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Appendix 1 Calculation of a Chemical Formula from a Mineral Analysis

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Appendix 1 Calculation of a Chemical Formula from a Mineral Analysis Appendix 1 Calculation of a chemical formula from a mineral analysis Appendix 1 Magnesiohornblende analysis 3 4 2 Atomic proportion No. of anions on 1 Molecular of oxygen from basis of 24 (O,OH) 5 Wt.% of oxides proportion of oxides each molecule i.e. col. 368.3735 No. of ions in formula SiO 51.63 0.8594 1.7188 14.392 Si 7.196 2 8.00 0.804 } Al2O3 7.39 0.0725 0.2175 1.821 Al 1.214 0.410 3+ Fe2O3 2.50 0.0157 0.0471 0.394 Fe 0.263 FeO 5.30 0.0738 0.0738 0.618 Fe2+ 0.618 5.07 MnO 0.17 0.0024 0.0024 0.020 Mn 0.020 } MgO 18.09 0.4489 0.4489 3.759 Mg 3.759 CaO 12.32 0.2197 0.2197 1.840 Ca 1.840 2.00 Na2O 0.61 0.0098 0.0098 0.082 Na 0.164 } H2O+ 2.31 0.1282 0.1282 1.073 OH 2.146 2.15 Total 100.32 2.8662 24 = 8.3735 2.8662 The procedure for calculating a chemical formula is Column 5 gives the number of cations associated described by means of the above example, a with the oxygens in column 4. Thus for SiO2 there is magnesiohornblende. one silicon for two oxygens so the column 4 entry is divided by 2. For A12O3 there are two aluminiums for Column 1 lists the composition of the mineral every three oxygens so the column 4 entry is multiplied expressed in the usual manner as weight percentages by ~˜. For divalent ions the column 5 value is the same ¯ of oxides. as that of column 4, and for monovalent ions (including hydrogen) the latter is doubled in column 5. Column 2 is derived by dividing each column 1 entry The numbers of ions on the basis of 24 oxygens by the molecular weight of the oxide concerned (see given in column 5 can be grouped as shown to conform Appendix 2). The figures so obtained therefore express to a structural formula. In the present example it is the molecular proportions of the various oxides. assumed that the tetrahedral sites which are not filled by Si are occupied by Al, and the remaining Al atoms are Column 3 is derived from column 2 by multiplying by in octahedral coordination. the number of oxygen atoms in the oxide concerned. It It should be noted that a chemical analysis in itself thus gives a set of numbers proportional to the numbers can give only the ratios of atoms in the formula, and of oxygen atoms associated with each of the elements that the actual numbers of atoms given depends on an concerned. At the foot of column 3 is its total (T). assumption about the actual number of one of them or If we require the amphibole formula based upon 24 of a group of them. A check of the correctness of the oxygen atoms (this represents half the content of the formula can be made if the cell volume and density are unit cell) we need to re-cast the oxygen atom accurately known, since a calculated density can then be proportions so that they total 24. This is done by compared with that measured. multiplying all of them by 24/T and the results are A check of charge balance, made by adding positive given in Column 4. and negative charges in the formula, is a check only on 485 Downloaded from http://pubs.geoscienceworld.org/books/book/chapter-pdf/3764459/9780903056434_backmatter.pdf by guest on 29 September 2021 Appendix 1 arithmetic and not on the quality of the analysis. This is The procedure for obtaining column 2 is as before, because any analysis expressed in terms of neutral the fluorine content being divided by 19, the atomic oxides must lead to numbers of cations and oxygens weight of fluorine. For column 3 the number of fluorine which balance electrically. atoms is inserted along with the oxygens and the total is In many silicates, as in the example above, the only again too high. It is necessary to subtract the oxygen anion in the mineral is oxygen (or OH). Each element is equivalent of the fluorine atoms, i.e. half their number, expressed (and generally directly determined) as a to give a true total. weight percentage of oxide, even though the oxides do In the case of fluorphlogopite, the number of anions not exist as such in the mineral. The calculation (O,OH,F) assumed is 24, so that the total of column 3 is procedure outlined is justifiable, as each element can divided into 24 to give the multiplying factor which is be thought of as associated with its appropriate share of applied to produce column 4. Column 5 is derived as the oxygen atoms in the crystal structure. before. If oxygen is not the only anion present the Many modern analyses are carried out by use of an calculation is somewhat more complicated, and an electron microprobe and this method does not yield 2+ example (a fluorphlogopite) is shown below. values for H2O and does not differentiate between Fe Here fluorine is shown, as well as the oxides of all and Fe3+. the cations, as a weight percentage. We may assume for If H2O has not been determined or is thought to be simplicity that the fluorine atoms in the structure are unreliable the mineral formula can be calculated on an bonded to magnesium atoms only, and yet the same anhydrous basis assuming the (OH) content to be ideal. atoms of magnesium are recorded as combined with Thus for the magnesiohornblende example above, oxygen in MgO. Thus an excess of oxygen is recorded instead of using 24(O,OH) to derive the scaling factor, and the total will exceed 100%. To obtain a real total 23(O) equivalents [i.e. 22(O) + 2(OH)] can be assumed (which is a measure to some extent of the accuracy of as associated with all the cations apart from hydrogen. the data), an oxygen equivalent of the fluorine atoms If Fe3+ has not been determined, total iron is usually must be subtracted. presented as an FeO equivalent. There are many ways of One excess oxygen atom is recorded for each two gaining an estimate of Fe3+ and Fe2+,mostofwhichare atoms of fluorine present, so that the oxygen equivalent referred to by Droop (1987). The most appropriate of a fluorine by weight is obtained by multiplying the method depends upon the mineral concerned. Some fluorine content by the factor methods re-assign Fe to Fe2O3 and FeO in the way which gives the ideal total for all the cations or for atomic weight of oxygen 16 ; i.e. particular groups. Some adjust Fe so that trivalent ions 2 Â atomic weight of fluorine 38 in octahedral sites balance the Al replacing Si in tetrahedra. All methods depend on assumptions which The oxygen equivalent of the fluorine weight is may or may not be warranted and all may give only an subtracted from the total of column 1 to give a true total. approximate result. Fluorphlogopite analysis 3 4 2 Atomic proportion No. of anions on 1 Molecular of oxygen from basis of 24 (O,OH) 5 Wt.% of oxides proportion of oxides each molecule i.e. col. (3)68.500 No. of ions in formula SiO2 41.18 0.6854 1.3708 11.651 Si 5.826 Al2O3 12.52 0.1228 0.3684 3.131 Al 2.088 TiO2 0.99 0.0124 0.0248 0.211 Ti 0.105 FeO 0.30 0.0041 0.0041 0.034 Fe 0.034 MnO 0.04 0.0005 0.0005 0.004 Mn 0.004 MgO 27.32 0.6779 0.6779 5.762 Mg 5.762 Na2O 0.88 0.0142 0.0142 0.121 Na 0.241 K2O 11.93 0.1266 0.1266 1.076 K 2.152 H2O+ 1.06 0.0588 0.0588 0.500 OH 1.000 F 6.74 0.3547 0.3547 3.113 F 3.113 102.96 3.015 –O:F 2.84 À0.1773 = 760.3547 Total 100.12 2.8235 24 = 8.500 2.8235 486 Downloaded from http://pubs.geoscienceworld.org/books/book/chapter-pdf/3764459/9780903056434_backmatter.pdf by guest on 29 September 2021 Appendix 2 Further reading Droop, G.T.R. (1987) A general equation for estimating Fe3+ concentrations in ferromagnesian silicates and oxides from microprobe analyses using stoichiometric criteria. Mineralogical Magazine, 51, 431À435. Appendix 2 Appendix 2 Atomic and molecular weights for use in calculations of mineral formulae from chemical analyses. Al2O3 101.96 H2O 18.015 S 32.06 B 10.81 HfO2 210.49 SO3 80.06 B2O3 69.62 K2O 94.20 Sc2O3 137.91 BaO 153.33 La2O3 325.81 SiO2 60.08 BeO 25.01 Li2O 29.88 SnO 134.71 CO2 44.01 MgO 40.30 SrO 103.62 CaO 56.08 MnO 70.94 Ta2O5 441.89 CeO2 172.12 MnO2 86.94 ThO2 264.04 Ce2O3 328.24 Mn3O4 228.81 TiO2 79.87 Cl 35.45 Na2O 61.98 UO2 270.03 CoO 74.93 NiO 74.69 U3O8 842.08 Cr2O3 151.99 Nb2O5 265.81 V2O5 181.88 CuO 79.55 P2O5 141.94 Y2O3 225.81 F 19.00 PbO 223.20 ZnO 81.38 FeO 71.84 Rb2O 186.94 ZrO2 123.22 Fe2O3 159.69 487 Downloaded from http://pubs.geoscienceworld.org/books/book/chapter-pdf/3764459/9780903056434_backmatter.pdf by guest on 29 September 2021 Appendix 3 End-member calculations Appendix 3 For some purposes it can be useful to express the and Fe2SiO4(Fa) may suffice, but the Mn2SiO4 (tephroite formula of a mineral which can show a range of end-member) could also be calculated.
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