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Units & norm calculation
Definitions
Whole-rock chemical data are commonly expressed in weight per cent of oxides. In order
to visualize chemical composition of rocks, petrologists use recalculations to obtain molar
ratios or directly the mineral proportions. All units have status of molar quantities (molar
masses), but differ in how individual oxides or mineral formulae are expressed:
Molar (molecular) units = molar quantities (molar masses) of oxides or minerals in their
common form (e.g. SiO2, MgO, CaO, Na2O, P2O5; Mg2SiO4, NaAlSi3O8 etc.).
Cation units = molar masses of oxides or minerals, expressed in “one-cation” form (e.g.
SiO2, MgO, CaO, NaO0.5, PO2.5; [Mg0.667Si0.333]1O1.333, [Na0.2Al0.2Si0.6]1O1.6).
Oxygen units = molar masses of oxides or minerals, expressed in “one-oxygen” form
(e.g. Si0.5O, MgO, CaO, Na2O, P0.4O; Mg0.5Si0.33O1, Na0.125Al0.125Si0.375O1)
Mineral recalculations or unit conversions can be performed in a straightforward chemical
manner: conversion factors are stoichiometric coefficients in balanced equations between
the above formulas. For petrologists, however, the use of fractional formulas is
inconvenient, and proportional numbers of cations or oxygens are rather used as
conversion factors. Our treatment will follow common petrological usage and a separate
reference to chemical approach will be made where necessary.
Molar (molecular) units
1 CaO
1 SiO2
1 Na2O
1 Al2O3
1 [Mg2Si]3O4
1 [NaAlSi3]5O8
Cation units
1 CaO
1 SiO2
2 NaO0.5
2 AlO1.5
3 [Mg0.67Si0.33]1O1.33
5 [Na0.2Al0.2Si0.6]1O1.6
Oxygen units
1 CaO
2 Si0.5O1
1 Na2O
3 Al0.67O1
4 Mg0.5Si0.33O1
8 Na0.125Al0.125Si0.375O1
Example 1: Convert 7.2 molar units of albite into cation and oxygen units:
Since one molecule of albite NaAlSi3O8 contains 5 cations and 8 anions (oxygens), the
corresponding quantities are:
7.2 molar Ab = 36 cation units Ab = 57.6 oxygen units Ab
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Weight - molar conversions
Molar quantities are obtained by dividing weight per cent by molecular weight of individual
oxides.
Example 2: Convert following wt-% concentrations into molar units and cations (SiO2
52.39 wt %, Al2O3 14.38, CaO 8.45, Na2O 2.78 wt %):
SiO2
Al2O3
CaO
Na2O
Weight per cent
52.39
14.38
8.45
2.78
Molecular weight
60.09
101.96
56.08
61.98
Molar units
52.39 / 60.09 = 0.8719
14.38 / 101.96 = 0.1410
8.45 / 56.08 = 0.1507
2.78 / 61.98 = 0.0449
SiO2
Al2O3
CaO
Na2O
Weight per cent
52.39
14.38
8.45
2.78
Molecular weight
60.09
101.96
56.08
61.98
Cations
52.39 / 60.09 = 0.8719
14.38 / 101.96 x 2 = 0.2820*)
8.45 / 56.08 = 0.1507
2.78 / 61.98 x 2 = 0.0897*)
If you think in terms of fractional “one-cation” formulas, then you can use the “halfmolecular” weights of AlO1.5 and NaO0.5 directly, i.e. 14.38 / 50.99 = 0.2820 and 2.78 /
30.99 = 0.0897 respectively.
*)
Mineral recalculation
Petrologists usually use one of the three recalculation procedures: oxides and minerals in
moles (e.g. CIPW norm), oxides in cations and minerals in moles (e.g. Morse’s recipes
and PetroLab), oxides and minerals in cations (e.g. Niggli values, Barth’s norms).
Example 3: What molar (molecular) amount of K-feldspar (KAlSi3O8) can be formed from
7.3 cations K?
One molecule of K-feldspar KAlSi3O8 contains 1 cation of K. Therefore, 7.3 cations K
correspond to 7.3 molar units of K-feldspar. From atomic proportions, the formation of Kfeldspar will also consume 7.3 cations Al and 21.6 cations Si from their available amounts
in the rock.
1 KAlSi3O8 (moles) = 1 KO0.5 (cat’s) + 1 AlO1.5 (cat’s) + 3 SiO2 (cat’s).
Example 4: How many cation units of K-feldspar can be formed from 7.3 cations K?
One molecule of K-feldspar KAlSi3O8 containts 5 cations (1 K + 1 Al + 3 Si). Therefore 5 x
7.3 = 36.5 cation units of K-feldspar is produced from 7.3 cations K. Again, 7.3 cat’s Al and
21.6 cat’s Si are also consumed by the formation of K-feldspar.
Chemists would get lost here, because the equation 5 KAlSi 3O8 = 1 KO0.5 + 1 AlO1.5 +
3SiO2 is not balanced chemically, they would rather write (see definitions of fractional
formulas in the introduction):
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5 [K0.2Al0.2Si0.6]1O1.6 = 1 KO0.5 + 1 AlO1.5 + 1 SiO2
which is stated correctly in cation units. The result is the same as before (36.5 cat’s of Kfeldspar come from 7.3 cat’s K).
Mineral – mineral conversions
Example 5: In a silica-undersaturated rock, we need to convert 20.5 cation units of Kfeldspar (KAlSi3O8) into leucite (KAlSi2O6) and quartz (SiO2). In a two-step manner:
balanced equation:
# cations in each molecule:
1 KAlSi3O8 = 1 KAlSi2O6 + 1 SiO2
5
4
1
we obtain the coefficients: 5 K-feldspar = 4 leucite + 1 quartz. Therefore, 20.5 cations of Kfeldspar will be transformed into 16.4 cat’s leucite and 5.1 cat’s quartz. Note that for
chemists, the equation 5 Kf = 4 Leu + 1 Qz is again not balanced for individual elements,
they would rather write (in accordance with our definitions at the beginning):
5 [K0.2Al0.2Si0.6]1O1.6 = 4 [K0.25Al0.25Si0.5]1O1.5 + 1 Si1O2
Example 6: Convert 20.5 molar units of K-feldspar into leucite and quartz:
balanced equation:
# moles on each side:
1 KAlSi3O8 = 1 KAlSi2O6 + 1 SiO2
1
1
1
Consequently, 20.5 moles of K-feldspar will be converted into 20.5 moles of leucite and
20.5 moles of quartz!
Unit – unit conversions
At the end of norm calculation, it is desirable to convert mineral proportions into the most
convenient units. Mineral abundances in volume per cent (as observed in the microscope)
are best comparable to proportions in oxygen units, whereas for plotting, one may
sometimes use cation units or weight per cent.
Example 7: Convert 4.9 molar units of albite (NaAlSi3O8) into cation units:
Since 1 molecule of albite containts 5 cations, 4.9 molar units of albite equals to 24.5
cation units of albite. Picky chemist would write:
1 NaAlSi3O8 (moles) = 5 [Na0.2Al0.2Si0.6]1O1.6 (cation units)
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Example 8: Convert the following norm from cation units into oxygen units: forsterite 17,
diopside 24, anorthite 37 and albite 22 cations.
Formula
Forsterite
Diopside
Anorthite
Albite
Forsterite
Diopside
Anorthite
Albite
TOTAL
Mg2SiO4
CaMgSi2O6
CaAl2Si2O8
NaAlSi3O8
cation units
17
24
37
22
100
# cations
# oxygens
3
4
5
5
4
6
8
8
conversion
17 x 1.333 = 22.7
24 x 1.5 = 36.0
37 x 1.6 = 59.2
22 x 1.6 = 35.2
153.1
conversion
factor
4/3 = 1.333
6/4 = 1.5
8/5 = 1.6
8/5 = 1.6
oxygen units
14.83
23.51
38.67
22.99
100
Example 9: Convert the same norm as in #7 from cation units to weight per cent:
formula
Forsterite
Diopside
Anorthite
Albite
Forsterite
Diopside
Anorthite
Albite
TOTAL
Mg2SiO4
CaMgSi2O6
CaAl2Si2O8
NaAlSi3O8
cation units
17
24
37
22
100
molecular
weight
140.71
216.57
278.22
262.24
# cations
4
4
5
5
conversion
17 x 35.18 = 598.06
24 x 54.14 = 1299.36
37 x 55.64 = 2058.68
22 x 52.45 = 1153.90
5110
conversion factor
(m.w. / cat’s)
140.71 / 4 = 35.18
216.57 / 4 = 54.14
278.22 / 5 = 55.64
262.24 / 5 = 52.45
weight per cent
11.70
25.43
40.29
22.58
100
Remarks
All molar (molecular), cation and oxygen units used in petrological calculations are molar
quantities (molar masses). Petrologists commonly calculate with mineral formulas (e.g.
Mg2SiO4, NaAlSi3O8, SiO2 etc.) and numbers of cations or oxygens are then used as
multiplication factors for conversion into cations or oxygen units. From chemical
perspective, the use of fractional formulas, as defined at the beginning, is probably easier
and very straightforward.
Note also that cation and oxygen units are mass-conservative (barycentric) but molar units
are not! During norm calculation, the total of oxides (in cation or oxygen units) at the
beginning must be equal to the total of minerals (in cation or oxygen units) at the end.
However, the total of oxides in molar units is not equal to the molar total of minerals at the
end!