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

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Presentation on theme: "Lecture 5 Crystal Chemistry Part 4: Compositional Variation of Minerals 1. Solid Solution 2. Mineral Formula Calculations."— 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) 2 SiO 4 ; Sphalerite (Fe,Zn)S Coupled substitution e.g. Plagioclase (Ca,Na)Al (1-2) Si (3-2) O 8 (Ca 2+ + Al 3+ = Na + + Si 4+ ) 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) Beryl, arguably considered a ring silicate (a Cyclosilicate) Yellow, green (SiO 4 ) -4 Purple Be tetrahedra Blue Al +3 in voids

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 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)  SiO 2 TiO 2 (+4)  Al 2 O 3 Cr 2 O 3 Fe 2 O 3 (+3)  MgO MnO FeO CaO(+2)  Na 2 O K 2 O H 2 O (+1)

8 Problem 1 Calculate a formula for these Weight Percents  Oxide Wt% MolWt Moles Moles Moles Oxide Oxide Cation Oxygen  SiO *  MgO  total 100%  Mole ratios Mg : Si : O = 1 : 1 : 3  Formula is: MgSiO 3 Enstatite Checked 9 Sept 2011 CLS *59.85/60.086

9 Problem 2 Formula to weight percents  Kyanite is Al 2 SiO 5  Calculate the weight percents of the oxides: – SiO 2 – Al 2 O 3

10 Problem 2 p2 Kyanite: Al 2 SiO 5  Oxide Moles MolWt Grams Wt% PFU Oxide Oxide SiO / Al2O / Formula weight % Checked 9 Sept 2011 CLS

11 Problem 3: Solid Solutions Weight percents to formula  Alkali Feldspars may exist with any composition between NaAlSi 3 O 8 ( Albite ) and KAlSi 3 O 8 ( Sanidine, Orthoclase and Microcline )  Formula has 8 oxygens: (Na,K)AlSi 3 O 8  The alkalis may substitute in any ratio, but total alkalis (Na + K) 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  SiO  Al2O  Na2O  K2O   Units: Wt% [g/FU] / MolWt [g/mole]  moles\FU  oxygens is wrong for this mineral. Multiply cations by 8.000/ 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 )AlSi 3 O 8

13 Various Simple Solid Solutions  Alkali Feldspars  NaAlSi 3 O 8 - KAlSi 3 O 8  Orthopyroxenes:  MgSiO 3 - FeSiO 3 Enstatite - Ferrosilite (opx)  MgCaSi 2 O 6 -FeCaSi 2 O 6 Diopside-Hedenbergite (cpx)  Olivines: Mg 2 SiO 4 - Fe 2 SiO 4 Forsterite - Fayalite  Garnets:  Mg 3 Al 2 Si 3 O 12 - Fe 3 Al 2 Si 3 O 12 Pyrope - Almandine

14 Problem 4: Orthopyroxenes Solid Solution Weight Percent Oxides from Formula  Given the formula En 70 Fs 30 for an Orthopyroxene, calculate the weight percent oxides.  En = Enstatite = Mg 2 Si 2 O 6  Fs = Ferrosilite = Fe 2 Si 2 O 6  Formula is (Mg 0.7 Fe 0.3 ) 2 Si 2 O 6 = (Mg 1.4 Fe 0.6 )Si 2 O 6

15 Problem 4 Weight Percent Oxides from Formula Recall formula was (Mg 1.4 Fe 0.6 ) Si 2 O 6  Oxide Moles MolWt Grams Wt% PFU Oxide Oxide  SiO2 2 x =  MgO 1.4 x =  FeO 0.6 x =  Formula weight tot % For example / =.5469 (i.e %) Checked 23 September 2011 CLS

16 Problem 5 Weight Percent Oxides from Formula  Consider a Pyroxene solid solution of 40% Jadeite (NaAlSi 2 O 6 ) and 60% Aegirine (NaFe +3 Si 2 O 6 ).  Calculate the weight percent oxides  Formula is Na(Al 0.4 Fe 0.6 )Si 2 O 6

17 Problem 5 continued Formula Unit is Na(Al 0.4 Fe 0.6 )Si 2 O 6 Calculate Weight Percent Oxides Oxide Moles MolWt Grams Wt% PFU Oxide Oxide  SiO  Al 2 O  Fe 2 O  Na 2 O  Formula weight Example: 2x = / =.5471 x 100 = 54.71% SiO 2 Checked Sept CLS Example: 0.4 moles Al given as Al 2 O 3 is 0.2 moles/per formula unit Al 2 O 3 0.2x = ; / =.0929 x 100 = 9.29%

18 Some Coupled Solid Substitutions  Plagioclase Feldspar CaAl 2 Si 2 O 8 - NaAlSi 3 O 8  Jadeite - Diopside NaAlSi 2 O 6 - CaMgSi 2 O 6

19 Problem 6 Coupled Substitution  Given 40% Anorthite; 60% Albite  Calculate Weight percent Oxides First write the formulas Anorthite is CaAl 2 Si 2 O 8 Albite is NaAlSi 3 O 8 An40 Ab60 is Ca 0.4 Na 0.6 Al 1.4 Si 2.6 O 8 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. 9 th 2011 CLS Ca same as Anorthite, Na Same as Albite

20 Problem 6 Coupled Substitution  An40 Ab60 formula is Ca.4 Na.6 Al 1.4 Si 2.6 O 8 Oxide Moles MolWt Grams Wt% PFU Oxide Oxide SiO Al2O CaO Na2O Formula weight 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 percents Jadeite is NaAlSi2O6 Diopside is CaMgSi2O6 We are given the following chemical analysis of a Px: Oxide Wt% MolWt Moles Moles Moles Prop. Cations to O 6 Oxide Oxide Cation Oxygen  SiO  Na 2 O  Al 2 O  MgO  CaO  But pyroxenes here have 6 moles oxygens/mole, not Multiply moles cation by 6/  As always, Moles Oxide = weight percentage divided by molec weight  Na.3 Ca.7 Al.3 Mg.7 Si 2 O 6 = 30% Jadeite 70% Diopside This page checked Sept CLS 

22 Next Lecture Thermodynamics


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