CIPW Norms Calculation Guide
Learn how to calculate CIPW norms for igneous rocks following specific steps involving oxide proportions and mineral allotments. This method, developed by Cross, Washington, Iddings, and Pirsson in 1931, helps convert rock compositions into ideal mineral compositions.
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CIPW Norms Calculation Rajnikant Patidar Department of Geology M.L. Sukhadia University, Udaipur
What is CIPW Norms The CIPW norm is named after the four petrologists, Cross, Washington, who devised it in 1931. Iddings, Pirsson and A norm is a means of converting the chemical composition of an igneous rock to an ideal mineral composition.
Steps for Norm Calculation Convert the weight % of oxides (except of H2O) into molecular proportion by dividing each value by their appropriate molecular weight. Amount of MnO is added to FeO. An An amount of CaO equal to 3 times that of P2O5 is alloted for apatite (ap).
An amount of FeO equal to that of TiO2 is alloted for ilmenite (il). If there is excess of TiO2 than equal amount of CaO is alloted to form sphene. An amount of Al2O3 equal to that of K2O with 6 times SiO2is alloted for orthoclase (or). An amount of Al2O3 equal to that of Na2O with 6 times SiO2is alloted for albite (ab).
If Al2O3remains after the allotment to orthoclase and albite then allot the amount of Al2O3equal amount of CaO and 2 times SiO2 to form anorthite (an). If there is insufficient Al2O3then the equal amount of Al2O3and Na2O will be taken for albite and remaining Na2O is reserved for acmite (ac). In this case anorthite will not form. If still excess Al2O3 remains, it will be allotted to corundum (c).
If there is excess of CaO over Al2O3then the remaining CaO is reserved for diopside and wollastonite (corundum will not form in such case). Form acmite (if excess Na2O is present) by using equal amount of Fe2O3and 4 times of SiO2. Remaining Fe2O3(after acmite) be allotted with equal amount of FeO to magnetite (mt).
If excess Fe2O3 present, allot to haematite (hm). Form diopside (di) by allotting remaining CaO with proportionate amount of MgO and FeO (if available) and SiO2. CaO (full amount), MgO & FeO in proportionate MgO = MgO/MgO+FeO X CaO FeO = FeO/MgO+FeO X CaO The reamining MgO & FeO will be used for hypersthene.
If there is excess of MgO and FeO remain after the allotment in diopside then allot it to hypersthene (hy). If the CaO ends after allotment to anorthite then the diopside will not form and remaining MgO and FeO will be allotted to hypersthene. It SiO2still available then form the quartz (q). If SiO2 become deficient then: Calculate the deficient amount of silica (D) If D is less than half of remaining MgO+FeO olivine (equal amount of D) is form and to hypersthene. then rest is allotted
Use right side of allotted values and finally calculate the norms by multiplying by the molecular weight of each normative mineral.
Exercises Analysis in wt% Value (1) Value (2) SiO2 TiO2 Al2O3 Fe2O3 FeO 72.82 61.21 0.28 0.70 13.27 16.96 1.48 2.99 1.11 2.29 MnO 0.06 0.15 MgO 0.39 0.93 CaO 1.14 2.34 Na2O K2O P2O5 3.53 5.47 4.30 4.98 0.07 0.21
Answers Norms Q Or Ab An C Di Hy Ol Mt il Ap Value in % (1) Value in % (2) 32.87 25.44 30.07 4.76 1.02 5.00 29.41 46.26 7.05 - - 2.14 2.06 1.34 - - 2.14 0.54 0.17 4.33 1.34 0.49
Exercise Harker Variation Diagram 1 2 3 4 5 6 49.20 57.94 61.21 65.55 68.55 72.82 SiO2 TiO2 Al2O3 Fe2O3 FeO 1.84 0.87 0.70 0.60 0.54 0.28 15.74 17.02 16.96 15.04 14.55 13.27 3.79 3.27 2.99 2.13 1.53 1.42 7.13 4.04 3.29 2.99 2.53 1.11 0.24 0.18 0.14 0.10 0.08 0.06 MnO 6.73 3.40 2.23 1.92 1.14 0.62 MgO 9.47 6.79 3.24 3.03 2.68 1.14 CaO 2.91 3.48 4.47 4.87 3.87 3.55 Na2O K2O P2O5 1.10 2.62 3.98 4.12 4.27 4.62 0.35 0.27 0.21 0.16 0.14 0.10
