Klein 322332, 3334
Nesse 169  180
Chemical Compositions  One of the aims of elemental / chemical analysis is to calculate a structural formula and determine the distribution of cations amongst several structural sites. The process of converting chemical analysis to structural formulas is known a chemical recalcuation.
Types of elemental analysis. The choice of type of elemental analysis depends upon the concentration of the elements in the sample, the level of accuracy required, and availability of analytical equipment.
Accuracy and precision is also dependent upon the the "skill" of the analyst.
What one chooses to analyze is the most important decision. The purity of the sample is an important factor. There are several methods to hygrade or beneficiate samples, such that you can be assured that the sample analysis is monomineralic.
These techniques include:
A common reference:
Physical methods in determinative mineralogy (1977) ed. J. Zussman, Academic Press, London.
Dissolution methods  These methods involve dissolving the entire sample and then analyzing the concentration of dissolved ions. Concentrations can be measured with instruments such as:
Pros  allows accurate analysis of lighter elements,
Cons  Must be a good chemist. (often dealing with dilute solutions or one must dilution solutions, which propogates error).
EDS  the sample is the target, and the intensity of the
characteristic radiation for each atom type is proportional to its
abundance.
XRF  Uses Bragg's Law (nl =
2dsinQ). By using crystals
with known dsapcings,
it is possible to analyze for a specific wavelength or energy that
is characteristic
of an element (e.g., l =
1.937Å
for FeKa). The energy of the diffracted
beam is proportional to the number of atoms present.
One can also excite Xrays with high energy electron source (as in the Electron Mircoprobe). Small sampling areas (10 microns) are possible.
Pros  Quick, relative easy
Cons  Light elements are not amenable ( H, Li, B, C, O)
FTIR, NMR, Mossbauer, Optical.
Pros  provide additional information about particular elements and/or functional groups and coordination state.
Cons  limited access and very specific to particular elements
and/or
functional groups
One goal of elemental analysis is to obtain the chemical formula of a mineral. Therefore, we need to obtain the atomic proportions of the elements present.
Probably the most important aspect of quantitative elemental analysis is make sure the total of the individual measures add up to 100%!
For the case of native elements the measurement is simple. For example, Cu = 100%. Therefore, the intensity measurement from an XRF analysis can be made directly proportional to the amount of Cu present.
For polyatomic minerals the situation is a little more complex. Quantitative analysis requires calibration.
We know that there is one sodium atom for each chlorine atom in halite .
Given the atomic weights of Na = 22.9898 and Cl = 35.453, it is
easy to
show the relative weight percent of each element in the compound.
For a recalculation (i.e., working the problem "forward") the general rule is to normalize the chemical formula by the dominant anionic group(s) (Cl^{} in the case for halite and O^{=} in the case of the rock forming silicates ). Assuming there is some way to measure the weight percent of Na and Cl in halite, we can recalculate the atomic ratios. As seen in the table below the second column contains the measured values (Note that they don't add up to exactly 100%) The atomic proportions are determined by dividing the wt% by the atomic weight. Normalization by the number of anions yields the atomic ratios, which are the subscript coefficients in the chemical formula. The resultant formula is Na_{0.995}Cl_{1.000}. Errors cause small problems, however upon rounding off the the numbers, one can see the final formula is reported as NaCl.

Weight % 

Atomic Proportions 














Calculation of unit cell contents
It can be shown that the density of mineral is related to the
mass of
the atoms and number of atoms in the unit cell. This relation is
given by
the following equation;
where, D = density g/cm^{3}
Z = number of formula units (atoms / unit cell)
M = Molecular weight (a.m.u or g/mole)
N = Avogadro's number = 6.022 x 10^{23} (atoms/mole)
V = Volume of unit cell Å^{3}
10^{24} = Å^{3} > cm^{3}
Dimensional analysis:
We can rearrange the equation in terms of the total mass of the unit cell,
or
The amount of each constituent (Q) (i.e., element or oxide) is usually given as a percentage (q%) of the total mineral. That is every 100 a.m.u. of mineral contains q a.m.u. of Q or q/M atoms of Q. Therefore, by simple proportion the number of Q atoms in the unit cell is,
Best shown by example 
Halite
Unit cell of halite is Na_{4}Cl_{4}
Quartz
By difference the percentage of oxygen = 53.2%, M = 15.9994
Unit cell of quartz is Si_{3}O_{6}