How Many Molecules Pyrite Cube weighs 778 g how many molecules is that About 4 000 000 000 000 000 000 000 000 Are they ALL Iron and Sulfur Stoichiometry Some minerals contain varying amounts of 2 elements which substitute for each other Solid solution specific elements substitute for each other in the mineral structure defined in terms of the end members species which contain 100 of one of the elements Chemical Formulas Subscripts represent relative numbers of elements present Parentheses separate complexes or substituted elements that go into a particular place Fe OH 3 Fe bonded to 3 separate OH groups Mg Fe SiO4 Olivine group mineral composed of 0 100 of Mg 100 Mg Fe Goldschmidt s rules of Substitution 1 The ions of one element can extensively replace those of another in ionic crystals if their radii differ by less than about 15 2 Ions whose charges differ by one may substitute readily if electrical neutrality is maintained if charge differs by more than one substitution is minimal Goldschmidt s rules of Substitution 3 When 2 ions can occupy a particular position in a lattice the ion with the higher charge density forms a stronger bond with the anions surrounding the site 4 Substitution may be limited when the electronegativities of competing ions are different forming bonds of different ionic character KMg3 AlSi3O10 OH 2 phlogopite K Li Al 2 3 AlSi3O10 OH 2 lepidolite KAl2 AlSi3O10 OH 2 muscovite Amphiboles Ca2Mg5Si8O22 OH 2 tremolite Ca2 Mg Fe 5Si8O22 OH 2 actinolite Actinolite series minerals K Na 0 1 Ca Na Fe Mg 2 Mg Fe Al 5 Si Al 8O22 OH 2 Hornblende Compositional diagrams FeO wustite Fe3O4 magnetite Fe2O3 hematite A Fe O A1B1C1 A1B2C3 x x B C Si fayalite forsterite enstatite Fe Mg fayalite Fe ferrosilite forsterite Mg Pyroxene solid solution MgSiO3 FeSiO3 Olivine solid solution Mg2SiO4 Fe2SiO4 Minor trace elements Because a lot of different ions get into any mineral s structure as minor or trace impurities strictly speaking a formula could look like Ca0 004Mg1 859Fe0 158Mn0 003Al0 006Zn0 002Cu0 001Pb0 000 01Si0 0985Se0 002O4 One of the ions is a determined integer the other numbers are all reported relative to that one Normalization Analyses of a mineral or rock can be reported in different ways Element weight Analysis yields x grams element in 100 grams sample Oxide weight because most analyses of minerals and rocks do not include oxygen and because oxygen is usually the dominant anion assume that charge imbalance from all known cations is balanced by some of oxygen Number of atoms need to establish in order to get to a mineral s chemical formula Technique of relating all ions to one often Oxygen is called normalization Normalization Be able to convert between element weight oxide weight and of atoms What do you need to know in order convert these Element s weight atomic mass Si 28 09 g mol O 15 99 g mol SiO2 60 08 g mol Original analysis Convention for relative oxides SiO2 Al2O3 Fe2O3 etc based on charge neutrality of complex with oxygen using dominant redox species Normalization example Start with data from quantitative analysis weight percent of oxide in the mineral Convert this to moles of oxide per 100 g of sample by dividing oxide weight percent by the oxide s molecular weight Fudge factor is process called normalization where we divide the number of moles of one thing by the total moles all species oxides then are presented relative to one another Feldspar analysis Ca Na K 1 Fe Al Si 4 O8 Oxide wt in the of moles mineral of oxide in mole of 2 cations in of O determined the oxides in oxide in oxide by analysis mineral the mineral 2 oxide Atomic weight of oxide g mol SiO2 60 08 1 2 65 90 1 09687 73 83 Si Al2 O3 Fe 2O3 CaO Na2O K2O 101 96 159 68 56 08 61 96 94 20 2 2 1 2 2 3 3 1 1 1 19 45 1 03 0 61 7 12 6 20 0 19076 0 00645 0 01088 0 11491 0 06582 12 84 0 43 0 73 7 73 4 43 Al 1 48569 100 SUM Cation moles of moles of O Number of cations contributed moles of in by each ion in the sample cation mineral 4 73 83 147 66 2 95 3 25 68 0 87 0 73 15 47 8 86 38 52 1 30 0 73 7 73 4 43 1 03 0 03 0 03 0 62 0 35 125 44 200 38 Fe3 Ca2 Na K of moles Oxygen choosen 8 Ca0 73 Na15 47 K8 86 Fe 0 87Al25 68 Si73 83O200 38 Ca0 03 Na0 62K0 35 Fe 0 03 Al1 03Si2 95 O8 to get here from formula above adjust by 8 200 38
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