1. Geology 2217 Lab. 1. Recalculation of chemical analyses.
For a number of purposes such as the study of igneous rocks, ore deposits and a number of other projects, it is necessary to recalculate chemical analyses of minerals in terms of the structural formulae of the minerals from which they were obtained. A structural formula is a representation of the chemical formula of the mineral in terms of the atomic positions occupied. The main purpose of this type of calculation is to show how minor elements substitute for the principal components of the mineral.
There are two kinds of calculations:
1. Analyses of minerals that do not contain oxygen in their formulae, such as native elements (e.g. gold, PGEs) and sulfides and similar minerals (e.g. pyrrhotite, arsenopyrite).
2. Analyses of oxygen-containing minerals, such as silicates, oxides and carbonates.
Case 1: Minerals not containing oxygen in the formula
In these minerals, all components are analyzed, typically by some type of microbeam technique. Consider an analysis of kesterite, which has an end-member composition of
Cu2ZnSnS4 but usually has substantial substitution of Fe for Zn giving Cu2(Zn,Fe)SnS4. The parentheses around Zn,Fe indicates that both atoms occupy the same structural position in the atomic structure, hence are a solid solution. The total number of atoms in the mineral is then eight.
element wt % element Zn Fe Cu Sn S
total
Note that this is a hypothetical analysis. Only by chance would a real analysis total exactly 100%.

2. Step 1: Divide the wt% element by the gram-atomic wt in each case.
wt% element/gram-atomic wt = atomic proportion
element wt % element atomic prop. Zn Fe Cu Sn S
total
Step 2: Calculate the stoichiometric factor = no. of atoms in formula/sum of atomic proportions
8 atoms/1.756 = 4.556
Step 3: Multiply each atomic prop. by the atomic factor
element wt % element atomic prop stoichiometric coefficients Zn Fe Cu Sn S
The structural formula is then Cu1.93(Zn0.72Fe0.24)Σ 0.96Sn1.25S3.85
Note several points:
1. Since stoichiometric coefficients for Zn and Fe have been calculated, they are not separated by a comma.
2. Decimals are preceded by a zero.
3. Rounding is carried out only in the final result, not in the intermediate steps.
4. The same number of decimal places for each element are given in the result.
In examining the result, we can see that it is not in good agreement with the general structural formula Cu2(Zn,Fe)SnS4.
This is an indication that there was a problem with the analysis, as seen in the errors in the proportions of the components.
Note also that if we had assumed that there were four sulfur atoms in carrying out the calculation, the errors in the
final proportions would have been different and perhaps obscured.

3. Case 2: Minerals containing oxygen in the formula
Oxygen cannot presently be determined by normally available means, nor with sufficient accuracy even with special techniques. In the past, minerals were analyzed by “wet chemical” methods in which the results were obtained in terms of constituent oxides. For example, enstatite MgSiO3 would be analyzed as MgO and SiO2. Now, oxygen-containing minerals are generally analyzed by microbeam methods that measure a signal from the heavier atoms, in the above example, from Mg and Si. By means of a computer program these are calculated as the equivalent percentage of oxides. So, as oxygen is not measured, in converting an analysis in terms of oxides to a structural formula, the number of oxygen atoms is taken from the ideal formula. In the case of enstatite, the number is three.
Another problem that arises is that many minerals contain hydrogen in the form of OH. It is possible to analyze for hydrogen, but a separate method is required. Hence, in hydrous minerals (those that contain OH), the analysis often omits this component, and an accurate analysis will yield a low total. This is illustrated in the example below for muscovite KAl2(AlSi3)O10(OH)2. This is an ideal structural formula, in which the Al in the Si site is indicated. But muscovite may contain minor element substitutions, specifically Na for K, Mg, Fe2+ and Fe3+ in the Al2 site and F and Cl for OH.

4. Step 1: Divide the wt% oxide by the gram-formula wt oxide for each component oxide to give the molecular proportion oxide.
oxide wt% oxide g-formula wt oxide mole prop.
SiO
Al2O
TiO
FeO
Fe2O
MgO
Na2O
K2O
H2O
total 99.28

5. Step 2: Multiply the molecular proportions by the number of oxygens in the oxide to give the oxygen proportion and by the number of cations in the oxide to give the cation proportion.
g-formula
oxide wt% oxide wt oxide mole prop oxygen prop.
SiO x =1.5140
Al2O x =1.1103
TiO x =0.0092
FeO x =0.0103
Fe2O x =0.0072
MgO x =0.0124
Na2O x =0.0121
K2O x =0.0954
H2O x =0.2397
total 99.28
Step 3: Total the oxygen proportion and then divide the number of oxygens in the formula by the sum of the oxygen proportions to give the oxygen factor. (This step has the effect of adjusting the number of cations to the number of oxygens in the formula.)
oxygen proportion cation proportion
SiO x = x =0.7570
Al2O x = x =0.7402
TiO x = x =0.0046
FeO x = x =0.0103
Fe2O x = x =0.0048
MgO x = x =0.0124
Na2O x = x =0.0242
K2O x = x =0.1908
H2O x = x =0.4794
12 oxygens/ = oxygen factor

6. Step 4: Multiply each of the cation proportions by the oxygen factor to give the number of cations. These are the stoichiometric coefficients in the mineral formula.
SiO x = x =
Al2O x = x =
TiO x = x =
FeO x = x =
Fe2O x = x =
MgO x = x =
Na2O x = x =
K2O x = x =
H2O x = x =
99.28
Step 5: Construct the structural formula using the stoichiometric coefficients calculated in Step 4.
Some intuition is needed in order to put all components in the correct position. As well, there are some conventions to be learned, e.g. for feldspars and micas.
(K0.761Na0.097)Σ0.858(Al1.987Mg0.049Fe Fe )Σ2.096(Al Si3.017)Σ4.000(OH)1.911
Looking at the formula, it is apparent that there is a deficiency in both the large cation site and in the hydroxl ion site. It may be a matter of some components not sought in the analysis such as Ca for the large cation site or Cl and F in the hydroxyl site. In the latter case, missing Cl and F could allow extra OH to enter the large cation site. This changes the optical properties of muscovite producing sericite, not a mineral but something we see in thin section.

7. If H2O had not been analyzed, as is often the case, the formula can be calculated on an anhydrous basis. In this case, instead of 12 oxygens in muscovite, one makes the calculation with 11. This is because the two negative charges on the missing hydroxyl ions must be compensated by the -2 charge on an oxygen ion. Following the steps:
Step 1: Divide the wt% oxide by the gram-formula wt oxide for each component oxide to give the molecular proportion oxide.
oxide wt% oxide g-formula wt oxide mole prop.
SiO
Al2O
TiO
FeO
Fe2O
MgO
Na2O
K2O
total 94.96
Note that the low total reflects the missing hydroxyl ions. All hydroxyl-containing minerals will have low total in a correct microbeam analysis.
Step 2: Multiply the molecular proportions by the number of oxygens in the oxide to give the oxygen proportion and by the number of cations in the oxide to give the cation proportion.
oxide wt% oxide wt oxide mole prop oxygen prop.
SiO x =1.5140
Al2O x =1.1103
TiO x =0.0092
FeO x =0.0103
Fe2O x =0.0072
MgO x =0.0124
Na2O x =0.0121
K2O x =0.0954

8. Step 3: Total the oxygen proportion and then divide the number of oxygens in the formula by the sum of the oxygen proportions to give the oxygen factor. (This step has the effect of adjusting the number of cations to the number of oxygens in the form
oxygen proportion cation proportion
SiO x = x =0.7570
Al2O x = x =0.7402
TiO x = x =0.0046
FeO x = x =0.0103
Fe2O x = x =0.0048
MgO x = x =0.0124
Na2O x = x =0.0242
K2O x = x =0.1908
11oxygens/ =
Step 4: Multiply each of the cation proportions by the oxygen factor to give the number of cations. These are the stoichiometric coefficients in the mineral formula.
SiO x = x =
Al2O x = x =
TiO x = x =
FeO x = x =
Fe2O x = x =
MgO x = x =
Na2O x = x =
K2O x = x =
The formula is then constructed as above and (OH)2 is added, without decimal places, as the coefficient is not calculated, but is assumed.
(K0.757Na0.096)Σ0.853(Al1.994Mg0.049Fe Fe )Σ2.053(Al Si3.005)Σ4.000(OH)2
Note that there are no decimal places given of the 2 for OH.
The result is only slightly different from the previous calculation, but, again, Cl and F may be missing.