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)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
Types of Solid Solution
Yellow, green (SiO4)-4
Purple Be tetrahedra
Blue Al+3 in voids
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)
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
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 1
Calculate a formula for these Weight Percents
 Oxide Wt% MolWt
Oxide
Moles Moles Moles
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 2
Formula to weight percents
 Kyanite is Al2SiO5
 Calculate the weight percents of the
oxides:
– SiO2
– Al2O3
Problem 2 p2
Kyanite: Al2SiO5
Oxide Moles MolWt Grams
PFU
Oxide Oxide

• SiO2
1
• Al2O3 1
•
60.086 60.086
101.963 101.963
Formula weight 162.049
Checked 9 Sept 2011 CLS
Wt%
60/162
102/162
37.08
62.92
100%
Problem 3: Solid Solutions
Weight 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 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 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 4
Weight Percent Oxides from Formula
Recall formula was (Mg





Oxide
Moles
PFU
1.4
Fe
MolWt
Oxide
0.6)
Si2O6
Grams
Oxide
Wt%
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 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
Problem 5 continued
Formula Unit is Na(Al0.4Fe0.6)Si2O6
Calculate Weight Percent Oxides
Oxide





Moles
PFU
SiO2
2.0
Al2O3
0.2
Fe2O3
0.3
Na2O
0.5
Formula weight
MolWt
Oxide
60.086
101.963
159.692
61.980
Grams
Oxide
120.172
20.393
47.908
30.990
219.463
Wt%
54.71
9.29
21.83
14.12
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
Some Coupled Solid Substitutions
 Plagioclase Feldspar CaAl2Si2O8 - NaAlSi3O8
 Jadeite - Diopside NaAlSi2O6 - CaMgSi2O6
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
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
Problem 6
Coupled Substitution
 An40 Ab60 formula is Ca.4 Na.6 Al1.4 Si2.6 O8
Oxide
SiO2
Al2O3
CaO
Na2O
Moles
PFU
MolWt
Oxide
Grams
Oxide
2.6
60.086
156.22
0.7
101.963
71.37
0.4
55.96
22.38
0.3
61.980
18.59
Formula weight 268.58
Wt%
58.17
26.57
8.33
6.92
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 CaMgSi2O6
We are given the following chemical analysis of a Px:
Oxide





Wt%
MolWt
Oxide
Moles
Oxide
SiO2 56.64 60.086 .9426
Na2O 4.38
61.99
.0707
Al2O3 7.21 101.963 .0707
MgO 13.30
40.312 .3299
CaO 18.46
55.96 .3299
Moles
Cation
.9426
.1414
.1414
.3299
.3299
Moles
Oxygen
1.8852
.0707
.2121
.3299
.3299
Prop. Cations to O6
2.00
.30
.30
.7
.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
Next Lecture
Thermodynamics