USING THE C.I.P.W. NORM METHOD (With Quartz in the Norm)

 

n  Convert all oxides to molecular amounts (moles), neglecting any less than 0.0001

n  Molecular amounts of MnO and NiO are added to that of FeO, and those of BaO and SrO to that of CaO

n  All molecular amounts for all oxides attributed to each mineral formed should be recorded on the spread sheet

n  Begin the allotment of oxides to the minerals
  1 An amount of CaO equal to 3.33 that of P2O5 (or 3.00 P2O5 and 0.33 F, if F is present in the
      norm) is allotted to apatite (got rid of P2O5 and F)
  2 An amount of FeO equal to that of Cr2O3 is allotted to chromite if Cr2O3 is present (got rid
      of Cr2O3)
  3 An amount of FeO equal to that of TiO2 is allotted for ilmenite (got rid of TiO2?)- any more
     TiO2 save it for # 9 below

       3b An equal amount of CaO with that of CO2, if present, to form calcite (got rid of CO2)
        4  Allot all the amount of ZrO2, if present, with an equal amount of SiO2 for zircon (got rid
            of ZrO2)
        5  Allot an equal amount of K2O with Al2O3 and 6 times that of SiO2 to form orthoclase (as is
             rarely the case, if there is an excess of K2O,over an equal amount of Al2O3, use an equal
             amount of it and SiO2 to form potassium metasilicate (got rid of K2O)
        6  Allot an equal amount of Na2O to that of Al2O3 and 6 times that of SiO2 to form albite

        6b If there is an excess of Na2O, above (not too often), use an equal amount of it with
             Fe2O3 and 4 times that amount of SiO2 to form acmite

      6c If there is still an excess of Na2O, it is used with an equal amount of SiO2 to form sodium
             metasilicate (got rid of the Na2O)

  7  If there is any remaining Al2O3 after forming orthoclase and albite above, allot an equal
       amount of it to CaO and 2 times the amount of SiO2 to form anorthite

           8 If there is an excess of Al2O3 after forming anorthite, it is used to form corundum (got rid of
                  Al2O3)

              9 If there is an excess of TiO2 # 3 above, allot an equal amount of it to an equal amount
                  of CaO, if any remains from 7 above, and an equal amount of SiO2 to form titanite
                  -use the remaining TiO2 to form rutile (got rid of the TiO2).  If there is any CaO
                   remaining, it is assigned to form diopside and/or wollastonite in #’s 12 and 13,
                   below.                    

           10  What usually happens, instead of forming acmite above in # 6b, an amount of Fe2O3 left
                    over from the formation of chromite and/or ilmenite in #’s 2 and 3 above equal to that of
                    FeO is used to form magnetite

           11 If there is any remaining Fe2O3, it is used to form hematite (got rid of Fe2O3)

                 AT THIS POINT, SUM ALL OF THE REMAINING AMOUNTS OF FeO AND
                 MgO LEFT FROM THE MINERAL ALLOTMENTS ABOVE AND
                 DETERMINE THE FRACTION OF EACH OF THE 2 OXIDES (DIVIDE

                     THE SUM INTO EACH INDIVIDUAL PORTION)

 

            12 Form diopside-- Allot an equal amount of FeO-MgO (sum) to that of CaO if remaining using
                    the proportionate amounts each of FeO and MgO calculated above. In other words, multiply
                    each fraction by the mole amount of CaO used to form diopside and place those numbers
                   each for FeO and MgO) in the column for diopside. The sum of FeO and MgO in the
                    column should equal the number of moles of CaO used. Multiply the amount of CaO by
                    2 and enter it in the SiO2 position

              13 If there is any CaO remaining from the formation of diopside it is used with an equal amount
                    of SiO2 to form wollastonite (got rid of CaO)

              14 If there is an excess of FeO-MgO remaining after forming diopside in # 12 above use it with
                    the same amount of SiO2 to form  hypersphene (got rid of FeO and MgO)

              15 Subtract the sum of SiO2 used for the formation of the minerals above from the total SiO2.
                     If there is an excess, the SiO2 is allotted to quartz (got rid of the SiO2)

 

             you are now finished with the allotments of oxide moles to specific minerals,
             now determine the weight % of each mineral in the rock which has
             quartz present

 

           16 Multiply the molecular weight of each balanced oxide of a mineral by the lowest (lower) mole
                    value of each oxide present in that mineral---do this for each mineral present. For instance,

                    the molecular weight of balanced orthoclase is 556 and magnetite is 232. The calculation for

                    diopside and hypersphene is more involved.  For diopside, one must add the total weight %

                    of CaO, FeO and MgO.  In other words, determine the molecular weight of CaO.SiO2
                          (116), FeO.SiO2 (132), MgO.SiO2 (100) and multiply each mol. weight by the appropriate
                    mole allotment of CaO, FeO, and MgO in the column for diopside. Add those 3 to obtain
                    the weight % of diopside.  For hypersphene, multiply 132 by the mole amount of FeO and
                    100 by the MgO amount and sum them for the weight % of hypersphene. For apatite
                    calculate the molecular weight from 3CaO.P2O5 and multiply it by the lower moles
                   (usually P2O5).     

 

 

                  You have just finished the calculation of the weight % of the mineral composition
               in the rock with quartz present in the Norm.