The main question that a person must know the answer to correctly understand the picture of the world is what is a substance in chemistry. This concept is formed at school age and guides the child in further development. When starting to study chemistry, it is important to find common ground with it at the everyday level, this allows you to clearly and easily explain certain processes, definitions, properties, etc.

Unfortunately, due to the imperfection of the education system, many people miss some fundamental basics. The concept of "substance in chemistry" is a kind of cornerstone, the timely assimilation of this definition gives a person the right start in the subsequent development in the field of natural science.

Concept formation

Before moving on to the concept of matter, it is necessary to define what the subject of chemistry is. Substances are what chemistry directly studies, their mutual transformations, structure and properties. In a general sense, matter is what physical bodies are made of.

So, in chemistry? Let us form a definition by passing from a general concept to a purely chemical one. A substance is a certain thing that necessarily has a mass that can be measured. This characteristic distinguishes matter from another type of matter - a field that has no mass (electric, magnetic, biofield, etc.). Matter, in turn, is what we are made of and everything that surrounds us.

A somewhat different characteristic of matter, which determines what exactly it consists of, is already the subject of chemistry. Substances are formed by atoms and molecules (some ions), which means that any substance consisting of these formula units is a substance.

Simple and complex substances

After mastering the basic definition, you can move on to complicating it. Substances come in different levels of organization, that is, simple and complex (or compounds) - this is the very first division into classes of substances, chemistry has many subsequent divisions, detailed and more complex. This classification, unlike many others, has strictly defined boundaries, each connection can be clearly attributed to one of the mutually exclusive species.

A simple substance in chemistry is a compound consisting of atoms of only one element from the periodic table of Mendeleev. As a rule, these are binary molecules, that is, consisting of two particles connected by a covalent non-polar bond - the formation of a common lone electron pair. So, atoms of the same chemical element have identical electronegativity, that is, the ability to hold a common electron density, so it is not shifted to any of the bond participants. Examples of simple substances (non-metals) are hydrogen and oxygen, chlorine, iodine, fluorine, nitrogen, sulfur, etc. A molecule of such a substance as ozone consists of three atoms, and all noble gases (argon, xenon, helium, etc.) consist of one. In metals (magnesium, calcium, copper, etc.) there is its own type of bond - metallic, which is carried out due to the socialization of free electrons inside the metal, and the formation of molecules as such is not observed. When recording a metal substance, simply the symbol of the chemical element is indicated without any indices.

A simple substance in chemistry, examples of which were given above, differs from a complex one in its qualitative composition. Chemical compounds are formed by atoms of different elements, from two or more. In such substances, covalent polar or ionic type of binding takes place. Since different atoms have different electronegativity, when a common electron pair is formed, it shifts towards a more electronegative element, which leads to a common polarization of the molecule. The ionic type is an extreme case of the polar one, when a pair of electrons completely passes to one of the binding participants, then the atoms (or groups of them) turn into ions. There is no clear boundary between these types, the ionic bond can be interpreted as a covalent strongly polar. Examples of complex substances are water, sand, glass, salts, oxides, etc.

Substance Modifications

Substances that are called simple actually have a unique feature that is not inherent in complex ones. Some chemical elements can form several forms of a simple substance. The basis is still one element, but the quantitative composition, structure and properties radically distinguish such formations. This feature is called allotropy.

Oxygen, sulfur, carbon and other elements have several For oxygen - this is O 2 and O 3, carbon gives four types of substances - carbine, diamond, graphite and fullerenes, the sulfur molecule can be rhombic, monoclinic and plastic modification. Such a simple substance in chemistry, examples of which are not limited to those listed above, is of great importance. In particular, fullerenes are used as semiconductors in engineering, photoresistors, additives for the growth of diamond films and for other purposes, and in medicine they are the most powerful antioxidants.

What happens to substances?

Every second there is a transformation of substances inside and around. Chemistry considers and explains those processes that go with a qualitative and / or quantitative change in the composition of the reacting molecules. In parallel, often interconnected, physical transformations also occur, which are characterized only by a change in the shape, color of substances or the state of aggregation, and some other characteristics.

Chemical phenomena are interaction reactions of various types, for example, compounds, substitutions, exchanges, decompositions, reversible, exothermic, redox, etc., depending on the change in the parameter of interest. These include: evaporation, condensation, sublimation, dissolution, freezing, electrical conductivity, etc. Often they accompany each other, for example, lightning during a thunderstorm is a physical process, and the release of ozone under its action is a chemical one.

Physical properties

In chemistry, a substance is matter that has certain physical properties. By their presence, absence, degree and intensity, one can predict how a substance will behave in certain conditions, as well as explain some chemical features of compounds. So, for example, high boiling points of organic compounds that contain hydrogen and an electronegative heteroatom (nitrogen, oxygen, etc.) indicate that such a chemical type of interaction as a hydrogen bond is manifested in a substance. Thanks to the knowledge of which substances have the best ability to conduct electric current, cables and wires of electrical wiring are made from certain metals.

Chemical properties

Chemistry is engaged in the establishment, research and study of the other side of the coin of properties. from her point of view, this is their reactivity to interaction. Some substances are extremely active in this sense, for example, metals or any oxidizing agents, while others, noble (inert) gases, practically do not enter into reactions under normal conditions. Chemical properties can be activated or passivated as needed, sometimes without much difficulty, and in some cases not easily. Scientists spend many hours in laboratories, by trial and error, achieving their goals, sometimes they do not achieve them. By changing the environmental parameters (temperature, pressure, etc.) or using special compounds - catalysts or inhibitors - it is possible to influence the chemical properties of substances, and hence the course of the reaction.

Classification of chemicals

All classifications are based on the division of compounds into organic and inorganic. The main element of organics is carbon, connecting with each other and hydrogen, carbon atoms form a hydrocarbon skeleton, which is then filled with other atoms (oxygen, nitrogen, phosphorus, sulfur, halogens, metals and others), closes in cycles or branches, thereby justifying a wide variety of organic compounds. To date, 20 million such substances are known to science. While there are only half a million mineral compounds.

Each compound is individual, but it also has many similar features with others in properties, structure and composition, on this basis there is a grouping into classes of substances. Chemistry has a high level of systematization and organization; it is an exact science.

inorganic substances

1. Oxides - binary compounds with oxygen:

a) acidic - when interacting with water, they give acid;

b) basic - when interacting with water, they give a base.

2. Acids - substances consisting of one or more hydrogen protons and an acid residue.

3. Bases (alkalis) - consist of one or more hydroxyl groups and a metal atom:

a) amphoteric hydroxides - exhibit the properties of both acids and bases.

4. Salts - the result between an acid and an alkali (soluble base), consist of a metal atom and one or more acidic residues:

a) acid salts - the anion of the acid residue contains a proton, the result of incomplete dissociation of the acid;

b) basic salts - a hydroxyl group is associated with the metal, the result of incomplete dissociation of the base.

organic compounds

There are a great many classes of substances in organic matter, it is difficult to remember such a volume of information at once. The main thing is to know the basic divisions into aliphatic and cyclic compounds, carbocyclic and heterocyclic, saturated and unsaturated. Also, hydrocarbons have many derivatives in which the hydrogen atom is replaced by halogen, oxygen, nitrogen and other atoms, as well as functional groups.

Substance in chemistry is the basis of existence. Thanks to organic synthesis, a person today has a huge amount of artificial substances that replace natural ones, and also have no analogues in their characteristics in nature.

Solving qualitative problems of identifying substances in bottles without labels involves a series of operations, the results of which can determine which substance is in a particular bottle.

The first stage of the decision is a thought experiment, which is a plan of action and their expected results. To record a thought experiment, a special matrix table is used, it indicates the formulas of the substances being determined horizontally and vertically. In places of intersection of the formulas of interacting substances, the expected results of observations are recorded: - gas evolution, - precipitation, changes in color, smell, or the absence of visible changes are indicated. If, according to the condition of the problem, it is possible to use additional reagents, then it is better to write down the results of their use before compiling the table - the number of substances to be determined in the table can thus be reduced.
The solution of the problem will therefore consist of the following steps:
- preliminary discussion of individual reactions and external characteristics of substances;
- recording formulas and expected results of pairwise reactions in a table,
- conducting an experiment in accordance with the table (in the case of an experimental task);
- analysis of the results of reactions and their correlation with specific substances;
- the formulation of the answer to the problem.

It must be emphasized that a thought experiment and reality do not always completely coincide, since real reactions are carried out at certain concentrations, temperatures, and lighting (for example, AgCl and AgBr are identical in electric light). A thought experiment often overlooks many small things. For example, Br 2 /aq is perfectly discolored by solutions of Na 2 CO 3, On 2 SiO 3, CH 3 COONa; the formation of a precipitate of Ag 3 PO 4 does not occur in a strongly acidic environment, since the acid itself does not give this reaction; glycerol forms a complex with Сu (OH) 2, but does not form with (CuOH) 2 SO 4 if there is no excess of alkali, etc. The real situation does not always agree with the theoretical prediction, and in this chapter the "ideal" matrix tables and "reality" will sometimes be different. And in order to understand what is really happening, look for every opportunity to work with your hands experimentally in a lesson or an elective (remember the safety requirements at the same time).

Example 1 The numbered vials contain solutions of the following substances: silver nitrate, hydrochloric acid, silver sulfate, lead nitrate, ammonia, and sodium hydroxide. Without using other reagents, determine in which bottle the solution of which substance is located.

Decision. To solve the problem, we will compile a matrix table, in which we will enter in the appropriate squares below the diagonal that intersects it, the observation data of the results of merging the substances of one test tubes with others.

Observation of the results of sequential pouring of the contents of some numbered test tubes to all others:

1 + 2 - a white precipitate precipitates; ;
1 + 3 - no visible changes are observed;

Substances	1. AgNO3,	2. HCl	3. Pb(NO 3) 2,	4.NH4OH	5.NaOH
1. AgNO3	X	AgCl white	-	precipitate is dissolved	Ag 2 O brown
2. HCl	white	X	PbCl 2 white,	-	_
3. Pb(NO 3) 2	-	white PbCl 2	X	Pb(OH) 2 turbidity)	Pb(OH) 2 white
4.NH4OH	-	-	(clouding)		-
S. NaOH	brown	-	white	-	X

1 + 4 - depending on the order of draining the solutions, a precipitate may form;
1 + 5 - a brown precipitate forms;
2 + 3 - a white precipitate precipitates;
2 + 4 - no visible changes are observed;
2+5 - no visible changes are observed;
3+4 - turbidity is observed;
3 + 5 - a white precipitate falls out;
4 + 5 - no visible changes are observed.

Let us write down further the equations of the occurring reactions in those cases when changes in the reaction system are observed (gas evolution, precipitation, color change) and enter the formula of the observed substance and the corresponding square of the matrix table above the diagonal that intersects it:

I. 1 + 2:	AgNO 3 + Hcl		AgCl + HNO 3 ;
II. 1+5:	2AgNO3 + 2NaOH		Ag 2 O + 2NaNO 3 + H 2 O;
			brown (2AgOH Ag 2 O + H 2 O)
III. 2+3:	2HCl + Pb (NO 3) 2		PbCl 2 + 2HNO 3;
			white
IV. 3+4:	Pb(NO 3) 2 + 2NH 4 OH		Pb (OH) 2 + 2NH 4 NO 3;
			turbidity
V.3 + 5:	Pb(NO 3) 2 + 2NaOH		Pb(OH) 2 + 2NaNO 3
			white

(when lead nitrate is added to an excess of alkali, the precipitate can immediately dissolve).
Thus, on the basis of five experiments, we distinguish between the substances in the numbered test tubes.

Example 2. Eight numbered test tubes (from 1 to 8) without inscriptions contain dry substances: silver nitrate (1), aluminum chloride (2), sodium sulfide (3), barium chloride (4), potassium nitrate (5), phosphate potassium (6), as well as solutions of sulfuric (7) and hydrochloric (8) acids. How, without any additional reagents, except water, to distinguish between these substances?

Decision. First of all, let's dissolve the solids in water and mark the test tubes where they ended up. Let's make a table-matrix (as in the previous example), in which we will enter the observational data of the results of merging the substances of some test tubes with others below and above the diagonal that intersects it. In the right part of the table, we introduce an additional column "general result of observation", which we will fill in after the end of all experiments and summing up the results of observations horizontally from left to right (see, for example, p. 178).

1+2:	3AgNO3 + A1C1,		3AgCl white	+ Al(NO 3) 3 ;
1 + 3:	2AgNO 3 + Na 2 S		Ag 2S black	+ 2NaNO 3 ;
1 + 4:	2AgNO 3 + BaCl 2		2AgCl white	+ Ba(NO 3) 2 ;
1 + 6:	3AgN0 3 + K 3 PO 4		Ag 3 PO 4 yellow	+ 3KNO 3 ;
1 + 7:	2AgNO 3 + H 2 SO 4		Ag,SO 4 white	+ 2HNOS ;
1 + 8:	AgNO 3 + HCl		AgCl white	+ HNO 3 ;
2 + 3:	2AlCl 3 + 3Na 2 S + 6H 2 O		2Al(OH)3,	+ 3H 2 S + 6NaCl;
		(Na 2 S + H 2 O NaOH + NaHS, hydrolysis);
2 + 6:	AlCl 3 + K 3 PO 4		A1PO 4 white	+ 3KCl;
3 + 7:	Na 2 S + H 2 SO 4		Na2SO4	+ H 2 S
3 + 8:	Na 2 S + 2HCl		-2NaCl	+H2S;
4 + 6:	3BaCl2 + 2K3PO4		Ba 3 (PO 4) 2 white	+ 6KC1;
4 + 7	BaCl 2 + H 2 SO 4		BaSO4 white	+ 2HC1.

Visible changes do not occur only with potassium nitrate.

By the number of times a precipitate precipitates and a gas is released, all reagents are uniquely determined. In addition, BaCl 2 and K 3 PO 4 are distinguished by the color of the precipitate with AgNO 3: AgCl is white, and Ag 3 PO 4 is yellow. In this problem, the solution can be simpler - any of the acid solutions allows you to immediately isolate sodium sulfide, it determines silver nitrate and aluminum chloride. Among the remaining three solids, barium chloride and potassium phosphate are determined by silver nitrate, hydrochloric and sulfuric acids are distinguished by barium chloride.

Example 3 Four unlabeled tubes contain benzene, chlorhexane, hexane and hexene. Using the minimum quantities and number of reagents, propose a method for the determination of each of the indicated substances.

Decision. The substances to be determined do not react with each other; it makes no sense to compile a table of pairwise reactions.
There are several methods for determining these substances, one of them is given below.
Bromine water decolorizes immediately only hexene:

C 6 H 12 + Br 2 \u003d C 6 H 12 Br 2.

Chlorhexane can be distinguished from hexane by passing their combustion products through a solution of silver nitrate (in the case of chlorhexane, a white precipitate of silver chloride precipitates, insoluble in nitric acid, unlike silver carbonate):

2C 6 H 14 + 19O 2 \u003d 12CO 2 + 14H 2 O;
C 6 H 13 Cl + 9O 2 \u003d 6CO 2 + 6H 2 O + HC1;
HCl + AgNO 3 \u003d AgCl + HNO 3.

Benzene differs from hexane in freezing in ice water (for C 6 H 6 mp = +5.5 ° C, and for C 6 H 14 mp = -95.3 ° C).

1. Equal volumes are poured into two identical beakers: one with water, the other with a dilute solution of sulfuric acid. How, without any chemical reagents at hand, to distinguish between these liquids (you can’t taste the solutions)?

2. Four test tubes contain powders of copper(II) oxide, iron(III) oxide, silver, and iron. How to recognize these substances using only one chemical reagent? Recognition by appearance is excluded.

3. Four numbered tubes contain dry copper(II) oxide, carbon black, sodium chloride, and barium chloride. How, using the minimum amount of reagents, to determine which of the test tubes contains which substance? Justify your answer and confirm with the equations of the corresponding chemical reactions.

4. Six unlabeled test tubes contain anhydrous compounds: phosphorus(V) oxide, sodium chloride, copper sulfate, aluminum chloride, aluminum sulfide, ammonium chloride. How can you determine the contents of each tube if there is only a set of empty tubes, water and a burner? Suggest an analysis plan.

5 . Four unlabeled tubes contain aqueous solutions of sodium hydroxide, hydrochloric acid, potash, and aluminum sulfate. Suggest a way to determine the contents of each tube without using additional reagents.

6 . Numbered tubes contain solutions of sodium hydroxide, sulfuric acid, sodium sulfate and phenolphthalein. How to distinguish between these solutions without using additional reagents?

7. Unlabeled jars contain the following individual substances: powders of iron, zinc, calcium carbonate, potassium carbonate, sodium sulfate, sodium chloride, sodium nitrate, as well as solutions of sodium hydroxide and barium hydroxide. There are no other chemical reagents at your disposal, including water. Make a plan to determine the contents of each jar.

8 . Four numbered jars without labels contain solid phosphorus oxide (V) (1), calcium oxide (2), lead nitrate (3), calcium chloride (4). Determine which jar contains each from of these compounds, if it is known that substances (1) and (2) react violently with water, and substances (3) and (4) dissolve in water, and the resulting solutions (1) and (3) can react with all other solutions with the formation of precipitation.

9 . Five test tubes without labels contain solutions of hydroxide, sulfide, chloride, sodium iodide and ammonia. How to determine these substances using one additional reagent? Give the equations of chemical reactions.

10. How to recognize solutions of sodium chloride, ammonium chloride, barium hydroxide, sodium hydroxide, which are in vessels without labels, using only these solutions?

11. . Eight numbered tubes contain aqueous solutions of hydrochloric acid, sodium hydroxide, sodium sulfate, sodium carbonate, ammonium chloride, lead nitrate, barium chloride, silver nitrate. Using indicator paper and carrying out any reactions between solutions in test tubes, determine which substance is contained in each of them.

12. Two test tubes contain solutions of sodium hydroxide and aluminum sulfate. How to distinguish between them, if possible, without the use of additional substances, with only one empty test tube or even without it?

13. Five numbered tubes contain solutions of potassium permanganate, sodium sulfide, bromine water, toluene and benzene. How, using only the named reagents, to distinguish between them? Use to detect each of the five substances their characteristic features (specify them); give a plan for the analysis. Write schemes of necessary reactions.

14. Six unnamed flasks contain glycerol, aqueous glucose solution, butyric aldehyde (butanal), hexene-1, aqueous sodium acetate solution and 1,2-dichloroethane. With only anhydrous sodium hydroxide and copper sulfate as additional chemicals, determine what is in each vial.

1. To determine water and sulfuric acid, you can use the difference in physical properties: boiling and freezing points, density, electrical conductivity, refractive index, etc. The strongest difference will be in electrical conductivity.

2. Pour hydrochloric acid to the powders in test tubes. Silver will not react. When dissolving iron, gas will be released: Fe + 2HCl \u003d FeCl 2 + H 2
Iron oxide (III) and copper oxide (II) dissolve without gas evolution, forming yellow-brown and blue-green solutions: Fe 2 O 3 + 6HCl \u003d 2FeCl 3 + 3H 2 O; CuO + 2HCl \u003d CuCl 2 + H 2 O.

3. CuO and C are black, NaCl and BaBr 2 are white. The only reagent can be, for example, dilute sulfuric acid H 2 SO 4:

CuO + H 2 SO 4 = CuSO 4 + H 2 O (blue solution); BaCl 2 + H 2 SO 4 = BaSO 4 + 2HCl (white precipitate).
Dilute sulfuric acid does not react with soot and NaCl.

4 . We put a small amount of each of the substances in water:

CuSO 4 + 5H 2 O \u003d CuSO 4 5H 2 O	(a blue solution and crystals are formed);
Al 2 S 3 + 6H 2 O \u003d 2Al (OH) 3 + 3H 2 S	(a precipitate precipitates and a gas with an unpleasant odor is released);
AlCl 3 + 6H 2 O \u003d A1C1 3 6H 2 O + Q AlCl 3 + H 2 O AlOHCl 2 + HCl AlOHC1 2 + H 2 0 \u003d Al (OH) 2 Cl + HCl A1 (OH) 2 C1 + H 2 O \u003d A1 (OH) 2 + HCl	(violent reaction proceeds, precipitation of basic salts and aluminum hydroxide is formed);
P 2 O 5 + H 2 O \u003d 2HPO 3 HPO 3 + H 2 O \u003d H 3 PO 4	(violent reaction with the release of a large amount of heat, a clear solution is formed).

Two substances - sodium chloride and ammonium chloride - dissolve without reacting with water; they can be distinguished by heating dry salts (ammonium chloride sublimes without residue): NH 4 Cl NH 3 + HCl; or by the color of the flame with solutions of these salts (sodium compounds color the flame yellow).

5. Compile a table of pairwise interactions of the indicated reagents

Substances	1.NaOH	2 HCl	3. K 2 CO 3	4. Al 2 (SO 4) 3	Overall observation result
1, NaOH		-	-	Al(OH)3	1 draft
2. HC1	_		CO2	__	1 gas
3. K 2 CO 3	-	CO2		Al(OH)3 CO2	1 sediment and 2 gases
4. Al 2 (S0 4) 3	A1(OH) 3	-	A1(OH) 3 CO2		2 draft and 1 gas
NaOH + HCl \u003d NaCl + H 2 O
K 2 CO 3 + 2HC1 \u003d 2KS1 + H 2 O + CO 2

3K 2 CO 3 + Al 2 (SO 4) 3 + 3H 2 O \u003d 2 Al (OH) 3 + 3CO 2 + 3K 2 SO 4;

Based on the presented table, all substances can be determined by the number of precipitation and gas evolution.

6. All solutions are mixed in pairs. A pair of solutions that gives a raspberry color - NaOH and phenolphthalein. The raspberry solution is added to the two remaining test tubes. Where the color disappears - sulfuric acid, in the other - sodium sulfate. It remains to distinguish between NaOH and phenolphthalein (tubes 1 and 2).
A. From tube 1, add a drop of solution to a large amount of solution 2.
B. From tube 2 - a drop of solution is added to a large amount of solution 1. In both cases, crimson staining.
To solutions A and B add 2 drops of sulfuric acid solution. Where the color disappears, a drop of NaOH was present. (If the color disappears in solution A, then NaOH is in tube 1).

Substances	Fe	Zn	CaCO 3	K 2 CO 3	Na2SO4	NaCl	NaNO 3
Va(OH) 2				sediment	sediment	solution	solution
NaOH		possible release of hydrogen		solution	solution	solution	solution
There is no precipitate in the case of two salts at Ba(OH) 2 and in the case of four salts of NaOH	dark powders (soluble in alkalis - Zn, insoluble in alkalis - Fe)		CaCO 3 gives a precipitate with both alkalis	give one sediment, differ in the color of the flame: K + - violet, Na + - yellow		do not give precipitation; differ in their behavior when heated (NaNO 3 melts, and then decomposes with the release of O 2, then NO 2

8 . React violently with water: P 2 O 5 and CaO to form, respectively, H 3 PO 4 and Ca (OH) 2:

P 2 O 5 + 3H 2 O \u003d 2H 3 RO 4, CaO + H 2 O \u003d Ca (OH) 2.
Substances (3) and (4) -Pb(NO 3) 2 and CaCl 2 - dissolve in water. Solutions can react with each other as follows:

Substances	1. H 3 RO 4	2. Ca (OH) 2,	3. Pb(NO 3) 2	4. CaCl2
1. H 3 RO 4		CaHPO 4	PbHPO 4	CaHPO 4
2. Ca (OH) 2	Sanro 4		Pb(OH)2	-
3. Pb(NO 3) 2	PbHPO 4	Pb(OH)2		PbCl 2
4. CaCl 2	CaHPO 4		PbCl2

Thus, solution 1 (H 3 PO 4) forms precipitates with all other solutions upon interaction. Solution 3 - Pb(NO 3) 2 also forms precipitates with all other solutions. Substances: I -P 2 O 5, II -CaO, III -Pb (NO 3) 2, IV-CaCl 2.
In the general case, most of the precipitation will depend on the order in which the solutions are drained and the excess of one of them (in a large excess of H 3 PO 4, lead and calcium phosphates are soluble).

9. The problem has several solutions, two of which are given below.
a. Add copper sulphate solution to all test tubes:
2NaOH + CuSO 4 = Na 2 SO 4 + Cu (OH) 2 (blue precipitate);
Na 2 S + CuSO 4 = Na 2 SO 4 + CuS (black precipitate);
NaCl + CuSO 4 (no changes in a dilute solution);
4NaI+2CuSO 4 = 2Na 2 SO 4 + 2CuI+I 2 (brown precipitate);
4NH 3 + CuSO 4 = Cu(NH 3) 4 SO 4 (blue solution or blue precipitate soluble in excess ammonia solution).

b. Add silver nitrate solution to all test tubes:
2NaOH + 2AgNO 3 \u003d 2NaNO 3 + H 2 O + Ag 2 O (brown precipitate);
Na 2 S + 2AgNO 3 = 2NaNO 3 + Ag 2 S (black precipitate);
NaCl + AgNO 3 = NaN0 3 + AgCl (white precipitate);
NaI + AgNO 3 = NaNO 3 + AgI (yellow precipitate);
2NH 3 + 2AgNO 3 + H 2 O = 2NH 4 NO 3 + Ag 2 O (brown precipitate).
Ag 2 O dissolves in an excess of ammonia solution: Ag 2 0 + 4NH 3 + H 2 O = 2OH.

10 . To recognize these substances, reactions of all solutions with each other should be carried out:

Substances	1.NaCl	2.NH4C1	3.Ba(OH),	4.NaOH	Overall observation result
1.NaCl		___	_	_	no interaction observed
2.NH4Cl	_	X	NH3	NH3	gas is released in two cases
3. Ba (OH) 2	-	NH3	X	-
4.NaOH	-	NH3	-	X	in one case gas is released

NaOH and Ba(OH) 2 can be distinguished by the different colors of the flame (Na+ is colored yellow, and Ba 2 + is green).

11. Determine the acidity of solutions using indicator paper:
1) acidic environment -Hcl, NH 4 C1, Pb (NO 3) 2;
2) neutral medium - Na 2 SO 4, ВаС1 2, AgNO 3;
3) alkaline environment - Na 2 CO 3, NaOH. We make a table.

Solving quality problems
course in organic chemistry

Elective course grade 11

Continuation. See No. 23/2006, 7/2007.

Section 2
Establishing the structure of substances
based on the data of physicochemical methods
and chemical properties (continuation)

Lesson 6. Calculation problems
to establish the structure of matter

Target. To teach schoolchildren to solve calculation problems to establish the structure of matter.

Exercise 1. Establish the structure of a hydrocarbon, the combustion of one volume of which produces six volumes of carbon dioxide, and when chlorinated in the light - only two monochlorine derivatives.

Decision

Task scheme:

Actually, there are two clues for solving the problem: this is the release of six volumes of CO 2 (which means that there are 6 carbon atoms in the molecule) and that chlorination takes place in the light (which means it is an alkane).

The formula of the hydrocarbon is C 6 H 14.

Set up the structure. Since this hydrocarbon has only two monochloro derivatives, its carbon chain is as follows:

This is 2,3-dimethylbutane. The frameworks of chlorohydrocarbons are as follows:

Task 2. To burn a portion of an alkane containing 1 10 23 molecules, a portion of oxygen containing 1.6 10 24 atoms is required. Set the composition and possible structure (all isomers) of the alkane.

Decision

When analyzing the solution, one should pay attention to the arrangement of the coefficients in a general form (through n), because without this, the problem cannot be solved:

With n H2 n+2 + (1,5n+ 0.5)O 2 \u003d n CO 2 + ( n+ 1) H 2 O.

(alkane) \u003d 1 10 23 / (6.02 10 23) \u003d 0.166 mol,

(O 2) \u003d 1.6 10 24 / (6.02 10 23 2) \u003d 1.33 mol.

Let's make a proportion:

1 mole of alkane - 1.5 n+ 0.5 oxygen,

0.166 mol of alkane - 1.33 mol of oxygen.

From here n = 5.

This is C 5 H 12 pentane, three isomers are possible for it:

Task 3. The mixture of alkane and oxygen, the volume ratio of which corresponds to the stoichiometric, after combustion, vapor condensation and reduction to the initial conditions, was reduced by half in volume. Establish the structure of the alkane that was part of the mixture.

Decision

When analyzing the solution, it is necessary to pay attention to the arrangement of coefficients in a general form through n, because without this, the problem cannot be solved:

With n H2 n+2 + (1,5n+ 0.5)O 2 \u003d n CO 2 + ( n+ 1) H 2 O.

Before the reaction, the total volume of gases was:

(1 + 1,5n+ 0.5) l.

After the reaction, we take into account only the volume of CO 2 - n l (water H 2 O at 20 ° C - liquid).

We make the equation: 1 + 1.5 n + 0,5 = 2n.

From here n = 3.

Answer. Propane C 3 H 8 .

Task 4. A mixture of alkane and oxygen, the volume ratio of which corresponds to the stoichiometric, after combustion, condensation of water vapor and reduction to standard. decreased in volume by 1.8 times. Determine the formula of the alkane that was part of the mixture if it is known that its molecule has four primary carbon atoms.

Answer. Neopentane (CH 3) 3 CCH 3.

Task 5. When a mixture of cis- and trans-isomers of alkene was passed through an excess of potassium permanganate solution, the mass of the precipitate that formed turned out to be greater than the mass of the initial alkene. Set the structure of the alkene.

Decision

Let us write the equation for the reaction of an alkene with a solution of potassium permanganate:

3C n H2 n+ 2KMnO 4 + 4H 2 O \u003d 3С n H2 n(OH)2 + 2MnO2 + 2KOH.

Let 1 mol of alkene enter into the reaction, then 0.6667 mol of manganese (IV) oxide is released.

M r(MnO 2) = 87, m(MnO 2) \u003d 87 0.6667 \u003d 58 g.

Therefore, taking into account the condition of the problem, the relative molecular weight of the alkene is less than 58. This condition is met by alkenes C 2 H 4 , C 3 H 6 , C 4 H 8 .

According to the condition of the problem, the alkene has cis- and trans-isomers. Then ethene and propene are definitely not suitable. Butene-2 ​​remains: only it has cis- and trans-isomers.

Answer. Buten-2.

Task 6. Nitration of one of the homologues of benzene with a mass of 31.8 g was obtained only one mononitro derivative with a mass of 45.3 g. Establish the structure of the initial substance of the reaction product.

Decision

According to the condition of the problem (C 6 H 5 R) \u003d (C 6 H 4 RNO 2). Using the formula = m/M, we get:

31.8 / (77 + R) = 45.3 / (77 - 1 + 46 + R).

Hence R = 29.

Since R = C n H2 n+1 , the ratio is correct:

12n + 2n + 1 = 29.

So n\u003d 2, the radical R is C 2 H 5.

However, according to the condition of the problem, only one nitro derivative is obtained. Therefore, the starting material cannot be ethylbenzene, since then ortho- and para-nitro derivatives would be formed. This means that the benzene homolog does not contain an ethyl radical, but two methyl radicals. They are located symmetrically pair-xylene). With this arrangement of substituents, only one nitro derivative is obtained.

Reaction equation:

Task 7. By heating a mixture of two saturated primary alcohols with a branched skeleton in the presence of sulfuric acid, a mixture of three organic substances belonging to the same class of compounds was obtained. Substances were obtained in equal molar ratios with a total mass of 21.6 g, while water with a mass of 2.7 g was released. Set all possible formulas of the starting compounds and calculate the mass of the initial mixture.

Decision

We analyze the condition of the problem to write the equation. In the presence of sulfuric acid, either intramolecular or intermolecular dehydration, or a combination of both, is possible. If the dehydration is intramolecular, then only two unsaturated hydrocarbons are obtained; if it is intermolecular, then a mixture of three esters is obtained. The combined option does not make sense to consider, because. by condition, substances of the same class are obtained. Reaction equation:

Calculate the amount of water substance:

(H 2 O) = m/M\u003d 2.7 / 18 \u003d 0.15 mol.

Since the reaction products were obtained in equal molar ratios, it means that each ether turned out to be: 0.15 / 3 \u003d 0.05 mol.

We compose the material balance equation:

0,05 (M(R) + ( M(R") + 16) + 0.05 (2 M(R) + 16) + 0.05 (2 M(R") + 16) = 21.6

From here ( M(R)+ M(R") = 128. Both radicals R and R" are limiting, so their total molar mass can be written as follows:

M(WITH n H2 n+1) = 128.

Substituting the values ​​of atomic masses, we find:

12n + 2n+ 1 = 128, n = 9.

The molecules of two alcohols contain 9 carbon atoms.

By the condition of the problem, alcohols are primary and have a branched carbon skeleton. This means that one alcohol contains 4 carbon atoms, and the other - 5.

Formula options:

Mass of the initial mixture: 21.6 + 2.7 = 24.3 g.

Section 3
Identification of organic substances
(qualitative reactions to different classes of compounds)

Lesson 7. Recognition of organic substances
with qualitative responses

Goals. To teach how to solve problems for the determination of substances, to consolidate knowledge of the qualitative reactions of organic compounds of different classes.

Exercise 1. Four test tubes contain the following substances: hexane, 2-methylpentene-1,
Pentin-2, Pentin-1. What chemical reactions can be used to distinguish between these substances?

Decision

This problem presents three classes of compounds: alkanes, alkenes and alkynes. For alkanes, there are no special qualitative reactions; for alkenes, this is the decolorization of bromine water and a solution of potassium permanganate. Alkynes are also characterized by decolorization of bromine water and potassium permanganate, but the reaction is slower (Table 1). The proposed two alkynes differ in the position of the triple bond. Alkynes, which have a triple bond at the edge, react with an ammonia solution of silver oxide and copper(I) oxide.

Table 1

tube number	Reagents			Conclusion - substance in vitro
	Oh	Br 2 (in H 2 O)	KMnO 4 (solution)
1	–	–	–	Hexane
2	–	Fast discoloration	Fast discoloration	2-Methylpentene-1
3	–	slow discoloration	slow discoloration	Pentin-2
4	Sediment	slow discoloration	slow discoloration	Pentin-1

First, a reaction is carried out for the detection of pentin-1:

CH 3 CH 2 CH 2 CCH + OH CH 3 CH 2 CH 2 CCAg + 2NH 3 + H 2 O.

Then, by the absence of reaction with bromine water, hexane is detected:

CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 + Br 2 (H 2 O) ....

Pentin-2 decolorizes bromine water slowly, and 2-methylpentene-2 ​​quickly:

The reaction with potassium permanganate can be omitted.

Task 2. Three test tubes without inscriptions contain liquids: n-propanol, 1-chlorobutane and glycerin. Distinguish these substances.

Decision

The test tubes contain substances of three classes: alcohol, polyhydric alcohol and halogen derivatives of alkanes. Glycerin has a viscosity, so we can already assume in which test tube it is. Qualitative reaction to polyhydric alcohols - interaction with copper (II) hydroxide to cornflower blue staining. Alcohol can be distinguished from a haloalkane by reacting with sodium without heating. In a test tube with alcohol, hydrogen gas bubbles will be observed (Table 2).

table 2

tube number	Reagent			Conclusion - substance in vitro
	By appearance	Cu(OH)2	Na
1	Viscosity	Cornflower blue coloring	Effervescence	Glycerol
2	–	–	Effervescence	Propanol
3	–	–	–	1-Chlorobutane

Reaction equations:

Task 3. The following liquids are poured into three test tubes: benzene, styrene, phenylacetylene. Determine which substance is which.

Decision

All substances contain an aromatic ring:

Reaction equations:

Let's make a table (Table 3).

Table 3

tube number	Reagent		Conclusion - a substance in a test tube
	Oh	Br 2 (in H 2 O)
1	–	–	C 6 H 6, benzene
2	–	Bromine water discoloration	C 6 H 5 CH \u003d CH 2, styrene
3	Precipitation	Bromine water discoloration	C 6 H 5 CCH, phenylacetylene

Task 4. Three test tubes without signatures contain the following substances: butanol-1, ethylene glycol, a solution of phenol in benzene. What reactions can be used to distinguish between these substances?

Decision

Let's make a table (Table 4).

Table 4

Reaction equations:

TASK FOR SELF SOLUTION

Exercise 1. Four unlabeled flasks contain the following organic substances: ethanol, acetaldehyde, ethylene glycol, and an aqueous solution of phenol. Suggest a way to distinguish between these substances.

Let's make a table - a solution scheme (Table 5).

Table 5

tube number	Reagents			Conclusion - the substance in the flask
	Cu(OH)2	Br 2 (in H 2 O)	Oh
1	–	–	–	ethanol
2	–	–	Sediment	Acetaldehyde
3	Cornflower blue coloring	–	–	ethylene glycol
4	–	Sediment	–	Phenol (in H 2 O)

Task 2. Four test tubes contain the following substances: formic acid, propionic acid, methanol, acetaldehyde. What reactions can be used to distinguish between these substances? Write equations for these reactions.

Let's make a table - a solution scheme (Table 6).

Table 6

tube number	Reagents		Conclusion - a substance in a test tube
	Litmus	Oh
1	Red	Sediment	Formic acid
2	Red	–	propionic acid
3	Violet	–	methanol
4	Violet	Sediment	Acetic aldehyde

Task 3. Write reaction equations that can be used to distinguish between the following organic solids: glucose, sucrose, sodium acetate, starch, and phenol.

Let's make a table - a solution scheme (Table 7).

Table 7

tube number	Reagents
	Solubility in cold water	Cu(OH)2		Iodine solution
1	Soluble	Cornflower blue coloring	color change to carrot	We do not conduct experience	Glucose
2	Soluble	Cornflower blue coloring	Virtually unchanged	We do not conduct experience	sucrose
3	Soluble	Without changes	Without changes	Without changes	sodium acetate
4	Insoluble	We do not conduct experience	We do not conduct experience	Blue staining	Starch
5	Slightly soluble	We do not conduct experience	We do not conduct experience	Without changes	Phenol

The reaction with a solution of bromine water for the determination of phenol can be omitted. Two substances remained unidentified - sodium acetate and phenol. Moreover, sodium acetate is highly soluble in cold water, and phenol is poorly. So they can be distinguished.

Task 4. How to distinguish between organic substances: phenylammonium chloride, sodium acetate, glucose, aminoacetic acid? Write the equations of the reactions that must be carried out to recognize substances.

Let's make a table - a solution scheme (Table 8).

Table 8

tube number	Reagents			Conclusion - analyte
	Cu(OH)2	Relation to heating solutions with cornflower blue staining	NaOH (sol.) when heated
1	Without changes	Without changes	Gas evolution, ammonia smell	Phenyl ammonium chloride
2	Without changes	Without changes	Methane gas release	sodium acetate
3	Cornflower blue coloring	color change to carrot	No visible changes	Glucose
4	Dark blue staining	Without changes	No visible changes	Aminoacetic acid

To be continued

Bodies, substances, particles

Any object, any living being can be called a body. A stone, a lump of sugar, a tree, a bird, a wire are bodies. It is impossible to list all the bodies, there are countless of them. The sun, the planets, the moon are also bodies. They are called celestial bodies.

The bodies can be divided into two groups.

Bodies created by nature itself are called natural bodies.
The bodies created by human hands are called artificial bodies.

Consider the drawings. Under the natural bodies, paint the circles in green, under the artificial ones in brown.

Bodies are made up of substances. The lump of sugar is the body, and the sugar itself is the substance. Aluminum wire is the body, aluminum is the substance. There are bodies that are formed not by one, but by several or many substances.

Substances is what bodies are made of.

Distinguish between solid, liquid and gaseous substances.
Sugar, aluminum are examples of solids. Water is a liquid substance. Air consists of several gaseous substances (gases).

Write down what substance the body is made of.

What body has a particular shape?
Answer: Solid bodies have a constant shape.

Fill in the table

Aluminum, silver, notebook, wood, TV, kettle, water, saw, cupboard, starch.

Substances, and therefore bodies, are made up of particles.
Each substance consists of special particles, which differ in size and shape from the particles of other substances.
Scientists have found that there are gaps between the particles. In solids, these gaps are very small, in liquids they are larger, and in gases even more. In any substance, all particles move.
Particles can be depicted using models, such as balls.