Mineral Chemistry Calculations |
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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 |
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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: |
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2. Electroneutrality -the total +(plus) charge and -(minus) charge component of elements in a mineral must be equal to give electoneutrality--the 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 - charges-----O(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 mineral--also, 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.
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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, CuFeS2 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- scripts--if 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 subscripts--this 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, CuFeS2 -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 cation--then 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 cation--factor 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/orYH2O should be converted toYH2O --2 examples of how formulas are converted to using a and b and to the YH2O are shown below in the following balanced formulas: Al2Si2O5.(OH)4 > Al2O3 + 2SiO2 + 2H2O and 2Al3(PO4)2(OH)3 .5H2O > 3Al2O3 + 2P2O5 + 13 H2O -calculate the weight % of the mineral, beryl (Be3Al2Si6O18) first follow steps a and b above: from step a: Be3Al2Si6O18 = BeO +Al2O3 +SiO2 from step b: Be3Al2Si6O18 = 3BeO+ Al2O3+6SiO2 --note in this case there is no water to shown as YH2O |
element oxide | molecular weight | # moles | molecular weight contribution | weight % |
BeO | 25 | 3 | 75 | (75/537)x100=13.97 |
Al2O3 | 102 | 1 | 102 | (102/537)x100=19.0 |
SiO2 | 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: SiO2 = silica CaO = lime Al2O3 = alumina MgO = magnesia Fe2O3 = ferric oxide MnO = manganous oxide FeO = ferrous oxide MnO2 = manganic oxide K2O = potash P2O5 = phosphate Na2O = soda TiO2 = 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 cation-O pair given in the problem- -need each of the cation-O weight % which is given- -calculate the molecular ratios of each cation oxide: you can obtain the weight % of each cation-oxide 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 |
Al2O3 | 102 | 19.0 | (19/102) = 0.186 | (.186/.186) = 1 |
SiO2 | 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: Be3Al2Si6O18 < 3BeO+1Al2O3+ 6SiO2 -the following treats a problem as above if the mineral formula contains water in the the form of YH2O or (OH)xor both -determine the molecular ratio of the water from the given weight % H2O as is done for any cation oxide alike in the table above--the resulting mineral formula containing YH2O may be the actual formula for the mineral, if not, then it must be altered to obtain the actual formula with only OHx or both water forms--an example of this can be explained in the calculation of a mineral yielding the mineral formula, Ca2B6O11.5H2O--this is not the actual formula of the mineral--a series of mineral formulas can be obtained by using a manipulation of the Y (number of moles) in YH2O to form a series of mineral formulas each expressing the number of x in (OH)x and the number for Y inYH2O--one of these resulting mineral formulas will be the correct one--the following is the process by which all possible mineral formulas can be obtained -start with the formula with water expressed only in YH2O--each successive mineral formula has one less non-water O, 2 more (OH) and one less H2O until all YH2O is changed to (OH)x--this procedure keeps the number of H and O balanced and the mineral formula electrically neutral--always factor subscripts when needed original formula: Ca2B6O11.5H2O next formulas: Ca2B6O10(OH)2.4H2O > 2(CaB3O5(OH).2H2O)factored Ca2B6O9(OH)4.3H2O Ca2B6O8(OH)6.2H2O > 2(CaB3O4(OH)3.H2O)factored Ca2B6O7(OH)8.H2O Ca2B6O6(OH)10 > 2(CaB3O3(OH)5)--factored and cannot go -further because there is no YH2O remaining--the 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, Al3(PO4)2(OH)3.5H2O- 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
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