Lecture 5 Crystal Chemistry

Part 4: Compositional Variation of Minerals 1. Solid Solution2. 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 SolutionSimple cationic or anionic substitution

e.g. Olivine (Mg,Fe)2SiO4; Sphalerite (Fe,Zn)SCoupled substitution

e.g. Plagioclase (Ca,Na)Al(1-2)Si(3-2)O8

(Ca2+ + Al3+ = Na+ + Si4+) neutrality preserved

Types of Solid Solution

2) Interstitial Solid SolutionOccurrence of ions and molecules within large voids within certain minerals (e.g., Beryl)

Beryl, arguably considered a ring silicate (a Cyclosilicate)

Yellow, green (SiO4)-4

Purple Be tetrahedraBlue Al+3 in voids

Types of Solid Solution

3) Omission Solid SolutionExchange 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

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) SiO2 TiO2 (+4)

Al2O3 Cr2O3 Fe2O3 (+3) MgO MnO FeO CaO(+2) Na2O K2O H2O (+1)

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

Problem 2Formula to weight percents

Kyanite is Al2SiO5

Calculate the weight percents of the oxides:

– SiO2

– Al2O3

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

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.

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

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

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

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

Problem 5Weight Percent Oxidesfrom 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

Problem 5 continuedFormula 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 219.463 100.00Example: 2x 60.086 = 120.172 120.172/219.463 = .5471 x 100 = 54.71% SiO2

Checked Sept 9 2011 CLS

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%

Some Coupled Solid Substitutions

Plagioclase Feldspar CaAl2Si2O8 - NaAlSi3O8

Jadeite - Diopside NaAlSi2O6 - CaMgSi2O6

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

Notice Silica (0.4 x 2 Silica in Anorthite) + (0.6 x 3 in Albite) = 2.6Aluminum (0.4 x 2 Aluminum in Anorthite) + (0.6 x 1 in Albite) = 1.4

Checked Sept. 9th 2011 CLS

Ca same as Anorthite, Na Same as Albite

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

Problem 7 Given Analysis Compute Mole percents Jadeite 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

This page checked Sept 2 2007 CLS

http://www.science.uwaterloo.ca/~cchieh/cact/c120/formula.html

Next Lecture

Thermodynamics