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 toYH2O --2 examples of how formulas                  are converted to using a and b and  to the YH2O are shown below in the following balanced                  formulas:                    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)xor 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                                                              Ca2B6O9(OH)4.3H2O                                                              Ca2B6O8(OH)6.2H2O > 2(CaB3O4(OH)3.H2O)factored                                                              Ca2B6O7(OH)8.H2O                                                              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(Ĺ), 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                    CLICK HERE TO SEE MORE 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