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Mineral Chemistry Calculations

Mineral Chemistry Calculations. Valences and Electroneutrality in Minerals. Refer to the table in the internet notes of typical valences in minerals Electroneutrality

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Mineral Chemistry Calculations

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  1. Mineral Chemistry Calculations

  2. 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

  3. 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,

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

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