Chemical formulas for dummies. Chemical calculations Introduction to nitrogen. Amines

On their basis, schemes and equations of chemical reactions are drawn up, as well as chemical classification and nomenclature of substances. One of the first to use them was the Russian chemist A. A. Iovsky.

A chemical formula may indicate or reflect:

1 molecule (as well as ion, radical...) or 1 mole of a specific substance;
qualitative composition: what chemical elements the substance consists of;
quantitative composition: how many atoms of each element does a molecule (ion, radical...) contain?

For example, the formula HNO 3 means:

1 molecule of nitric acid or 1 mole of nitric acid;
qualitative composition: the nitric acid molecule consists of hydrogen, nitrogen and oxygen;
quantitative composition: a nitric acid molecule contains one hydrogen atom, one nitrogen atom and three oxygen atoms.

Kinds

Currently, the following types of chemical formulas are distinguished:

The simplest formula . It can be obtained experimentally by determining the ratio of chemical elements in a substance using the atomic mass values of the elements. So, the simplest formula of water is H 2 O, and the simplest formula of benzene is CH (unlike C 6 H 6 - true). Atoms in formulas are indicated by the signs of chemical elements, and their relative quantities are indicated by numbers in subscript format.
True Formula . Molecular formula - can be obtained if the molecular mass of a substance is known. The true formula of water is H 2 O, which coincides with the simplest one. The true formula of benzene is C 6 H 6, which differs from the simplest one. True formulas are also called gross formulas . They reflect the composition, but not the structure, of the molecules of a substance. The true formula shows the exact number of atoms of each element in one molecule. This quantity corresponds to a [lower] index - a small number after the symbol of the corresponding element. If the index is 1, that is, there is only one atom of a given element in the molecule, then such an index is not indicated.
Rational formula . Rational formulas highlight groups of atoms characteristic of classes of chemical compounds. For example, for alcohols the group -OH is allocated. When writing a rational formula, such groups of atoms are enclosed in parentheses (OH). The number of repeating groups is indicated by numbers in subscript format, which are placed immediately after the closing bracket. Square brackets are used to reflect the structure of complex compounds. For example, K4 is potassium hexacyanocobaltate. Rational formulas are often found in a semi-expanded form, when some of the same atoms are shown separately for better reflection structure of the molecule of a substance.
Markush formula represent a formula in which the active nucleus and a number of substituent options are isolated, combined into a group of alternative structures. It is a convenient way to designate chemical structures in a generalized form. The formula refers to the description of an entire class of substances. The use of “broad” Markush formulas in chemical patents leads to a lot of problems and discussions.
Empirical formula. Different authors may use this term to mean the simplest , true or rational formulas.
Structural formula. Graphically shows the relative arrangement of atoms in a molecule. Chemical bonds between atoms are indicated by lines (dashes). There are two-dimensional (2D) and three-dimensional (3D) formulas. Two-dimensional are a reflection of the structure of matter on a plane (also skeletal formula- attempts to approximate a 3D structure on a 2D plane). Three-dimensional [spatial models] make it possible to represent its composition most closely to theoretical models of the structure of a substance, and, often (but not always), a more complete (true) relative arrangement of atoms, bond angles and distances between atoms.

The simplest formula: C 2 H 6 O
True, empirical, or gross formula: C 2 H 6 O
Rational formula: C 2 H 5 OH
Rational formula in semi-expanded form: CH 3 CH 2 OH

N N │ │ N-S-S-O-N │ │ N N

Structural formula (3D):

Option 1: Option 2:

The simplest formula C 2 H 6 O can equally correspond to dimethyl ether (rational formula; structural isomerism): CH 3 -O-CH 3.

There are other ways to write chemical formulas. New methods appeared in the late 1980s with the development of personal computer technology (SMILES, WLN, ROSDAL, SLN, etc.). IN personal computers Special software tools called molecular editors are also used to work with chemical formulas.

Notes

Basic concepts of chemistry (undefined) (unavailable link). Retrieved November 23, 2009. Archived November 21, 2009.
Distinguish empirical And true formulas. Empirical formula expresses the simplest formula substance (chemical compound), which is determined by elemental analysis. Thus, the analysis shows that simplest, or empirical, the formula of some compound corresponds to CH. True Formula shows how many of these simplest CH groups are contained in the molecule. Let's imagine true formula in the form (CH) x, then at x = 2 we have acetylene C 2 H 2, at x = 6 we have benzene C 6 H 6.
Strictly speaking, you cannot use the terms “ molecular formula" And " molecular mass"salts, since salts do not contain molecules, but only ordered lattices consisting of ions. None of the sodium ions [cations] in the sodium chloride structure “belong” to any particular chloride ion [anion]. It's correct to talk about chemical formula salt & its corresponding formula mass. Because the chemical formula (true) sodium chloride - NaCl, formula mass sodium chloride is defined as the sum of the atomic masses of one sodium atom and one chlorine atom: 1 sodium atom: 22.990 a. eat.
1 chlorine atom: 35.453 a. eat.
-----------
Total: 58,443 a. eat.
It is customary to call this quantity “

Well, to complete our acquaintance with alcohols, I will also give the formula of another well-known substance - cholesterol. Not everyone knows that it is a monohydric alcohol!

|`/`\\`|<`|w>`\`/|<`/w$color(red)HO$color()>\/`|0/`|/\<`|w>|_q_q_q<-dH>:a_q|0<|dH>`/<`|wH>`\|dH; #a_(A-72)<_(A-120,d+)>-/-/<->`\

I marked the hydroxyl group in it in red.

Carboxylic acids

Any winemaker knows that wine should be stored without access to air. Otherwise it will turn sour. But chemists know the reason - if you add another oxygen atom to an alcohol, you get an acid.
Let's look at the formulas of acids that are obtained from alcohols already familiar to us:

Substance			Skeletal formula
Methane acid (formic acid)	H/C`\|O\|\OH	HCOOH	O//\OH
Ethanoic acid (acetic acid)	H-C-C\O-H; H\|#C\|H	CH3-COOH	/`\|O\|\OH
Propanic acid (methylacetic acid)	H-C-C-C\O-H; H\|#2\|H; H\|#3\|H	CH3-CH2-COOH	\/`\|O\|\OH
Butanoic acid (butyric acid)	H-C-C-C-C\O-H; H\|#2\|H; H\|#3\|H; H\|#4\|H	CH3-CH2-CH2-COOH	/\/`\|O\|\OH
Generalized formula	(R)-C\O-H	(R)-COOH or (R)-CO2H	(R)/`\|O\|\OH

A distinctive feature of organic acids is the presence of a carboxyl group (COOH), which gives such substances acidic properties.

Anyone who has tried vinegar knows that it is very sour. The reason for this is the presence of acetic acid in it. Typically table vinegar contains between 3 and 15% acetic acid, with the rest (mostly) water. Consumption of acetic acid in undiluted form poses a danger to life.

Carboxylic acids can have multiple carboxyl groups. In this case they are called: dibasic, tribasic etc...

Food products contain many other organic acids. Here are just a few of them:

The name of these acids corresponds to the food products in which they are contained. By the way, please note that here there are acids that also have a hydroxyl group, characteristic of alcohols. Such substances are called hydroxycarboxylic acids(or hydroxy acids).
Below, under each of the acids, there is a sign specifying the name of the group of organic substances to which it belongs.

Radicals

Radicals are another concept that has influenced chemical formulas. The word itself is probably known to everyone, but in chemistry radicals have nothing in common with politicians, rebels and other citizens with an active position.
Here these are just fragments of molecules. And now we will figure out what makes them special and get acquainted with a new way of writing chemical formulas.

Generalized formulas have already been mentioned several times in the text: alcohols - (R)-OH and carboxylic acids - (R)-COOH. Let me remind you that -OH and -COOH are functional groups. But R is a radical. It’s not for nothing that he is depicted as the letter R.

To be more specific, a monovalent radical is a part of a molecule lacking one hydrogen atom. Well, if you subtract two hydrogen atoms, you get a divalent radical.

Radicals in chemistry received proper names. Some of them even received Latin designations similar to the designations of the elements. And besides, sometimes in formulas radicals can be indicated in abbreviated form, more reminiscent of gross formulas.
All this is demonstrated in the following table.

Name	Structural formula	Designation	Brief formula	Example of alcohol
Methyl	CH3-()	Me	CH3	(Me)-OH	CH3OH
Ethyl	CH3-CH2-()	Et	C2H5	(Et)-OH	C2H5OH
I cut through	CH3-CH2-CH2-()	Pr	C3H7	(Pr)-OH	C3H7OH
Isopropyl	H3C\CH(`/H3C)-()	i-Pr	C3H7	(i-Pr)-OH	(CH3)2CHOH
Phenyl	`/`=`\//-\\-{}	Ph	C6H5	(Ph)-OH	C6H5OH

I think everything is clear here. I just want to draw your attention to the column where examples of alcohols are given. Some radicals are written in a form that resembles the gross formula, but the functional group is written separately. For example, CH3-CH2-OH turns into C2H5OH.
And for branched chains like isopropyl, structures with brackets are used.

There is also such a phenomenon as free radicals. These are radicals that, for some reason, have separated from functional groups. In this case, one of the rules with which we began studying the formulas is violated: the number of chemical bonds no longer corresponds to the valence of one of the atoms. Well, or we can say that one of the connections becomes open at one end. Free radicals usually live for a short time as the molecules tend to return to a stable state.

Introduction to nitrogen. Amines

I propose to get acquainted with another element that is part of many organic compounds. This nitrogen.
It is denoted by the Latin letter N and has a valency of three.

Let's see what substances are obtained if nitrogen is added to the familiar hydrocarbons:

Substance	Expanded structural formula	Simplified structural formula	Skeletal formula
Aminomethane (methylamine)	H-C-N\H;H\|#C\|H	CH3-NH2	\NH2
Aminoethane (ethylamine)	H-C-C-N\H;H\|#C\|H;H\|#3\|H	CH3-CH2-NH2	/\NH2
Dimethylamine	H-C-N<`\|H>-C-H; H\|#-3\|H; H\|#2\|H	$L(1.3)H/N<_(A80,w+)CH3>\dCH3	/N<_(y-.5)H>\
Aminobenzene (Aniline)	H\N\|C\\C\|C<\H>`//C<\|H>`\C<`/H>`\|\|C<`\H>/	NH2\|C\\CH\|CH`//C<_(y.5)H>`\HC`\|\|HC/	NH2\|\\|`/`\`\|/_o
Triethylamine	$slope(45)H-C-C/N\C-C-H;H\|#2\|H; H\|#3\|H; H\|#5\|H;H\|#6\|H; #N`\|C<`-H><-H>`\|C<`-H><-H>`\|H	CH3-CH2-N<`\|CH2-CH3>-CH2-CH3	\/N<`\|/>\\|

As you probably already guessed from the names, all these substances are united under the general name amines. The functional group ()-NH2 is called amino group. Here are some general formulas of amines:

In general, there are no special innovations here. If these formulas are clear to you, then you can safely engage in further study of organic chemistry using a textbook or the Internet.
But I would also like to talk about formulas in inorganic chemistry. You will see how easy it will be to understand them after studying the structure of organic molecules.

Rational formulas

It should not be concluded that inorganic chemistry is easier than organic chemistry. Of course, inorganic molecules usually look much simpler because they don't tend to form such complex structures like hydrocarbons. But then we have to study more than a hundred elements that make up the periodic table. And these elements tend to combine according to their chemical properties, but with numerous exceptions.

So, I won’t tell you any of this. The topic of my article is chemical formulas. And with them everything is relatively simple.
Most often used in inorganic chemistry rational formulas. And now we’ll figure out how they differ from those already familiar to us.

First, let's get acquainted with another element - calcium. This is also a very common element.
It is designated Ca and has a valency of two. Let's see what compounds it forms with the carbon, oxygen and hydrogen we know.

Substance	Structural formula	Rational formula
Calcium oxide	Ca=O	CaO
Calcium hydroxide	H-O-Ca-O-H	Ca(OH)2
Calcium carbonate	$slope(45)Ca`/O\C\|O`\|/O`\#1	CaCO3
Calcium bicarbonate	HO/`\|O\|\O/Ca\O/`\|O\|\OH	Ca(HCO3)2
Carbonic acid	H\|O\C\|O`\|/O`\|H	H2CO3

At first glance, you can see that the rational formula is something between a structural and a gross formula. But it is not yet very clear how they are obtained. To understand the meaning of these formulas, you need to consider the chemical reactions in which substances participate.

Calcium in its pure form is a soft white metal. It does not occur in nature. But it is quite possible to buy it at a chemical store. It is usually stored in special jars without access to air. Because in air it reacts with oxygen. Actually, that’s why it doesn’t occur in nature.
So, the reaction of calcium with oxygen:

2Ca + O2 -> 2CaO

The number 2 before the formula of a substance means that 2 molecules are involved in the reaction.
Calcium and oxygen produce calcium oxide. This substance also does not occur in nature because it reacts with water:

CaO + H2O -> Ca(OH2)

The result is calcium hydroxide. If you look closely at its structural formula (in the previous table), you can see that it is formed by one calcium atom and two hydroxyl groups, with which we are already familiar.
These are the laws of chemistry: if a hydroxyl group is added to an organic substance, an alcohol is obtained, and if it is added to a metal, a hydroxide is obtained.

But calcium hydroxide does not occur in nature due to the presence of carbon dioxide in the air. I think everyone has heard about this gas. It is formed during the respiration of people and animals, the combustion of coal and petroleum products, during fires and volcanic eruptions. Therefore, it is always present in the air. But it also dissolves quite well in water, forming carbonic acid:

CO2 + H2O<=>H2CO3

Sign<=>indicates that the reaction can proceed in both directions under the same conditions.

Thus, calcium hydroxide, dissolved in water, reacts with carbonic acid and turns into slightly soluble calcium carbonate:

Ca(OH)2 + H2CO3 -> CaCO3"|v" + 2H2O

A down arrow means that as a result of the reaction the substance precipitates.
With further contact of calcium carbonate with carbon dioxide in the presence of water, a reversible reaction occurs to form an acidic salt - calcium bicarbonate, which is highly soluble in water

CaCO3 + CO2 + H2O<=>Ca(HCO3)2

This process affects the hardness of the water. When the temperature rises, bicarbonate turns back into carbonate. Therefore, in regions with hard water, scale forms in kettles.

Chalk, limestone, marble, tuff and many other minerals are largely composed of calcium carbonate. It is also found in corals, mollusk shells, animal bones, etc...
But if calcium carbonate is heated very hot high fire, then it will turn into calcium oxide and carbon dioxide.

This short story about the calcium cycle in nature should explain why rational formulas are needed. So, rational formulas are written so that the functional groups are visible. In our case it is:

In addition, individual elements - Ca, H, O (in oxides) - are also independent groups.

Ions

I think it's time to get acquainted with ions. This word is probably familiar to everyone. And after studying the functional groups, it doesn’t cost us anything to figure out what these ions are.

In general, the nature of chemical bonds is usually that some elements give up electrons while others gain them. Electrons are particles with a negative charge. An element with a full complement of electrons has zero charge. If he gave away an electron, then its charge becomes positive, and if he accepted it, then it becomes negative. For example, hydrogen has only one electron, which it gives up quite easily, turning into a positive ion. There is a special entry for this in chemical formulas:

H2O<=>H^+ + OH^-

Here we see that as a result electrolytic dissociation water breaks down into a positively charged hydrogen ion and a negatively charged OH group. The OH^- ion is called hydroxide ion. It should not be confused with the hydroxyl group, which is not an ion, but part of some kind of molecule. The + or - sign in the upper right corner shows the charge of the ion.
But carbonic acid never exists as an independent substance. In fact, it is a mixture of hydrogen ions and carbonate ions (or bicarbonate ions):

H2CO3 = H^+ + HCO3^-<=>2H^+ + CO3^2-

The carbonate ion has a charge of 2-. This means that two electrons have been added to it.

Negatively charged ions are called anions. Typically these include acidic residues.
Positively charged ions - cations. Most often these are hydrogen and metals.

And here you can probably fully understand the meaning of rational formulas. The cation is written in them first, followed by the anion. Even if the formula does not contain any charges.

You probably already guess that ions can be described not only by rational formulas. Here is the skeletal formula of the bicarbonate anion:

Here the charge is indicated directly next to the oxygen atom, which received an extra electron and therefore lost one line. Simply put, each extra electron reduces the number of chemical bonds depicted in the structural formula. On the other hand, if some node of the structural formula has a + sign, then it has an additional stick. As always, this fact needs to be demonstrated with an example. But among the substances familiar to us there is not a single cation that consists of several atoms.
And such a substance is ammonia. Its aqueous solution is often called ammonia and is included in any first aid kit. Ammonia is a compound of hydrogen and nitrogen and has the rational formula NH3. Consider the chemical reaction that occurs when ammonia is dissolved in water:

NH3 + H2O<=>NH4^+ + OH^-

The same thing, but using structural formulas:

H|N<`/H>\H + H-O-H<=>H|N^+<_(A75,w+)H><_(A15,d+)H>`/H + O`^-# -H

On the right side we see two ions. They were formed as a result of one hydrogen atom moving from a water molecule to an ammonia molecule. But this atom moved without its electron. The anion is already familiar to us - it is a hydroxide ion. And the cation is called ammonium. It exhibits properties similar to metals. For example, it may combine with an acidic residue. The substance formed by combining ammonium with a carbonate anion is called ammonium carbonate: (NH4)2CO3.
Here is the reaction equation for the interaction of ammonium with a carbonate anion, written in the form of structural formulas:

2H|N^+<`/H><_(A75,w+)H>_(A15,d+)H + O^-\C|O`|/O^-<=>H|N^+<`/H><_(A75,w+)H>_(A15,d+)H`|0O^-\C|O`|/O^-|0H_(A-15,d-)N^+<_(A105,w+)H><\H>`|H

But in this form the reaction equation is given for demonstration purposes. Typically equations use rational formulas:

2NH4^+ + CO3^2-<=>(NH4)2CO3

Hill system

So, we can assume that we have already studied structural and rational formulas. But there is another issue that is worth considering in more detail. How do gross formulas differ from rational ones?
We know why the rational formula of carbonic acid is written H2CO3, and not some other way. (The two hydrogen cations come first, followed by the carbonate anion.) But why is the gross formula written CH2O3?

In principle, the rational formula of carbonic acid may well be considered a true formula, because it has no repeating elements. Unlike NH4OH or Ca(OH)2.
But an additional rule is very often applied to gross formulas, which determines the order of elements. The rule is quite simple: carbon is placed first, then hydrogen, and then the remaining elements in alphabetical order.
So CH2O3 comes out - carbon, hydrogen, oxygen. This is called the Hill system. It is used in almost all chemical reference books. And in this article too.

A little about the easyChem system

Instead of a conclusion, I would like to talk about the easyChem system. It is designed so that all the formulas that we discussed here can be easily inserted into the text. Actually, all the formulas in this article are drawn using easyChem.

Why do we even need some kind of system for deriving formulas? The whole point is that standard way display of information in Internet browsers is hypertext markup language (HTML). It is focused on processing text information.

Rational and gross formulas can be depicted using text. Even some simplified structural formulas can also be written in text, for example alcohol CH3-CH2-OH. Although for this you would have to use the following entry in HTML: CH ₃-CH ₂-OH.
This of course creates some difficulties, but you can live with them. But how to depict the structural formula? In principle, you can use a monospace font:

H H | | H-C-C-O-H | | H H Of course it doesn’t look very nice, but it’s also doable.

The real problem comes when trying to draw benzene rings and when using skeletal formulas. There is no other way left except connecting a raster image. Rasters are stored in separate files. Browsers can include images in gif, png or jpeg format.
To create such files, a graphic editor is required. For example, Photoshop. But I have been familiar with Photoshop for more than 10 years and I can say for sure that it is very poorly suited for depicting chemical formulas.
Molecular editors cope with this task much better. But with a large number of formulas, each of which is stored in a separate file, it is quite easy to get confused in them.
For example, the number of formulas in this article is . They are displayed in the form of graphic images (the rest using HTML tools).

The easyChem system allows you to store all formulas directly in an HTML document in text form. In my opinion, this is very convenient.
In addition, the gross formulas in this article are calculated automatically. Because easyChem works in two stages: first the text description is converted into an information structure (graph), and then various actions can be performed on this structure. Among them, the following functions can be noted: calculation of molecular weight, conversion to a gross formula, checking for the possibility of output as text, graphic and text rendering.

Thus, to prepare this article, I only used a text editor. Moreover, I didn’t have to think about which of the formulas would be graphic and which would be text.

Here are a few examples that reveal the secret of preparing the text of an article: Descriptions from the left column are automatically turned into formulas in the second column.
In the first line, the description of the rational formula is very similar to the displayed result. The only difference is that the numerical coefficients are displayed interlinearly.
In the second line, the expanded formula is given in the form of three separate chains separated by a symbol; I think it is easy to see that the textual description is in many ways reminiscent of the actions that would be required to depict the formula with a pencil on paper.
The third line demonstrates the use of slanted lines using the \ and / symbols. The ` (backtick) sign means the line is drawn from right to left (or bottom to top).

There is much more detailed documentation on using the easyChem system here.

Let me finish this article and wish you good luck in studying chemistry.

A brief explanatory dictionary of terms used in the article

Hydrocarbons Substances consisting of carbon and hydrogen. They differ from each other in the structure of their molecules. Structural formulas are schematic images of molecules, where atoms are denoted by Latin letters and chemical bonds by dashes. Structural formulas are expanded, simplified and skeletal. Expanded structural formulas are structural formulas where each atom is represented as a separate node. Simplified structural formulas are those structural formulas where hydrogen atoms are written next to the element with which they are associated. And if more than one hydrogen is attached to one atom, then the amount is written as a number. We can also say that groups act as nodes in simplified formulas. Skeletal formulas are structural formulas where carbon atoms are depicted as empty nodes. The number of hydrogen atoms bonded to each carbon atom is equal to 4 minus the number of bonds that converge at the site. For knots formed not by carbon, the rules of simplified formulas apply. Gross formula (aka true formula) - a list of all chemical elements that make up the molecule, indicating the number of atoms in the form of a number (if there is one atom, then the unit is not written) Hill system - a rule that determines the order of atoms in the gross formula formula: carbon is placed first, then hydrogen, and then the remaining elements in alphabetical order. This is a system that is used very often. And all the gross formulas in this article are written according to the Hill system. Functional groups Stable combinations of atoms that are conserved during chemical reactions. Often functional groups have their own names and influence Chemical properties and scientific name of the substance

This approach is also prompted by the similarity of the gross formulas of sulfuric, selenic and so-called telluric acids - H2SO4, H2SeO4 and H2TeO4, respectively. However, if the first two compounds fully correspond to the structural concepts of acids, since they contain isolated tetrahedral complex radicals 2- or 2- with a coordination number of S and Se equal to 4, which gives grounds for writing their structural formulas in the form of H2 and H2, this cannot be said about "telluric acid". The study of this compound did not reveal any anionic groups 2- with CN Te = 4 in its structure. Instead, it was found that Te6+ ions have CN = 6, i.e. correspond to the CN of amphoteric or weakly acidic anion formers. The structure of this compound turned out to consist of chains of TeO4(OH)2 - octahedra, in two opposite vertices of which there are OH ions, connected to each other by common O atoms of the equatorial vertices of the octahedra. It is easy to see that by cutting out the repeatability element of such a structure, we obtain a structural formula in the form Te(OH)2O2. Thus, this compound is a Te6+ hydroxide oxide with very slightly acidic properties, sharply distinguishing it from sulfuric and selenic acids.

Slide 109 from the presentation “Systematics of Minerals” for chemistry lessons on the topic “Minerals”

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Minerals

"Chemistry of Minerals" - Minerals and mineralogy are of extremely great interest. Minerals. Minerals in nature. Among industrially valuable minerals, it is customary to distinguish two groups. Properties of minerals. Minerals in foods. Precious minerals. The importance of minerals in human life. Minerals have played an important role in human development.

“Systematics of minerals” - Metals, the elements of which occupy the largest left part. Cenosymmetric element. A family of zeolites that includes subfamilies. Basic requirements for the taxonomy of minerals. There are incommensurably more connections of different elements with each other. Assignment of a mineral to a specific class of oxysols. Minerals are predominantly of the covalent-ionic and ionic type.

“Classification of minerals” - Cosmic body. Quartz. Opal. Classification of minerals. Sphalerite. Class of native elements. Halite. Silicates are characterized by complex chemical composition. Dolomite. Coloring. Silicates. Minerals of the sulfate class. Minerals. Quartz and chalcedony. Silicate class. The most common mineral of the first class is sulfur.

“Ural Gems” - But especially prized: green patterned malachite and pink eagle. Often in the form of crystals or their fragments. Products with DIAMONDS. Diamond. Precious stones are found in nature in a wide variety of shapes and forms. Emerald (obsolete: Smaragd) - gem 1st class. Emerald.

“Ores of ferrous and non-ferrous metals” - Get acquainted with educational material. Flaw. Use of steel and cast iron. Ore. Rust. Basic properties of metals. Material about ore. How to determine which metal is ferrous and which is non-ferrous. Iron. Expected results.

“Gold deposit” - Radioactive elements. Coal. Mineral resource base. Antimony. Deposits of tin and tungsten. Non-metallic minerals. Oil and gas. Fossil fuels. Colored and rare metals. Dynamics of annual gold production. Gold. Antimony deposits. Gold deposits. Tin and tungsten. Improving legislation in the mining sector.

Simple chemical calculations. Basic concepts and laws of chemistry Chemical symbolism A chemical symbol (symbol of a chemical element) is used as an abbreviation for the name of an element. As a sign they usually take one or two letters from Latin name elements. Si - copper (cuprum), Au - gold (Aurum), etc. The system of chemical signs was proposed in 1811 by the Swedish scientist J. Verzelius. The chemical sign means: [D) the name of the element; 1 mole of its atoms; |T] atomic number; [b] relative atomic mass of the element. Chemical calculations A chemical formula is an expression of the composition of a substance using chemical symbols. From the chemical formula you can find out: (TJ the name of the substance; (2] one of its molecules; how many moles of atoms of each element contains one mole of the substance. Characterizing the relationship between the mass amounts of elements that make up the substance, the formula makes it possible to calculate the mass of each element in the compound and its mass fraction. Example 1 Calculate the mass fraction of hydrogen in ammonia. Given: M(N) = 14 g/mol M(H) = 1 g/mol Find: w(H) Solution: 1) Determine the xmolar mass of NH3: M( NH3) = 14 + 1- 3 = 17 g/mol 2) Determine the mass of ammonia in the amount of substance 1 mol: m(NH3) = 1 mol 17 g/mol = 17 g 3) From the formula of ammonia it follows that the amount of substance atomic hydrogen is 3 times more than the amount of substance NH3: v(H) - 3v(NH3), v(H) = 3-1 = 3 mol. 4) Calculate the mass of hydrogen: m = v M; t(H) = 3 1 = 3 g. 5) Find the mass fraction of hydrogen in ammonia: c;(H) = - = 0.176 or 17.6%. 17 Answer: w(H) = 17.6%. CG Example 2 Calculate the mass of phosphorus that can be obtained from 620 kg of calcium orthophosphate. Given: m(Ca3(P04)2) = 620 kg Find: t(P Solution: 1) Determine the molar mass of Ca3 (P04)2: M(Ca3 (P04)2) = 40 3 + 31 2 + 16 8 = 310 g/mol. 2) Calculate the amount of calcium orthophosphate: = 2 03 mol. 3) From the formula of calcium orthophosphate it follows that the amount of atomic phosphorus substance is 2 times greater than the amount of substance Ca3(P04)2: v(P) = 2v(Ca3(P04)2), v(P) = 2 2 103 - 4 103 mole. 4) Find the mass of phosphorus; t(P) - 4 103 31 = 124 kg. Answer: t(P) = 124kg. There are simple and true (molecular) formulas. The simplest formula expresses the smallest ratio between the numbers of atoms of the elements included in the molecule. The true formula shows the actual number of atoms in the molecule corresponding to the smallest ratio. To establish the true formula, you need to know not only the mass composition of the substance, but also its molecular weight. w(C) = 75% Find: Solution: 1) Select the mass of the unknown compound per 100 g. Then the masses of the elements H and C are equal: E Example 3 Derive the formula for a compound containing 25% hydrogen and 75% carbon. /n(H) = 100-0.25 = 25 g, m(C) = 100 0.75 = 75 g. 2) Determine the amounts of substances of the atomic elements H and C: 25 75 v(H) = -- = 25 mol, v(C) = -- = 6.25 mol. 1 A/ 3) We make up the quantitative ratio of substances: v(H): v(C) - 25: 6.25. 4) Divide the right side of the proportion by a smaller number (6.25) and obtain the ratio of atoms in the formula of the unknown compound: *(C): y(H) = 1:4. The simplest formula of the compound is CH4. Answer: CH4. Example 4 Upon complete combustion of 2.66 g of a certain substance, 1.54 g of carbon monoxide (IV) and 4.48 g of sulfur oxide (IV) were formed. The vapor density of this substance in air is 2.62. Derive the true formula of this substance. Given: m(C02) = 1.54 g m(S02) = 4.48 g Find: the true formula of the substance Solution: 1) Calculate the amounts of carbon monoxide (IV) and sulfur oxide (IV): 1.54 v(C02 ) = --- = 0.035 mol, 44 4.48 v(S02) ~ -GT~ = 0.07 mol. 64 2) Determine the amounts of atomic carbon and sulfur substances: v(C) = v(C02) - 0.035 mol, v(S) = v(S02) = 0.07 mol. 3) Find the masses of carbon and sulfur: /72(C) = 0.035-12 "0.42 g, m(S) = 0.07 32 "2.24 g. The total mass of these elements is 2.66 g and is equal to the mass burning substance. Therefore, it consists only of carbon and sulfur. 4) Find the simplest formula of the substance: v(C): v(S) - 0.035: 0.07 - 1:2. The simplest formula is CS2. 5) Determine the molar mass of CS2: M(CS2) = 12 + 32 2 = 76 g/mol. 6) Calculate the true formula of the substance: Af = 29 1> = 29 2.62 - 76 g/mol. IST. VOED. * "Thus, the true formula of the substance coincides with the simplest one. Answer: L/ist - 76 g/mol. Example 5 Derive the true formula of an organic compound containing 40.03% C, 6.67% H and 53.30% O. The molar mass of this compound is 180 g/mol. Given: u>(C) = 40.03% w(H) - 6.67% w(0) = 53.30% t(CxH02) = 180 g/mol Find : схяуог Solution: 1) Let us denote the number of carbon atoms by x, the number of hydrogen atoms by y, the number of oxygen atoms by z. 2) We divide the percentage of elements accordingly by the values of their relative atomic masses and find the ratios between atoms in the molecule of a given compound: 40, 03 6.67 53.30 x:y: z = 3.33: 6.67: 3.33 3) We bring the found values to integer values: x: y: z = 1: 2: 1. The simplest formula of an organic compound will be CH20. The molar mass is: (12 + 2 + 16)-30 g/mol. The molar mass of the simplest formula is 6 times 180:30 = 6 less than the molar mass of the true formula of this compound. Therefore, to derive the true formula of an organic compound, the number of atoms is needed multiply by 6. Then we get C6H1206. Answer: SbN12Ob. Example 6 Establish the formula of calcium chloride crystalline hydrate if upon calcination of 6.57 g of it, 3.24 g of condensed water was released. Given: /l(CaC12 *H20) = 6.57 g m(H20) = 3.24 g Find: crystalline hydrate formula Solution: 1) Calculate the mass of the anhydrous CaC12 salt contained in the crystalline hydrate: t(CaC12) - 6.57 - 3.2 = 3.33 g. 2) Determine the amounts of substances CaCl2 and H20: 3 33 v(CaCL) - ---- 0.03 mol, 111 3.24 v(H90) --- 0.18 mol. 2 18 3) Find the formula of the crystalline hydrate: v(CaCl2): v(H20) = 0.03: 0.18 = 1:6. The formula of the crystalline hydrate is CaC12 6H20. Answer: CaC12 6H20. Chemical equation is an image chemical reaction using chemical symbols and formulas. The equation characterizes both the qualitative side of the reaction (which substances entered into a chemical reaction and which were obtained during it) and the quantitative side (what are the quantitative relationships between the masses or volumes of the gases of the starting substances and the reaction products). Reflection of the quantitative side of chemical processes by equations allows for various calculations to be made on their basis: finding the mass or volume of starting substances to obtain a given amount of reaction products, mass or volume of new substances that can be obtained from a given amount of starting substances, etc. Example 7 What What mass of aluminum must be taken to reduce iron from 464 g of iron scale? Given: m(Fe304) = 464 g Find: m(A1) Solution: 1) Write down the reaction equation and indicate the quantitative ratios of the required substances: 8A1 + 3Fe304 - 9Fe + 4A1203. 8 mol 3 mol 2) Determine the molar mass of Fe304: M(Fe304) - 56 3 + 16 4 = 232 g/mol. 3) Calculate the amount of iron scale substance (Fe304): 464 v(Fe304) - -- = 2 mol. Using the equation of a chemical reaction, you can calculate which substance and in what quantity is taken in excess (or deficiency) during the interaction of given amounts of reacting substances. Example 9 Iron filings weighing 5.6 g were added to a solution containing 37.6 g of copper nitrate. Calculate whether copper nitrate will remain in the solution after the end of the chemical reaction. Given: /n(Cu(N03)3) = 37.6 g m(Fe) - 5.6 g Find: will copper nitrate remain in the solution Solution: 1) Write the reaction equation: Cu(N03)2 + Fe = Fe (N03)2 + Si. 2) Find the molar mass of Cu(N03)2: M(Cu(N03)2) = 64 + 14 2 + 16 6 - 188 g/mol. 3) Determine the amounts of substances Cu(N03)2 and Fe: 37.6 v(Cu(N03)2) = -n- = 0.2 mol, v(Fe) = - " =0.1 mol. 56 4) We calculate the amount of substance Cu (N03)2 according to the reaction equation according to the proportion: 1 mol Cu(N03)2 - 1 mol Fe v mol Cu(N03)2 - 0.1 mol Fe v(Cu(N03)2) = 0.1 mol. Comparing the initial amount of Cu(N03)2 and required for the reaction, we conclude that the amount of Cu(N03)2 was taken in excess. Calculation of the amount of reactants and reaction products must be carried out based on the amount of the substance taken in deficiency. In our case - for Fe. We calculate the amount of substance and mass of Cu(N03)2 in solution after the reaction: v(Cu(N03)2) = 0.2 - 0.1 = 0.1 mol, m(Cu(N03)2) = 0 ,1 188 = 18.8 g. Answer: m(Cu(N03)2) = 18.8 g. Using the chemical equation, calculations can also be made in the case when the starting substance contains a certain specified amount of impurities. Example 10 Calculate the amount of sodium nitrite that is formed when 1 kg of Chilean saltpeter containing 85% NaN03 is calcined. Given: /n(nitrate) = 1 kg to(NaN03) = 85% Find: m(NaN02) Solution: 1) Write the reaction equation: 2NaN03 = 2NaN02 + 02|. 2) Determine the mass of NaN03: t(nitrate) to(NaN03) m(NaNOo) = 37 100% 1 103- 85% m(NaN03) = = 850 g. v(NaN03) = = 10 mol. 3) Determine the amount of substance NaN03: 850 85 4) Calculate the amount of substance NaN02 according to the reaction equation according to the proportion: 2 mol NaN03 - 2 mol NaN02 10 mol NaNO. - v mol NaNO., Find the mass of NaN02: m(NaN02) = 10 69 = 690 g. Answer: m(NaN02) = 690 g. Based on the chemical reaction equation (or chemical formula), problems on product yield are solved. Example 11 Sand weighing 2 kg was fused with an excess of potassium hydroxide, resulting in a reaction that produced potassium silicate weighing 3.82 kg. Determine the yield of the reaction product if the mass fraction of silicon (IV) oxide in the sand is 90%. Given: t(sand) = 2 kg 0)(Si02) = 90% m(K2Si03) = 3.82 kg Find: 4(K2Si03) Solution: 1) Write the reaction equation: Si02 + 2KOH = K2Si03 + H20. 2) Determine the mass of Si02: t(sand) 90% 2>"wG%-2 90% t(810^)=-tshg=1"8kg- 3) Determine the amount of substance Si02: 1.8-103 v(Si02) ---- = 30 mol. E-Ll 4) We calculate the amount of substance K2Si03 according to the reaction equation according to the proportion: 1 mol Si02 - 1 mol K2Si03 30 mol Si02 - v mol K2Si03 v(K2Si03) = 30 mol. 5) Find the mass of K2Si03, which should be formed in accordance with the theoretical calculation: m(K2Si03) - 30 154 - 4620 g or 4.62 kg. 6) We calculate the yield of the reaction product: 3.82 100% llgtl/ L-Shch - 82.7%. Answer: Ti(K2Si03) - 82.7%. Problems for independent solution 1. Calculate the mass fraction of each element in the following chromium compounds: a) Fe(Cr02)2; b) Cr2(S04)3; c) (NH4)2Cr04. 2. Calculate the mass of copper contained in 444 g of basic copper carbonate. Answer: 256 g. 3. Calculate the mass of iron that can be obtained from 320 g of red iron ore. Answer: 224 g. 4. How many moles of lead nitrate are contained in: a) 414 g of lead; b) 560 g of nitrogen; c) 768 g of oxygen. Answer: a) 2 mol; b) 20 mol; c) 8 mol. 5. Calculate the mass of phosphorus that can be obtained from 1 ton of phosphorite containing 31% calcium orthophosphate. Answer: 62 kg. 6. Unpurified Glauber's salt contains 94% crystalline hydrate. Calculate the mass of anhydrous sodium sulfate that can be obtained from 6.85 tons of this raw material. Answer: 2.84 t. 7. Derive the simplest formula of a compound containing 44.89% potassium, 18.37% sulfur and 36.74% oxygen. Answer: K2S04. 8. The mineral copper luster contains 79.87% copper and 20.13% sulfur. Find the formula of the mineral. Answer: Cu2S. 9. Calcium or magnesium, burning in a nitrogen atmosphere, form compounds containing 18.92% and 27.75% nitrogen, respectively. Find the formulas of these compounds. Answer: Ca3N2; Mg3N2. 10. Hydrocarbon contains 85.72% carbon and 14.28% hydrogen. Find its formula and determine which homological series it belongs to. Answer: C2H4. 11. The molar mass of the compound is 98 g/mol. Determine the formula of this compound containing 3.03% H, 31.62% P and 65.35% O. Answer: H3P04. 12. When organic matter consisting of carbon, hydrogen and sulfur is burned, 2.64 g of carbon monoxide (IV), 1.62 g of water and 1.92 g of sulfur oxide (IV) are obtained. Find the formula of this substance. Answer: C2H6S. 13. Establish the true formula of an organic substance if, when burning 2.4 g of it, 5.28 g of carbon monoxide (IV) and 2.86 g of water were obtained. The vapor density of this substance for hydrogen is 30. Answer: C3H80. 14. Establish the formula of one of the crystalline hydrates of sodium sulfate if, during its dehydration, the weight loss is 20.22% of the mass of the crystalline hydrate. Answer: Na2S04 2H20. 15. 0.327 g of zinc was dissolved in sulfuric acid and 1.438 g of zinc salt crystalline hydrate was crystallized from the resulting solution. Determine the formula of crystalline hydrate. Answer: ZnS04 7N20. 16. When tungsten (VI) oxide was reduced with hydrogen, 27 g of water was formed. What mass of tungsten can be obtained in this way? Answer: 92 g. 17. An iron plate was immersed in a solution of copper sulfate. After some time, the mass of the plate increased by 1g. What mass of copper is deposited on the plate? Answer: 8 g. 18. Determine which substance and in what quantity will remain in excess as a result of the reaction between 4 g of magnesium oxide and 10 g of sulfuric acid. Answer: 0.20 g H2S04. 19. What volume of carbon dioxide will be required to convert 50 g of calcium carbonate into bicarbonate? Answer: 11.2l C02. 20. What composition and in what quantity is the salt formed by the interaction of a solution containing 9 g of sodium hydroxide with carbon dioxide formed by burning 2.24 liters of methane? Answer: 11.9 g Na2C03. 21. The decomposition of 44.4 g of malachite yielded 4.44 liters of carbon monoxide (IV) (n.o.). Determine the mass fraction (%) of impurities in malachite. Answer: 0.9%. 22. When a mixture of magnesium and magnesium oxide weighing 5 g was treated with hydrochloric acid, 4 liters of hydrogen were released. Calculate the mass fraction of magnesium in the mixture. Answer: 85.7%. 23. What volume of ammonia (n.a.) will be obtained by heating a mixture of 5.35 g of ammonium chloride with 10 g of calcium hydroxide? Answer: 2.24 l. 24. What mass of silicon, containing 8% impurities, reacted with a solution of sodium hydroxide if 5.6 liters of hydrogen (n.e.) were released? Answer: 3.8 g. 25. Phosphoric acid weighing 195 kg was obtained from natural phosphorite weighing 310 kg. Calculate the mass fraction of Ca3(P04)2 in natural phosphorite. Answer: 99.5%.

In mineralogy, it is important to be able to calculate the formula of a mineral based on the results of its chemical analysis. This section provides a number of examples of such calculations for different minerals. When the calculations are made and the structural formula is obtained, it becomes clear whether it coincides with the crystal chemical data for the mineral. It should be noted that even if the total sum of components in the analysis turns out to be equal to 100%, this does not always mean that the composition of the mineral is determined correctly and accurately.

5.7.1 Calculation of sulfide analysis

In the case of sulfide minerals, analytical results are usually expressed in mass percentages.

Table 5.1 Results of chemical analysis of iron-containing sphalerite from the Renström deposit, North. Sweden (byR. S. Duckworth and D. Richard,Mineral. Mag. 57:83-91, 1993)

Element	Mac.%	Atomic	Atomic
		quantities	ratios
			at S = 1
	57,93	0,886	0,858
	8,21	0,1407	0,136
	33,09	1,032	1,000
Sum	99,23

max (wt.%) of elements. Calculating a formula based on data from such analyzes is a simple arithmetic problem. In the example of iron-bearing sphalerite below (Table 5.1), the first step is to divide the mass percentage of each element by its atomic mass to obtain the mole fraction of that element. The structural formula of iron-bearing sphalerite looks like (Zn, Fe)S, and therefore, in order for the results to have the correct relationships, it is necessary to reduce either the sum of the mole fractions of Zn and Fe, or the mole fraction of S, to unity. The formula used, allowing for both completely cationic and completely anionic lattice, is valid for the case under consideration, and if the results of the analysis are correct, then the formulas calculated by both methods should coincide. Thus, bringing S to unity and rounding the resulting values to the second decimal place, we obtain the formula (Zn 086 Fe 014) 100 S. Some sulfide minerals (for example, pyrrhotite Fe 1-x S) have a non-stoichiometric content of cations. In such cases, analyzes should be calculated based on the amount of sulfur ions.

5.7.2 Calculation of silicate analysis

The results of analyzes of rock-forming minerals (see, for example, the analysis of garnet in Table 5.2) are usually expressed in mass percentage of oxides. The calculation of the analysis presented in this form is somewhat more complex and includes a number of additional operations.

molecular weight, which gives the relative content of oxide molecules (column 2).

2. Calculate the atomic quantities of oxygen. To do this, each value in column 2 is multiplied by the number of oxygen atoms in the corresponding oxides, which gives the relative content of oxygen atoms contributed to the formula by each element (column 3).

At the bottom of column 3 is the total number of oxygen atoms (2.7133).

3. If we want to obtain a garnet formula based on 12 oxygen atoms, then it is necessary to recalculate the ratios of oxygen atoms so that their total number is 12. To do this, the numbers in column 3 for each oxide are multiplied by 12/T, where T is the total amount of oxygen from column 3. The results are shown in column 4.

4. Calculate the atomic ratios for various cations. For this purpose, the numbers in column 4 must be multiplied or divided by the values of these ratios, determined by stoichiometry. So, for example, SiO 2 has one silicon per two oxygen. Therefore, the corresponding number in column 4 is divided by 2. In A1 2 0 3, for every three oxygen atoms there are two aluminum atoms, in which case the number in column 4 is multiplied by 2/3. For divalent cations, the numbers in columns 4 and 5 are the same.

Table 5.2 Results of chemical analysis of garnet, Wesselton mine, Kimberley, South Africa (according toA.D. Edgar and N.E. Charbonneau,Am.Mineral. 78: 132-142, 1993)

Oxide	Mmass% oxides	Molecular quantities oxides	Atomic amount of oxygen in a molecule	Number of anions per 12 O atoms, i.e. column (3) x 4.422	Number of cations in the formula
Si0 2	40,34	0,6714	1,3426	5,937	Si 2.968
A1 2 0 3	18,25	0,1790	0,537	2,374	Al 1.582
	4,84	0,0674	0,0674	0,298	Fe 0.298
	0,25	0,0035	0,0035	0,015	Mn 0.015
Ti0 2	2,10	0,0263	0,0526	0,232	Ti 0.116
Cr 2 0 3	2,22	0,0146	0,0438	0,194	Cr 0.129
	18,77	0,3347	0,3347	1,480	Ca 1.480
	13,37	0,3317	0,3317	1,467	Mg 1.467
Sum	100,14		2,7133
			12/2,7133 = 4,422

The quantities of cations in the formula corresponding to the established number of oxygen atoms (12) and given in column 5 can be grouped as shown in the table in accordance with the structural formula of garnet A 3 B 2 [(Si, Al)0 4 ], where A - divalent cations (Ca, Mg, Fe, Mn), and B - trivalent cations (Al, Cr), as well as Ti 4 +. The deficiency of Si is compensated by Al, which is taken in such quantity as to completely fill the tetrahedral positions. The remaining aluminum atoms are to position B,

To quickly assess the correctness of the arithmetic operations performed, you need to check the balance of valences by summing up the positive and negative charges.

5.7.3 Calculation of analysis in the presence of different anions

In the last example, we will briefly consider the calculation of the formula based on the results of analysis in the presence of different anions in the mineral composition (Table 5.3). In our case, the mineral is represented by fluorine-apatite Ca 5 (PO 4) 3 ^,0,OH), which, in addition to

Table 5.3 Results of chemical analysis of apatite

Oxides	(!) ~	(2.)		Ch 4) Ka number
	Wt.%	Molek	Molek	Ch 4) Ka number
		lar	lar	tions in
		if	quantity	based on
Na2O K2O P2O5 H2O Sum O=FjCl Sum	55,08 0,32 0,02 0,05 0,03 0,04 0,0! 42,40 1,63 0,20 1,06 100,84 -0,72 100,12	quality 0,9822 0,0020 0,0003 0,0012 0,0003 0,0006 0,0001 0,2987 0,0858 0,0056 0,0567 0,0914 3/2, 5409 =	va oxygen 0,9822 0,0060 0,0003 0,0012 0,0003 0,0006 0,0001 1,4935 0,0858 0,0056 0,0567 0,0914 2,5409 4, 9386	13 anions (4.9386) 4,85 0,02 0,01 0,01 2,95 0,42 0,03 0,56

oxygen contains F and Cl. The results of the analysis are again expressed in mass percentage of oxides, although in fact some of them are halogens. In such cases, it is necessary to correct the total amount of oxygen by taking into account the number of moles of oxygen equivalent to the halides present.

So, the calculation includes the following steps.

To do this, the number of moles indicated in column 2 must be multiplied by the stoichiometric

anion number. Remember to subtract the oxygen equivalent (in this case 0.0914 moles) of F and Cl present in the mineral (table bets 3).

3. Sum up the number of anions, remembering to subtract 0.0914 moles of oxygen associated with the F and Cl present (2.5409).

4. If we want to obtain an apatite formula based on 13 anions, then we need to recalculate the ratios of the anions so that their total number is 13. To do this, each of them is multiplied by 13/2.5409, those. at 4.9386.

5. Calculate the ratios of atoms of various cations. To do this, you need to multiply the molecular quantities given in column 2 by 4.9386, and then multiply or divide the resulting values by the values of these ratios, determined by the stoichiometry of the oxides. For example, at P 2 O 5 on There are two phosphorus atoms per mole of oxide. The final results are shown in column 4.

Literature for further study

1. Goldstein, J. L., Newbury, D. E., Echhn, P., Joy, D. S., FiOTi, C. and Lifshm, E. Scanning Electron Microscopy and X-ray Microanalysis. New York, Plenum, 1984.

2. Marfunin, A. S. (ed.]. Methods and Instrumentation: Results and Recent Developments, vol. 2 of Advanced Mineralogy Berlin, Springer-Verlag, 1985.

3. Willard, H. H., Merntt, L. L., Dean, J. A. and Settle, F. A. Instrumental Methods of Analysis, 7th edn. Belmont, CA, Wadsworth, 1988.

Editor's Addition

1. Garanin V.K., Kudryavtseva G.P.The use of electron probe instruments for the study of mineral matter. M, Nedra, 1983, 216 p.

2. Laputina I.P. Microprobe in mineralogy. M., on Science, 1991, 139 p.

The physical properties of minerals are determined by the interaction between structure and chemical composition. These properties include those that affect appearance mineral, such as its luster and color. Other properties affect the physical characteristics of minerals - hardness, piezoelectricity, magnetism. We will first look at the density of minerals, since this property is directly related to their structure and composition.