In this lab you will research the chemical formulas and structure of the 10 common rock forming minerals, along with some new minerals introduced here.
You will need to find out:
· What is the chemical formula of the mineral?
· You should write this like this – Water = H2O
· The % oxides in each one (see below for methodology)
Calculating % oxides for elements in a mineral
Chemical analysis of minerals very often presents the composition of elements as weight percent (or wt%) of oxides rather than of individual elements. The reason for this appears to mostly be tradition, but it is standard across geology, so you should be able to understand what the numbers mean and how they are calculated.
Take diopside as an example - a pyroxene with the chemical formula MgCaSi2O6. Its chemical analysis is as follows:
SiO2 55.49
MgO 18.61
CaO 25.90
Total 100
This is how the chemical compositions of minerals are presented. How are these numbers calculated? Or, if we have these numbers, how is the chemical formula calculated?
First, some basic chemistry. Since all elements differ in size, we cannot use how much weight there is of each element, as for bigger elements this will mean fewer atoms. Instead we will use a standard measurement for the amount of the element known as a mole (notified as mol). A mole is the weight of 6.02214086 x 1023 atoms of a certain element. This number is also known as Avogadro’s Constant, and is defined by the number of atoms in 12 grams of Carbon-12 (the most common isotope of carbon). Larger elements (i.e. with more protons and neutrons) will have heavier molar weights than smaller ones; the same applies for molar weights of compounds or molecules.
We will start with calculating the wt% oxides from our chemical formula. The non-oxygen elements in the formula have a ratio of Si : Mg : Ca of 2 : 1 : 1. We will assume that the chemical formula is completely balanced (it does not have a residual positive or negative charge) and so the molar ration of Si : Mg : Ca must also be 2 : 1 : 1. Since we are using oxides however, we cannot use these ratios as they are. We need to convert the number of moles of the element into a mole weight oxide. A mole of an element is simply the atomic weight in grams. For silicon, this is 28.084g / mol, while for oxygen this is 15.999 g / mol. A mole of SiO2 therefore weights 28.084 + 15.999 + 15.999 = 60.082 g / mol.
The molar weight and ratios of diopside are:
SiO2 = 60.082 g / mol = 120.164 g in one mol of diopside (since there are two Si atoms)
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MgO = 40.3044 g / mol = 40.3044 g in one mol of diopside
CaO = 56.0774 g / mol = 56.0774 g in one mol of diopside
The total weight of a diopside mol is the sum of the weight of oxides in one mol – in this case it is:
Total formula molecular weight = 120.164 + 40.3044 + 56.0774 = 216.5458 g
To calculate wt% from this, all we need to do is show how much % of each individual oxide is the total formula weight. For SiO2 this is calculated so:
Wt% oxide = (120.164 / 216.5458) x 100
Wt% oxide = 55.49
The other components are as follows:
MgO Wt% oxide = (40.3044 / 216.5458) x 100 = 18.61
CaO Wt% oxide = (56.0774 / 216.5458) x 100 = 25.90
Total = 55.49 + 18.61 + 25.90 = 100
Now we will work backwards to find the chemical formula from some chemical analysis. We will use the same mineral as before to keep things simple.
The first step is to convert the oxide weight % into moles –
SiO2 wt % = 55.49
SiO2 molar weight = 60.082 g / mol
Moles of oxide = 55.49 / 60.082 = 0.923 mols
The other components are therefore as follows
Moles of oxide of MgO = 18.61 / 40.3044 = 0.462 mols
Moles of oxide of CaO = 25.90 / 56.0774 = 0.462 mols
The total number of moles of oxide that this mineral contains is 0.923 + 0.462 + 0.462 = 1.847. As a component of this, SiO2 is 50% (0.923 ˜ 1.847 / 2), and MgO and CaO are both 25% (0.462 ˜ 1.847 / 4). So the ratio of moles of oxide of SiO2 : MgO : CaO is 2 : 1 : 1.
In SiO2, there are 2 moles of oxygen for every 1 mole of Silicon (MgO and CaO both have one of each). So total moles of elements in this mineral are:
0.923 + 0.462 + 0.462 + (0.923 x 2) + 0.462 + 0.462
(2)Si + Mg + Ca + O (2xSiO2) + O (MgO) + O (CaO)
So our total ratio of Si : Mg : Ca : O is 2 : 1 : 1 : 6
We just need to fit this together into: MgCaSi2O6 – Diopside
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Why in this particular order? For silicate minerals, the Si and O components always go at the end of the formula (sometimes followed by OH or H2O if the mineral contains water in its crystal structure). The other components go in order of their atomic number. In this case, Mg = 12 while Ca = 20, so Mg occurs first in the formula.
Ideally you would want to balance the equation out to get rid of any decimal places, but some minerals will have a very small fraction of one element, and so you will occasionally see something like “F0.2” in a chemical formula for a mineral.
Part 1
Research the chemical formulas for the 10 rock forming minerals, as well as the new minerals listed below (a fantastic resources to use is www.mindat.org), and write them below. Indicate whether or not the mineral is hydrous (i.e. does it contain water in the chemical formula), and if the mineral contains multiple cations, indicate the end members of the series, as well as their names. To help you out, olivine has been completed for you.
Hydrous / Anhydrous? Chemical formula
Olivine: Anhydrous Fe2SiO4 (fayalite) to Mg2SiO4 (forsterite)
Quartz:
Plagioclase (use
the Ca and Na
end members):
Alkali Feldspar:
Muscovite:
Biotite*1:
Amphibole*2:
Pyroxene*3:
Garnet*4:
Calcite:
6
New minerals:
Hydrous / Anhydrous? Chemical formula
Chlorite (as
clinochlore):
Zircon:
Andalusite /
Kyanite /
Sillimanite:
*1 Biotite has a large range of types. Use Phlogopite and Annite as the two examples.
*2 Amphiboles have a very varied chemical composition – 76 end member minerals! For this exercise, please use the calcic amphibole ‘tremolite’, the iron amphibole ‘grunerite’ and the alkali amphibole ‘glaucophane’ as examples of amphibole chemical compositions (note that the table in Part 2 uses different amphiboles to these)
*3 Pyroxenes are a little less complicated, but still have many end members. In this exercise, use diopside, enstatite, jadeite and pigeonite as the pyroxene examples.
*4 Garnet too has many end members – but pyrope, almandine and grossular garnets can be used here as end member examples.
A note on iron:
Iron can be incorporated into minerals as either Fe2+ or Fe3+. It is not always easily to see which type of iron a mineral might contain, but the total number of O ions must balance out the rest of the cations so that the overall charge of the mineral is zero. In the example of fayalite, it contains 4 O atoms, giving it a total valance of -8. It also contains a single Si (+4) and two Fe. To balance out the equation perfectly, the Fe must be present as Fe2+. Mineral analysis will often put iron in as FeO (T) and not differentiate between them – in this case you will need to take this extra step to calculate the mineral formula.
A note on water and fluorine:
Water in minerals can be bound up as OH- ions or H2O molecules (and in some cases, a bit of both). Similar to Fe, you may need to play around with using OH- or H2O to balance your formula. Some molecules can substitute F- ions instead of OH- ions. For simplicities sake, we will assume none of the minerals we will use for these exercises contains fluorine.
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