**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 SiO _{2}, quartz cannot form and a
redistribution or reallocation of the SiO_{2} 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. **

** Step 1. Redistribution
of SiO_{2} between new hypersphene and
olivine **

**a.
****Add all the
SiO _{2} mole amounts used in the formation of minerals (general sum) **

**b.
****Subtract the
moles of SiO _{2} obtained in a from the total moles of SiO_{2}
at the beginning of mineral formation. This difference is the available SiO_{2}
for the redistribution.**

**a.
**** 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 SiO _{2} and there is further
redistribution in 2 below.**

** **

b. __Form the number of new moles of hypersphene (x__): x = 2S-M, where S is the moles of SiO_{2}
in c above and M is the sum of FeO, MgO moles allotted to hypersphene
initially. Place this number in the SiO_{2} 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 SiO_{2} for olivine.
Multiply this mole number by the FeO fraction
and the MgO fraction and then each by 2 since olivine
= 2(Fe,Mg)O. SiO_{2}
and place the respective mole amounts in the olivine column. At this point the
sum of the SiO_{2} used to form hypersphene
and olivine should equal the available SiO_{2} obtained from c above
and the sum of each FeO and MgO
to form hypersphene and olivine should equal the
original amounts of each from the original hypersphene. Determine the weight % hypersphene,
the same way you did initially.
Determine the weight % olivine: FeO. (**1/2
SiO _{2}) times the mole amount of FeO. MgO. (**

**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 SiO_{2}. There is now no hypersphene
present in the norm.**

** **

**
a. Add the mole amounts of SiO _{2} 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 SiO _{2} present initially in the rock.
This represents the amount of SiO_{2} moles available to continue.**

**c.
****If the mole
amount of SiO _{2} in b is greater than twice the amount of moles of
Na_{2}O allotted to albite initially, and it
is 6 times less than the Na_{2}O amount used initially, new albite and nephelite will form
according to d below. If not, there is
still a deficiency of SiO_{2}--- go to Step 3.**

**d.
**__Form new albite moles (x)__:
x = (S-2N)/4; where S is the amount of SiO_{2} moles available
in b; N is the mole amount of Na_{2}O_{ }used for initial albite. Place the calculated mole amount in the Na_{2}O
row for new albite.
Allot an equal amount of Al_{2}O_{3} and 6 times the Na_{2}O
for the amount of SiO_{2} moles.
Determine the weight % albite just like
before. __Form nephelite
moles (y): __y = N-x. Place this calculated amount in the Na_{2}O
row for nephelite, and allot an equal amount of this
for Al_{2}O_{3} and 2 times the Na_{2}O mole amount to
SiO_{2}. Determine the weight %
for nephelite by multiplying its
balanced molecular weight times the mole amount of Na_{2}O. At this point there is no deficiency of SiO_{2}
and all of the Na_{2}O and Al_{2}O_{3} are used. **AT THIS POINT THE NORM CALCULATION IS FINISHED.**

**Step 3.**** Form nephelite,
new orthoclase and leucite: All the Na_{2}O moles allotted to **

albite is used to form nephelite. Place this amount in the row of Na_{2}O for

nephelite, and an amount equal to Al_{2}O_{3} and 2 times the Na_{2}O amount for

SiO_{2}. Determine the weight% of nephelite.

in the norm.

will be for new orthoclase and leucite.

** a. Add all the mole amounts of SiO _{2}
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 SiO _{2} present.
This represents
the mole amount
available to continue.**

** c. If the amount of SiO _{2} in b is
greater than 4 times the K_{2}O 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 SiO_{2} in **

b; K is the mole amount of K_{2}O used to form initial orthoclase. Place this

calculated mole amount in the row of K_{2}O for new orthoclase, an equal

amount in that for Al_{2}O_{3}, and 6 times the amount of K_{2}O amount for SiO_{2}.

Determine the orthoclase weight % as before. Form leucite moles (y): y = K-x.

Place this mole amount in the row for K_{2}O

for leucite, an equal amount for Al_{2}O_{3}, and 4 times the K_{2}O amount for SiO_{2}.

Determine the weight % for leucite by multiplying the mole amount of K_{2}O

by the molecular balanced weight.

CALCULATION IS FINISHED.

** Step 4. Form leucite,
new diopside, new olivine and Ca orthosilicate:
All the K_{2}O moles **

allotted to form orthoclase initially is used to form leucite with an equal amount

of Al_{2}O_{3}, and 4 times the amount of K_{2}O for SiO_{2}. Determine the weight % of

leucite. There is now no orthoclase in the norm.

deficiency, so the next allotment will be for new diopside, new olivine,

and Ca orthosilicate.

** a.
Add the SiO _{2} 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 SiO _{2} initially in the rock. This
represents the mole amount available to
continue.**

** c. If **

** **