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GEOS 3010Homework #3

Name: ____________________

Due Date: _________________

Mineral Chemistry:

Formula Normalization, Solid Solution, Polymorphs, and Isomorphs

Objective

In this exercise, you will practice the skill of normalizing a mineral formula from chemical

analysis. You will also identify aspects of bonding, solid solution, polymorphism, and

isomorphism in example minerals.

Introduction

The composition of minerals was originally determined by various wet chemical analysis

techniques. These days analysis is done mainly by beam analytical instruments: electron

microprobes and energy dispersive x-ray analyzers. These instruments work by using an energy

beam (high energy electrons) fired at a mineral grain. The mineral emits lots of things when

struck by this beam including low energy x-rays. By measuring the energy levels or the wave

length of the emitted x-rays, the geologist can determine what kinds of elements are present,

and how much of each element is present.

The resulting chemical analysis of a mineral shows the elements that are present, and how much

of each, but not how they are combined in the structure of the mineral. The results of chemical

analyses are given in weight percent (wt%). In order to determine the formula for the mineral,

wt % must be converted to atomic proportions (or moles). Dividing the wt % of each element by

its atomic weight (the average weight of the element, as given on the periodic table) gives the

atomic proportions. Here’s an example for a sample of the mineral marcasite:

Element

Wt %

Fe

S

Total

46.55

53.05

99.6

Atomic Weight

(g/mol)

55.85

32.07

—

Atomic Proportions

(moles)

(46.55 / 55.85) = 0.834

(53.05 / 32.07) = 1.655

—

But we have a little problem. Note that the proportions are 0.83 moles of Fe to 1.65 moles of S.

However, formulas are written as proportions of whole numbers (with the exception of

elements in solid solution, coming later). Therefore, the next step is to convert the atomic

proportions into whole number, relative proportions.

How can we do this? The magic lies in understanding proportions – that they are always relative

to each other. For example, each of the following sets of numbers is a 1:2 proportion (the

second number is twice as large as the first number): 1:2, 2:4, 5:10, 0.5:1. You can maintain the

1:2 proportion as long as you do the same thing to both sides, in other words, multiply both

sides by two, divide them by 8.776, etc. This preserves the proportional value.

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Back to our problem – how do we turn the atomic proportions into whole numbers? Divide any

number by itself to get the number 1. But since this is a proportion, we have to remember to

divide both sides by the same number. Therefore, the simplest thing to do is to divide both sides

by the smaller number, yielding whole number proportions. So,

0.834 / 0.834 = 1 mole of Fe

1.655 / 0.834 = 1.98, or 2 moles of S

This means that the mineral formula contains 2 S atoms for every 1 Fe atom. We can therefore

write the formula for marcasite as FeS2. But wait! Why wasn’t the proportion of Fe to S exactly 1

to 2? Because no analysis is perfect – some analytical error can be expected. You can see in the

table that the total weight is not 100% (it is 99.6%). That is OK, a total wt% between 98.0 and

102.0 is considered to be within analytical error.

What about elements that share solid solution? In this case it is perfectly acceptable to have

decimal fractions in the formula for the elements in solid solution (but note that the total of

these elements must be a whole number). The formula is written to reflect the exact

proportions of each element in the solid solution relationship. For example, the mineral zincite’s

formula is written as (Zn,Mn)O. This means that there can be any proportion of Zn to Mn as long

as the total Zn + Mn is 1:1 with O. So a specimen with the formula (Zn0.80Mn0.20)O has 80% Zn in

the solid solution, whereas a specimen with the formula (Zn0.25Mn0.75) has 25% Zn. In both cases,

however, the total Zn + Mn is 1, which maintains the 1:1 ratio with oxygen.

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Formula Normalization and Recalculation Problem

1. The analyses for several other samples of marcasite are presented below.

Element

Sample 1 (wt%) Sample 2 (wt%)

Fe

46.54

46.55

Sample 3

(wt%)

46.53

Sample 4

(wt%)

47.22

S

53.46

53.05

53.30

52.61

Total

100.00

99.60

99.83

99.83

a. What are the atomic proportions for each sample of marcasite? Show your work for

each calculation in the table, and round your answer to the 0.001 decimal place.

Element

Fe

Atomic Weight

(g/mol)

55.85

S

32.07

Sample 1

Atomic %

Sample 2

Atomic %

Sample 3

Atomic %

Sample 4

Atomic %

b. What are the relative proportions for each sample of marcasite? Show your work for

each calculation in the table, and report the answer as a whole number.

Element

Sample 1

Rel Proportions

Sample 2

Rel Proportions

Sample 3

Rel Proportions

Sample 4

Rel Proportions

Fe

S

c. What are the relative proportions of Fe and S in marcasite? What is the formula for each

sample of marcasite?

d. What do you think is the reason for variation between these sample of marcasite? Do

you think there is any atomic substitution, or is the composition of marcasite fixed?

[HINT: go back and re-read the introduction to the homework.]

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e. Iron occurs as either the ferrous (Fe+2) or ferric (Fe+3) ion in nature. Is marcasite a simple

ionically bonded mineral? What type of bonding do you think occurs in this sulfide

mineral? [HINT: balance the valences, think about marcasite’s luster, and review the

Chemistry Review lecture slides (atoms, elements, periodic table, and bonds).]

f. What other common sulfide mineral has an identical chemical composition to marcasite?

[HINT: look up in your R&M lab guidebook.] What is that relationship called?

Atomic Substitution and Solid Solution Problem

2. Let’s take a look at chemical composition in a more complicated material system, the zinciron sulfide mineral, sphalerite.

Element

Sample 1 (wt%)

.015

–

Sample 2

(wt%)

7.99

–

Sample 3

(wt%)

11.05

–

Sample 4

(wt%)

18.25

2.66

Fe

Mn

Cd

Zn

66.98

1.23

57.38

0.3

55.89

0.28

44.67

S

32.78

32.99

32.63

33.57

99.91

99.59

99.87

99.43

As the analyses indicate, sphalerite may be almost pure zinc sulfide or may contain a lot of iron

and small amounts of other transition metals.

a. What is the chemical formula shown for sphalerite in your R&M lab guidebook? What do

the parentheses in the formula mean?

b. What is the valence of sulfur in sphalerite?

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c. What are the most probable valences of each cation? Which cations can substitute in

sphalerite’s mineral structure?

d. What type of bonding do you think exists in sphalerite? [HINT: like pyrite and marcasite,

it’s also a sulfide mineral.]

e. In the table below, calculate the atomic proportions (round off to the 0.001 place) for

each element in each analysis. Show your work in the table. You will need to look up the

gram atomic weight for each element in the periodic table, and enter this in the first

column.

Element

Atomic Weight

(g/mol)

Sample 1

Atomic %

Sample 2

Atomic %

Sample 3

Atomic %

Sample 4

Atomic %

Fe

Mn

Cd

Zn

Total metals

(add Fe, Mn,

Cd, and Zn)

S

d. What are the relative proportions of total metals to sulfur for each sample? Note that we

do not need to normalize these to whole numbers to get the relative proportions.

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f. For Sample #4, calculate the normalized proportion of each metal relative to total metals

(for example, the normalized proportion of Fe in Sample 4 = atomic % Fe in Sample 4 /

atomic % total metals in Sample 4). Show your work below.

g. Write a chemical formula for Sample #4 of sphalerite that shows the atomic proportions

of each metal. [HINT: group solid solution elements together. Remember that in this case

it is OK to have decimal fractions in the formula, since the solid solution elements are

sharing the same site and we want to know exactly what proportion of each element is in

the solid solution. The rest of the formula should be in whole numbers.]

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