Mineral recalculation

Mineral Recalculation

Methodology

The calculation of mineral chemistry from probe analyses entails the conversion of oxide or element data to moles of cations and oxygen and then renormalization on the basis of a given number of cations or oxygens. In analyses by electron microprobe, iron is assumed to be Fe2+ (FeO), but many minerals (e.g., amphibole, pyroxene) contain Fe3+ also. One can make an attempt at determining the proportion of the two oxidation states of Fe by using cation formulation; however, the validity of the results depend upon the overall quality of the analysis. Minor errors in the determination of SiO2, for example, can produce large changes in the calculated Fe2O3. This will be covered in more detail below.

To determine the mole of cations present, one divides the measured oxide weight percents by the cation-equivalent molecular weights given in the table below. The numbers in the table take into account the fact that some oxides contain two or more cations. Moles of oxygen are calculated assuming that all cations are bonded to oxygen. Thus, we may simply multiply the moles of cations by the oxygen factor to get moles of oxygen.

Cation-Mole and Oxygen Factors

The results of the molar calculations must next be recalculated. This is usually done by assuming a number of oxygen atoms in the mineral formula. for some minerals we can use charge balance and cation normalization to estimate Fe3+. Appropriate normalization numbers using oxygen (O) and cations for some minerals are given in the following table, which also gives site occupancies.

Mineral Normalizations and Sites

Using the normalized cation amounts, we fill the cation sites in the minerals. Cations are assigned in the order shown in the table for the structural formulae. For example, for amphibole, one first fills the T site, then the C site, then the B site, and finally, the A site. As an example, feldspar analyses are determined on an 8-oxygen basis: the cation conversion factor is 8 divided by the total moles of oxygen determined by calculation. All the cation (or oxygen) values are multiplied by this conversion factor to yield normalized molar amounts. In addition, if the concentrations of F and/or Cl have been determined, their abundances are divided by atomic weight (18.9984 and 35.453, respectively) and multiplied by the same conversion factor. Note! Site assignments are a subject of research and debate, and you may wish to use a different site filling scheme.

For analyses with F and/or Cl, we subtract the F and Cl values from the number for the OH-site to determine the OH content of the site. We can calculate the H2O content of minerals by multiplying the OH value by 9.0076 to get moles of H in the OH-site and dividing by the conversion factor. This number is added to analytical original total. The oxygen equivalents of F and Cl are calculated by multiplying moles of F by 0.4211 and moles of Cl by 0.2256, and the oxygen-equivalents for the F and Cl are subtracted from the analytical total.

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Oxygen normalization examples

Phlogopite mica

As an example of the recalculation procedure, consider the following microprobe data for a phlogopite (biotite) from the Miocene volcanic rocks of the Hickey Formation of central Arizona. Note that more digits than are significant are retained throughout the calculations and rounding only occurs at the end.

MicaDataTable

For mica, we want to recalculate atom concentrations on a 11-oxygen basis (10 oxygen atoms and 2 OH); the factor needed to convert 2.5083 into 11 is:

C.F. = 11 / 2.5083 = 4.3854.
Next we multiply all the cations by this conversion factor:

Mice Normalized Cations

Now we can fill the cation sites. First, we fill the T-site with Si and sufficient Al to bring its occupancy to 4.000. If the amount of Si exceeds 4.00 or if Si + Al are insufficient to fill the site (within a 2% error), the analysis is bad and should be discarded. We get 3.994 at the T-site occupancy in our mica.

(Si2.752Al1.243)T
Next, we fill the M-site with the remaining Al, Ti, and Fe, adding Cr, Mg, and Mn to bring the total to 3.000. The acceptable range is about 2.94 to 3.06 (±2% error); the data sum to 2.965.

(Ti0.124Fe0.617Cr0.035Mg2.220Mn0.005)M
Finally, we fill the A-site with Na, K, Ca, and Ba. The A-site should total to near 1.00; the data yield 0.967. It is possible that there are vacancies in the A-site, or that the mica is slightly altered, so we will not reject the analysis on the basis of A-site occupancy.

(Na0.100K0.853Ba0.014)A
We can now calculate the occupancy of the (OH) site. First, divide F by its atomic weight (18.9984) and multiply by the conversion factor. Subtracting the result from 2.000 determines the OH occupancy in the OH-site:

F = (3.02 / 18.9984) x 4.3854 = 0.697
OH = 2.000 – 0.697 = 1.303

So the site occupancy is

(F0.697Cl0.000OH1.303)OH
We can back-calculate the H2O content of the mica by multiplying the OH value by 9.0076 to get moles of H and dividing by the conversion factor:

H2O = (1.3029 * 9.0076) / 4.3854 = 2.68 wt. %
Finally, lets add this add this to the total above and subtract the oxygen-equivalents for the F:

(OH = F) = 3.02 * 0.4211 = 1.27
Total = 98.9 + 2.68 – 1.27 = 100.3

The site occupancies should probably not be reported to more than 3 significant digits:

(Na0.10K0.85Ba0.01)A(Ti0.12Fe0.62Cr0.03Mg2.22Mn0.01)M(Si2.75Al1.24)TO10(F0.70OH1.30)OH
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Amphibole

As another example, consider the following microprobe data for an an amphibole from K-rich dikes in the Mountain Pass area of eastern California.

Amphibole Data

For amphibole, we want to recalculate atom concentrations on a 23-oxygen basis (22 oxygens and 2 OH); the factor needed to convert 2.6541 into 23 is:

C.F. = 23 / 2.6541 = 8.6658.

Next we multiply all the cations by this conversion factor:

Amphibole Normalized Cations

Now we fill the cation sites. First, we fill the T-site with Si and sufficient Al to bring its occupancy to 8.000. If the amount of Si exceeds 8.00 or if Si + Al are insufficient to fill the site (within a 2% error), the analysis is bad and should be discarded. We get 8.000 for the T-site occupancy in our amphibole.

(Si7.995Al0.005)T

Next, we fill the B and C sites. It is not really possible to determine the independent occupancies in amphiboles (generally, larger cations, such as Na, are in the B site), so we fill them together. We use the remaining Al, Ti, Fe, Mg, Mn, Ca, and Na (in this order) to bring the B+C total to 7.000. The acceptable range is about 7.14 to 8.86 (±2% error); we can completely fill the site, so there is no problem.

(Al0.134Cr0.004Fe1.444Mg3.621Mn0.032Ca1.212Na0.553)B+C

Finally, we fill the A-site with the remaining Na, and K. The A-site should total between 0 and 1; the data yield 0.414.

(Na0.267K0.146)A

We can now calculate the occupancy of the (OH) site. First, divide F by its atomic weight (18.9984) and multiply by the conversion factor. Subtracting the result from 2.000 determines the OH occupancy in the OH-site:

F = (0.47 / 18.9984) * 8.6658 = 0.214
OH = 2.000 – 0.214 = 1.786

So the site occupancy is

(F0.214OH1.786)OH

We can back-calculate the H2O content of the mica by multiplying the OH value by 9.0076 to get moles of H and dividing by the conversion factor:

H2O = (1.786 * 9.0076) / 8.6658 = 1.86 wt. %

Finally, lets add this add this to the total above and subtract the oxygen-equivalents for the F:

(OH = F) = 0.47 * 0.42108 = 0.20
Total = 97.38 + 1.86 – 0.20 = 99.04

The total is acceptable (between 98% and 102%), It should recognized that the calculations above were done assuming that all iron is Fe2+. Amphiboles can contain significant amounts of Fe3+, which would change the calculation of oxygen, but it is not possible to determine the amount of Fe3+ using microprobe.

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Alkali Feldspar

As another example, consider an alkali feldspar (#1 in Table 37, The Rock-Forming Minerals, 2nd ed., Deer, Howie & Zussman, 1992). This is a wet-chemical analysis and we will ignore the water, because it probably reflects the fact that the analysis was done on a mineral separate and water occurred on the grain surfaces. Additionally, we will ignore the elements that are not normally found in a feldspar structure (TiO2, MgO) because these probably result from tiny inclusions in the feldspar grains analyzed. Here is a summary table for the calculation:

Feldspar Data Reduction Table

The correction factor is: 8 / 2.984 = 2.681. The corrected cations are given in the last column. The resulting formula is:

(Na0.73K0.19Ca0.06Ba0.01)A(Si2.93Al1.06Fe0.01)TO8
With feldspar, there are several internal checks we can make:

  1. The A site should total 1.00 … it’s 0.99
  2. The T site should total 4.00 … it’s 4.00
  3. Because of charge balance constraints, Al should equal (1.0 + Ba + Ca) = 1.07 … it’s 1.06
  4. For the same reason, Si should equal (2.0 + Na + K) = 2.92 … it’s 2.92.

We can conclude that this is a very good feldspar analysis.

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Cation normalization example

Augite pyroxene

In some cases, it may be appropriate to use cation normalization to determine the valence state of Fe. This will only work if the analysis is very good. As an example, we will use an augite microprobe standard, a material that was analysed very carefully by wet chemical analysis. We will combine FeO and Fe2O3 from the wet chemical analysis (2.66 and 5.70 wt. %, respectively) as would be the case with a microprobe analysis into total FeO. Using this analysis, let’s see if charge balance constraints and cation normalization will yield the same Fe2+/Fe3+ ratio as the wet chemical analysis. Our “microprobe” analysis of the this material is given in the table below.

In this case, we’ll normalize the cations to a total of 4.00. Our correction factor is:

4 / 1.7982 = 2.2244
We will multiply the cation values and the oxgyen values by this factor:

Augite Normalized Cations and Oxygen

Notice that the oxygen atoms do not total to 6 as would be expected from the pyroxene formula. Some oxygen is “missing,” indicating that some of the Fe is in the 3+ state. We can estimate how much by converting Fe2+ into Fe3+:

2FeO + O = Fe2O3
Notice that for every two Fe2+ cations we convert into Fe3+ with get to add an oxygen. We need to add 0.0170 more oxygens (6 – 5.9830) to bring the total to six, which indicates that we need to convert twice as much Fe2+ into Fe3+. This results in:

Fe3+ = 2 x 0.0170 = 0.0340
Fe2+ = 0.2269 – 0.0340 = 0.1929
The resulting ratio Fe2+/(Fe2+ + Fe3+) = 0.8502, so this fraction of the total iron should be expressed as FeO:

FeO = 0.8502 x 7.33 = 6.23 wt. %.
The remainder of the iron will be Fe2O3. We convert the remaining FeO into Fe2O3:

Fe2O3 = 1.1113 * (7.33 – 6.23) = 1.22 wt. %
These two numbers are close to the analytical values: 6.23 vs. 5.42 for FeO and 1.22 vs. 2.12 for Fe2O3. The reason that this result is not closer probably is because one or more of the other elements is not determined with sufficient accuracy. For example, if CaO were 22.0 wt. %, the results would be FeO = 5.70 and Fe2O3= 1.81… Conclusion? It is probably not a wise idea to trust microprobe analyses to determine the abundances of the two species of iron! Similar procedures can be used to recalculate spinels, for example, magnetite.

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