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UT Arlington GEOL 2313 - Minerals and Rocks Homework 3

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GEOL2313 – Minerals and Rocks G. Mattioli and D. Zachry, Dept. of Geosciences, Univ. of Arkansas, Fall 2008 Homework Assignment 3 – Calculation of CIPW Norm Due in Class November 17, 2008 Problem 1 There are several different ways to calculate the normative mineralogy of an igneous rock. This type of petrologic analysis tool was first developed by W. Cross, J.P. Iddings, L.V. Pirsson, and H.S. Washington, and hence is referred to as the CIPW norm. The method outlined below ignores mineral solid solutions and is taken from Appendix B of Hyndman’s Petrology of Igneous and Metamorphic Rocks. Other methods have been developed that include minor and trace elements not normally reported in older texts or standard XRF whole rock major element analyses. Required data for the normative minerals are shown in Table 1 and the data for this example are shown in Table 2. Using the whole rock compositions given in the spreadsheet on the web (http://comp.uark.edu/~mattioli/geol_2313/CIPW_norm.xls), calculate the CIPW norm for the Granite, Diorite, Gabbro, and Ijolite compositions. Please show all work and be as neat as possible. Follow the step by step instructions below. Problem 2 Discuss the utility of the CIPW norm. Why was it devised and how is used today? Why is it impossible for olivine and quartz to appear in the norm of the same rock? Table 1 Chemical formulae for normative mineralsInstructions A step-by-step calculation of the CIPW norm from a chemical analysis of a granite sample from the Idaho batholith (num. 134.15) is shown below. Oxide Wt. % Mol. Wt. Mol. % (with most calculated values subtracted as an example) SiO2 70.66 60.09 1.1759 – 0.2262 – 0.4608 – 0.0612 = 0.4277* TiO2 0.24 79.90 0.0030 – 0.0030 = 0 Al2O3 15.99 101.96 0.1568 – 0.0377 – 0.0768 – 0.0306 = 0.0117 Fe2O3 0.76 159.69 0.0048 – 0.0047 = 0 FeO 0.70 71.85 0.0097 + 0.0004 – 0.0030 – 0.0047 = 0.0024 MnO 0.03 70.94 0.0004 – 0.0004 = 0 MgO 0.47 40.30 0.0117 CaO 1.81 56.08 0.0323 – 0.0017 = 0.0306 – 0.0306 = 0 Na2O 4.76 61.98 0.0768 – 0.0768 = 0 K2O 3.55 94.20 0.0377 – 0.0377 = 0 P2O5 0.07 141.95 0.0005 – 0.0005 = 0 H2O 0.37 Table 2 Oxide weight percentages, molecular weights, and mols for example (* does not include subtracted value beyond step 7; final value = 0.4137) 1) Divide the weight percentage (wt. %) of each oxide by its molecular weight (Table 2) to obtain the number of molecules (mols or molecular proportion) of each oxide. 2) Add mol. prop. of MnO to that of FeO (e.g. 0.0101 here). 3) Calculate normative apatite, which equals mol. prop. P2O5 (e.g., ap = 0.0005 here). Note that Ca is also in apatite, you must subtract the equivalent number of moles of CaO (3.33 times that of P2O5) from the original CaO mol. prop. (3.33 x 0.0005 = 0.0017). 4) Calculate normative ilmenite, which equals the mol. prop. of TiO2 that does not exceed the available amount of FeO, as it also is required to form stoichiometric ilmenite (e.g., il = 0.0030). You must also subtract the equivalent mol. prop. of FeO from the original FeO mol. prop. If the mol. prop. of TiO2 is greater than that of FeO, excess TiO2 will remain. This will be used to calculate normative sphene or “titanite” (e.g. tn = 0 in this example). Again, since CaO (see Table 2) is also in sphene, you must subtract an amount of CaO equal to the TiO2 used to make sphene. All remaining TiO2 is assigned to make rutile.5) Calculate normative orthoclase, which equals the mol. prop. of K2O ( or = 0.0377 here). Subtract an equal amount of Al2O3 and six times the mol. prop. of K2O from the original amount of SiO2 as these are the appropriate stoichiometric proportions of these components to make orthoclase. In rare cases of excess K2O over the amount of Al2O3, assign this to normative potassium metasilicate or kalsilite (ks = 0 in this example). Subtract the equivalent amount of mols from the mol. prop. of SiO2. 6) Calculate normative albite, which equals the mol. prop. of Na2O (ab = 0.0768). Similar to step (5) above, subtract a similar amount of Al2O3 and six times the mol. prop. of Na2O from the original amount of SiO2 as these are the appropriate stoichiometric proportions of these components to make albite. If there is excess of Na2O over Al2O3, assign it to normative acmite, which equals the excess mol. prop. of Na2O or Fe2O3, whichever is greater (ac = 0 here; note that acmite is the same as aegirine, NaFeSi2O6). Subtract this amount of Na2O and mol. prop. Fe2O3 and 4 times this amount from the mol. prop. of SiO2. Excess Na2O is assigned to normative sodium metasilicate (ns = 0 here). Subtract this amount from mol. prop. SiO2. Any excess Fe2O3 should be assigned to normative magnetite (see step 8 below). 7) Calculate normative anorthite, which equals the excess CaO left over after forming titanite in step 4 (here an = 0.0306). Subtract CaO used to make anorthite from the remaining CaO; subtract the same amount of Al2O3 and 2 times that amount of SiO2, corresponding to the appropriate stoichiometry for anorthite. If there is any remaining excess Al2O3, assign it to normative corundum (c = 0.0117 here). Any excess CaO is used to make normative diopside and wollastonite in step 10 below (both equal 0 here). 8) Calculate normative magnetite, which corresponds to the mo, prop. of Fe2O3 or any excess Fe2O3 left over after formation of acmite in step 6 above (mt = 0.0047 here). Subtract an equal amount of FeO to yield the appropriate stoichiometry for magnetite. If there any excess Fe2O3 remains after forming magnetite and using all the FeO, then it should be assigned to form normative hematite (hm = 0 here). 9) Calculate the mol. prop. MgO/(MgO + remaining FeO). This is referred to as the Mg# and for the example here is 0.0116/(0.0116 + 0.0024), which equals 0.8286.10) Calculate normative diopside, which is equal to the mol. prop. of CaO remaining after making anorthite in step 7 (di = 0 here). An equal amount of MgO + FeO is allocated to diopside, maintaining the appropriate Mg# as calculated in step 9. Subtract amounts of MgO and FeO from their totals. Subtract 2 times the amount of CaO from SiO2. If CaO remains after


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