Keywords: Chemistry grade 8. All formulas and definitions, symbols physical quantities, units of measurement, prefixes for designating units of measurement, relationships between units, chemical formulas, basic definitions, briefly, tables, diagrams.

1. Symbols, names and units of measurement
some physical quantities used in chemistry

Physical quantity	Designation	unit of measurement
Time	t	from
Pressure	p	Pa, kPa
Amount of substance	ν	mole
Mass of matter	m	kg, g
Mass fraction	ω	Dimensionless
Molar mass	M	kg/mol, g/mol
Molar volume	V n	m 3 / mol, l / mol
Volume of matter	V	m 3, l
Volume fraction		Dimensionless
Relative atomic mass	A r	Dimensionless
	M r	Dimensionless
Relative density of gas A over gas B	D B (A)	Dimensionless
Matter density	R	kg / m 3, g / cm 3, g / ml
Avogadro constant	N A	1/mol
Temperature absolute	T	K (Kelvin)
Celsius temperature	t	°С (degree Celsius)
thermal effect chemical reaction	Q	kJ/mol

2. Relations between units of physical quantities

3. Chemical formulas in grade 8

4. Basic definitions in grade 8

• Atom- the smallest chemically indivisible particle of a substance.
• Chemical element a certain type of atom.
• Molecule- the smallest particle of a substance that retains its composition and chemical properties and consists of atoms.
• Simple substances Substances whose molecules are made up of atoms of the same type.
• Complex Substances Substances whose molecules are made up of different types of atoms.
• The qualitative composition of the substance shows what atoms it consists of.
• The quantitative composition of the substance shows the number of atoms of each element in its composition.
• Chemical formula- conditional record of the qualitative and quantitative composition of a substance by means of chemical symbols and indices.
• Atomic mass unit(amu) - a unit of measurement of the mass of an atom, equal to the mass of 1/12 of a carbon atom 12 C.
• mole- the amount of a substance that contains the number of particles equal to the number of atoms in 0.012 kg of carbon 12 C.
• Avogadro constant (Na \u003d 6 * 10 23 mol -1) - the number of particles contained in one mole.
• Molar mass of a substance (M ) is the mass of a substance taken in an amount of 1 mol.
• Relative atomic mass element BUT r - the ratio of the mass of an atom of a given element m 0 to 1/12 of the mass of a carbon atom 12 C.
• Relative molecular mass substances M r - the ratio of the mass of a molecule of a given substance to 1/12 of the mass of a carbon atom 12 C. The relative molecular mass is equal to the sum of the relative atomic masses chemical elements, forming a compound, taking into account the number of atoms of a given element.
• Mass fraction chemical element ω(X) shows what part of the relative molecular weight of substance X is accounted for by this element.

ATOMIC-MOLECULAR STUDIES
1. There are substances with molecular and not molecular structure.
2. There are gaps between the molecules, the dimensions of which depend on the state of aggregation of the substance and temperature.
3. Molecules are in continuous motion.
4. Molecules are made up of atoms.
6. Atoms are characterized by a certain mass and size.
At physical phenomena Molecules are preserved, while chemical ones, as a rule, are destroyed. Atoms in chemical phenomena rearrange, forming molecules of new substances.

THE LAW OF CONSTANT COMPOSITION OF A SUBSTANCE
Each chemically pure substance molecular structure, regardless of the method of preparation, has a constant qualitative and quantitative composition.

VALENCE
Valency is the property of an atom of a chemical element to attach or replace a certain number of atoms of another element.

CHEMICAL REACTION
A chemical reaction is a process in which another substance is formed from one substance. Reagents are substances that enter into a chemical reaction. Reaction products are substances that are formed as a result of a reaction.
Signs of chemical reactions:
1. Release of heat (light).
2. Color change.
3. The appearance of a smell.
4. Precipitation.
5. Gas release.

• chemical equation- recording a chemical reaction using chemical formulas. Shows what substances and in what quantity react and are obtained as a result of the reaction.

LAW OF CONSERVATION OF MASS
The mass of substances that entered into a chemical reaction is equal to the mass of substances formed as a result of the reaction. As a result of chemical reactions, atoms do not disappear and do not appear, but their rearrangement occurs.

The most important classes of inorganic substances

Summary of the lesson “Chemistry Grade 8. All formulas and definitions.

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The chemical formula reflects the composition of a substance. For example, H 2 O - two hydrogen atoms are connected to an oxygen atom. Chemical formulas also contain some information about the structure of the substance: for example, Fe (OH) 3, Al 2 (SO 4) 3 - these formulas indicate some stable groups (OH, SO 4) that are part of the substance - its molecule or formula units.

Molecular formula indicates the number of atoms of each element in the molecule. The molecular formula describes substances with a molecular structure (gases, liquids, and some solids). The composition of a substance with an atomic or ionic structure can only be described by a formula unit.

formula unit indicates the simplest ratio between the number of atoms of different elements in a substance. For example, the formula unit of benzene is CH, the molecular formula is C 6 H 6.

Structural (graphical) formula indicates the order of connection of atoms in a molecule and in a formula unit and the number of bonds between atoms.

Valence

Correct writing such formulas is based on the idea of valency(valentia - strength) as the ability of an atom of a given element to attach to itself a certain number of other atoms. In modern chemistry, three types of valency are considered: stoichiometric, electronic and structural.

Stoichiometric valency chemical element - is the number of equivalents that a given atom can attach to itself, or is the number of equivalents in an atom. Equivalents are determined by the number of attached or substituted hydrogen atoms, so the stoichiometric valence is equal to the number of hydrogen atoms with which a given atom interacts. But not all elements interact with hydrogen, and almost everything interacts with oxygen, so the stoichiometric valency can be defined as twice the number of attached oxygen atoms.

For example, the stoichiometric valence of sulfur in hydrogen sulfide H 2 S is 2, in oxide SO 2 - 4, in oxide SO 3 -6.

When determining the stoichiometric valency of an element according to the formula of a binary compound, one should be guided by the rule: the total valency of all atoms of one element must be equal to the total valency of all atoms of another element.

Knowing the valency of the elements and this rule, it is possible to compose the chemical formula of the compound. When compiling formulas, the following procedure should be observed.

1. Write, in ascending order of electronegativity, the chemical symbols of the elements that make up the compound, for example:

2. Above the symbols of chemical elements put down their valency (it is customary to denote it with Roman numerals):

I II III I III II

3. Using the above rule, determine the least common multiple of the numbers expressing the stoichiometric valency of both elements (2, 3 and 6, respectively).

4) Dividing the least common multiple by the valency of the corresponding element, find the number of atoms in the formula of the compounds:

I II III I III II

K 2 O AlCl 3 Al 2 O 3

Example 15 Write the formula for chlorine oxide, knowing that chlorine in it is heptavalent and oxygen is divalent.

Solution. We find the smallest multiple of the numbers 2 and 7 - it is equal to 14. Dividing the least common multiple by the stoichiometric valency of the corresponding element, we find the number of atoms: chlorine 14 : 7=2, oxygen 14 : 2=7. Thus, the oxide formula is Cl 2 O 7.

Oxidation state also characterizes the composition of the substance and is equal to the stoichiometric valence with a plus sign (for a metal or a more electropositive element in a molecule) or minus.

1. In simple substances ax the oxidation state of the elements is zero.

2. The oxidation state of fluorine in all compounds is -1. The remaining halogens (chlorine, bromine, iodine) with metals, hydrogen and other more electropositive elements also have an oxidation state of -1, but in combination with more electronegative elements they have positive oxidation states.

3. Oxygen in compounds has an oxidation state of -2; the exceptions are hydrogen peroxide H 2 O 2 and its derivatives (Na 2 O 2, BaO 2, etc., in which oxygen has an oxidation state of -1, as well as oxygen fluoride OF 2, in which the oxidation state of oxygen is +2.

4. Alkaline elements (Li, Na, K, etc.) and elements of the main subgroup of the second group Periodic system(Be, Mg, Ca, etc.) always have an oxidation state equal to the group number, that is, +1 and +2, respectively.

5. All elements of the third group, except for thallium, have a constant oxidation state equal to the group number, i.e. +3.

6. The highest oxidation state of an element is equal to the group number of the Periodic system, and the lowest is the difference: group number is 8. For example, the highest oxidation state of nitrogen (it is located in the fifth group) is +5 (in nitric acid and its salts), and the lowest is -3 (in ammonia and ammonium salts).

7. The oxidation states of the elements in the compound compensate each other so that their sum for all atoms in a molecule or a neutral formula unit is zero, and for an ion - its charge.

These rules can be used to determine the unknown oxidation state of an element in a compound, if the oxidation states of the rest are known, and to formulate multi-element compounds.

Example 16 Determine the oxidation state of chromium in the K 2 CrO 4 salt and in the Cr 2 O 7 2 - ion.

Solution. The oxidation state of potassium is +1 (rule 4) and oxygen -2 (rule 3). The oxidation state of chromium is denoted by X. For the formula unit K 2 CrO 4 we have:

2∙(+1) + X + 4∙(-2) = 0,

therefore, the oxidation state of chromium is X = +6.

For the ion Cr 2 O 7 2 - we have: 2∙X + 7∙(-2) = -2, X = +6.

We see that the oxidation state of chromium is the same in both cases.

Example 17. Determine the oxidation state of phosphorus in the compounds P 2 O 3 and PH 3.

Solution. In the P 2 O 3 compound, the oxidation state of oxygen is -2. Based on the fact that algebraic sum oxidation states of the molecule should be equal to zero, we find the oxidation state of phosphorus: 2∙X + 3∙(-2) = 0, hence X = +3.

In the PH 3 compound, the oxidation state of hydrogen is +1, hence X + 3 ∙ (+1) \u003d 0, X \u003d -3.

Example 18. Write the formulas of the oxides that can be obtained by thermal decomposition of the following hydroxides (bases and acids): Fe(OH) 3 , Cu(OH) 2 , H 2 SiO 3 , H 3 AsO 4 , H 2 WO 4 .

Solution. Fe (OH) 3 - the charge of the hydroxide ion is -1, therefore, the oxidation state of iron is +3 and the formula of the corresponding oxide is Fe 2 O 3.

Cu (OH) 2 - since there are two hydroxide ions, the total charge of which is -2, the oxidation state of copper is +2 and the oxide formula is CuO.

H2SiO3. The oxidation state of hydrogen is +1, oxygen -2, silicon - X. Algebraic equation: 2∙(+1) + X + 3∙(-2) = 0. X = +4. The oxide formula is SiO 2.

H 3 AsO 4 - the degree of oxidation of arsenic in acid is calculated by the equation:

3 . (+1) + X + 4 (-2) = 0; X = +5.

Thus, the oxide formula is As 2 O 5.

H2WO4. The oxidation state of tungsten, calculated in the same way (check!) is +6. Therefore, the formula of the corresponding oxide is WO 3 .

Chemical elements are divided into elements of constant and variable valency; accordingly, the former have a constant degree of oxidation in any compounds, and the latter have a different one, which depends on the composition of the compound /

Consider how, using the Periodic system of D.I. Mendeleev, it is possible to determine the oxidation states of elements.

For stable oxidation states of elements main subgroups the following regularities are observed.

1. The elements of groups I-III have the only oxidation states - positive, and equal in magnitude to the group numbers, except for thallium, which has oxidation states +1 and +3.

2. The elements of IV-VI groups, in addition to the maximum positive oxidation state corresponding to the group number, and negative, equal to the difference between the number 8 and the group number, there are also intermediate oxidation states, usually differing by 2 units. For group IV, the oxidation states are +4, +2, -4, -2; for group V +5, +3, -3, -1; for VI group - +6, +4, -2.

3. The elements of group VII have all oxidation states from +7 to -1, differing by two units, i.e. +7,+5, +3, +1 and -1. But in this group (halogens) fluorine is released, which does not have positive oxidation states and in compounds with other elements exists only in one oxidation state -1.

Note. There are several unstable compounds of chlorine, bromine and iodine with even oxidation states +2, +4 and +6 (ClO, ClO 2 , ClO 3 and others).

Elements side subgroups there is no simple relationship between stable oxidation states and group number. For the most common elements of the elements of the secondary subgroups, stable oxidation states should simply be remembered. These elements include: chromium Cr (+3 and +6), manganese Mn (+7, +6, +4 and +2), iron Fe, cobalt Co and nickel Ni (+3 and +2), copper Cu ( +2 and +1), silver Ag (+1), gold Au (+3 and +1), zinc Zn and cadmium Cd (+2), mercury Hg (+2 and +1).

To draw up formulas for three- and multi-element compounds, it is necessary to know the oxidation states of all elements. In this case, the number of element atoms in the formula is determined from the condition that the sum of the oxidation states of all atoms is equal to zero (in a formula unit) or charge (in an ion). For example, if it is known that in the formula unit there are K, Cr and O atoms with oxidation states equal to +1, +6 and -2, respectively, then this condition will be satisfied by the formulas K 2 CrO 4 , K 2 Cr 2 O 7 , K 2 Cr 3 O 10 and many others; similarly to this ion with a charge of -2, containing Cr +6 and O - 2, the formulas CrO 4 2 -, Cr 2 O 7 2 -, Cr 3 O 10 2 - , Cr 4 O 13 2 -, etc. will correspond.

Electronic valence element is equal to the number of chemical bonds formed by an atom of this element.

In most compounds, the electronic valency of the elements is equal to the stoichiometric. But there are many exceptions. For example, in hydrogen peroxide H 2 O 2, the stoichiometric valency of oxygen is one (one hydrogen atom per oxygen atom), and the electronic one is two, which follows from the structural formula, which shows the chemical bonds of atoms: H–O–O–H . The discrepancy between the values ​​of the stoichiometric and electronic valences is explained in this case by the fact that the oxygen atoms are bonded not only to the hydrogen atoms, but also to each other.

Thus, there are chemical compounds, in which the stoichiometric and electronic valencies do not coincide. These include, for example, complex compounds.

Structural (coordination) valence, or the coordination number is determined by the number of neighboring atoms. For example, in the SO 3 molecule, sulfur has 3 neighboring oxygen atoms and a structural valence and coordination number of 3, while the stoichiometric valence is 6.

Electronic and coordination valencies are discussed in more detail in the chapters "Chemical bond" and "Complex compounds".

Formula definition tasks organic matter there are several types. Usually, the solution of these problems is not particularly difficult, but often graduates lose points on this problem. There are several reasons:

1. Incorrect design;
2. The solution is not mathematical, but by enumeration;
3. Incorrectly compiled general formula of a substance;
4. Errors in the reaction equation involving a substance written in general view.

Types of tasks in task C5.

1. Determination of the formula of a substance by mass fractions of chemical elements or by the general formula of a substance;
2. Determination of the formula of a substance by combustion products;
3. Determination of the formula of a substance by chemical properties.

Necessary theoretical information.

1. Mass fraction of an element in a substance.
   The mass fraction of an element is its content in a substance as a percentage by mass.
   For example, a substance of composition C 2 H 4 contains 2 carbon atoms and 4 hydrogen atoms. If we take 1 molecule of such a substance, then its molecular weight will be equal to:
   Mr (C 2 H 4) \u003d 2 12 + 4 1 \u003d 28 a.m.u. and it contains 2 12 a.m.u. carbon.

   To find the mass fraction of carbon in this substance, it is necessary to divide its mass by the mass of the entire substance:
   ω(C) = 12 2 / 28 = 0.857 or 85.7%.
   If a substance has the general formula C x H y O z, then the mass fractions of each of their atoms are also equal to the ratio of their mass to the mass of the entire substance. The mass x of C atoms is -12x, the mass y of H atoms is y, the mass z of oxygen atoms is 16z.
   Then
   ω(C) = 12 x / (12x + y + 16z)

   If we write this formula in general form, we get the following expression:

2. Molecular and simplest formula of a substance.

   Molecular (true) formula - a formula that reflects the real number of atoms of each type included in the molecule of a substance.
   For example, C 6 H 6 is the true formula of benzene.
   The simplest (empirical) formula - shows the ratio of atoms in a substance.
   For example, for benzene, the ratio C:H = 1:1, i.e. The simplest formula for benzene is CH.
   The molecular formula may coincide with the simplest or be a multiple of it.

   Examples.

   If only the mass fractions of elements are given in the problem, then in the process of solving the problem, only the simplest formula of a substance can be calculated. To obtain the true formula in the problem, additional data is usually given - the molar mass, the relative or absolute density of the substance, or other data that can be used to determine the molar mass of the substance.

3. Relative density of gas X by gas Y - D by Y (X).
   Relative density D is a value that shows how many times gas X is heavier than gas Y. It is calculated as the ratio of the molar masses of gases X and Y:
   D by Y (X) \u003d M (X) / M (Y)
   Often used for calculations relative densities of gases for hydrogen and for air.
   Relative gas density X for hydrogen:
   D by H 2 \u003d M (gas X) / M (H 2) \u003d M (gas X) / 2
   Air is a mixture of gases, so only the average molar mass can be calculated for it. Its value is taken as 29 g/mol (based on the approximate average composition).
   That's why:
   D by air. = M (gas X) / 29
4. The absolute density of a gas under normal conditions.

   The absolute density of a gas is the mass of 1 liter of gas under normal conditions. Usually for gases it is measured in g / l.
   ρ = m (gas) / V (gas)
   If we take 1 mole of gas, then:
   ρ \u003d M / V m,
   and the molar mass of a gas can be found by multiplying the density by the molar volume.

5. General formulas of substances of different classes.
   Often, to solve problems with chemical reactions, it is convenient to use an unusual general formula, but by a formula in which a multiple bond or a functional group is singled out separately.

   Class of organic substances	General molecular formula	Formula with highlighted multiple bond and functional group
   Alkanes	C n H 2n+2	—
   Alkenes	C n H 2n	C n H 2n+1 -CH=CH 2
   Alkynes	C n H 2n−2	C n H 2n+1 -C≡CH
   dienes	C n H 2n−2	—
   Benzene homologues	C n H 2n−6	C 6 H 5 -C n H 2n+1
   Limit monohydric alcohols	C n H 2n+2 O	C n H 2n+1 -OH
   Polyhydric alcohols	C n H 2n+2 O x	C n H 2n+2−x (OH) x
   Limit aldehydes	C n H 2n O
   Esters	C n H 2n O 2

Determination of formulas of substances by mass fractions of atoms that make up its composition.

The solution to these problems consists of two parts:

• first, the molar ratio of atoms in a substance is found - it corresponds to its simplest formula. For example, for a substance of composition A x B y, the ratio of the amounts of substances A and B corresponds to the ratio of the number of their atoms in the molecule:
  x: y = n(A) : n(B);
• then, using the molar mass of the substance, determine its true formula.

Example 1
Determine the formula of a substance if it contains 84.21% C and 15.79% H and has a relative density in air of 3.93.

Example 1 solution.

1. Let the mass of the substance be 100 g. Then the mass C will be 84.21 g, and the mass H will be 15.79 g.
2. Find the amount of matter of each atom:
   ν(C) \u003d m / M \u003d 84.21 / 12 \u003d 7.0175 mol,
   ν(H) = 15.79 / 1 = 15.79 mol.
3. We determine the molar ratio of C and H atoms:
   C: H \u003d 7.0175: 15.79 (we reduce both numbers by a smaller one) \u003d 1: 2.25 (we multiply by 4) \u003d 4: 9.
   Thus, the simplest formula is C 4 H 9.
4. Calculate the molar mass from the relative density:
   M \u003d D (air.) 29 \u003d 114 g / mol.
   The molar mass corresponding to the simplest formula C 4 H 9 is 57 g / mol, which is 2 times less than the true molar mass.
   Hence, the true formula is C 8 H 18.

There is a much simpler way to solve this problem, but, unfortunately, they will not give a full score for it. But it is suitable for checking the true formula, i.e. with it you can check your solution.

Method 2: We find the true molar mass (114 g / mol), and then we find the masses of carbon and hydrogen atoms in this substance by their mass fractions.
m(C) = 114 0.8421 = 96; those. number of C atoms 96/12 = 8
m(H) = 114 0.1579 = 18; i.e., the number of H atoms 18/1 = 18.
The formula of the substance is C 8 H 18.

Answer: C 8 H 18.

Example 2
Determine the formula of alkyne with a density of 2.41 g/l under normal conditions.

Example 2 solution.

General formula of alkyne С n H 2n−2
How, given the density of a gaseous alkyne, to find its molar mass? Density ρ is the mass of 1 liter of gas under normal conditions.
Since 1 mole of a substance occupies a volume of 22.4 liters, it is necessary to find out how much 22.4 liters of such a gas weigh:
M \u003d (density ρ) (molar volume V m) \u003d 2.41 g / l 22.4 l / mol \u003d 54 g / mol.
Next, we write an equation relating the molar mass and n:

14n − 2 = 54, n = 4.
Hence, alkyne has the formula C 4 H 6.

Answer: C 4 H 6.

Example 3
Determine the formula of the limiting aldehyde if it is known that 3 10 22 molecules of this aldehyde weigh 4.3 g.

Example 3 solution.

In this problem, the number of molecules and the corresponding mass are given. Based on these data, we need to find again the value of the molar mass of the substance.
To do this, you need to remember how many molecules are contained in 1 mol of a substance.
This is Avogadro's number: N a = 6.02 10 23 (molecules).
So, you can find the amount of aldehyde substance:
ν \u003d N / Na \u003d 3 10 22 / 6.02 10 23 \u003d 0.05 mol,
and molar mass:
M \u003d m / n \u003d 4.3 / 0.05 \u003d 86 g / mol.
Further, as in the previous example, we make an equation and find n.
The general formula of the limiting aldehyde is C n H 2n O, that is, M \u003d 14n + 16 \u003d 86, n \u003d 5.

Answer: C 5 H 10 O, pentanal.

Example 4
Determine the formula of dichloroalkane containing 31.86% carbon.

Example 4 solution.

The general formula of dichloroalkane is: C n H 2n Cl 2, there are 2 chlorine atoms and n carbon atoms.
Then the mass fraction of carbon is equal to:
ω(C) = (number of C atoms per molecule) (atomic mass of C) / (molecular mass of dichloroalkane)
0.3186 = n 12 / (14n + 71)
n = 3, the substance is dichloropropane.

Answer: C 3 H 6 Cl 2, dichloropropane.

Determination of formulas of substances by combustion products.

In tasks for combustion, the amount of substances of elements included in the substance under study is determined by the volumes and masses of combustion products - carbon dioxide, water, nitrogen and others. The rest of the solution is the same as in the first type of problems.

Example 5
448 ml (n.a.) gaseous saturated non-cyclic hydrocarbon was burned, and the reaction products were passed through an excess of lime water, while forming 8 g of a precipitate. What hydrocarbon was taken?

Example 5 solution.

1. The general formula of a gaseous saturated non-cyclic hydrocarbon (alkane) is C n H 2n + 2
   Then the combustion reaction scheme looks like this:

   C n H 2n+2 + O 2 → CO 2 + H 2 O
   It is easy to see that the combustion of 1 mole of alkane will release n moles of carbon dioxide.

   The amount of alkane substance is found by its volume (do not forget to convert milliliters to liters!):

   ν(C n H 2n+2) = 0.488 / 22.4 = 0.02 mol.

2. When carbon dioxide is passed through Ca (OH) 2 lime water, calcium carbonate precipitates:

   CO 2 + Ca (OH) 2 \u003d CaCO 3 + H 2 O

   The mass of calcium carbonate precipitate is 8 g, the molar mass of calcium carbonate is 100 g/mol.

   So its amount of matter
   ν (CaCO 3) \u003d 8/100 \u003d 0.08 mol.
   The amount of carbon dioxide substance is also 0.08 mol.

3. The amount of carbon dioxide is 4 times more than alkane, so the formula of alkane is C 4 H 10.

Answer: C 4 H 10.

Example 6
Relative vapor density organic compound nitrogen is 2. When burning 9.8 g of this compound, 15.68 liters of carbon dioxide (n.a.) and 12.6 g of water are formed. Derive the molecular formula of the organic compound.

Example 6 solution.

Since the substance turns into carbon dioxide and water during combustion, it means that it consists of atoms C, H and, possibly, O. Therefore, its general formula can be written as C x H y O z.

1. We can write the combustion reaction scheme (without placing the coefficients):

   C x H y O z + O 2 → CO 2 + H 2 O

   All the carbon from the original substance goes into carbon dioxide, and all the hydrogen goes into water.

2. We find the amounts of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain:
   ν (CO 2) \u003d V / V m \u003d 15.68 / 22.4 \u003d 0.7 mol.
   One molecule of CO 2 accounts for one atom C, which means that there are as many moles of carbon as CO 2.

   ν(C) = 0.7 mol
   

   One molecule of water contains two atom H, means the amount of hydrogen twice as much than water.
   ν(H) \u003d 0.7 2 \u003d 1.4 mol.

3. We check the presence of oxygen in the substance. To do this, the masses C and H must be subtracted from the mass of the entire initial substance.
   m(C) = 0.7 12 = 8.4 g, m(H) = 1.4 1 = 1.4 g
   The mass of the entire substance is 9.8 g.
   m(O) = 9.8 - 8.4 - 1.4 = 0, i.e. there are no oxygen atoms in this substance.
   If oxygen were present in a given substance, then by its mass it would be possible to find the amount of the substance and calculate the simplest formula based on the presence of three different atoms.
4. The next steps are already familiar to you: the search for the simplest and true formulas.
   C: H = 0.7: 1.4 = 1: 2
   The simplest formula of CH 2.
5. We are looking for the true molar mass by the relative density of the gas with respect to nitrogen (do not forget that nitrogen consists of diatomic N 2 molecules and its molar mass is 28 g / mol):
   M ist. \u003d D by N 2 M (N 2) \u003d 2 28 \u003d 56 g / mol.
   The true formula is CH 2, its molar mass is 14.
   56 / 14 = 4.
   The true formula is C 4 H 8.

Answer: C 4 H 8.

Example 7
Determine the molecular formula of the substance, during the combustion of 9 g of which 17.6 g of CO 2, 12.6 g of water and nitrogen were formed. The relative density of this substance with respect to hydrogen is 22.5. Determine the molecular formula of the substance.

Example 7 solution.

1. Substance contains C,H atoms and N. Since the mass of nitrogen in the combustion products is not given, it will have to be calculated based on the mass of all organic matter.
   Combustion reaction scheme:
   C x H y N z + O 2 → CO 2 + H 2 O + N 2
2. We find the amounts of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain:

   ν (CO 2) \u003d m / M \u003d 17.6 / 44 \u003d 0.4 mol.
   ν(C) = 0.4 mol.
   ν (H 2 O) \u003d m / M \u003d 12.6 / 18 \u003d 0.7 mol.
   ν(H) \u003d 0.7 2 \u003d 1.4 mol.

3. Find the mass of nitrogen in the original substance.
   To do this, the masses C and H must be subtracted from the mass of the entire initial substance.

   M(C) = 0.4 12 = 4.8 g,
   m(H) = 1.4 1 = 1.4 g

   The mass of the entire substance is 9.8 g.

   M(N) = 9 - 4.8 - 1.4 = 2.8 g,
   ν(N) \u003d m / M \u003d 2.8 / 14 \u003d 0.2 mol.

4. C:H:N=0.4:1.4:0.2=2:7:1
   The simplest formula is C 2 H 7 N.
   True molar mass
   M \u003d D according to H 2 M (H 2) \u003d 22.5 2 \u003d 45 g / mol.
   It coincides with the molar mass calculated for the simplest formula. That is, this is the true formula of the substance.

Answer: C 2 H 7 N.

Example 8
The substance contains C, H, O and S. When burning 11 g of it, 8.8 g of CO 2, 5.4 g of H 2 O were released, and sulfur was completely converted into barium sulfate, the mass of which turned out to be 23.3 g. Determine substance formula.

Example 8 solution.

The formula of a given substance can be represented as C x H y S z O k . When it is burned, carbon dioxide, water and sulfur dioxide are produced, which are then converted into barium sulfate. Accordingly, all sulfur from the original substance is converted to barium sulfate.

1. We find the amounts of substances of carbon dioxide, water and barium sulfate and the corresponding chemical elements from the substance under study:

   ν (CO 2) \u003d m / M \u003d 8.8 / 44 \u003d 0.2 mol.
   ν(C) = 0.2 mol.
   ν (H 2 O) \u003d m / M \u003d 5.4 / 18 \u003d 0.3 mol.
   ν(H) = 0.6 mol.
   ν (BaSO 4) \u003d 23.3 / 233 \u003d 0.1 mol.
   ν(S) = 0.1 mol.

2. We calculate the estimated mass of oxygen in the initial substance:

   M(C) = 0.2 12 = 2.4 g
   m(H) = 0.6 1 = 0.6 g
   m(S) = 0.1 32 = 3.2 g
   m(O) = m substances − m(C) − m(H) − m(S) = 11 − 2.4 − 0.6 − 3.2 = 4.8 g,
   ν(O) = m / M = 4.8 / 16 = 0.3 mol

3. We find the molar ratio of elements in a substance:
   C:H:S:O=0.2:0.6:0.1:0.3=2:6:1:3
   Substance formula C 2 H 6 SO 3.
   It should be noted that in this way we obtained only the simplest formula.
   However, the resulting formula is true, because when you try to double this formula (C 4 H 12 S 2 O 6), it turns out that 4 carbon atoms, in addition to sulfur and oxygen, have 12 H atoms, and this is impossible.

Answer: C 2 H 6 SO 3.

Determination of formulas of substances by chemical properties.

Example 9
Determine the formula of alkadiene if 80 g of a 2% bromine solution can decolorize it.

Example 9 solution.

1. The general formula for alkadienes is С n H 2n−2.
   Let us write the reaction equation for the addition of bromine to alkadiene, not forgetting that in the diene molecule two double bonds and, accordingly, 2 moles of bromine will react with 1 mole of the diene:
   С n H 2n−2 + 2Br 2 → С n H 2n−2 Br 4
2. Since the mass and percentage concentration of the bromine solution that reacted with the diene are given in the problem, it is possible to calculate the amount of substance of the reacted bromine:

   M (Br 2) \u003d m solution ω \u003d 80 0.02 \u003d 1.6 g
   ν (Br 2) \u003d m / M \u003d 1.6 / 160 \u003d 0.01 mol.

3. Since the amount of bromine that reacted is 2 times more than the alkadiene, you can find the amount of the diene and (since its mass is known) its molar mass:

   0,005		0,01
   C n H 2n−2	+ 2Br2 →	C n H 2n−2 Br 4

   M diene \u003d m / ν \u003d 3.4 / 0.05 \u003d 68 g / mol.

4. We find the formula of alkadiene according to its general formulas, expressing the molar mass in terms of n:

   14n - 2 = 68
   n = 5.

   This is C 5 H 8 pentadiene.

Answer: C 5 H 8.

Example 10
In the interaction of 0.74 g of saturated monohydric alcohol with metallic sodium, hydrogen was released in an amount sufficient to hydrogenate 112 ml of propene (n.a.). What is this alcohol?

Example 10 solution.

1. The formula for limiting monohydric alcohol is C n H 2n + 1 OH. Here it is convenient to write the alcohol in a form in which it is easy to formulate the reaction equation - i.e. with a separate OH group.
2. Let's compose the reaction equations (we must not forget about the need to equalize the reactions):

   2C n H 2n+1 OH + 2Na → 2C n H 2n+1 ONa + H 2
   C 3 H 6 + H 2 → C 3 H 8

3. You can find the amount of propene, and from it - the amount of hydrogen. Knowing the amount of hydrogen, by the reaction we find the amount of alcohol substance:

   ν (C 3 H 6) \u003d V / V m \u003d 0.112 / 22.4 \u003d 0.005 mol => ν (H 2) \u003d 0.005 mol,
   ν alcohol \u003d 0.005 2 \u003d 0.01 mol.

4. Find the molar mass of alcohol and n:

   M alcohol \u003d m / ν \u003d 0.74 / 0.01 \u003d 74 g / mol,
   14n + 18 = 74
   14n = 56
   n = 4.

   Alcohol - butanol C 4 H 7 OH.

Answer: C 4 H 7 OH.

Example 11.
Define Formula ester, upon hydrolysis of 2.64 g of which 1.38 g of alcohol and 1.8 g of monobasic carboxylic acid.

Example 11 solution.

1. The general formula of an ester consisting of an alcohol and an acid with different number carbon atoms can be represented as follows:
   C n H 2n+1 COOC m H 2m+1
   Accordingly, alcohol will have the formula
   C m H 2m+1 OH,
   and acid
   C n H 2n+1 COOH .
   Ester hydrolysis equation:
   C n H 2n+1 COOC m H 2m+1 + H 2 O → C m H 2m+1 OH + C n H 2n+1 COOH
2. According to the law of conservation of mass of substances, the sum of the masses of the initial substances and the sum of the masses of the reaction products are equal.
   Therefore, from the data of the problem, you can find the mass of water:

   M H 2 O \u003d (mass of acid) + (mass of alcohol) - (mass of ether) \u003d 1.38 + 1.8 - 2.64 \u003d 0.54 g
   ν H 2 O \u003d m / M \u003d 0.54 / 18 \u003d 0.03 mol

   Accordingly, the amounts of acid and alcohol substances are also equal to a mole.
   You can find their molar masses:

   M acid \u003d m / ν \u003d 1.8 / 0.03 \u003d 60 g / mol,
   M alcohol \u003d 1.38 / 0.03 \u003d 46 g / mol.

   We get two equations, from which we find m and n:

   M C n H 2n + 1 COOH \u003d 14n + 46 \u003d 60, n \u003d 1 - acetic acid
   M C m H 2m + 1 OH = 14m + 18 = 46, m = 2 - ethanol.

   Thus, the desired ester is the ethyl ester of acetic acid, ethyl acetate.

Answer: CH 3 COOC 2 H 5 .

Example 12.
Determine the formula of an amino acid if, by treating 8.9 g of it with an excess of sodium hydroxide, 11.1 g of the sodium salt of this acid can be obtained.

Example 12 solution.

1. The general formula of an amino acid (assuming that it does not contain any other functional groups, except for one amino group and one carboxyl group):
   NH2-CH(R)-COOH.
   It could be written in different ways, but for the convenience of writing the reaction equation, it is better to isolate the functional groups separately in the amino acid formula.
2. You can write an equation for the reaction of this amino acid with sodium hydroxide:
   NH 2 -CH(R)-COOH + NaOH → NH 2 -CH(R)-COONa + H 2 O
   The amounts of the substance of the amino acid and its sodium salt are equal. However, we cannot find the mass of any of the substances in the reaction equation. Therefore, in such problems it is necessary to express the amounts of substances of an amino acid and its salt in terms of molar masses and equate them:

   M (amino acids NH 2 -CH (R) -COOH) \u003d 74 + M R
   M(salts NH 2 -CH(R)-COONa) \u003d 96 + M R
   ν amino acids = 8.9 / (74 + M R),
   ν salt = 11.1 / (96 + M R)
   8.9 / (74 + M R) = 11.1 / (96 + M R)
   M R = 15

   It is easy to see that R = CH 3 .
   This can be done mathematically if we assume that R - C n H 2n+1 .
   14n + 1 = 15, n = 1 . Set the formula of the limiting monobasic carboxylic acid, the calcium salt of which contains 30.77% calcium.

   Part 2. Determination of the formula of a substance by combustion products.

   2-1. The relative vapor density of an organic compound in terms of sulfur dioxide is 2. When 19.2 g of this substance is burned, 52.8 g of carbon dioxide (N.O.) and 21.6 g of water are formed. Derive the molecular formula of the organic compound.

   2-2. When burning organic matter weighing 1.78 g in excess oxygen, 0.28 g of nitrogen, 1.344 l (n.o.) CO 2 and 1.26 g of water were obtained. Determine the molecular formula of the substance, knowing that the indicated sample of the substance contains 1.204 10 22 molecules.

   2-3. Carbon dioxide obtained from the combustion of 3.4 g of hydrocarbon was passed through an excess of calcium hydroxide solution and 25 g of precipitate was obtained. Derive the simplest formula for a hydrocarbon.

   2-4. During the combustion of organic matter containing C, H and chlorine, 6.72 l (N.O.) of carbon dioxide, 5.4 g of water, 3.65 g of hydrogen chloride were released. Set the molecular formula of the burned substance.

   2-5. (USE-2011) During the combustion of the amine, 0.448 l (n.o.) of carbon dioxide, 0.495 g of water and 0.056 l of nitrogen were released. Determine the molecular formula of this amine.

   Part 3. Determination of the formula of a substance by chemical properties.

   3-1. Determine the formula of an alkene if it is known that 5.6 g of it, when added to water, form 7.4 g of alcohol.

   3-2. For the oxidation of 2.9 g of saturated aldehyde to acid, 9.8 g of copper (II) hydroxide was required. Determine the formula of the aldehyde.

   3-3. Monobasic monoamino acid weighing 3 g with an excess of hydrogen bromide forms 6.24 g of salt. Determine the amino acid formula.

   3-4. In the interaction of the limiting dihydric alcohol weighing 2.7 g with an excess of potassium, 0.672 liters of hydrogen were released. Determine the formula of alcohol.

   3-5. (USE-2011) When saturated monohydric alcohol was oxidized with copper (II) oxide, 9.73 g of aldehyde, 8.65 g of copper and water were obtained. Determine the molecular formula of this alcohol.

   Answers and comments to tasks for independent solution.

   1-2. C 3 H 6 (NH 2) 2

   1-3. C 2 H 4 (COOH) 2

   1-5. (HCOO) 2 Ca - calcium formate, salt of formic acid

   2-1. C 8 H 16 O

   2-2. C 3 H 7 NO

   2-3. C 5 H 8 (we find the mass of hydrogen by subtracting the mass of carbon from the mass of hydrocarbon)

   2-4. C 3 H 7 Cl (do not forget that hydrogen atoms are found not only in water, but also in HCl)

   3-2. C 3 H 6 O

   3-3. C 2 H 5 NO 2

   
   

   Classification inorganic substances and their nomenclature are based on the simplest and most constant characteristic over time -chemical composition, which shows the atoms of the elements that form a given substance, in their numerical ratio. If a substance is made up of atoms of one chemical element, i.e. is a form of existence of this element in a free form, then it is called a simple substance ; if the substance is made up of atoms of two or more elements, it is calledcomplex substance. All simple substances (except monatomic) and all complex substances are calledchemical compounds, since in them the atoms of one or different elements are interconnected by chemical bonds.
   

   The nomenclature of inorganic substances consists of formulas and names.Chemical formula- depiction of the composition of a substance with the help of symbols of chemical elements, numerical indices and some other signs.chemical name- a representation of the composition of a substance using a word or group of words. The construction of chemical formulas and names is determined by the systemnomenclature rules.
   

   Symbols and names of chemical elements are given in the Periodic system of elements of D.I. Mendeleev. Elements are conditionally divided into metals and non-metals . Non-metals include all elements of the VIIIA group (noble gases) and VIIA group (halogens), elements of the VIA group (except polonium), elements nitrogen, phosphorus, arsenic (VA group); carbon, silicon (IVA-group); boron (IIIA-group), as well as hydrogen. The remaining elements are classified as metals.
   

   When compiling the names of substances, Russian names of elements are usually used, for example, dioxygen, xenon difluoride, potassium selenate. By tradition, for some elements, the roots of their Latin names are introduced into derivative terms:
   

   The followingnumerical prefixes:
   

   1 - mono

   7 - hepta

   2 - di

   3 - three

   9 - nona

   4 - tetra

   5 - penta

   6 - hexa

   An indefinite number is indicated by a numerical prefix n - poly.
   

   For some simple substances also use special names such as 3 - ozone, R 4 - white phosphorus.
   

   Chemical formulascomplex substances are made up of the designationelectropositive(conditional and real cations) andelectronegative(conditional and real anions) components, for example, CuSO 4 (here Cu 2+ - real cation, SO 4 2- - real anion) and PCl 3 (here P +III - conditional cation, Cl-I - conditional anion).
   

   Names of complex substances make up the chemical formulas from right to left. They consist of two words - the names of the electronegative components (in the nominative case) and the electropositive components (in genitive case), for example:
   

   CuSO4 - copper(II) sulfate
   PCl 3 - phosphorus trichloride
   LaCl 3 - lanthanum(III) chloride
   CO - carbon monoxide
   

   The number of electropositive and electronegative components in the names is indicated by the numerical prefixes given above (universal method), or by the oxidation states (if they can be determined by the formula) using Roman numerals in parentheses (the plus sign is omitted). In some cases, the ion charge is given (for complex cations and anions), using Arabic numerals with the corresponding sign.
   

   The following special names are used for common multielement cations and anions:
   

   NH 4 + - ammonium

   HF2- - hydrodifluoride

   For a small number of well-known substances also use special titles:
   

   AsH 3 - arsine

   HN 3 - hydrogen azidide

   B 2 H 6 - borane

   H2 S - hydrogen sulfide

   1. Acid and basic hydroxides. salt
   

   Hydroxides - a type of complex substances, which include atoms of a certain element E (except for fluorine and oxygen) and the hydroxo group OH; general formula of hydroxides E (OH) n , where n = 1÷6. Hydroxide form E(OH) n is called ortho-form; for n > 2 hydroxide can also be found in meta -form, including, in addition to E atoms and OH groups, O oxygen atoms, for example, E (OH) 3 and EO(OH), E(OH) 4 and E(OH) 6 and EO 2 (OH) 2 .
   

   Hydroxides are divided into two chemically opposite groups: acidic and basic hydroxides.
   

   Acid hydroxidescontain hydrogen atoms, which can be replaced by metal atoms, subject to the rule of stoichiometric valence. Most acid hydroxides are found in meta -form, with hydrogen atoms in the formulas of acid hydroxides put in first place, for example H 2 SO 4, HNO 3 and H 2 CO 3, not SO 2 (OH) 2, NO 2 (OH) and CO (OH) 2 . The general formula of acid hydroxides is H x EO y , where the electronegative component EO at x- called an acid residue. If not all hydrogen atoms are replaced by a metal, then they remain in the composition of the acid residue.
   

   The names of common acid hydroxides consist of two words: their own name with the ending "aya" and the group word "acid". Here are the formulas and proper names of common acid hydroxides and their acid residues (a dash means that the hydroxide is not known in free form or in acidic form). aqueous solution):
   

   HAso 2 - metaarsenic

   AsO 2 - - metaarsenite

   H 3 AsO 3 - orthoarsenic

   AsO 3 3- - orthoarsenite

   H 3 AsO 4 - arsenic

   AsO 4 3- - arsenate

   -	B 4 O 7 2- - tetraborate

   -	ВiО 3 - - bismuthate

   H 2 CrO 4 - chrome

   CrO 4 2- - chromate

   -	НCrO 4 - - hydrochromate

   H 2 Cr 2 O 7 - dichromic

   Cr 2 O 7 2- - dichromate

   -	FeO 4 2- - ferrate

   HIO 3 - iodine

   IO 3 - - iodate

   HIO 4 - metaiodine

   IO 4 - - metaperiodat

   H 5 IO 6 - orthoiodic

   IO 6 5- - orthoperiodate

   HMnO 4 - manganese

   MnO 4 - - permanganate

   HNO 2 - nitrogenous

   NO 2 - - nitrite

   HNO 3 - nitrogen

   NO 3 - - nitrate

   HPO 3 - metaphosphoric

   PO 3 - - metaphosphate

   H3PO4 - orthophosphoric

   PO 4 3- - orthophosphate

   HPO 4 2- - hydrogen orthophosphate

   H 2 PO 4 - - dihydrootophosphate

   H 4 P 2 O 7 - diphosphoric

   P 2 O 7 4- - diphosphate

   Less common acid hydroxides are named according to the nomenclature rules for complex compounds, for example:
   

   The names of acid residues are used in the construction of the names of salts.
   

   Basic hydroxidescontain hydroxide ions, which can be replaced by acidic residues, subject to the rule of stoichiometric valency. All basic hydroxides are found in ortho -form; their general formula is M(OH) n , where n = 1.2 (rarely 3.4) and M n +- metal cation. Examples of formulas and names of basic hydroxides:
   

   The most important chemical property of basic and acid hydroxides is their interaction with each other with the formation of salts (salt formation reaction), for example:
   

   Ca (OH) 2 + H 2 SO 4 \u003d CaSO 4 + 2H 2 O
   

   Ca (OH) 2 + 2H 2 SO 4 \u003d Ca (HSO 4) 2 + 2H 2 O
   

   2Ca(OH) 2 + H 2 SO 4 = Ca 2 SO 4 (OH) 2 + 2H 2 O
   

   Salts - a type of complex substances, which include cations M n+ and acid residues*.
   

   Salts with the general formula M x (EO y) n are called average salts, and salts with unsubstituted hydrogen atoms - sour salts. Sometimes salts also contain hydroxide and/or oxide ions; such salts are called main salts. Here are examples and names of salts:
   

   CuCO3

   Copper(II) carbonate

   Ti(NO 3 ) 2 O

   Titanium oxide dinitrate

   Acid and basic salts can be converted to medium salts by reaction with the corresponding basic and acidic hydroxide, for example:
   

   Ca(HSO 4 ) 2 + Ca(OH) = CaSO 4 + 2H2 O
   

   Ca2 SO4 (OH)2 + H2 SO4 = Ca2 SO4 + 2H2 O
   

   There are also salts containing two different cations: they are often calleddouble salts, for example:
   

   2. Acid and basic oxides
   

   Oxides EXABOUTat- products of complete dehydration of hydroxides:
   

   Acid hydroxides (H2 SO4 , H2 CO3 ) answer acid oxides (SO3 , CO2 ), and basic hydroxides (NaOH, Ca(OH)2 ) - basic oxides(Na2 O, CaO), and the oxidation state of the element E does not change when moving from hydroxide to oxide. An example of formulas and names of oxides:
   

   SO3 - sulfur trioxide

   Na2 O - sodium oxide

   P4 O10 - tetraphosphorus decaoxide

   THO2 - thorium(IV) oxide

   Acid and basic oxides retain the salt-forming properties of the corresponding hydroxides when interacting with hydroxides of opposite properties or with each other:
   

   N2 O5 + 2NaOH = 2NaNO3 + H2 O
   

   3CaO + 2H3 PO4 = Ca3 (PO4 ) 2 + 3H2 O
   

   La2 O3 + 3SO3 = La2 (SO4 ) 3
   

   3. Amphoteric oxides and hydroxides
   

   Amphoterichydroxides and oxides - a chemical property consisting in the formation of two rows of salts by them, for example, for hydroxide and aluminum oxide:
   

   (a) 2Al(OH)3 + 3SO3 = Al2 (SO4 ) 3 + 3H2 O
   

   Al2 O3 + 3H2 SO4 = Al2 (SO4 ) 3 + 3H2 O
   

   (b) 2Al(OH)3 + Na2 O = 2NaAlO2 + 3H2 O
   

   Al2 O3 + 2NaOH = 2NaAlO2 + H2 O
   

   Thus, hydroxide and aluminum oxide in reactions (a) exhibit the propertiesmajorhydroxides and oxides, i.e. react with acid hydroxides and oxide, forming the corresponding salt - aluminum sulfate Al2 (SO4 ) 3 , while in reactions (b) they also exhibit the propertiesacidichydroxides and oxides, i.e. react with basic hydroxide and oxide, forming a salt - sodium dioxoaluminate (III) NaAlO2 . In the first case, the aluminum element exhibits the property of a metal and is part of the electropositive component (Al3+ ), in the second - the property of a non-metal and is part of the electronegative component of the salt formula (AlO2 - ).
   

   If these reactions proceed in an aqueous solution, then the composition of the resulting salts changes, but the presence of aluminum in the cation and anion remains:
   

   2Al(OH)3 + 3H2 SO4 = 2 (SO4 ) 3
   

   Al(OH)3 + NaOH = Na
   

   Here square brackets denote complex ions3+ - hexaaquaaluminum(III) cation,- - tetrahydroxoaluminate(III)-ion.
   

   Elements that exhibit metallic and non-metallic properties in compounds are called amphoteric, these include elements of the A-groups of the Periodic system - Be, Al, Ga, Ge, Sn, Pb, Sb, Bi, Po, etc., as well as most elements of B- groups - Cr, Mn, Fe, Zn, Cd, Au, etc. Amphoteric oxides are called the same as the main ones, for example:
   

   If several oxidation states correspond to an amphoteric element in compounds, then the amphotericity of the corresponding oxides and hydroxides (and, consequently, the amphotericity of the element itself) will be expressed differently. For low oxidation states, hydroxides and oxides have a predominance of basic properties, and the element itself has metallic properties, so it is almost always a part of cations. For high degrees oxidation, on the contrary, in hydroxides and oxides there is a predominance acid properties, and the element itself has non-metallic properties, so it is almost always included in the composition of anions. For example, manganese(II) oxide and hydroxide are dominated by basic properties, while manganese itself enters into the composition of cations of the type2+ , while manganese(VII) oxide and hydroxide are dominated by acidic properties, and manganese itself is a part of an anion of the MnO type4 - . Amphoteric hydroxides with a large predominance of acidic properties, formulas and names are attributed according to the model of acid hydroxides, for example HMnVIIO4 - permanganic acid.
   

   Thus, the division of elements into metals and non-metals is conditional; between elements (Na, K, Ca, Ba, etc.) with purely metallic and elements (F, O, N, Cl, S, C, etc.) with purely non-metallic properties there is a large group of elements with amphoteric properties.
   

   4. Binary connections
   

   An extensive type of inorganic complex substances is binary compounds. These include, first of all, all two-element compounds (except basic, acidic and amphoteric oxides), for example H2 O, KBr, H2 S, Cs2 (S2 ), N2 O, NH3 , HN3 , CaC2 , SiH4 . The electropositive and electronegative components of the formulas of these compounds include single atoms or bonded groups of atoms of the same element.
   

   Multi-element substances, in the formulas of which one of the components contains atoms of several elements that are not interconnected, as well as single-element or multi-element groups of atoms (except hydroxides and salts), are considered as binary compounds, for example CSO, IO2 F3 , SBrO2 F, CrO(O2 ) 2 , PSI3 , (CaTi)O3 , (FeCu)S2

   Pb(N3 ) 2 - lead(II) azide
   

   For some binary compounds, special names are used, the list of which was given earlier.
   

   Chemical properties binary compounds are quite diverse, so they are often divided into groups according to the name of the anions, i.e. halides, chalcogenides, nitrides, carbides, hydrides, etc. are considered separately. Among binary compounds, there are also those that have some signs of other types of inorganic substances. So, compounds CO, NO, NO2 , and (FeIIFe2 III)O4 , whose names are built using the word oxide, cannot be attributed to the type of oxides (acidic, basic, amphoteric). Carbon monoxide CO, nitrogen monoxide NO and nitrogen dioxide NO2 do not have the corresponding acid hydroxides (although these oxides are formed by non-metals C and N), they do not form salts, the anions of which would include C atomsII, NIIand NIV. Double oxide (FeIIFe2 III)O4 - oxide of diiron (III) - iron (II), although it contains atoms of the amphoteric element - iron, in the composition of the electropositive component, but in two varying degrees oxidation, as a result of which, when interacting with acid hydroxides, it forms not one, but two different salts.
   

   Binary compounds such as AgF, KBr, Na2 S, Ba(HS)2 , NaCN, NH4 Cl, and Pb(N3 ) 2 , are built, like salts, from real cations and anions, which is why they are calledsalinebinary compounds (or just salts). They can be considered as products of substitution of hydrogen atoms in the compounds HF, HCl, HBr, H2 S, HCN and HN3 . The latter in an aqueous solution have an acidic function, and therefore their solutions are called acids, for example HF (aqua) - hydrofluoric acid, H2 S(aqua) - hydrosulfide acid. However, they do not belong to the type of acid hydroxides, and their derivatives do not belong to the salts within the classification of inorganic substances.

   Value and its dimension	Ratio
   Atomic mass of element X (relative)
   Element number	Z= N(e –) = N(R +)
   Mass fraction of element E in substance X, in fractions of a unit, in%)
   Amount of substance X, mol
   Amount of gas substance, mol	V m= 22.4 l/mol (n.o.) well. - R= 101 325 Pa, T= 273 K
   Molar mass of substance X, g/mol, kg/mol
   Mass of substance X, g, kg	m(X)= n(X) M(X)
   Molar volume of gas, l / mol, m 3 / mol	V m= 22.4 l / mol at n.o.
   Gas volume, m 3	V = V m × n
   Product yield
   Substance density X, g / l, g / ml, kg / m 3
   Density gaseous substance X for hydrogen
   Density of a gaseous substance X in air	M(air) = 29 g/mol
   Combined gas law
   Mendeleev-Clapeyron equation	PV = nRT, R= 8.314 J/mol×K
   Volume fraction of a gaseous substance in a mixture of gases, in fractions of a unit or in%
   Molar mass of a mixture of gases
   Mole fraction of substance (X) in the mixture
   The amount of heat, J, kJ	Q = n(X) Q(X)
   Thermal effect of the reaction	Q =–H
   Heat of formation of substance X, J/mol, kJ/mol
   Chemical reaction rate (mol/lsec)
   Mass action law (for a simple reaction)	a A+ in B= from C + d D u = k from a(A) from in(B)
   Van't Hoff's rule
   Solubility of substance (X) (g/100 g solvent)
   Mass fraction of substance X in a mixture A + X, in fractions of a unit, in%
   Mass of solution, g, kg	m(rr) = m(X) + m(H2O) m(rr) = V(rr)  (rr)
   Mass fraction of the dissolved substance in the solution, in fractions of a unit, in %
   Solution Density
   The volume of the solution, cm 3, l, m 3
   Molar concentration, mol/l
   The degree of dissociation of the electrolyte (X), in fractions of a unit or%
   Ionic product of water	K(H 2 O) =
   Hydrogen indicator	pH = –lg

   Main:

   Kuznetsova N.E. and etc. Chemistry. 8 cells-10 cells .. - M .: Ventana-Graf, 2005-2007.

   Kuznetsova N.E., Litvinova T.N., Levkin A.N. Chemistry. Grade 11 in 2 parts, 2005-2007.

   Egorov A.S. Chemistry. A new textbook for preparing for universities. Rostov n/a: Phoenix, 2004.– 640 p.

   Egorov A.S. Chemistry: a modern course to prepare for the exam. Rostov n / a: Phoenix, 2011. (2012) - 699 p.

   Egorov A.S. Self-instruction manual for solving chemical problems. - Rostov-on-Don: Phoenix, 2000. - 352 p.

   Chemistry / manual-tutor for university applicants. Rostov-n/D, Phoenix, 2005– 536 p.

   Khomchenko G.P., Khomchenko I.G. Tasks in chemistry for university students. M.: high school. 2007.–302p.

   Additional:

   Vrublevsky A.I.. Educational and training materials for preparation for centralized testing in chemistry / A.I. Vrublevsky - Mn .: Unipress LLC, 2004. - 368 p.

   Vrublevsky A.I.. 1000 tasks in chemistry with chains of transformations and control tests for schoolchildren and university entrants.– Mn.: Unipress LLC, 2003.– 400 p.

   Egorov A.S.. All types of computational tasks in chemistry for preparing for the Unified State Examination.–Rostov n/D: Phoenix, 2003.–320p.

   Egorov A.S., Aminova G.Kh. Typical tasks and exercises to prepare for the exam in chemistry. - Rostov n / D: Phoenix, 2005. - 448 p.

   Unified state exam 2007. Chemistry. Educational and training materials for the preparation of students / FIPI - M .: Intellect-Center, 2007. - 272 p.

   USE-2011. Chemistry. Training kit, ed. A.A. Kaverina. - M .: National Education, 2011.

   The only real options for tasks to prepare for the unified state exam. USE.2007. Chemistry/V.Yu. Mishina, E.N. Strelnikov. M.: Federal Testing Center, 2007.–151p.

   Kaverina A.A.. The optimal bank of tasks for preparing students. Unified State Exam 2012. Chemistry. Tutorial./ A.A. Kaverina, D.Yu. Dobrotin, Yu.N. Medvedev, M.G. Snastina. - M .: Intellect-Center, 2012. - 256 p.

   Litvinova T.N., Vyskubova N.K., Azhipa L.T., Solovieva M.V.. Test tasks in addition to tests for students of 10-month correspondence preparatory courses (guidelines). Krasnodar, 2004. - S. 18 - 70.

   Litvinova T.N.. Chemistry. USE-2011. Training tests. Rostov n/a: Phoenix, 2011.– 349 p.

   Litvinova T.N.. Chemistry. Tests for the exam. Rostov n / D .: Phoenix, 2012. - 284 p.

   Litvinova T.N.. Chemistry. Laws, properties of elements and their compounds. Rostov n / D .: Phoenix, 2012. - 156 p.

   Litvinova T.N., Melnikova E.D., Solovieva M.V.., Azhipa L.T., Vyskubova N.K. Chemistry in tasks for applicants to universities. - M .: LLC "Publishing House Onyx": LLC "Publishing House "World and Education", 2009.- 832 p.

   Educational and methodological complex in chemistry for students of medical and biological classes, ed. T.N. Litvinova. - Krasnodar: KSMU, - 2008.

   Chemistry. USE-2008. Entrance tests, teaching aid / ed. V.N. Doronkin. - Rostov n / a: Legion, 2008. - 271 p.

   List of sites on chemistry:

   1. Alchemist. http:// www. alchemist. en

   2. Chemistry for everyone. Electronic guide for full course chemistry.

   http:// www. informika. en/ text/ database/ chemy/ START. html

   3. School chemistry - a reference book. http:// www. school chemistry. by. en

   4. Tutor in chemistry. http://www. chemistry.nm.ru

   Internet resources

   Alchemist. http:// www. alchemist. en

   Chemistry for everyone. Electronic reference book for a complete course of chemistry.

   http:// www. informika. en/ text/ database/ chemy/ START. html

   School chemistry - a reference book. http:// www. school chemistry. by. en

   http://www.classchem.narod.ru

   Chemistry tutor. http://www. chemistry.nm.ru

   http://www.alleng.ru/edu/chem.htm- Internet educational resources in chemistry

   http://schoolchemistry.by.ru/- school chemistry. On this site there is an opportunity to pass On-line testing on various topics, as well as demo options Unified State Exam

   Chemistry and life–XX1st century: popular scientific journal. http:// www. hij. en