**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 SiO_{2} - ratio of 1:2 -Si = cation
(+4), O = anion (-2). Although SiO_{2} 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,
MgSiO_{3}, reported as weight percents of the component oxides,
SiO_{2} 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

SiO_{2}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 - MgSiO_{3}

**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 CO_{2}43.95 44.01 0.9986 0.9986=1 1.997399.91 2.9952=3

Answer: CaCO_{3} 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, Al_{2
}SiO_{5}, compute the weight percents of the
component oxides:

Oxide Moles Molec. Wt Grams Weight Form.Unit oxide oxide Percent SiO_{2}1 60.086 60.086 37.08 Al_{2}O_{3}1 101.963 101.963 67.92 Form.Wt. 162.049 100.00

**Example 4.2.1**. Alkali feldspars may exist with any
composition between NaAlSi_{3}O_{8} and
KAlSi_{3}O_{8}. 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 SiOMole Ratios: (Na_{2}68.20 60.086 1.1350 1.1350 2.2701 Al_{2}O_{3}19.29 101.963 .1892 3784 .5676 Na_{2}O 10.20 61.9796 .1646 .3291 .1646 K_{2}O 2.32 94.2037 .0246 .0493 .0246 Total Alk. .3784 3.0269 x 2.6430 = 8.000

Other examples of solid solution are:
MgSiO_{3} - FeSiO_{3} Enstatite - Ferrosilite (pyroxenes)
(Mg_{0.537}Fe_{0.463})SiO_{3} = MSiO_{3} MgCaSi_{2}O_{6} - FeCaSi_{2
}O_{6} Diopside - hedenbergite (clinopyroxene)
Mg_{2}SiO_{4} - Fe_{2}SiO
_{4} Forsterite - fayalite (olivines)
Mg_{3}Al_{2}Si_{3}O_{12} - Fe_{3}Al_{2}Si_{3}O_{12} 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 = NaAlSi_{3}O_{8} = Ab;
Orthoclase = KAlSi_{3}O_{8} = Or,
In example 4.2.1., feldspar may be denoted Ab_{87}Or_{13}

Given the formula En_{70}Fs_{30}, calculate
the weight percent oxides. En = Mg_{2}Si_{2}O
_{6}, Fs =
Fe_{2}Si_{2}O_{6} First,
write out the complete formula (i.e.(Mg_{.7}Fe_{.3
})_{2}Si_{2}O_{6}); 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 SiO_{2}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
(NaAlSi_{2}O_{6}) and Acmite (NaFeSi_{2
}O_{6})
Jd_{40}Ac_{60}, i.e. Na(Fe_{.6}Al
_{.4})Si_{2}O_{6}

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

SiO_{2}2.0 60.086 120.172 54.76 Al_{2}O_{3}0.2 101.963 20.393 9.29 Fe_{2}O_{3}0.3 159.692 47.908 21.83 Na_{2}O 0.5 61.980 30.990 14.12 219.463 100.00

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 (NaAlSi_{3}O_{8})
and anorthite (CaAl_{2}Si_{2}O_{8})
form a **coupled substitution series** that is also isomorphous.

Some other examples of coupled isomorphous series are:
MgCaSi_{2}O_{6} - CaAlAlSiO_{
6} (Diopside - Calcium Tschermaks pyroxene)
NaAlSi_{2}O_{6} - CaMgSi_{
2}O_{6} (Jadeite - diopside pyroxene)

**Example 4.3.1.** You should be able to manipulate these formulae backwards
and forwards. For example, given formula An_{40}Ab_{60}
= Ca_{.4}Na_{.6}Al_{1.4}Si_{ 2.6}O_{8},
find the weight percents of component oxides.

Oxide Moles Molec. Wt Grams Weight Form.Unit oxide oxide Percent SiO_{2}2.60 60.086 156.22 58.17 Al_{2}O_{3}0.70 101.963 71.37 26.57 CaO 0.40 55.96 22.38 8.33 Na_{2}O 0.30 61.980 18.59 6.92268.58 100.00

**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

SiO_{2}56.64 60.086 0.9426 0.9426=2.0 1.8852 Al_{2}O_{3}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 Na_{2}O 4.38 61.980 0.0707 0.1414=0.3 0.0707Totals 99.99 2.8278

6.00/2.8278 = 2.1218

Na_{.3}Ca_{.7}Al_{.3}Mg_{.
7}Si_{2}O_{6} or, Jadeite NaAlSi_{2
}O_{6} = 30%, Diopside CaMgSi_{2}O_{6
} = 70%, or
Jd_{30}Di_{70}.

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 AV_{c
}, where V_{c }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 a^{3} or

[6.023 x 10^{23}][5.640 x 10^{-8}]^{3} = 108.06 cm^{3}

Density = rho = 233.77gm/108.06cm^{3}
= 2.16 gm/cm^{3}

Fluorite has a cell edge of 5.46 Å. The chemical formula for fluorite
is CaF_{2}. 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 CaF_{2} 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 a^{3} = 162.77Å^{3}
(1 Å^{3} = 10^{-24} cm^{3}). Avogadro's number
of cells is A x a^{3} = (6.022 x 10^{23})
x (162.77 x 10^{-24}) = 98.04 cm^{3}.
So, 312.30 g / 98.04 cm^{3} = 3.19 g/cm^{3}
= density.

Let's work cuprite (Cu_{2}O) backwards (i.e. given the
density find the cell edge).

density = 6.1 g/cm^{3} atomic weight Cu = 63.546 g
Z = 2 atomic weight O = 15.9994 g
Formula weight Cu_{2}O = 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 = V^{1/3} = [286.16 g / (6.022 x 10^{23}
x 6.1 g/cm^{3})]^{1/3} = 4.27 Å

**Example 4.4.4.** Calculate the density of ferberite,
FeWO_{4}, 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 Å^{3}
V =1.33 x 10^{-22} cm^{3}

rho = Z(FW)/AV = 2 (303.695) / (6.02 x 10^{23})(1.33 x 10^{-22})
rho = 7.56 g/cm^{3}

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.

- Tetragonal V =
*a*^{2}*c* - Hexagonal V =
*a*^{2}*c*sin 120º - Orthorhombic V =
*abc* - Monoclinic V =
*abc*sin beta

V = *abc*(1 + 2*cos alpha*cos beta*cos gamma - cos^{2}alpha- cos^{2}beta- cos^{2}gamma)^{1/2}

Mineral Structures and Properties Data Base