Units & Norm Calculation
DD, Jan 2000
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) unitsCation unitsOxygen units
1 CaO1 CaO1 CaO
1 SiO21 SiO22 Si0.5O1
1 Na2O2 NaO0.51 Na2O
1 Al2O32 AlO1.53 Al0.67O1
1 [Mg2Si]3O43 [Mg0.67Si0.33]1O1.334 Mg0.5Si0.33O1
1 [NaAlSi3]5O85 [Na0.2Al0.2Si0.6]1O1.68 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
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 %):
Weight per centMolecular weightMolar units
SiO252.3960.0952.39 / 60.09 = 0.8719
Al2O314.38 101.9614.38 / 101.96 = 0.1410
CaO 8.4556.088.45 / 56.08 = 0.1507
Na2O 2.7861.98 2.78 / 61.98 = 0.0449
Weight per centMolecular weightCations
SiO252.3960.0952.39 / 60.09 = 0.8719
Al2O314.38 101.9614.38 / 101.96 x 2 = 0.2820*)
CaO 8.4556.088.45 / 56.08 = 0.1507
Na2O 2.7861.98 2.78 / 61.98 x 2 = 0.0897*)
*) If you think in terms of fractional “one-cation” formulas, then you can use the “half-molecular” 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 K-feldspar 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 KAlSi3O8 = 1 KO0.5 + 1 AlO1.5 + 3SiO2 is not balanced chemically, they would rather write (see definitions of fractional formulas in the introduction):
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 K-feldspar 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 K-feldspar (KAlSi3O8) into leucite (KAlSi2O6) and quartz (SiO2). In a two-step manner:
balanced equation: 1 KAlSi3O8 = 1 KAlSi2O6 + 1 SiO2
# cations in each molecule: 5 4 1
we obtain the coefficients: 5 K-feldspar = 4 leucite + 1 quartz. Therefore, 20.5 cations of K-feldspar 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:1 KAlSi3O8 = 1 KAlSi2O6 + 1 SiO2
# moles on each side: 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)
Example 8: Convert the following norm from cation units into oxygen units: forsterite 17, diopside 24, anorthite 37 and albite 22 cations.
Formula # cations # oxygensconversion
factor
ForsteriteMg2SiO434 4/3 = 1.333
DiopsideCaMgSi2O6466/4 = 1.5
AnorthiteCaAl2Si2O8588/5 = 1.6
AlbiteNaAlSi3O8588/5 = 1.6
cation unitsconversion oxygen units
Forsterite1717 x 1.333 = 22.714.83
Diopside2424 x 1.5 = 36.023.51
Anorthite3737 x 1.6 = 59.2 38.67
Albite2222 x 1.6 = 35.222.99
TOTAL 100 153.1100
Example 9: Convert the same norm as in #7 from cation units to weight per cent:
formula molecular # cationsconversion factor
weight (m.w. / cat’s)
ForsteriteMg2SiO4 140.714 140.71 / 4 = 35.18
DiopsideCaMgSi2O6 216.574216.57 / 4 = 54.14
AnorthiteCaAl2Si2O8 278.225278.22 / 5 = 55.64
AlbiteNaAlSi3O8 262.245262.24 / 5 = 52.45
cation unitsconversion weight per cent
Forsterite1717 x 35.18 = 598.0611.70
Diopside2424 x 54.14 = 1299.3625.43
Anorthite3737 x 55.64 = 2058.6840.29
Albite2222 x 52.45 = 1153.9022.58
TOTAL 1005110 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!