Download 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
• 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
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