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Mineral Chemistry Calculations • Valences and Electroneutrality in Minerals Refer to the table in the internet notes of typical valences in minerals • Electroneutrality • is the equality of plus and minus attribute in a mineral contributed by the product of the valence and subscript of each element • in order for a mineral to be stable, it must be electroneutral • an example is orthoclase, KAlSi3O8--the plus and minus attributes are: • + charges---K(1x +1) + Al(1x+3) + Si(3x+4) = + 16 • - charges--- O(8x-2) = - 16------ +16 + -16 = 0 Mineral Formula Calculations • The following section treats the determination of weight percent of elements and cation oxides in minerals as well as the determination of the chemical formula of the mineral • Determination of the weight % of elements in a mineral • this portion does not include oxygen based minerals--we will treat this later • need formula of mineral • need atomic weights of individual elements in mineral • chemical formulas are usually written in a specific order • from left to right:cations, anions, then yH2O,if present • cations increase in valence or if equal valence, • by alphabetic order of the chemical symbols • calculation of weight % of elements in chalcopyrite (CuFeS2) • CuFeS2 = 1Cu + 1Fe + 2S • Cu (63.54x1/183.5)x100 = 34.62% • Fe (55.8x1/183.5)x100 = 30.43 % • S (32.06x2/183.5)x100 = 34.94% • Determination of the formula of a mineral • for minerals without oxygen in formula • need weight % of each element in mineral • need atomic weights of each element • in the calculation of the formula of a mineral with Cu, Fe and S--we will use the example of chalcopyrite and work the problem above backwards • divide weight % of each element in the mineral by the atomic weight to obtain atomic proportion • Cu (34.62/63.54) = 0.54 (atomic proportion) • Fe (30.43/55.85) = 0.54 • S (34.94/32.1) = 1.08 • establish subscripts of elements by dividing each atomic proportion by the lowest atomic proportion • subscripts = Cu (0.54/0.54) = 1; Fe (0.54/0.54) = 1; S (1.08/0.54) = 2 • mineral formula is 1Cu + 1Fe + 2S = CuFeS2 • if the decimal portion of a subscript is too high to simply eliminate, all subscripts must be multiplied by same whole number to obtain insignificant decimal numbers which then can be eliminated • Determination of the weight % of element oxides in a mineral • need mineral’s chemical formula • need molecular weights of element oxides • need to establish balanced element oxides from formula • Be3Al2Si6O18 = BeO + Al2O3 + SiO2 • = 3 BeO + 1 Al2O3 + 6 SiO2 • BeO (25 x 3) = (75/537)x100 = 13.97% • Al2O3 (102 x 1) = (102/537)x100 = 19.0% • SiO2 (60 x 6) = (360/537)x100 = 67.03% • you should be familiar with names of elements oxides appearing in the internet notes--know ‘em • convert any OHx and/or H2O to YH2O and proceed treat the YH2O as any cation oxide as above • Determination of the formula of a mineral with oxygen in the formula • need molecular weights of each element oxide • need each element oxide weight % • calculate the subscripts for each element in beryl by first establishing molecular proportions • BeO (13.97/25) = 0.559 • Al2O3 (19/102) = 0.186 • SiO2 (67.03/60) = 1.11 • next let us determine the molecular ratios for each element (cation) oxide • BeO (0.559/0.186) = 3 • Al2O3 (0.186/0.186) = 1 • SiO2 (1.11/0.186) = 6 • Next, associate the molecular ratios with the appropriate element oxide and determine the subscript for each element in formula • 3BeO + 1Al2O3 + 6SiO2 = Be3Al2Si6O18 • if water appears in formula, it may be in the H2O or (OH)x form or both • an altered mineral formula involves steps to manipulate yH2O to determine the mineral formula with the correct distribution of water form(s) • Ca2B6O11.5H2O is a formula for this mineral but the water is not in this specific form in the actual chemical formula for the mineral • hence the following procedure explains how the formula with water in the yH2O can be manipulated to create a series of formulas, one of which will be the correct chemical formula for the mineral • start with the formula with the yH2O, create a new formula by subtracting 1 H2O and 1 from the subscript associated with the non water oxygen in the formula • next add 2 (OH) waters = (OH)2 to the formula • these steps create a new electroneutral formula • continue to form new formulas using the same steps above on each created formula until no H2O water form exists • one of the created formulas is the correct formula • all formula subscripts should be factored correctly • let’s try an example • Ca2B6O11.5H2O is the formula expressed with yH2O • Ca2B6O10(OH)2 .4H2O = CaB3O5(OH) .2H2O • Ca2B6O9(OH)4 .3H2O • Ca2B6O8(OH)6 .2H2O = CaB3O4(OH)3 .1H2O • Ca2B6O7(OH)8 .1H2O • Ca2B6O6(OH)10 = CaB3O3(OH)5 • cannot proceed further--all out of H2O • the above red formula is the correct chemical formula for the mineral colemanite Specific Gravity Calculation • S.G. of a substance is the density number without the associated unit of density (grams/cc) and is obtained by dividing the density of a substance by the density of water (1 gram/cc--this cancels out the unit of density) • S. G depends on • the kind of atoms comprising a substance (atomic weight of atoms) • the manner in which these atoms are packed (closely or loosely packed) • S. G. = (Z•M)/(N•V) • Z = number of formula weights per unit cell; N = Avagadro’s Number (6.023•1023); V = volume of unit cell; M = molecular weight of formula of mineral • The following is a specific gravity calculation for wavelite, Al3(PO4)2(OH)3•5(H2O) • Z= 4; M = 412; V = (a = 9.62 x 10-8; b = 17.34 x 10-8; c= 6.99 x 10-8) • S. G. =((4•412)/(6.023 x 1023)•(116.6 x 10-23)) = 2.34 • NOW TRY OR FINISH PROBLEM SET #1