B. Mineral Formula Calculations
-we will review the calculation of weight percent of elements in minerals and the determination 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.
1. Determination of weight % of elements in a mineral
-need chemical formula of mineral
-need atomic weights of individual elements comprising mineral formula
-chemical formulas are traditionally written with cations preceding anions--cations
appear in order from left to right according to increasing valence--if the same
valence exists for two or more cations, they can be written left to right according
to alphabetic order of their chemical symbols--the following is an example of the
determination of weight %
-calculate the weight % of the elements in the mineral, chalcopyrite (CuFeS2)molecular weight
Element
atomic weight
atoms/formul 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
(32.06/183.5)x100=34.94
2.
element
Cu
Fe
S
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 subscripts--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-atomic weights
63.54
55.85
32.06
weight %
34.62
30.43
34.94
atomic proportion
(34.6/63.5)=0.54
(30.4/55.9)=0.54
(34.9/32.1)=1.08
subscript
(0.54/0.54)=1
(0.54/0.54)=1
(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
1
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
BeO
Al2O3
SiO2
molecular weight
25
102
60
# moles
3
1
6
molecular weight
contribution
75
102
360
weight %
(75/537)x100=13.97
(102/537)x100=19.0
(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
2
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
BeO
Al2O3
SiO2
molecular weight
25
102
60
weight %
13.97
19.0
67.03
molecular proportion
(13.97/25) = 0.559
(19/102) = 0.186
(67.03/60) = 1.11
molecular ratios
(.559/.186) = 3
(.186/.186) = 1
(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 ay 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
3
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
How do you calculate the weight % of elements from a mineral (molecular) formula?
To do so, you will first need to determine the molecular weight of the mineral.
For example: KAlSi3O8
1. Set up a table with 5 columns, placing the element name in the first column.
2. In column 2, enter the number of atoms of the element in the mineral formula.
3. In column 3, enter the atomic weight of each of the elements in the mineral (if you don't know
these, consult a periodic table).
4. Multiply the value in column 2, by the atomic weight in column 3, and enter the results in
column 4 (in the case of O, 8 x 15.999 = 127.992).
5. Sum the values in the column 4 (and enter the sum at the bottom; e.g. 278.330)
6. To calculate the weight % of the element in the mineral, divide the value for the contribution to
the molecular weight (for example, O 127.992), divided by the sum of the contributions (278.330),
and multiply by 100 (to arrive at 45.99 weight % for O)
Element
Number of
Atomic
Contribution to
Weight %
Atoms in
Weight
Molecular
Element in
Formula
Weight
Mineral
K
1
39.098
39.098
14.05
Al
1
26.982
26.982
9.69
Si
3
28.086
84.258
30.27
O
8
15.999
127.992
45.99
278.330
4