Mineral Chemistry Calculations 

In
this section we will treat chemical calculations involving mineral formulas and specific gravity CLICK HERE FOR A CHEMICAL TABLE CLICK HERE FOR ANOTHER CHEMICAL TABLE 

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: 



2. Electroneutrality the total +(plus) charge and (minus) charge component of elements in a mineral must be equal to give electoneutralitythe 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  chargesO(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 mineralalso, 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.


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, CuFeS_{2} 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 scriptsif 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 subscriptsthis 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, CuFeS_{2 } 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 cationthen 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 cationfactor 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/orYH_{2}O should be converted toYH_{2}O 2 examples of how formulas are converted to using a and b and to the YH_{2}O are shown below in the following balanced formulas: Al_{2}Si_{2}O_{5}.(OH)_{4} > Al_{2}O_{3} + 2SiO_{2} + 2H_{2}O and 2Al_{3}(PO_{4})_{2}(OH)_{3} .5H_{2}O > 3Al_{2}O_{3} + 2P_{2}O_{5} + 13 H_{2}O calculate the weight % of the mineral, beryl (Be_{3}Al_{2}Si_{6}O_{18}) first follow steps a and b above: from step a: Be_{3}Al_{2}Si_{6}O_{18} = BeO +Al_{2}O_{3} +SiO_{2} from step b: Be_{3}Al_{2}Si_{6}O_{18} = 3BeO+ Al_{2}O_{3}+6SiO_{2 } note in this case there is no water to shown as YH_{2}O 
element oxide  molecular weight  # moles  molecular weight contribution  weight % 
BeO  25  3  75  (75/537)x100=13.97 
Al_{2}O_{3}  102  1  102  (102/537)x100=19.0 
SiO_{2}  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: SiO_{2} = silica CaO = lime Al_{2}O_{3 }= alumina MgO = magnesia Fe_{2}O_{3 }= ferric oxide MnO = manganous oxide FeO = ferrous oxide MnO_{2} = manganic oxide K_{2}O = potash P_{2}O_{5} = phosphate Na_{2}O = soda TiO_{2} = 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 cationO pair given in the problem need each of the cationO weight % which is given calculate the molecular ratios of each cation oxide: you can obtain the weight % of each cationoxide 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 
Al_{2}O_{3}  102  19.0  (19/102) = 0.186  (.186/.186) = 1 
SiO_{2}  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: Be_{3}Al_{2}Si_{6}O_{18 }< 3BeO+1Al_{2}O_{3}+ 6SiO_{2} 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 % H_{2}O as is done for any cation oxide alike in the table abovethe resulting mineral formula containing YH_{2}O may be the actual formula for the mineral, if not, then it must be altered to obtain the actual formula with only OH_{x} or both water formsan example of this can be explained in the calculation of a mineral yielding the mineral formula, Ca_{2}B_{6}O_{11}.5H_{2}Othis is not the actual formula of the minerala series of mineral formulas can be obtained by using a manipulation of the Y (number of moles) in YH_{2}O to form a series of mineral formulas each expressing the number of x in (OH)_{x} and the number for Y inYH_{2}Oone of these resulting mineral formulas will be the correct onethe following is the process by which all possible mineral formulas can be obtained start with the formula with water expressed only in YH_{2}Oeach successive mineral formula has one less nonwater O, 2 more (OH) and one less H_{2}O until all YH_{2}O is changed to (OH)_{x}this procedure keeps the number of H and O balanced and the mineral formula electrically neutralalways factor subscripts when needed original formula: Ca_{2}B_{6}O_{11}.5H_{2}O next formulas: Ca_{2}B_{6}O_{10}(OH)_{2}.4H_{2}O > 2(CaB_{3}O_{5}(OH).2H_{2}O)factored Ca_{2}B_{6}O_{9}(OH)_{4}.3H_{2}O Ca_{2}B_{6}O_{8}(OH)_{6}.2H_{2}O > 2(CaB_{3}O_{4}(OH)_{3}.H_{2}O)factored Ca_{2}B_{6}O_{7}(OH)_{8}.H_{2}O Ca_{2}B_{6}O_{6}(OH)_{10} > 2(CaB_{3}O_{3}(OH)_{5})factored and cannot go further because there is no YH_{2}O remainingthe 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(Å), 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, Al_{3}(PO_{4})_{2}(OH)_{3}.5H_{2}O 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 