1 / 21

Lecture 5 Crystal Chemistry Part 4: Compositional Variation of Minerals 1. Solid Solution 2. Mineral Formula Calcula

Solid Solution in Minerals. Where atomic sites are occupied by variable proportions of two or more different ionsDependent on: Similar ionic size (differ by less than 15-30%)Must have electrostatic neutralityAtomic sites are more accommodating at higher temperatures ? BUT as temperatures cool ex

norris
Download Presentation

Lecture 5 Crystal Chemistry Part 4: Compositional Variation of Minerals 1. Solid Solution 2. Mineral Formula Calcula

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

    1. Lecture 5 Crystal Chemistry Part 4: Compositional Variation of Minerals 1. Solid Solution 2. 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. Types of Solid Solution 2) Interstitial Solid Solution Occurrence of ions and molecules within large voids within certain minerals (e.g., Beryl, Zeolite)

    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 ranging 0-0.2 Where Fe+2 absent from some octohedral sites, some Iron probably Fe+3 to restore electrical neutrality

    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 1 Calculate a formula for these Weight Percents Oxide Wt% MolWt Moles Moles Moles Oxide Oxide Cation Oxygen SiO2 59.85 60.086 .9960 .9960 1.9920 MgO 40.15 40.312 .9960 .9960 .9960 100.0 2.9980 Mole ratios Mg : Si : O = 1 : 1 : 3 Formula is: MgSiO3 Checked 17 July 2007 CLS

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

    10. Kyanite: Al2SiO5 Oxide Moles MolWt Grams Wt% PFU Oxide Oxide

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

    12. Problem 3 (cont’) Solid Solutions Weight 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 Sept 2nd 2007 CLS

    13. Various Simple Solid Solutions Alkali Feldspars NaAlSi3O8 - KAlSi3O8 Orthopyroxenes: MgSiO3- FeSiO3 Enstatite-Ferrosilite MgCaSi2O6-FeCaSi2O6 Diopside-Hedenbergite 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 4 Weight Percent Oxides from Formula Recall formula was (Mg 1.4 Fe 0.6) Si2O6 Oxide Moles MolWt Grams Wt% PFU Oxide Oxide SiO2 2 60.086 120.172 54.69 MgO 1.4 40.312 56.437 25.69 FeO 0.6 71.846 43.108 19.62 Formula weight 219.717 100.00% For example 120.172/219.717 = .5469 x 100 = 54.69%

    16. Problem 5 Weight Percent Oxides from Formula 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 continued Formula Unit is Na(Al0.4Fe0.6)Si2O6 Calculate 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

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

    19. Problem 6 Coupled 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

    20. Problem 6 Coupled Substitution An40 Ab60 is Ca .4 Na .6 Al 1.4 Si 2.6 O 8 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

    21. Problem 7 Given Analysis Compute Mole percents Jadeite is NaAlSi2O6 Diopside is CaMgSi2O6 Oxide Wt% MolWt Moles Moles Moles Prop. Cations 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

    22. Next Lecture Thermodynamics

More Related