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:  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

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

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

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

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

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)

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

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

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

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.

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

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

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

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

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

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%

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

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

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

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 

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