Recalculation of Silicate and other oxide analyses.
Chemical analyses of silicates or any other mineral with oxygen in the anionic radical are
reported as charge balanced oxide components such as SiO2 (not SiO4 –2), Al2O3, etc. These
are determined in weight percent by the chemist and report that way. You can also look at the
weight percent as the number of grams of each component. These should total to 100% or 100
grams, when they don’t it means there was either an element present that was not analyzed for
or there was error in the analysis.
Once we have the chemical analysis we want to calculate the formula from this analysis.
This is especially important for solid solution series minerals like olivine and the feldspars and
other minerals where there is substitution of elements. The recalculation is particularly useful in
determining how much of the Al present is partitioned between the tetrahedral and octahedral
sites.
RECALCULATION
 The first step in this process is to calculate the moles of oxide. This is easily done by
dividing the wt.% by the molecular weight of the oxide.
 Next calculate the moles of O. For SiO2 you must multiply the moles of SiO2 by 2 to get the
moles of O in a mole of SiO2. For CaO the moles of O are the same.
 Then sum up the moles of O. We want the moles of O to equal the moles of O in the
stoichiometric formula so we must apply a correction according to the following formula
 Sum O x Correction Factor = Formula O or Correction Factor = Formula O/Sum O.
 Next multiply the moles of O column by this correction factor. The sum of the this new
column (moles of anion or O on the formula basis—which is 8 in this example).
 The final step requires requires calculating the number of cations in formula units. To do this
go back to the original oxide component and ratio the amount of cation to the amount of O.
For SiO2 this equals ½ , for CaO this is 1, for Na2O this number is 2. Simply multiply the
formula O by this to get formula cations.
Interesting things to notice here. See how Al and Si occupy the tetrahedral site and their
formula cations add up to 4. In a pure alkali feldspar without Ca present Si should be 3 and Al
1. However, remember there is a 1 for 1 coupled solid substitution of Ca and Al for (K, Na) and
Si. Notice in the example formula that the Al greater than 1 is equal to the moles of Ca. Also
notice that the moles of Ca, K, and Na are ready for plotting on a ternary diagram in this
example. That is because they add up to a formula value of 1. For some minerals like olivine
where the formula is (Mg, Fe, Ca)2 this will require renormalizing to 100% as is shown on the
right hand part of the example.
This also gives us a way to partition Al between tetrahedral and octahedral sites in
minerals like micas and amphiboles where it occurs in both. You simply make sure that the
tetrahedral site is filled (so that Si and Al add up to the formula amount) and then assign the
leftover to the octahedral site. This is really important in hornblendes as the amount of
tetrahedral Al is pressure dependent so it serves as a geobarometer.
SHORTCUT METHOD FOR PLOTTING ENDMEMBER COMPOSITIONS
If you just want to plot a mineral on a binary or ternary diagram, you don’t have to go through
the entire recalculation. Just take the end members you are interested in and find the number of
oxide component moles. Then determine the moles of cations by multiplying this by the number
of cation moles in the oxide formula (Ca is CaO times 1, K is K2O times 2). Then sum up the
result and normalize to 100% and you’re ready to plot.