Chapter 4. Mineral Stoichiometry: Chemical Formulae, End-Members, Simple Solid Solu tions, Coupled Substitutions, Mineral Densities

4.1. Chemical Formulae

Example 4.1.1. Halite NaCl - Na = cation (+1), Cl = anion (-1) NaCl is an ionic crystal with a ratio of 1:1 for the charge on each atom. An ionic compound is neutral with a net charge of 0.

Example 4.1.2. Quartz SiO2 - ratio of 1:2 -Si = cation (+4), O = anion (-2). Although SiO2 may not be totally ionic, the atomic ratios are rigidly fixed by the structure and the requirement for electroneutrality.

Chemical analyses of minerals are customarily reported as weight percents of component oxides. This is an unfortunate relic of wet chemical analysis, but is so firmly entrenched in the science that it is important that you be able to manipulate these and convert them to atom ratios. In the table below is a typical analysis of a simple end-member silicate, enstatite, MgSiO3, reported as weight percents of the component oxides, SiO2 and MgO. Also given are the molecular weights of these oxides, and the relative mole numbers of the oxide, the cations, and the oxygen to illustrate the simple ratios that exist.

Example 4.1.3. - Enstatite, a member of the pyroxene group.


Oxide       Wt%         Molec. Wt      Moles      Moles      Moles
                        oxide          oxide      cations    oxygen

SiO2 59.85 60.086 0.9960 0.9960 1.9920 MgO 40.15 40.312 0.9960 0.9960 0.9960 100.00 2.9880

Mole ratios -Mg : Si : O = 1 : 1 : 3 - MgSiO3

Example 4.1.4. Given the following analysis in weight percents of component oxides, compute a formula.


Oxide	Wt%	Molec. Wt 	Moles	Moles	Moles
		oxide	oxide	cations	oxygen


CaO	55.96	56.08	0.9979	0.9979=1	0.997
CO2	43.95	44.01	0.9986	0.9986=1	1.9973
	99.91				2.9952=3

Answer: CaCO3 This could be either calcite or aragonite, two different minerals with different structures but the same formula (polymorphs).

Example 4.1.5. Given the formula for kyanite, Al2 SiO5, compute the weight percents of the component oxides:


Oxide	Moles	Molec. Wt 	Grams	Weight
	Form.Unit	oxide	
oxide	Percent


SiO2	1	 60.086	60.086	37.08
Al2O3	1	101.963	101.963	67.92
Form.Wt.			162.049	100.00

4.2. Simple Solid Solutions

Example 4.2.1. Alkali feldspars may exist with any composition between NaAlSi3O8 and KAlSi3O8. The alkalis may substitute for each other in any ratio, but the ratio of alkalis to aluminum is fixed at 1.00.

Given in the table below is a chemical analysis in weight percent oxides of a typical alkali feldspar illustrating the free substitution of K and Na and the fixed ratios of other cations and anions.


Oxide	Wt%	Molec. Wt 	Moles	Moles	Moles
		oxide	oxide	cations	oxygen


SiO2	68.20	60.086	1.1350	1.1350	2.2701
Al2O3	19.29	101.963	.1892	3784	.5676
Na2O	10.20	61.9796	.1646	.3291	.1646
K2O	2.32	94.2037	.0246	.0493	.0246
Total Alk.				.3784	3.0269
					x 2.6430 =
					8.000
Mole Ratios: (Na0.87K0.13)Al1Si3O8

Other examples of solid solution are: MgSiO3 - FeSiO3 Enstatite - Ferrosilite (pyroxenes) (Mg0.537Fe0.463)SiO3 = MSiO3 MgCaSi2O6 - FeCaSi2 O6 Diopside - hedenbergite (clinopyroxene) Mg2SiO4 - Fe2SiO 4 Forsterite - fayalite (olivines) Mg3Al2Si3O12 - Fe3Al2Si3O12 pyrope - almandine (garnets)

These are said to be isomorphous series because the whole range of compositions has the same crystal structure, that is, the same space group and nearly the same atom positions and coordinations.

The extremes of chemical chemical composition in such isomorphous series are known as end- members Examples: Albite = NaAlSi3O8 = Ab; Orthoclase = KAlSi3O8 = Or, In example 4.2.1., feldspar may be denoted Ab87Or13

Example 4.2.2. Compute weight percent oxides from end-members.

Given the formula En70Fs30, calculate the weight percent oxides. En = Mg2Si2O 6, Fs = Fe2Si2O6 First, write out the complete formula (i.e.(Mg.7Fe.3 )2Si2O6); compute the the numbers of moles of the component oxides and their weights together with the gram formula weight of your formula.


Oxide	Moles	Molec. Wt 	Grams	Weight
	Form.Unit	oxide	
oxide	Percent


SiO2	2.00	60.086	120.172	54.69
FeO	0.60	71.846	43.108	19.62
MgO	1.40	40.312	56.437	25.69
Form.Wt.			219.717	100.00

Example 4.2.3. Crystalline solution between Jadeite (NaAlSi2O6) and Acmite (NaFeSi2 O6) Jd40Ac60, i.e. Na(Fe.6Al .4)Si2O6


Oxide Moles Molec. Wt Grams Weight Form.Unit oxide oxide Percent
SiO2 2.0 60.086 120.172 54.76 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 14.12 219.463 100.00

4.3. Coupled Substitution

If the atoms substituting for one another are of different valences, there must be an accompanying substitution to compensate the charge to maintain neutrality.

In the plagioclase feldspars, Ca and Na have the same radius but different charges, so that if Ca+2 substitutes for Na+1 there must be a compensating substitution of Al+3 for Si+4

Thus the end-members, albite (NaAlSi3O8) and anorthite (CaAl2Si2O8) form a coupled substitution series that is also isomorphous.

Some other examples of coupled isomorphous series are: MgCaSi2O6 - CaAlAlSiO 6 (Diopside - Calcium Tschermaks pyroxene) NaAlSi2O6 - CaMgSi 2O6 (Jadeite - diopside pyroxene)

Example 4.3.1. You should be able to manipulate these formulae backwards and forwards. For example, given formula An40Ab60 = Ca.4Na.6Al1.4Si 2.6O8, find the weight percents of component oxides.


Oxide	Moles	Molec. Wt 	Grams	Weight
	Form.Unit	oxide	
oxide	Percent


SiO2	2.60	60.086	156.22	58.17 
Al2O3	0.70	101.963	71.37	26.57
CaO	0.40	55.96	22.38	8.33
Na2O	0.30	61.980	18.59	6.92

			 268.58	100.00 

Example 4.3.2. Given the following weight percents oxides, compute mole percents of end- members Jadeite - Diopside


Oxide	Wt%	Molec. Wt 	Moles	Moles	Moles
		oxide	oxide	cations	oxygen

SiO2 56.64 60.086 0.9426 0.9426=2.0 1.8852 Al2O3 7.21 101.963 0.0 707 0.1414=0.3 0.2121 MgO 13.30 40.312 0.3299 0.3299=0.7 0.3299 CaO 18.46 55.96 0.3299 0.3299=0.7 0.3299 Na2O 4.38 61.980 0.0707 0.1414=0.3 0.0707 Totals 99.99 2.8278

6.00/2.8278 = 2.1218

Na.3Ca.7Al.3Mg. 7Si2O6 or, Jadeite NaAlSi2 O6 = 30%, Diopside CaMgSi2O6 = 70%, or Jd30Di70.

4.4. Unit Cells and Mineral Density

It is useful at this time to review the notion of the unit cell. The unit cell is the smallest physical unit of a crystal's structure which, when repeated by translation symmetry operations, will generate a whole crystal. The unit cell may contain several formula units of atoms. The number of formula units per unit cell is an integer called Z. Avogadro's number of unit cells, then, will have a weight, in grams, equal to Z formula weights and have a volume equal to AVc , where Vc is the cell volume. This permits us to calculate the theoretical density of any mineral from its cell volume, its formula, and its Z number.

Example 4.4.1.: Halite NaCl

Na 22.9898 Cl 35.453 Formula weight 58.44 grams

Z = 4

Avogadro's Number of unit cells contains Z (i.e. 4) formula weights or 233.77 grams and has a volume of A x a3 or

[6.023 x 1023][5.640 x 10-8]3 = 108.06 cm3

Density = rho = 233.77gm/108.06cm3 = 2.16 gm/cm3

Example 4.4.2. Fluorite CaF2

Fluorite has a cell edge of 5.46 Å. The chemical formula for fluorite is CaF2. The number of formula units per unit cell is 4. This means that there are 4 Ca atoms and 8 fluorine atoms. Fluorite has a molecular weight of 40.08(Ca) + 2(18.998)(F) = 78.08. This is the weight of Avogadro's number of CaF2 groups. If Z = 4, what is the weight of this many cells? 4 x 78.08 = 312.30 g. What is the volume of this many cells? One cell is a3 = 162.77Å3 (1 Å3 = 10-24 cm3). Avogadro's number of cells is A x a3 = (6.022 x 1023) x (162.77 x 10-24) = 98.04 cm3. So, 312.30 g / 98.04 cm3 = 3.19 g/cm3 = density.

Example 4.4.3. Cuprite Cu2O

Let's work cuprite (Cu2O) backwards (i.e. given the density find the cell edge).

density = 6.1 g/cm3 atomic weight Cu = 63.546 g Z = 2 atomic weight O = 15.9994 g Formula weight Cu2O = 143.08 g

If density = (Z)(FW) / (A)(V), where Z = # of formula units, FW = Formula weight, A = Avogadro's number, and V = volume, then to solve for the cell edge, a , we have a = V1/3 = [286.16 g / (6.022 x 1023 x 6.1 g/cm3)]1/3 = 4.27 Å

Example 4.4.4. Calculate the density of ferberite, FeWO4, which is monoclinic with a = 4.73, b = 5.70, c = 4.95, beta = 90.01º and Z = 2.


First find formula weight:
	Fe	55.847	1	55.847
	W	183.85	1	183.85
	O	15.9995	4	63.998
	Total			303.695

Then the cell volume. V = abc sin beta = (4.73)(5.70)(4.95)(sin beta) V = 133.457 Å3 V =1.33 x 10-22 cm3

rho = Z(FW)/AV = 2 (303.695) / (6.02 x 1023)(1.33 x 10-22) rho = 7.56 g/cm3

Example 4 requires that you be able to compute unit cell volumes for monoclinic cells. You should be able to do this for all but triclinic cells.

For the triclinic case, the general formula applicable to all cases is

V = abc(1 + 2*cos alpha*cos beta*cos gamma - cos2alpha- cos2beta- cos2gamma)1/2


GEOL 3010 Syllabus

Chapter 3 Mineralogy Notes

Chapter 5 Mineralogy Notes

Mineral Structures and Properties Data Base

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