2. Mineral Chemistry Calculations

3. 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, KAlSi 3 O 8 --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

4. 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 yH 2 O,if present cations increase in valence or if equal valence,

5. by alphabetic order of the chemical symbols calculation of weight % of elements in chalcopyrite (CuFeS 2 ) CuFeS 2 = 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

6. 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 = CuFeS 2 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

7. 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 Be 3 Al 2 Si 6 O 18 = BeO + Al 2 O 3 + SiO 2 = 3 BeO + 1 Al 2 O 3 + 6 SiO 2 BeO (25 x 3) = (75/537)x100 = 13.97% Al 2 O 3 (102 x 1) = (102/537)x100 = 19.0% SiO 2 (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 OH x and/or H 2 O to YH 2 O and proceed treat the YH 2 O as any cation oxide as above

8. 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 Al 2 O 3 (19/102) = 0.186 SiO 2 (67.03/60) = 1.11 next let us determine the molecular ratios for each element (cation) oxide BeO (0.559/0.186) = 3 Al 2 O 3 (0.186/0.186) = 1 SiO 2 (1.11/0.186) = 6

9. Next, associate the molecular ratios with the appropriate element oxide and determine the subscript for each element in formula 3BeO + 1Al 2 O 3 + 6SiO 2 = Be 3 Al 2 Si 6 O 18 if water appears in formula, it may be in the H 2 O or (OH) x form or both an altered mineral formula involves steps to manipulate yH 2 O to determine the mineral formula with the correct distribution of water form(s) Ca 2 B 6 O 11.5H 2 O is a formula for this mineral but the water is not in this specific form in the actual chemical formula for the mineral

10. hence the following procedure explains how the formula with water in the yH 2 O 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 yH 2 O, create a new formula by subtracting 1 H 2 O 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 H 2 O water form exists one of the created formulas is the correct formula all formula subscripts should be factored correctly

11. let’s try an example Ca 2 B 6 O 11. 5H 2 O is the formula expressed with yH 2 O Ca 2 B 6 O 10 (OH) 2. 4H 2 O = CaB 3 O 5 (OH). 2H 2 O Ca 2 B 6 O 9 (OH) 4. 3H 2 O Ca 2 B 6 O 8 (OH) 6. 2H 2 O = CaB 3 O 4 (OH) 3. 1H 2 O Ca 2 B 6 O 7 (OH) 8. 1H 2 O Ca 2 B 6 O 6 (OH) 10 = CaB 3 O 3 (OH) 5 cannot proceed further--all out of H 2 O the above red formula is the correct chemical formula for the mineral colemanite

12. 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. = (ZM)/(NV) Z = number of formula weights per unit cell; N = Avagadro’s Number (6.02310 23 ); V = volume of unit cell; M = molecular weight of formula of mineral

13. The following is a specific gravity calculation for wavelite, Al 3 (PO 4 ) 2 (OH) 3 5(H 2 O) 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. =((4412)/(6.023 x 10 23 )(116.6 x 10 -23 )) = 2.34 NOW TRY OR FINISH PROBLEM SET # 1