Transcript Lecture 4 - Stoichiometry

How Many Molecules?
• Pyrite Cube weighs 778 g – how many
molecules is that??
About 4,000,000,000,000,000,000,000,000
Are they ALL Iron and Sulfur?
Stoichiometry
• Some minerals contain varying amounts of
2+ elements which substitute for each
other
• Solid solution – specific elements
substitute for each other in the mineral
structure, defined in terms of the end
members – species which contain 100% of
one of the elements
Chemical Formulas
• Subscripts represent relative numbers of
elements present
• (Parentheses) separate complexes or
substituted elements that go into a
particular place
– Fe(OH)3 – Fe bonded to 3 separate OH
groups
– (Mg, Fe)SiO4 – Olivine group – mineral
composed of 0-100 % of Mg, 100-Mg% Fe
Goldschmidt’s rules of Substitution
1. The ions of one element can extensively
replace those of another in ionic crystals
if their radii differ by less than about 15%
2. Ions whose charges differ by one may
substitute readily if electrical neutrality is
maintained – if charge differs by more
than one, substitution is minimal
Goldschmidt’s rules of Substitution
3. When 2 ions can occupy a particular
position in a lattice, the ion with the
higher charge density forms a stronger
bond with the anions surrounding the site
4. Substitution may be limited when the
electronegativities of competing ions are
different, forming bonds of different ionic
character
• KMg3(AlSi3O10)(OH)2 - phlogopite
• K(Li,Al)2-3(AlSi3O10)(OH)2 – lepidolite
• KAl2(AlSi3O10)(OH)2 – muscovite
• Amphiboles:
• Ca2Mg5Si8O22(OH)2 – tremolite
• Ca2(Mg,Fe)5Si8O22(OH)2 –actinolite
Actinolite series
minerals
• (K,Na)0-1(Ca,Na,Fe,Mg)2(Mg,Fe,Al)5(Si,Al)8O22(OH)2
- Hornblende
Compositional diagrams
FeO
wustite
Fe3O4
magnetite
Fe2O3
hematite
A
Fe
O
A1B1C1
A1B2C3
x
x
B
C
Si
fayalite
forsterite
enstatite
Fe
ferrosilite
Mg
fayalite
Fe
forsterite
Mg
Pyroxene solid solution  MgSiO3 – FeSiO3
Olivine solid solution  Mg2SiO4 – Fe2SiO4
Minor, trace elements
• Because a lot of different ions get into any
mineral’s structure as minor or trace
impurities, strictly speaking, a formula
could look like:
• Ca0.004Mg1.859Fe0.158Mn0.003Al0.006Zn0.002Cu0.001Pb
0.00001Si0.0985Se0.002O4
• One of the ions is a determined integer, the
other numbers are all reported relative to that
one.
Normalization
• Analyses of a mineral or rock can be reported in
different ways:
– Element weight %- Analysis yields x grams element in
100 grams sample
– Oxide weight % because most analyses of minerals and
rocks do not include oxygen, and because oxygen is
usually the dominant anion - assume that charge
imbalance from all known cations is balanced by some %
of oxygen
– Number of atoms – need to establish in order to get to a
mineral’s chemical formula
• Technique of relating all ions to one (often Oxygen)
is called normalization
Normalization
• Be able to convert between element weight
%, oxide weight %, and # of atoms
• What do you need to know in order convert
these?
– Element’s weight  atomic mass (Si=28.09
g/mol; O=15.99 g/mol; SiO2=60.08 g/mol)
– Original analysis
– Convention for relative oxides (SiO2, Al2O3,
Fe2O3 etc)  based on charge neutrality of
complex with oxygen (using dominant redox
species)
Normalization example
• Start with data from quantitative analysis: weight
percent of oxide in the mineral
• Convert this to moles of oxide per 100 g of
sample by dividing oxide weight percent by the
oxide’s molecular weight
• ‘Fudge factor’ is process called normalization –
where we divide the number of moles of one
thing by the total moles  all species/oxides
then are presented relative to one another
Feldspar analysis
(Ca, Na, K)1 (Fe, Al, Si)4 O8
Oxide wt %
in the
# of moles
mineral of oxide in mole % of
2# cations in # of O (determined
the
oxides in
oxide
in oxide by analysis) mineral the mineral
2-
oxide
Atomic
weight
of oxide
(g/mol)
SiO2
60.08
1
2
65.90
1.09687
73.83
Cation
4+
Si
Al2 O3
Fe 2 O3
CaO
Na2 O
K2 O
101.96
159.68
56.08
61.96
94.20
2
2
1
2
2
3
3
1
1
1
19.45
1.03
0.61
7.12
6.20
0.19076
0.00645
0.01088
0.11491
0.06582
12.84
0.43
0.73
7.73
4.43
Al3+
Fe3+
Ca2+
Na+
K+
1.48569
100
SUM
moles of moles of O Number of
cations contributed moles of
in
by each ion in the
sample
cation
mineral
73.83
147.66
2.95
25.68
0.87
0.73
15.47
8.86
38.52
1.30
0.73
7.73
4.43
1.03
0.03
0.03
0.62
0.35
125.44
200.38
# of moles Oxygen choosen:
8
Ca0.73 Na15.47 K8.86 Fe 0.87 Al25.68 Si73.83 O200.38
Ca0.03 Na0.62 K0.35 Fe 0.03 Al1.03 Si2.95 O8
to get here from formula above, adjust by 8 / 200.38