Mineral Chemistry Calculations

        In this section we will treat chemical calculations involving mineral formulas and
                                                   specific gravity



    A.  Valences of Typical Elements Comprising Minerals and Electroneutrality
a familiarity with these two topics will help in solving chemical calculation problems given
          during the semester-

         1. Valences in minerals
   -the following table summarizes the valence states of representative elements found in minerals:
element valence state(s) element valence state(s)
O -2 N +5
Si +4 F -1
Al +3 P +5
Fe +2, +3 S -2, +6
Ca +2 V +5
Na +1 Cr +3,+6
K +1 Pb +2
Mg +2 Mn +2, +4
Ti +4 B +3
C +4 Ba +2
W +6 Mo +6
Li +1 Be +2
As -3,+3,+5


          2. Electroneutrality
             -the total +(plus) charge and -(minus) charge component of elements in a mineral 
              must be equal to give electoneutrality--the component of each charge is obtained by
              multiplying the subscript of each element by its valence number and adding like signs
             -an example is the electroneutrality in orthoclase (KAlSi3O8):
                    + charges---- K(1x +1)+Al(1x +3)+Si(3x +4) = +16
                    - charges-----O(8x -2) = -16

    B.  Mineral Formula Calculations
we will review the calculation of weight percent of elements in minerals and the determ-
          ination of the chemical formula of a mineral--also, we will determine the same
          for the element oxide composition of a mineral which is quite different than what you did
          in your chemistry course but very important in geology.

          1.  Determination of weight % of elements in a mineral
                -need chemical formula of mineral
                -need atomic weights of individual elements comprising mineral formula
                -chemical formulas are traditionally written with cations preceding anions--cations
                  appear in order from left to right according to increasing valence--if the same
                  valence exists for two or more cations, they can be written left to right according
                  to alphabetic order of their chemical symbols--the following is an example of the
                  determination of weight %
                       -calculate the weight % of the elements in the mineral, chalcopyrite (CuFeS2)-

element atomic weight # atoms/formula molecular weight contribution weight % of element
           Cu         63.54          1      63.54 (63.5/183.5)x100=34.62
           Fe         55.85          1      55.85 (55.9/183.5)x100=30.43
            S         32.06          2      64.12 (64.1/183.5)x100=34.94

          2.   Determination of the chemical formula of a mineral                          
                -need weight % of each element in the mineral
                -need atomic weights of elements in the mineral
                -the following is an example of the determination of the formula of the mineral,
                  chalcopyrite, CuFeS2 above-

                       -calculate the subscripts of Cu, Fe, and S in the mineral --the
                        atomic proportions must be normalized and rounded off to obtain the sub-
                        scripts--if a decimal portion of a subscript is high, all subscripts must be
                        multiplied by the same whole number and rounded off to generate only
                        whole number subscripts--this condition is not too abundant--
element atomic weights weight % atomic proportion subscript
Cu 63.54 34.62 (34.6/63.5)=0.54 (0.54/0.54)=1
Fe 55.85 30.43 (30.4/55.9)=0.54 (0.54/0.54)=1
S 32.06 34.94 (34.9/32.1)=1.08 (1.08/0.54)=2

                 -placing each appropriate subscript below the corresponding element in the formula
                  will result in the chemical formula of the mineral, CuFeS2
                 -notice in this case the decimal portion of subscript is small, (actually 0) and they do not   
                  have to be multiplied by the same whole number and rounded off to generate the whole 
                  number subscripts
         3.  Determination of the weight % of  element oxides in a mineral
               -this is a problem very similar to #1 above except for an additional step below since the
                 mineral contains O as an important anion in the formula
               -need chemical formula of mineral-
               -need  to determine the molecular weights of each element (cation) oxide-
               -break down chemical formula into balanced cation oxides
                    a. draw an arrow to the right of mineral formula and place an O (oxygen) after each
                       cation--then place the appropriate subscript for each cation below that of the
                       associated oxygen and 2 (-2, the valence of oxygen) for the subscript of the
                       cation--factor subscripts of each cation and O to  the lowest whole number if
                       necessary and keep only the factored subscripts
                    b. balance the number of each element on both sides of the arrow by placing a
                        whole number in front of each cation oxide pair and in front of the mineral
                        formula if necessary
               -any (OH)x water and/orYH2O should be converted toYH2--2 examples of how formulas 
                are converted to using a and b and  to the YH2O are shown below in the following balanced 

                   Al2Si2O5.(OH)4  > Al2O3 + 2SiO2 + 2H2O   and

                  2Al3(PO4)2(OH)3 .5H2O > 3Al2O3 + 2P2O5 + 13 H2O
                     -calculate the weight % of the mineral, beryl (Be3Al2Si6O18)
                        first follow steps a and b above:
from step a: Be3Al2Si6O18 = BeO +Al2O3 +SiO2
from step b: Be3Al2Si6O18 = 3BeO+ Al2O3+6SiO2  
          --note in this case there is no water to shown as YH2O
element oxide molecular weight # moles molecular weight contribution weight %
BeO 25 3 75 (75/537)x100=13.97
Al2O3 102 1 102 (102/537)x100=19.0
SiO2 60 6 360 (360/537x100=67.03

                       There are specific names given to cation oxides in Mineralogy
                        Know the names below given to some element (cation) oxides:
                            SiO2 = silica                        CaO = lime
                            Al2O3 = alumina                  MgO = magnesia
                            Fe2O3 = ferric oxide            MnO = manganous oxide
                            FeO = ferrous oxide            MnO2 = manganic oxide
                            K2O = potash                      P2O5 = phosphate
                            Na2O = soda                       TiO2 = titania

          4.  Determination of the chemical formula of a mineral with oxygen in 
            the formula      

               -this problem is very similar to problem # 2 above
               -need to determine molecular weights of  each cation-O pair given in the problem-
               -need each of the cation-O weight % which is given-
               -calculate the molecular ratios of each cation oxide:
                     you can obtain the weight % of each cation-oxide from the previous problem -

element oxide molecular weight weight % molecular proportion molecular ratios
BeO 25 13.97 (13.97/25) = 0.559 (.559/.186) = 3
Al2O3 102 19.0 (19/102) = 0.186 (.186/.186) = 1
SiO2 60 67.03 (67.03/60) = 1.11 (1.11/.186) = 6

                  -see the immediate formula below for the following steps to determine the formula
                  -place each mole number from the molecular ratio in front of each cation oxide (element oxide)
                  -the appropriate subscripts can then be determined for the mineral formula by balancing
                   the number of each element on both sides of the arrow, and will result in the following:
                   remember the order of element presentation in the formula as given in B 1 above:
                           Be3Al2Si6O18 < 3BeO+1Al2O3+ 6SiO2
                       -the following treats a problem as above if the mineral formula contains water in the 
                   the form of YH2O or (OH)x
or both 
-determine the molecular ratio of the water from the given weight % H2O as is done for
                        any cation oxide alike in the table above--the resulting mineral formula containing YH2O     
                        may be the actual formula for the mineral, if not, then it must be altered to obtain the actual
                        formula with only OHx or both water forms--an example of this can be explained in the 
                        calculation of a mineral yielding the
mineral formula, Ca2B6O11.5H2O--this is not the 
                        actual formula of the mineral--a series of mineral formulas can be obtained by using a
                        manipulation of the Y (number of moles) in YH2O to form a series of mineral formulas
                        each expressing the number of x in (OH)x and the number for Y inYH2O--one of these
                        resulting mineral formulas will be the correct one--the following is the process by which
                        all possible mineral formulas can be obtained
                           -start with the
formula with water expressed only in YH2O--each successive  
                            mineral formula has one less non-water O, 2 more (OH) and one less H2O until all  
                            YH2O is changed to (OH)x--this procedure keeps the number of H and O balanced 
                             and the mineral formula electrically neutral--always factor subscripts
when needed 
original formula: Ca2B6O11.5H2O
                                    next formulas: Ca2B6O10(OH)2.4H2O > 2(CaB3O5(OH).2H2O)factored
                                                             Ca2B6O8(OH)6.2H2O > 2(CaB3O4(OH)3.H2O)factored
                                                             Ca2B6O6(OH)10 > 2(CaB3O3(OH)5)--factored and cannot go
                              -further because there is no YH2O remaining--the 4th formula (underlined) 
                                represents the correct formula for the mineral, colemanite
-the above procedure helps review many concepts you already were taught in basic
                  chemistry as well as more
   C.  Specific Gravity Calculation
-specific gravity depends on (1) the kind of atoms of which the mineral is composed and
          (2) the manner in which these atoms are packed together-
         -specific gravity is density divided by the density of water (1 gram/cc)--hence the specific
          gravity number has no associated units 
         -S.G.=(ZxM)/(NxV) where Z is the number of formula weights or units per unit cell; We will
           treat the nature of the Z number in a later topic
          N is Avagadro's Number, 6.023x10+23; V is the Volume of the unit cell measured in ang-
          stroms(), V representing the product of 3 crystallographic axes lengths we will treat axes later; 
          M is the molecular weight of the chemical formula of the mineral
                -lets calculate the S.G. for wavelite, Al3(PO4)2(OH)3.5H2O-
                       Z = 4;  M = 412;  V = (a = 9.62A, b =17.34A, c = 6.99A) = (116.6x10-23)

                   S.G. =( (4x412)/(6.023x10+23)x(116.6x10-23)) = 2.34

     Now let us practice some problems concerning mineral chemical calculations

                      CLICK HERE TO SEE PROBLEMS

Let us now study the combining principles of atoms which will aid in understanding
     the nature of the classification of minerals  into groups, series, and silicate subclasses--
   - the section deals with atomic structure of minerals according to Pauling' s rules of
     stable crystal or atomic structures