Recalculations of Mineral Compositions

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Recalculations of Mineral Compositions

• Klein, 2002, read pages 94 - 98

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Recalculations of Mineral Compositions

• Native element minerals have an easy chemical
  formula, indicated by their elemental content
• Examples
• Gold Au
• Silver Ag
• Sulfur S
• Copper Cu
• Arsenic Sb
• BUT most minerals are compounds
• Examples 1/ Galena PbS (lead sulphide)
  2/ Chalcopyrite CuFeS2 (copper iron sulphide)

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Recalculations of Mineral Compositions

• Very few minerals have an exact composition (even
  gold, Au has admixtures)
• Only a few minerals are pure substances
• eg Quartz SiO2 and kyanite Al2SiO5
• In most minerals some substitution occurs
  (Remember - a mineral has a definite, but not
  entirely fixed chemical composition )
• Example Fe can substitute for Zn in structure of
  sphalerite, ZnS. In this case, the total
  anion-cation ratio stays fixed 11 but Zn Fe is
  variable

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Quantitative analyses are the basis for the
formulae of elements or oxides (in weight )
NOTE Total must be (close to) 100 in the
analysis
Recalculations of Mineral Compositions
A quantitative analysis tells us what elements
how much of each element but not where in the
mineral structure the reported percentages do
not represent the ratios of the different atoms
so we must recalculate by dividing by the
atomic weights of the elements (see Table 3.3, p
43)
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Recalculations of Mineral Compositions
To Determine Atomic Ratio from Mineral Analysis
instrumental analysis
1Cu,1Fe,2S
or CuFeS2 chalco-pyrite
1/ List atomic weights from table
2/ Divide weight by atomic weight to get atomic
proportions
3/ Divide proportions by smallest number to
obtain ratios
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Recalculations of Mineral Compositions
Reverse Procedure

• Calculate atomic percentages of chalcopyrite from
  chemical formula
• which is CuFeS2

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Recalculations of Mineral Compositions

• ChalcopyriteCuFeS2

Atomic weight
183.53
1/ List atomic weights and multiply by number of
elements in formula
3/ Divide atomic weight by total
4/ Multiply by 100 to obtain percentage
2/ Total atomic weights
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Recalculations of Mineral Compositions
Example Some sphalerite analyses (where Zn is
partially substituted by variable amounts of Fe,
Mn Cd)

• Weight recalculated into atomic proportions for
  sphalerite, ZnS

ANALYSIS
Atomic weight Zn65.37 S32.065
Divide by atomic weight Zn 67.10/65.371.026 S
32.90/32.0651.026
Divide by smallest ZnS11
(ZnFeMnCd)S Total 1.026 and 1.026 11
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Recalculations of Mineral Compositions

• Weight percentage recalculated into atomic
  proportions (2nd analysis)

Divide by atomic weights
Zn 66.98/65.371.024 Fe 0.15/55.850.003 S
32.78/32.0641.022
Total metals 1.027
Zn1.024/1.027 x 10099.7 Fe0.003/1.027 x
1000.3
(Zn99.7Fe0.3)S
(Zn99.7Fe0.3)S
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Recalculations of Mineral Compositions

• Weight recalculated into atomic proportions
  (3rd analysis)

Divide by atomic weight 18.25/55.850.327 2.66/54
.940.048 0.28/112.400.002 44.67/65.370.683

1.060
(ZnFeMnCd)S1.061.05 11
Percentage of metals
Fe0.327/1.060x10030.8 Zn0.683/1.060x10064.4
Mn0.048/1.060x1004.5 Cd0.002/1.060x1000.2

(Zn64.4Fe30.8Mn4.5Cd0.2)S
(Zn64Fe31Mn5Cd)S
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Recalculations of Mineral Compositions

• Majority of minerals are compounds containing
  oxygen (e.g. carbonates, sulfates, silicates,
  etc).
• By convention, all these analyses are reported as
  oxides
• e.g. CaO in gypsum

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Recalculations of Mineral Compositions

• Recalculation of gypsum analysis

Molecular proportions
Molecular ratios
Molecular weight
Weight
gypsum
CaO
SO3
H2O
Total
1/ Divide weight by molecular weight
2/ Divide by smallest number
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Recalculations of Mineral Compositions

• 1 CaO 1 SO3, and 2 H2O
• written as CaO.SO3.2H2O
• or CaSO4. 2H2O Gypsum

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Example of Olivine Analysis
Divide wt/mol weight gives
Olivine (Fe,Mg)2SiO4
Weight
Molecular weight
Mol. proportions
On the basis Of 4 Oxygen
Atomic ratio
Atomic proportions
Oxygen proportion
2.02
We know from the formula this should be 4
So multiply every factor with 4/2.3535 the
oxygen factor
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Recalculations of Mineral Compositions
is the general formula
Olivine (Fe,Mg,Mn)2SiO4
This analysis specifically gives the
formula (Fe0.9Mg1.4Mn0.1) SiO4
MgFe or Mg Fe ratio
2.02
Mg1.140 Fe 0.870 MgFe2.010
Mg1.140/2.010x10057 Fe 0.870/2.010x10043
Or 57 Forsterite and 43 Fayalite
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Recalculations of Mineral Compositions

• Olivine is a simple silicate, where Fe and Mg can
  substitute each other for one place
• In pyroxene and amphibole less simple.
• General pyroxene formula
• (Mg,Fe,Ca)(Si,Al)2O6 various sites
  substituted
• Endmembers
• Pure MgO is enstatite (En)
• Pure FeO is ferrosillite (Fs)
• Pure CaO is wollastonite (Wo)

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HOMEWORK
Read page 94 to 98 and go over follow the
Recalculation for the pyroxene analysis
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Multiply by oxygen factor
Oxygen factor is 6/2.68282.236469
Pyroxene (Mg,Fe,Ca)(Si,Al)2O6

2.0
1.033 1.0

1.006 1.0
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Recalculations of Mineral Compositions

• Recalculation steps
• Divide column(1) weight percent by molecular
  weight (Col 2)molecular proportions
• Cation proportions (Col 3) by multiplying mol.
  proportions (Col 2) with the number of cations in
  the molecule (e.g. Al2O3 2x).
• Number of oxygens (Col 4) by multiplying
  mol.props.(2) by number of oxygens in molecule
  (e.g. Al2O3 3x).
• Total number of oxygens (Col 4) should be 6
  oxygen factor is 6/total oxygens.
• Multiply cations (Col 3) by oxygen factor
• Assign cations to spaces

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Recalculations of Mineral Compositions

• Pyroxene end member calculation

• MgO as MgSiO3 or enstatite

• FeO as FeSiO3 or ferrosilite

• CaO as CaSiO3 or wollastonite

Total end member molecules
Multiply every factor by 1/total x 100
Gives end member percentages