgSiO3- FeSiO3 Enstatite - Ferrosilite (opx) • MgCaSi2O6-FeCaSi2O6 Diopside-Hedenbergite (cpx) • Olivines: Mg2SiO4- Fe2SiO4 Forsterite - Fayalite • Garnets: • Mg3Al2Si3O12- Fe3Al2Si3O12 Pyrope - Almandine

  14. Problem 4: Orthopyroxenes Solid Solution Weight Percent Oxides from Formula • Given the formula En70Fs30 for an Orthopyroxene, calculate the weight percent oxides. • En = Enstatite = Mg2Si2O6 • Fs = Ferrosilite = Fe2Si2O6 • Formula is (Mg0.7Fe0.3)2Si2O6 = (Mg1.4Fe0.6)Si2O6

  15. Problem 4Weight Percent Oxides from Formula Recall formula was (Mg 1.4 Fe 0.6) Si2O6 • Oxide Moles MolWt Grams Wt% PFU Oxide Oxide • SiO2 2 x 60.086 = 120.172 54.69 • MgO 1.4 x 40.312 = 56.437 25.69 • FeO 0.6 x 71.846 = 43.108 19.62 • Formula weight tot. 219.717 100.00% For example 120.172/219.717 = .5469 (i.e. 54.69%) Checked 23 September 2011 CLS

  16. Problem 5Weight Percent OxidesfromFormula • Consider a Pyroxene solid solution of 40% Jadeite (NaAlSi2O6) and 60% Aegirine (NaFe+3Si2O6). • Calculate the weight percent oxides • Formula is Na(Al0.4Fe0.6)Si2O6

  17. Problem 5 continuedFormula Unit is Na(Al0.4Fe0.6)Si2O6Calculate Weight Percent Oxides Oxide Moles MolWt Grams Wt% PFU Oxide Oxide • SiO2 2.0 60.086 120.172 54.71 • Al2O3 0.2 101.963 20.393 9.29 • Fe2O3 0.3 159.692 47.908 21.83 • Na2O 0.5 61.980 30.990 14.12 • Formula weight 219.463 100.00 Example: 2x 60.086 = 120.172 120.172/219.463 = .5471 x 100 = 54.71% SiO2 Example: 0.4 moles Al given as Al2O3 is 0.2 moles/per formula unit Al2O3 0.2x101.963 = 20.393; 20.393/219.463 = .0929 x 100 = 9.29% Checked Sept 9 2011 CLS

  18. Some Coupled Solid Substitutions • Plagioclase Feldspar CaAl2Si2O8 - NaAlSi3O8 • Jadeite - Diopside NaAlSi2O6 - CaMgSi2O6

  19. Problem 6Coupled Substitution • Given 40% Anorthite; 60% Albite • Calculate Weight percent Oxides • First write the formulas • Anorthite is CaAl2Si2O8 • Albite is NaAlSi3O8 • An40 Ab60 is Ca0.4Na0.6Al1.4Si2.6O8 Ca same as Anorthite, Na Same as Albite Notice Silica (0.4 x 2 Silica in Anorthite) + (0.6 x 3 in Albite) = 2.6 Aluminum (0.4 x 2 Aluminum in Anorthite) + (0.6 x 1 in Albite) = 1.4 Checked Sept. 9th 2011 CLS

  20. Problem 6Coupled Substitution • An40 Ab60 formula is Ca.4 Na.6 Al1.4 Si2.6 O8 Oxide Moles MolWt Grams Wt% PFU Oxide Oxide SiO2 2.6 60.086 156.22 58.17 Al2O3 0.7 101.963 71.37 26.57 CaO 0.4 55.96 22.38 8.33 Na2O 0.3 61.980 18.59 6.92 Formula weight 268.58 100.00 Example: Notice Al 1.4 moles/PFU reported as Al2O3 is 0.7 PFU Checked 9 August 2007 CLS

  21. Problem 7 Given Analysis Compute Mole percentsJadeite is NaAlSi2O6 Diopside is CaMgSi2O6We are given the following chemical analysis of a Px: Oxide Wt% MolWt Moles Moles Moles Prop. Cations to O6 Oxide Oxide Cation Oxygen • SiO2 56.64 60.086 .9426 .9426 1.8852 2.00 • Na2O 4.38 61.99 .0707 .1414 .0707 .30 • Al2O3 7.21 101.963 .0707 .1414 .2121 .30 • MgO 13.30 40.312 .3299 .3299 .3299 .7 • CaO 18.46 55.96 .3299 .3299 .3299 .7 2.8278 • But pyroxenes here have 6 moles oxygens/mole, not 2.8278. Multiply moles cation by 6/2.8278 • As always, Moles Oxide = weight percentage divided by molec weight • Na .3 Ca.7 Al.3 Mg .7 Si2O6= 30% Jadeite 70% Diopside • http://www.science.uwaterloo.ca/~cchieh/cact/c120/formula.html This page checked Sept 2 2007 CLS

  22. Next Lecture Thermodynamics

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