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 re-calcuation.
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 mono-mineralic.
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
XRF - Uses Bragg's Law (n λ = 2 d sin Θ). By using crystals with known d-spacings, it
is possible to analyze for a specific wavelength or energy that is
characteristic of an element (e.g., λ
= 1.937Å for FeKα). The
energy of the diffracted beam is proportional to the number of
One can also excite X-rays 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
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 Na0.995Cl1.000. Errors cause small problems, however upon rounding off the the numbers, one can see the final formula is reported as NaCl.
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/cm3
Z = number of formula units (atoms / unit cell)
M = Molecular weight (a.m.u or g/mole)
N = Avogadro's number = 6.022 x 1023 (atoms/mole)
V = Volume of unit cell Å3
10-24 = Å3 --> cm3
We can rearrange the equation in terms of the total mass of the unit cell,
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 -
Unit cell of halite is Na4Cl4
By difference the percentage of oxygen = 53.2%, M = 15.9994
Unit cell of quartz is Si3O6