C.I.P.W. NORM CALCULATION IF THERE IS A DEFICIENCY OF QUARTZ

After all the minerals have been formed, and there is a deficiency of SiO2, quartz cannot form and a redistribution or reallocation of the SiO2 needs to be accomplished.  This means that after the reallocation either some minerals formed will not be present, or if they are, the weight % will be different, or new minerals will form. In each of the separate redistribution steps below, if there is no SiO2 deficiency (sum of SiO2 used to form all minerals equals the total SiO2), the process of redistribution stops. This is true if the “d” path is used for a given step(1-3). If the “d” step is not used in a step (1-3), the process of redistribution will continue until a “d” step is used and there is no deficiency in SiO2.  The redistribution steps are given below.

Step 1.  Redistribution of SiO2 between new hypersphene and olivine

a.      Add all the SiO2 mole amounts used in the formation of minerals (general sum) excluding  that for hypersphene.

b.      Subtract the moles of SiO2 obtained in “a” from the total moles of SiO2 at the beginning of mineral formation. This difference is the available SiO2 for the redistribution.

c.        If this mole amount in “b” is greater than half the mole amount of the sum of FeO and MgO used for hypersphene initially, both hypersphene and olivine can be formed in “d” below and there is no further distribution.  If not, no hypersphene forms and only olivine forms. In this case there is a deficiency in SiO2 and there is further redistribution in “2” below.

d.      Form the number of new moles of hypersphene (x): x = 2S-M, where S is the moles of SiO2 in “c” above and M is the sum of FeO, MgO moles allotted to hypersphene initially. Place this number in the SiO2 column for hypersphene.  Multiply this number by the fraction of FeO and MgO and place these amounts in the respective rows for hypersphene.  Form the number of olivine moles: S-x and place this number in the SiO2 for olivine.  Subtract the number of FeO moles used in all minerals formed including the new hypersphene but not the initial hypersphene and subtract this from the beginning FeO mole amount.  Place this value in the FeO under Ol.  Do the same for MgO.  The sum of the MgO and FeO should equal twice the amount of silica alloted for Ol.  AT THIS POINT THE REALLOCATION STOPS BECAUSE THERE IS NOT A SILCA DEFICIENCY.  THE NORM CALCULATION IS FINISHED.

Step 2.   Form olivine, new albite and nephelite: The FeO and MgO used for initial
hypersphene is distributed for olivine, and ½ of the sum amounts of FeO and
MgO is allotted for SiO2. There is now no hypersphene present in the norm.

a. Add the mole amounts of SiO2 for all minerals formed up to this point
including olivine but not that for hypersphene and albite

b.      Subtract the mole amount in “a” from the total SiO2 present initially in the rock. This represents the amount of SiO2 moles available to continue.

c.       If the mole amount of SiO2 in “b” is greater than twice the amount of moles of Na2O allotted to albite initially, and it is 6 times less than the Na2O amount used initially, new albite and nephelite will form according to “d” below.  If not, there is still a deficiency of SiO2--- go to Step 3.

d.      Form new albite moles (x):  x = (S-2N)/4; where S is the amount of SiO2 moles available in “b”; N is the mole amount of Na2O used for initial albite. Place the calculated mole amount in the Na2O row for new albite.  Allot an equal amount of Al2O3 and 6 times the Na2O for the amount of SiO2 moles.  Determine the weight % albite just like before.  Form nephelite moles (y):  y = N-x.  Place this calculated amount in the Na2O row for nephelite, and allot an equal amount of this for Al2O3 and 2 times the Na2O mole amount to SiO2.  Determine the weight % for nephelite by multiplying it’s balanced molecular weight times the mole amount of Na2O.  At this point there is no deficiency of SiO2 and all of the Na2O and Al2O3 are used.  AT THIS POINT THE NORM CALCULATION IS FINISHED.

Step 3.  Form nephelite, new orthoclase and leucite:  All the Na2O moles allotted to
albite is used to form nephelite.  Place this amount in the row of Na2O for
nephelite, and an amount equal to Al2O3 and 2 times the Na2O amount for
SiO2. Determine  the weight% of nephelite.
There is now no albite present
in the norm.
There is still a SiO2 deficiency, so the next allotment
will be for new orthoclase and leucite
.

a.  Add all the mole amounts of SiO2 which you have allotted to minerals present
up to now ( olivine, new albite, and all those original minerals formed) except
that for orthoclase.

b. Subtract the amount in “a” from the total initial SiO2 present.  This represents
the mole amount available to continue.

c. If  the amount of SiO2 in “b” is greater than 4 times the K2O used for original
orthoclase and less than 6 times that amount, form new orthoclase and leucite
in “d” below.  If not, go to step 4 below.

d. Form new orthoclase moles (x): x = (S-4K)/2, where S is the amount of SiO2 in
‘b”; K is the mole amount of K2O used to form initial orthoclase.  Place this
calculated mole amount in the row of  K2O for new orthoclase, an equal
amount in that for Al2O3, and 6 times the amount of K2O amount for SiO2.
Determine the orthoclase weight % as before.  Form leucite moles (y): y = K-x.
Place this mole amount in the row for K2O
for leucite, an equal amount for Al2O3, and 4 times the K2O amount for SiO2.
Determine the weight % for leucite by multiplying the mole amount of K2O
by the molecular balanced weight.
AT THIS POINT THE NORM
CALCULATION IS FINISHED.

Step 4.  Form leucite, new diopside, new olivine and Ca orthosilicate: All the K2O moles
allotted to form orthoclase initially is used to form leucite with an equal amount
of Al2O3, and 4 times the amount of K2O for SiO2.  Determine the weight % of
leucite.  There is now no orthoclase in the norm.
There is still a SiO2
deficiency, so the next allotment will be for new diopside, new olivine,
and Ca orthosilicate.

a. Add the SiO2 mole amounts used for nephelite (in step 3, above), initial
anorthite, initial acmite (if present), olivine (in step 2 above), zircon and Na
metasilicate ( if either or both formed initially).

b. Subtract the mole amount in “a” from the total SiO2 initially in the rock. This
represents the mole amount available to continue.

c. If