Chemistry– the science of the composition, structure, properties and transformations of substances.

Atomic-molecular science. Substances consist of chemical particles (molecules, atoms, ions), which have a complex structure and consist of elementary particles (protons, neutrons, electrons).

Atom– a neutral particle consisting of a positive nucleus and electrons.

Molecule– a stable group of atoms connected by chemical bonds.

Chemical element– a type of atoms with the same nuclear charge. Element denote

where X is the symbol of the element, Z– serial number of the element in the Periodic Table of Elements D.I. Mendeleev, A– mass number. Serial number Z equal to the charge of the atomic nucleus, the number of protons in the atomic nucleus and the number of electrons in the atom. Mass number A equal to the sum of the numbers of protons and neutrons in an atom. The number of neutrons is equal to the difference A–Z.

Isotopes– atoms of the same element having different mass numbers.

Relative atomic mass(A r) is the ratio of the average mass of an atom of an element of natural isotopic composition to 1/12 of the mass of an atom of the carbon isotope 12 C.

Relative molecular weight(M r) is the ratio of the average mass of a molecule of a substance of natural isotopic composition to 1/12 of the mass of an atom of the 12 C carbon isotope.

Atomic mass unit(a.u.m) – 1/12 of the mass of an atom of the carbon isotope 12 C. 1 a.u. m = 1.66? 10 -24 years

Mole– the amount of a substance containing as many structural units (atoms, molecules, ions) as there are atoms in 0.012 kg of the carbon isotope 12 C. Mole– the amount of a substance containing 6.02 10 23 structural units (atoms, molecules, ions).

n = N/N A, Where n– amount of substance (mol), N– number of particles, a N A– Avogadro’s constant. The amount of a substance can also be denoted by the symbol v.

Avogadro's constant N A = 6.02 10 23 particles/mol.

Molar massM(g/mol) – ratio of the mass of the substance m(d) to the amount of substance n(mol):

M = m/n, where: m = M n And n = m/M.

Molar volume of gasV M(l/mol) – gas volume ratio V(l) to the amount of substance of this gas n(mol). Under normal conditions V M = 22.4 l/mol.

Normal conditions: temperature t = 0°C, or T = 273 K, pressure p = 1 atm = 760 mm. rt. Art. = 101,325 Pa = 101.325 kPa.

V M = V/n, where: V = V M n And n = V/V M .

The result is a general formula:

n = m/M = V/V M = N/N A .

Equivalent- a real or fictitious particle that interacts with one hydrogen atom, or replaces it, or is equivalent to it in some other way.

Molar mass equivalents M e– the ratio of the mass of a substance to the number of equivalents of this substance: M e = m/n (eq) .

In charge exchange reactions, the molar mass of substance equivalents is

with molar mass M equal to: M e = M/(n ? m).

In redox reactions, the molar mass of equivalents of a substance with molar mass M equal to: M e = M/n(e), Where n(e)– number of transferred electrons.

Law of equivalents– the masses of reactants 1 and 2 are proportional to the molar masses of their equivalents. m 1 / m 2= M E1/M E2, or m 1 /M E1 = m 2 /M E2, or n 1 = n 2, Where m 1 And m 2– masses of two substances, M E1 And M E2– molar masses of equivalents, n 1 And n 2– the number of equivalents of these substances.

For solutions, the law of equivalents can be written as follows:

c E1 V 1 = c E2 V 2, Where with E1, with E2, V 1 And V 2– molar concentrations of equivalents and volumes of solutions of these two substances.

United gas law: pV = nRT, Where p– pressure (Pa, kPa), V– volume (m 3, l), n– amount of gas substance (mol), T – temperature (K), T(K) = t(°C) + 273, R– constant, R= 8.314 J/(K? mol), with J = Pa m 3 = kPa l.

2. Atomic structure and Periodic Law

Wave-particle duality matter - the idea that every object can have both wave and corpuscular properties. Louis de Broglie proposed a formula connecting the wave and corpuscular properties of objects: ? = h/(mV), Where h– Planck’s constant, ? – wavelength that corresponds to each body with mass m and speed V. Although wave properties exist for all objects, they can be observed only for micro-objects with masses on the order of the mass of an atom and an electron.

Heisenberg Uncertainty Principle: ?(mV x) ?х > h/2n or ?V x ?x > h/(2?m), Where m– particle mass, x– its coordinate, Vx– speed in direction x, ?– uncertainty, error of determination. The uncertainty principle means that it is impossible to simultaneously indicate the position (coordinate) x) and speed (V x) particles.

Particles with small masses (atoms, nuclei, electrons, molecules) are not particles in the sense of Newtonian mechanics and cannot be studied by classical physics. They are studied by quantum physics.

Principal quantum numbern takes values ​​1, 2, 3, 4, 5, 6 and 7, corresponding to the electronic levels (layers) K, L, M, N, O, P and Q.

Level– the space where electrons with the same number are located n. Electrons of different levels are spatially and energetically separated from each other, since the number n determines electron energy E(the more n, the more E) and distance R between electrons and nucleus (the more n, the more R).

Orbital (side, azimuthal) quantum numberl takes values ​​depending on the number n:l= 0, 1,…(n- 1). For example, if n= 2, then l = 0, 1; If n= 3, then l = 0, 1, 2. Number l characterizes the sublevel (sublayer).

Sublevel– the space where electrons with certain n And l. Sublevels of a given level are designated depending on the number l:s- If l = 0, p- If l = 1, d- If l = 2, f- If l = 3. The sublevels of a given atom are designated depending on the numbers n And l, for example: 2s (n = 2, l = 0), 3d(n= 3, l = 2), etc. Sublevels of a given level have different energies (the more l, the more E): E s< E < Е А < … and the different shapes of the orbitals that make up these sublevels: the s-orbital has the shape of a ball, p-the orbital is shaped like a dumbbell, etc.

Magnetic quantum numberm 1 characterizes the orientation of the orbital magnetic moment, equal to l, in space relative to the external magnetic field and takes the following values: – l,…-1, 0, 1,…l, i.e. total (2l + 1) value. For example, if l = 2, then m 1 =-2, -1, 0, 1, 2.

Orbital(part of a sublevel) – the space where electrons (no more than two) are located with certain n, l, m 1. Sublevel contains 2l+1 orbital. For example, d– the sublevel contains five d-orbitals. Orbitals of the same sublevel having different numbers m 1, have the same energy.

Magnetic spin numberm s characterizes the orientation of the electron’s own magnetic moment s, equal to?, relative to the external magnetic field and takes two values: +? And _ ?.

Electrons in an atom occupy levels, sublevels and orbitals according to the following rules.

Pauli's Rule: In one atom, two electrons cannot have four identical quantum numbers. They must differ in at least one quantum number.

From the Pauli rule it follows that an orbital can contain no more than two electrons, a sublevel can contain no more than 2(2l + 1) electrons, a level can contain no more 2n 2 electrons.

Klechkovsky's rule: electronic sublevels are filled in in order of increasing amount (n + l), and in case of the same amount (n+l)– in ascending order of number n.

Graphic form of Klechkovsky's rule.

According to Klechkovsky’s rule, sublevels are filled in in the following order: 1s, 2s, 2р, 3s, Зр, 4s, 3d, 4р, 5s, 4d, 5р, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s,…

Although the filling of sublevels occurs according to the Klechkovsky rule, in the electronic formula the sublevels are written sequentially by level: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f etc. Thus, the electronic formula of the bromine atom is written as follows: Br(35e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 .

The electronic configurations of a number of atoms differ from those predicted by Klechkovsky's rule. So, for Cr and Cu:

Сr(24e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 and Cu(29e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1.

Rule of Hunda (Gunda): The filling of the orbitals of a given sublevel is carried out so that the total spin is maximum. The orbitals of a given sublevel are filled first with one electron at a time.

Electronic configurations of atoms can be written by levels, sublevels, orbitals. For example, the electronic formula P(15e) can be written:

a) by levels)2)8)5;

b) by sublevels 1s 2 2s 2 2p 6 3s 2 3p 3;

c) by orbital

Examples of electronic formulas of some atoms and ions:

V(23e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2;

V 3+ (20e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 0.

3. Chemical bond

3.1. Valence bond method

According to the valence bond method, a bond between atoms A and B is formed by sharing a pair of electrons.

Covalent bond. Donor-acceptor connection.

Valence characterizes the ability of atoms to form chemical bonds and is equal to the number of chemical bonds formed by an atom. According to the valence bond method, valence is equal to the number of shared pairs of electrons, and in the case of a covalent bond, valence is equal to the number of unpaired electrons in the outer level of an atom in its ground or excited states.

Valence of atoms

For example, for carbon and sulfur:

Saturability covalent bond: atoms form a limited number of bonds equal to their valence.

Hybridization of atomic orbitals– mixing of atomic orbitals (AO) of different sublevels of the atom, the electrons of which participate in the formation of equivalent?-bonds. Hybrid orbital (HO) equivalence explains the equivalence of the chemical bonds formed. For example, in the case of a tetravalent carbon atom there is one 2s– and three 2p-electron. To explain the equivalence of the four?-bonds formed by carbon in the molecules CH 4, CF 4, etc., atomic one s- and three R- orbitals are replaced by four equivalent hybrid ones sp 3-orbitals:

Focus A covalent bond is that it is formed in the direction of maximum overlap of the orbitals that form a common pair of electrons.

Depending on the type of hybridization, hybrid orbitals have a specific location in space:

sp– linear, the angle between the axes of the orbitals is 180°;

sp 2– triangular, angles between the axes of the orbitals are 120°;

sp 3– tetrahedral, angles between the axes of the orbitals are 109°;

sp 3 d 1– trigonal-bipyramidal, angles 90° and 120°;

sp 2 d 1– square, angles between the axes of the orbitals are 90°;

sp 3 d 2– octahedral, the angles between the axes of the orbitals are 90°.

3.2. Molecular orbital theory

According to the theory of molecular orbitals, a molecule consists of nuclei and electrons. In molecules, electrons are located in molecular orbitals (MO). The MOs of the outer electrons have a complex structure and are considered as a linear combination of the outer orbitals of the atoms that make up the molecule. The number of formed MOs is equal to the number of AOs involved in their formation. The energies of MOs can be lower (bonding MOs), equal (non-bonding MOs) or higher (antibonding MOs), than the energies of the AOs that form them.

Terms of interaction of JSC

1. AO interact if they have similar energies.

2. AOs interact if they overlap.

3. AO interact if they have the appropriate symmetry.

For a diatomic molecule AB (or any linear molecule), the symmetry of MO can be:

If a given MO has an axis of symmetry,

If a given MO has a plane of symmetry,

If the MO has two perpendicular planes of symmetry.

The presence of electrons on the bonding MOs stabilizes the system, as it reduces the energy of the molecule compared to the energy of the atoms. The stability of the molecule is characterized bond order n, equal to: n = (n light – n size)/2, Where n light and n size - number of electrons in bonding and antibonding orbitals.

The filling of MOs with electrons occurs according to the same rules as the filling of AOs in an atom, namely: Pauli’s rule (there cannot be more than two electrons on a MO), Hund’s rule (the total spin must be maximum), etc.

The interaction of 1s-AO atoms of the first period (H and He) leads to the formation of bonding?-MO and antibonding?*-MO:

Electronic formulas of molecules, bond orders n, experimental bond energies E and intermolecular distances R for diatomic molecules from atoms of the first period are given in the following table:

Other atoms of the second period contain, in addition to 2s-AO, also 2p x -, 2p y – and 2p z -AO, which upon interaction can form?– and?-MO. For O, F and Ne atoms, the energies of 2s- and 2p-AOs are significantly different, and the interaction between the 2s-AO of one atom and the 2p-AO of another atom can be neglected, considering the interaction between the 2s-AO of two atoms separately from the interaction of their 2p-AO. The MO scheme for molecules O 2, F 2, Ne 2 has the following form:

For atoms B, C, N, the energies of 2s– and 2p-AO are close in their energies, and the 2s-AO of one atom interacts with the 2p z-AO of another atom. Therefore, the order of MOs in molecules B 2, C 2 and N 2 differs from the order of MOs in molecules O 2, F 2 and Ne 2. Below is the MO scheme for molecules B 2, C 2 and N 2:

Based on the given MO schemes, it is possible, for example, to write down the electronic formulas of the molecules O 2 , O 2 + and O 2 ?:

O 2 + (11e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *1 ? y *0)

n = 2 R = 0.121 nm;

O 2 (12e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *1 ? y *1)

n = 2.5 R = 0.112 nm;

O 2 ?(13e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *2 ? y *1)

n = 1.5 R = 0.126 nm.

In the case of the O 2 molecule, MO theory allows us to foresee greater strength of this molecule, since n = 2, the nature of changes in binding energies and internuclear distances in the series O 2 + – O 2 – O 2 ?, as well as the paramagnetism of the O 2 molecule, the upper MOs of which have two unpaired electrons.

3.3. Some types of connections

Ionic bond– electrostatic bond between ions of opposite charges. An ionic bond can be considered an extreme case of a polar covalent bond. An ionic bond is formed if the difference in electronegativity of the atoms? X is greater than 1.5–2.0.

An ionic bond is non-directional non-saturable communication In a NaCl crystal, the Na+ ion is attracted by all the Cl ions? and is repelled by all other Na + ions, regardless of the direction of interaction and the number of ions. This determines the greater stability of ionic crystals compared to ionic molecules.

Hydrogen bond– a bond between a hydrogen atom of one molecule and an electronegative atom (F, CI, N) of another molecule.

The existence of a hydrogen bond explains the anomalous properties of water: the boiling point of water is much higher than that of its chemical analogues: t kip (H 2 O) = 100 °C, and t kip (H 2 S) = -61 ° C. No hydrogen bonds are formed between H 2 S molecules.

4. Patterns of chemical processes

4.1. Thermochemistry

Energy(E)- ability to produce work. Mechanical work (A) is performed, for example, by gas during its expansion: A = p?V.

Reactions that occur with the absorption of energy are: endothermic.

Reactions that involve the release of energy are: exothermic.

Types of energy: heat, light, electrical, chemical, nuclear energy, etc.

Energy types: kinetic and potential.

Kinetic energy– the energy of a moving body, this is the work that a body can do before it reaches rest.

Heat (Q)– a type of kinetic energy – associated with the movement of atoms and molecules. When communicating to a body of mass (m) and specific heat capacity (c) of heat? Q its temperature increases by? t: ?Q = m with ?t, where? t = ?Q/(c t).

Potential energy- energy acquired by a body as a result of a change in position in space by it or its component parts. The energy of chemical bonds is a type of potential energy.

First law of thermodynamics: energy can pass from one type to another, but cannot disappear or arise.

Internal energy (U) – the sum of the kinetic and potential energies of the particles that make up the body. The heat absorbed in the reaction is equal to the difference in the internal energy of the reaction products and reagents (Q = ?U = U 2 – U 1), provided that the system has not done any work on the environment. If the reaction occurs at constant pressure, then the released gases do work against external pressure forces, and the heat absorbed during the reaction is equal to the sum of the changes in internal energy ?U and work A = p?V. This heat absorbed at constant pressure is called the enthalpy change: ? Н = ?U + p?V, defining enthalpy How H = U + pV. Reactions of liquid and solid substances occur without significant changes in volume (?V = 0), so what about these reactions? N close to ?U (?Н = ?U). For reactions with a change in volume we have ?Н > ?U, if expansion is in progress, and ?N< ?U , if there is compression.

The change in enthalpy is usually referred to the standard state of a substance: that is, for a pure substance in a certain state (solid, liquid or gaseous), at a pressure of 1 atm = 101,325 Pa, a temperature of 298 K and a concentration of substances of 1 mol/l.

Standard enthalpy of formation?– heat released or absorbed during the formation of 1 mole of a substance from the simple substances that constitute it, under standard conditions. For example, ?N arr.(NaCl) = -411 kJ/mol. This means that in the reaction Na(s) + ?Cl 2 (g) = NaCl(s) when 1 mole of NaCl is formed, 411 kJ of energy is released.

Standard enthalpy of reaction?H– change in enthalpy during a chemical reaction, determined by the formula: ?N = ?N arr.(products) – ?N arr.(reagents).

So for the reaction NH 3 (g) + HCl (g) = NH 4 Cl (tv), knowing? H o 6 p (NH 3) = -46 kJ/mol, ? H o 6 p (HCl) = -92 kJ /mol and?H o 6 p (NH 4 Cl) = -315 kJ/mol we have:

H = ?H o 6 p (NH 4 Cl) – ?H o 6 p (NH 3) – ?H o 6 p (HCl) = -315 – (-46) – (-92) = -177 kJ.

If? N< 0, then the reaction is exothermic. If? N> 0, then the reaction is endothermic.

Law Hess: The standard enthalpy of a reaction depends on the standard enthalpies of the reactants and products and does not depend on the path of the reaction.

Spontaneous processes can be not only exothermic, i.e. processes with a decrease in energy (?N< 0), but can also be endothermic processes, i.e. processes with increasing energy (?N> 0). In all these processes, the “disorder” of the system increases.

EntropyS – a physical quantity characterizing the degree of disorder of the system. S – standard entropy, ?S – change in standard entropy. If?S > 0, disorder increases if AS< 0, то беспорядок системы уменьшается. Для процессов в которых растет число частиц, ?S >0. For processes in which the number of particles decreases, ?S< 0. Например, энтропия меняется в ходе реакций:

CaO(solid) + H 2 O(l) = Ca(OH) 2 (solid), ?S< 0;

CaCO 3 (tv) = CaO (tv) + CO 2 (g), ?S > 0.

Processes occur spontaneously with the release of energy, i.e. for which? N< 0, and with increasing entropy, i.e. for which?S > 0. Taking both factors into account leads to the expression for Gibbs energy: G = H – TS or? G = ?H – T?S. Reactions in which the Gibbs energy decreases, i.e. ?G< 0, могут идти самопроизвольно. Реакции, в ходе которых энергия Гиббса увеличивается, т. е. ?G >0, do not go spontaneously. The condition?G = 0 means that equilibrium has been established between the products and reactants.

At low temperatures, when the value T is close to zero, only exothermic reactions occur, since T?S– little and?G = ? N< 0. At high temperatures the values T?S great, and, neglecting the size? N, we have?G = – T?S, i.e., processes with increasing entropy will occur spontaneously, for which?S > 0, a?G< 0. При этом чем больше по абсолютной величине значение?G, тем более полно проходит данный процесс.

The value of AG for a particular reaction can be determined by the formula:

G = ?С arr (products) – ?G o b p (reagents).

In this case, the values ​​of ?G o br, as well as? N arr. and?S o br for a large number of substances are given in special tables.

4.2. Chemical kinetics

Chemical reaction rate(v) is determined by the change in the molar concentration of reactants per unit time:

Where v– reaction rate, s – molar concentration of the reagent, t- time.

The rate of a chemical reaction depends on the nature of the reactants and the reaction conditions (temperature, concentration, presence of a catalyst, etc.)

Effect of concentration. IN In the case of simple reactions, the reaction rate is proportional to the product of the concentrations of the reactants, taken in powers equal to their stoichiometric coefficients.

For reaction

where 1 and 2 are the directions of the forward and reverse reactions, respectively:

v 1 = k 1 ? [A] m ? [B]n and

v 2 = k 2 ? [C]p ? [D]q

Where v- speed reaction, k– rate constant, [A] – molar concentration of substance A.

Molecularity of the reaction– the number of molecules participating in the elementary act of the reaction. For simple reactions, for example: mA + nB> рС + qD, molecularity is equal to the sum of the coefficients (m + n). Reactions can be single-molecule, double-molecule, and rarely triple-molecule. Reactions of higher molecular weight do not occur.

Reaction order is equal to the sum of the exponents of the degrees of concentration in the experimental expression of the rate of a chemical reaction. So, for a complex reaction

mA + nB > рС + qD the experimental expression for the reaction rate is

v 1 = k 1 ? [A] ? ? [IN] ? and the reaction order is (? + ?). Wherein? And? are found experimentally and may not coincide with m And n accordingly, since the equation of a complex reaction is the result of several simple reactions.

Effect of temperature. The rate of a reaction depends on the number of effective collisions between molecules. An increase in temperature increases the number of active molecules, giving them the necessary energy for the reaction to occur. activation energy E act and increases the rate of a chemical reaction.

Van't Hoff's rule. When the temperature increases by 10°, the reaction rate increases by 2–4 times. Mathematically this is written as:

v 2 = v 1 ? ?(t 2 – t 1)/10

where v 1 and v 2 are the reaction rates at the initial (t 1) and final (t 2) temperatures, ? – temperature coefficient of reaction rate, which shows how many times the reaction rate increases with an increase in temperature by 10°.

More precisely, the dependence of the reaction rate on temperature is expressed Arrhenius equation:

k = A? e - E/(RT)

Where k– rate constant, A– constant independent of temperature, e = 2.71828, E– activation energy, R= 8.314 J/(K? mol) – gas constant; T– temperature (K). It can be seen that the rate constant increases with increasing temperature and decreasing activation energy.

4.3. Chemical equilibrium

A system is in equilibrium if its state does not change over time. Equality of the rates of forward and reverse reactions is a condition for maintaining the equilibrium of the system.

An example of a reversible reaction is the reaction

N 2 + 3H 2 - 2NH 3 .

Law of mass action: the ratio of the product of concentrations of reaction products to the product of concentrations of starting substances (all concentrations are indicated in powers equal to their stoichiometric coefficients) is a constant called equilibrium constant.

The equilibrium constant is a measure of the progress of a forward reaction.

K = O – direct reaction does not occur;

K =? – the direct reaction goes to completion;

K > 1 – balance shifted to the right;

TO< 1 – balance is shifted to the left.

Reaction equilibrium constant TO is related to the magnitude of the change in the standard Gibbs energy?G for the same reaction:

G= – RT ln K, or?G = -2.3RT lg K, or K= 10 -0.435?G/RT

If K > 1, then lg K> 0 and?G< 0, т. е. если равновесие сдвинуто вправо, то реакция – переход от исходного состояния к равновесному – идет самопроизвольно.

If TO< 1, then lg K < 0 и?G >0, i.e. if the equilibrium is shifted to the left, then the reaction does not spontaneously go to the right.

Law of equilibrium shift: If an external influence is exerted on a system in equilibrium, a process arises in the system that counteracts the external influence.

5. Redox reactions

Redox reactions– reactions that occur with a change in the oxidation states of elements.

Oxidation– process of electron donation.

Recovery– the process of adding electrons.

Oxidizer– an atom, molecule, or ion that accepts electrons.

Reducing agent– an atom, molecule, or ion that donates electrons.

Oxidizing agents, accepting electrons, go into a reduced form:

F 2 [approx. ] + 2e > 2F? [restored].

Reductants, giving up electrons, go into the oxidized form:

Na 0 [recovery ] – 1e > Na + [approx.].

The equilibrium between the oxidized and reduced forms is characterized by Nernst equations for redox potential:

Where E 0– standard value of redox potential; n– number of transferred electrons; [restored ] and [approx. ] are the molar concentrations of the compound in reduced and oxidized forms, respectively.

Values ​​of standard electrode potentials E 0 are given in tables and characterize the oxidative and reduction properties of compounds: the more positive the value E 0, the stronger the oxidizing properties, and the more negative the value E 0, the stronger the restorative properties.

For example, for F 2 + 2e - 2F? E 0 = 2.87 volts, and for Na + + 1e - Na 0 E 0 =-2.71 volts (the process is always recorded for reduction reactions).

A redox reaction is a combination of two half-reactions, oxidation and reduction, and is characterized by an electromotive force (emf) ? E 0:?E 0= ?E 0 ok – ?E 0 restore, Where E 0 ok And? E 0 restore– standard potentials of the oxidizing agent and reducing agent for this reaction.

E.m.f. reactions? E 0 is related to the change in the Gibbs free energy?G and the equilibrium constant of the reaction TO:

?G = – nF?E 0 or? E = (RT/nF) ln K.

E.m.f. reactions at non-standard concentrations? E equal to: ? E =?E 0 – (RT/nF) ? Ig K or? E =?E 0 –(0,059/n)lg K.

In the case of equilibrium?G = 0 and?E = 0, where does it come from? E =(0.059/n)lg K And K = 10 n?E/0.059 .

For the reaction to proceed spontaneously, the following relations must be satisfied: ?G< 0 или K >> 1, to which the condition corresponds? E 0> 0. Therefore, to determine the possibility of a given redox reaction, it is necessary to calculate the value? E 0. If? E 0 > 0, the reaction is in progress. If? E 0< 0, no response.

Chemical current sources

Galvanic cells– devices that convert the energy of a chemical reaction into electrical energy.

Daniel's galvanic cell consists of zinc and copper electrodes immersed in solutions of ZnSO 4 and CuSO 4, respectively. Electrolyte solutions communicate through a porous partition. In this case, oxidation occurs on the zinc electrode: Zn > Zn 2+ + 2e, and reduction occurs on the copper electrode: Cu 2+ + 2e > Cu. In general, the reaction goes: Zn + CuSO 4 = ZnSO 4 + Cu.

Anode– electrode on which oxidation occurs. Cathode– the electrode on which the reduction takes place. In galvanic cells, the anode is negatively charged and the cathode is positively charged. On element diagrams, metal and mortar are separated by a vertical line, and two mortars are separated by a double vertical line.

So, for the reaction Zn + CuSO 4 = ZnSO 4 + Cu, the circuit diagram of the galvanic cell is written: (-)Zn | ZnSO 4 || CuSO 4 | Cu(+).

The electromotive force (emf) of the reaction is? E 0 = E 0 ok – E 0 restore= E 0(Cu 2+ /Cu) – E 0(Zn 2+ /Zn) = 0.34 – (-0.76) = 1.10 V. Due to losses, the voltage created by the element will be slightly less than? E 0. If the concentrations of solutions differ from the standard ones, equal to 1 mol/l, then E 0 ok And E 0 restore are calculated using the Nernst equation, and then the emf is calculated. corresponding galvanic cell.

Dry element consists of a zinc body, NH 4 Cl paste with starch or flour, a mixture of MnO 2 with graphite and a graphite electrode. During its operation, the following reaction occurs: Zn + 2NH 4 Cl + 2MnO 2 = Cl + 2MnOOH.

Element diagram: (-)Zn | NH4Cl | MnO 2 , C(+). E.m.f. element - 1.5 V.

Batteries. A lead battery consists of two lead plates immersed in a 30% sulfuric acid solution and coated with a layer of insoluble PbSO 4 . When charging a battery, the following processes occur on the electrodes:

PbSO 4 (tv) + 2e > Pb (tv) + SO 4 2-

PbSO 4 (tv) + 2H 2 O > PbO 2 (tv) + 4H + + SO 4 2- + 2e

When the battery is discharged, the following processes occur on the electrodes:

Pb(tv) + SO 4 2- > PbSO 4 (tv) + 2e

PbO 2 (tv) + 4H + + SO 4 2- + 2e > PbSO 4 (tv) + 2H 2 O

The total reaction can be written as:

To operate, the battery requires regular charging and monitoring of the concentration of sulfuric acid, which may decrease slightly during battery operation.

6. Solutions

6.1. Concentration of solutions

Mass fraction of substance in solution w equal to the ratio of the mass of the solute to the mass of the solution: w = m water / m solution or w = m in-va /(V ? ?), because m solution = V p-pa ? ?r-ra.

Molar concentration With equal to the ratio of the number of moles of solute to the volume of solution: c = n(mol)/ V(l) or c = m/(M? V( l )).

Molar concentration of equivalents (normal or equivalent concentration) with e is equal to the ratio of the number of equivalents of a dissolved substance to the volume of solution: with e = n(mol eq.)/ V(l) or with e = m/(M e? V(l)).

6.2. Electrolytic dissociation

Electrolytic dissociation– decomposition of the electrolyte into cations and anions under the influence of polar solvent molecules.

Degree of dissociation?– ratio of the concentration of dissociated molecules (with diss) to the total concentration of dissolved molecules (with vol): ? = with diss / with ob.

Electrolytes can be divided into strong(? ~ 1) and weak.

Strong electrolytes(for them? ~ 1) – salts and bases soluble in water, as well as some acids: HNO 3, HCl, H 2 SO 4, HI, HBr, HClO 4 and others.

Weak electrolytes(for them?<< 1) – Н 2 O, NH 4 OH, малорастворимые основания и соли и многие кислоты: HF, H 2 SO 3 , H 2 CO 3 , H 2 S, CH 3 COOH и другие.

Ionic reaction equations. IN In ionic equations of reactions, strong electrolytes are written in the form of ions, and weak electrolytes, poorly soluble substances and gases are written in the form of molecules. For example:

CaCO 3 v + 2HCl = CaCl 2 + H 2 O + CO 2 ^

CaCO 3 v + 2H + + 2Cl? = Ca 2+ + 2Cl? + H 2 O + CO 2 ^

CaCO 3 v + 2H + = Ca 2+ + H 2 O + CO 2 ^

Reactions between ions go towards the formation of a substance that produces fewer ions, i.e. towards a weaker electrolyte or a less soluble substance.

6.3. Dissociation of weak electrolytes

Let us apply the law of mass action to the equilibrium between ions and molecules in a solution of a weak electrolyte, for example acetic acid:

CH 3 COOH - CH 3 COO? +H+

The equilibrium constants for dissociation reactions are called dissociation constants. Dissociation constants characterize the dissociation of weak electrolytes: the lower the constant, the less the weak electrolyte dissociates, the weaker it is.

Polybasic acids dissociate stepwise:

H 3 PO 4 - H + + H 2 PO 4 ?

The equilibrium constant of the total dissociation reaction is equal to the product of the constants of the individual stages of dissociation:

N 3 PO 4 - ZN + + PO 4 3-

Ostwald's dilution law: the degree of dissociation of a weak electrolyte (a) increases with decreasing its concentration, i.e., with dilution:

Effect of a common ion on the dissociation of a weak electrolyte: the addition of a common ion reduces the dissociation of the weak electrolyte. So, when adding CH 3 COOH to a solution of a weak electrolyte

CH 3 COOH - CH 3 COO? +H+ ?<< 1

a strong electrolyte containing an ion common to CH 3 COOH, i.e. an acetate ion, for example CH 3 COONa

CH 3 COOna - CH 3 COO? + Na + ? = 1

the concentration of acetate ion increases, and the CH 3 COOH dissociation equilibrium shifts to the left, i.e., acid dissociation decreases.

6.4. Dissociation of strong electrolytes

Ion activity A – concentration of an ion, manifested in its properties.

Activity factorf– ion activity ratio A to concentration with: f= a/c or A = fc.

If f = 1, then the ions are free and do not interact with each other. This occurs in very dilute solutions, in solutions of weak electrolytes, etc.

If f< 1, то ионы взаимодействуют между собой. Чем меньше f, тем больше взаимодействие между ионами.

The activity coefficient depends on the ionic strength of solution I: the higher the ionic strength, the lower the activity coefficient.

Ionic strength of solution I depends on charges z and concentrations from ions:

I = 0.52?s z2.

The activity coefficient depends on the charge of the ion: the greater the charge of the ion, the lower the activity coefficient. Mathematically, the dependence of the activity coefficient f on ionic strength I and ion charge z written using the Debye-Hückel formula:

Ion activity coefficients can be determined using the following table:

6.5 Ionic product of water. pH value

Water, a weak electrolyte, dissociates, forming H+ and OH? ions. These ions are hydrated, that is, connected to several water molecules, but for simplicity they are written in non-hydrated form

H 2 O - H + + OH?.

Based on the law of mass action, for this equilibrium:

The concentration of water molecules [H 2 O], i.e. the number of moles in 1 liter of water, can be considered constant and equal to [H 2 O] = 1000 g/l: 18 g/mol = 55.6 mol/l. From here:

TO[H 2 O] = TO(H 2 O ) = [H + ] = 10 -14 (22°C).

Ionic product of water– the product of concentrations [H + ] and – is a constant value at a constant temperature and equal to 10 -14 at 22°C.

The ionic product of water increases with increasing temperature.

pH value– negative logarithm of the concentration of hydrogen ions: pH = – log. Similarly: pOH = – log.

Taking the logarithm of the ionic product of water gives: pH + pHOH = 14.

The pH value characterizes the reaction of the medium.

If pH = 7, then [H + ] = is a neutral medium.

If pH< 7, то [Н + ] >– acidic environment.

If pH > 7, then [H + ]< – щелочная среда.

6.6. Buffer solutions

Buffer solutions are solutions that have a certain concentration of hydrogen ions. The pH of these solutions does not change when diluted and changes little when small amounts of acids and alkalis are added.

I. A solution of the weak acid HA, concentration – from the acid, and its salt with the strong base BA, concentration – from the salt. For example, an acetate buffer is a solution of acetic acid and sodium acetate: CH 3 COOH + CHgCOONa.

pH = pK acidic + log(salt/s sour).

II. A solution of the weak base BOH, concentration - from basic, and its salt with a strong acid BA, concentration - from salt. For example, an ammonia buffer is a solution of ammonium hydroxide and ammonium chloride NH 4 OH + NH 4 Cl.

pH = 14 – рК basic – log(with salt/with basic).

6.7. Hydrolysis of salts

Hydrolysis of salts– interaction of salt ions with water to form a weak electrolyte.

Examples of hydrolysis reaction equations.

I. A salt is formed by a strong base and a weak acid:

Na 2 CO 3 + H 2 O - NaHCO 3 + NaOH

2Na + + CO 3 2- + H 2 O - 2Na + + HCO 3 ? +OH?

CO 3 2- + H 2 O - HCO 3 ? + OH?, pH > 7, alkaline environment.

In the second stage, hydrolysis practically does not occur.

II. A salt is formed by a weak base and a strong acid:

AlCl 3 + H 2 O - (AlOH)Cl 2 + HCl

Al 3+ + 3Cl? + H 2 O - AlOH 2+ + 2Cl? + H + + Cl?

Al 3+ + H 2 O - AlOH 2+ + H +, pH< 7.

In the second stage, hydrolysis occurs less, and in the third stage there is practically no hydrolysis.

III. A salt is formed by a strong base and a strong acid:

K + + NO 3 ? + H 2 O ? no hydrolysis, pH? 7.

IV. A salt is formed by a weak base and a weak acid:

CH 3 COONH 4 + H 2 O - CH 3 COOH + NH 4 OH

CH 3 COO? + NH 4 + + H 2 O - CH 3 COOH + NH 4 OH, pH = 7.

In some cases, when the salt is formed by very weak bases and acids, complete hydrolysis occurs. In the solubility table for such salts the symbol is “decomposed by water”:

Al 2 S 3 + 6H 2 O = 2Al(OH) 3 v + 3H 2 S^

The possibility of complete hydrolysis should be taken into account in exchange reactions:

Al 2 (SO 4) 3 + 3Na 2 CO 3 + 3H 2 O = 2Al(OH) 3 v + 3Na 2 SO 4 + 3CO 2 ^

Degree of hydrolysish – the ratio of the concentration of hydrolyzed molecules to the total concentration of dissolved molecules.

For salts formed by a strong base and a weak acid:

= chрOH = – log, рН = 14 – рOH.

From the expression it follows that the degree of hydrolysis h(i.e. hydrolysis) increases:

a) with increasing temperature, as K(H 2 O) increases;

b) with a decrease in the dissociation of the acid forming the salt: the weaker the acid, the greater the hydrolysis;

c) with dilution: the smaller the c, the greater the hydrolysis.

For salts formed by a weak base and a strong acid

[H + ] = ch pH = – log.

For salts formed by a weak base and a weak acid

6.8. Protolytic theory of acids and bases

Protolysis– proton transfer process.

Protoliths– acids and bases that donate and accept protons.

Acid– a molecule or ion capable of donating a proton. Each acid has a corresponding conjugate base. The strength of acids is characterized by the acid constant K k.

H 2 CO 3 + H 2 O - H 3 O + + HCO 3 ?

K k = 4 ? 10 -7

3+ + H 2 O - 2+ + H 3 O +

K k = 9 ? 10 -6

Base– a molecule or ion that can accept a proton. Each base has a corresponding conjugate acid. The strength of bases is characterized by the base constant K 0.

NH3? H 2 O (H 2 O) - NH 4 + + OH?

K 0 = 1,8 ?10 -5

Ampholytes– protoliths capable of releasing and acquiring a proton.

HCO3? + H 2 O - H 3 O + + CO 3 2-

HCO3? – acid.

HCO3? + H 2 O - H 2 CO 3 + OH?

HCO3? – foundation.

For water: H 2 O+ H 2 O - H 3 O + + OH?

K(H 2 O) = [H 3 O + ] = 10 -14 and pH = – log.

Constants K k And K 0 for conjugate acids and bases are linked.

HA + H 2 O - H 3 O + + A?,

A? + H 2 O - HA + OH?,

7. Solubility constant. Solubility

In a system consisting of a solution and a precipitate, two processes take place - dissolution of the precipitate and precipitation. The equality of the rates of these two processes is a condition of equilibrium.

Saturated solution– a solution that is in equilibrium with the precipitate.

The law of mass action applied to the equilibrium between precipitate and solution gives:

Since = const,

TO = K s (AgCl) = .

In general we have:

A m B n(TV) - m A +n+n B -m

K s ( A m B n)= [A +n ] m[IN -m ] n .

Solubility constantK s(or solubility product PR) - the product of ion concentrations in a saturated solution of a slightly soluble electrolyte - is a constant value and depends only on temperature.

Solubility of a sparingly soluble substance s can be expressed in moles per liter. Depending on the size s substances can be divided into poorly soluble – s< 10 -4 моль/л, среднерастворимые – 10 -4 моль/л? s? 10 -2 mol/l and highly soluble s>10 -2 mol/l.

The solubility of compounds is related to their solubility product.

Condition for precipitation and dissolution of sediment

In the case of AgCl: AgCl - Ag + + Cl?

K s= :

a) condition of equilibrium between precipitate and solution: = Ks.

b) deposition condition: > Ks; during precipitation, ion concentrations decrease until equilibrium is established;

c) the condition for the dissolution of the precipitate or the existence of a saturated solution:< Ks; As the precipitate dissolves, the ion concentration increases until equilibrium is established.

8. Coordination compounds

Coordination (complex) compounds are compounds with a donor-acceptor bond.

For K 3:

ions of the outer sphere – 3K +,

inner sphere ion – 3-,

complexing agent – ​​Fe 3+,

ligands – 6CN?, their dentation – 1,

coordination number – 6.

Examples of complexing agents: Ag +, Cu 2+, Hg 2+, Zn 2+, Ni 2+, Fe 3+, Pt 4+, etc.

Examples of ligands: polar molecules H 2 O, NH 3, CO and anions CN?, Cl?, OH? and etc.

Coordination numbers: usually 4 or 6, less often 2, 3, etc.

Nomenclature. The anion is named first (in the nominative case), then the cation (in the genitive case). Names of some ligands: NH 3 - ammin, H 2 O - aquo, CN? – cyano, Cl? – chloro, OH? – hydroxo. Names of coordination numbers: 2 – di, 3 – three, 4 – tetra, 5 – penta, 6 – hexa. The oxidation state of the complexing agent is indicated:

Cl—diamminesilver(I) chloride;

SO 4 – tetrammine copper(II) sulfate;

K 3 – potassium hexacyanoferrate(III).

Chemical connection.

Valence bond theory assumes hybridization of the orbitals of the central atom. The location of the resulting hybrid orbitals determines the geometry of the complexes.

Diamagnetic complex ion Fe(CN) 6 4-.

Cyanide ion – donor

The iron ion Fe 2+ – acceptor – has the formula 3d 6 4s 0 4p 0. Taking into account the diamagnetic nature of the complex (all electrons are paired) and the coordination number (6 free orbitals are needed), we have d 2 sp 3-hybridization:

The complex is diamagnetic, low-spin, intraorbital, stable (no external electrons are used), octahedral ( d 2 sp 3-hybridization).

Paramagnetic complex ion FeF 6 3-.

Fluoride ion is a donor.

The iron ion Fe 3+ – acceptor – has the formula 3d 5 4s 0 4p 0 . Taking into account the paramagneticity of the complex (electrons are coupled) and the coordination number (6 free orbitals are needed), we have sp 3 d 2-hybridization:

The complex is paramagnetic, high-spin, outer-orbital, unstable (outer 4d orbitals are used), octahedral ( sp 3 d 2-hybridization).

Dissociation of coordination compounds.

Coordination compounds in solution completely dissociate into ions of the inner and outer spheres.

NO 3 > Ag(NH 3) 2 + + NO 3 ?, ? = 1.

The ions of the inner sphere, i.e., complex ions, dissociate into metal ions and ligands, like weak electrolytes, in stages.

Where K 1 , TO 2 , TO 1 _ 2 are called instability constants and characterize the dissociation of complexes: the lower the instability constant, the less the complex dissociates, the more stable it is.

several basic concepts and formulas.

All substances have different mass, density and volume. A piece of metal from one element can weigh many times more than an exactly the same size piece of another metal.

Mole(number of moles)

designation: mole, international: mol- a unit of measurement for the amount of a substance. Corresponds to the amount of substance that contains N.A. particles (molecules, atoms, ions) Therefore, a universal quantity was introduced - number of moles. A frequently encountered phrase in tasks is “received... mole of substance"

N.A.= 6.02 1023

N.A.- Avogadro's number. Also “a number by agreement.” How many atoms are there in the tip of a pencil? About a thousand. It is not convenient to operate with such quantities. Therefore, chemists and physicists around the world agreed - let’s designate 6.02 × 1023 particles (atoms, molecules, ions) as 1 mole substances.

1 mole = 6.02 1023 particles

This was the first of the basic formulas for solving problems.

Molar mass of a substance

Molar mass substance is the mass of one mole of substance.

Denoted as Mr. It is found according to the periodic table - it is simply the sum of the atomic masses of a substance.

For example, we are given sulfuric acid - H2SO4. Let's calculate the molar mass of a substance: atomic mass H = 1, S-32, O-16.
Mr(H2SO4)=1 2+32+16 4=98 g\mol.

The second necessary formula for solving problems is

substance mass formula:

That is, to find the mass of a substance, you need to know the number of moles (n), and we find the molar mass from the Periodic Table.

Law of conservation of mass - The mass of substances that enter into a chemical reaction is always equal to the mass of the resulting substances.

If we know the mass(es) of the substances that reacted, we can find the mass(es) of the products of that reaction. And vice versa.

The third formula for solving chemistry problems is

volume of substance:

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Where did the number 22.4 come from? From Avogadro's law:

equal volumes of different gases taken at the same temperature and pressure contain the same number of molecules.

According to Avogadro's law, 1 mole of an ideal gas under normal conditions (n.s.) has the same volume Vm= 22.413 996(39) l

That is, if in the problem we are given normal conditions, then, knowing the number of moles (n), we can find the volume of the substance.

So, basic formulas for solving problems in chemistry

Avogadro's numberN.A.

6.02 1023 particles

Quantity of substance n (mol)

n=V\22.4 (l\mol)

Mass of substance m (g)

Volume of substance V(l)

V=n 22.4 (l\mol)

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These are formulas. Often, to solve problems, you first need to write the reaction equation and (required!) arrange the coefficients - their ratio determines the ratio of moles in the process.

Magnitude and its dimension	Ratio
Atomic mass of element X (relative)
Element serial 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.s.) 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.S.
Gas volume, m 3	V = V m × n
Product yield
Density of substance X, g/l, g/ml, kg/m3
Density of gaseous substance X by hydrogen
Density of gaseous substance X in air	M(air) = 29 g/mol
United 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 a substance (X) in a mixture
Amount of heat, J, kJ	Q = n(X) Q(X)
Thermal effect of reaction	Q =–H
Heat of formation of substance X, J/mol, kJ/mol
Chemical reaction rate (mol/lsec)
Law of Mass Action (for a simple reaction)	a A+ V B= With C + d D u = k With a(A) With V(B)
Van't Hoff's rule
Solubility of the substance (X) (g/100 g solvent)
Mass fraction of substance X in mixture A + X, in fractions of a unit, in %
Weight of solution, g, kg	m(rr) = m(X)+ m(H2O) m(rr) = V(rr)  (rr)
Mass fraction of dissolved substance in solution, in fractions of a unit, in %
Solution density
Volume of solution, cm 3, l, m 3
Molar concentration, mol/l
Degree of electrolyte dissociation (X), in fractions of a unit or %
Ionic product of water	K(H2O) =
pH value	pH = –lg

Main:

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

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

Egorov A.S. Chemistry. A new textbook for preparing for higher education. Rostov n/d: Phoenix, 2004.– 640 p.

Egorov A.S. Chemistry: a modern course for preparing for the Unified State 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/tutor manual for applicants to universities. Rostov-n/D, Phoenix, 2005– 536 p.

Khomchenko G.P., Khomchenko I.G.. Problems in chemistry for applicants to universities. M.: Higher school. 2007.–302p.

Additional:

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

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

Egorov A.S.. All types of calculation problems in chemistry for preparation for the Unified State Exam. – Rostov n/D: Phoenix, 2003. – 320 p.

Egorov A.S., Aminova G.Kh.. Typical tasks and exercises for preparing for the chemistry exam. – Rostov n/d: Phoenix, 2005. – 448 p.

Unified State Exam 2007. Chemistry. Educational and training materials for preparing students / FIPI - M.: Intellect-Center, 2007. – 272 p.

Unified State Exam 2011. Chemistry. Educational and training kit ed. A.A. Kaverina. – M.: National Education, 2011.

The only real options for tasks to prepare for the Unified State Exam. Unified State Examination 2007. Chemistry/V.Yu. Mishina, E.N. Strelnikova. M.: Federal Testing Center, 2007.–151 p.

Kaverina A.A. The optimal bank of tasks for preparing students. Unified State Exam 2012. Chemistry. Textbook./ 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., Solovyova M.V.. Test tasks in addition to tests for students of 10-month correspondence preparatory courses (methodological instructions). Krasnodar, 2004. – P. 18 – 70.

Litvinova T.N.. Chemistry. Unified State Exam 2011. Training tests. Rostov n/d: Phoenix, 2011.– 349 p.

Litvinova T.N.. Chemistry. Tests for the Unified State 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., Solovyova M.V.., Azhipa L.T., Vyskubova N.K. Chemistry in tasks for applicants to universities. – M.: Onyx Publishing House LLC: Mir and Education Publishing House LLC, 2009. – 832 p.

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

Chemistry. Unified State Exam 2008. Entrance tests, teaching aid / ed. V.N. Doronkina. – Rostov n/a: Legion, 2008.– 271 p.

List of websites on chemistry:

1. Alhimik. http:// www. alhimik. ru

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

http:// www. informika. ru/ text/ database/ chemistry/ START. html

3. School chemistry - reference book. http:// www. schoolchemistry. by. ru

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

Internet resources

Alhimik. http:// www. alhimik. ru

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

http:// www. informika. ru/ text/ database/ chemistry/ START. html

School chemistry - reference book. http:// www. schoolchemistry. by. ru

http://www.classchem.narod.ru

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

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

http://schoolchemistry.by.ru/- school chemistry. This site has the opportunity to take On-line testing on various topics, as well as demo versions of the Unified State Exam

Chemistry and life—XXI century: popular science magazine. http:// www. hij. ru

Check information. It is necessary to check the accuracy of the facts and reliability of the information presented in this article. On the talk page there is a discussion on the topic: Doubts regarding terminology. Chemical formula ... Wikipedia

A chemical formula reflects information about the composition and structure of substances using chemical symbols, numbers and dividing symbols of brackets. Currently, the following types of chemical formulas are distinguished: The simplest formula. Can be obtained by experienced... ... Wikipedia

A chemical formula reflects information about the composition and structure of substances using chemical symbols, numbers and dividing symbols of brackets. Currently, the following types of chemical formulas are distinguished: The simplest formula. Can be obtained by experienced... ... Wikipedia

A chemical formula reflects information about the composition and structure of substances using chemical symbols, numbers and dividing symbols of brackets. Currently, the following types of chemical formulas are distinguished: The simplest formula. Can be obtained by experienced... ... Wikipedia

A chemical formula reflects information about the composition and structure of substances using chemical symbols, numbers and dividing symbols of brackets. Currently, the following types of chemical formulas are distinguished: The simplest formula. Can be obtained by experienced... ... Wikipedia

Main article: Inorganic compounds List of inorganic compounds by element informational list of inorganic compounds presented in alphabetical order (by formula) for each substance, hydrogen acids of the elements (if ... ... Wikipedia

This article or section needs revision. Please improve the article in accordance with the rules for writing articles... Wikipedia

A chemical equation (equation of a chemical reaction) is a conventional representation of a chemical reaction using chemical formulas, numerical coefficients and mathematical symbols. The equation of a chemical reaction gives qualitative and quantitative... ... Wikipedia

Chemical software are computer programs used in the field of chemistry. Contents 1 Chemical editors 2 Platforms 3 Literature ... Wikipedia

Books

• Japanese-English-Russian dictionary for installation of industrial equipment. About 8,000 terms, Popova I.S.. The dictionary is intended for a wide range of users and primarily for translators and technical specialists involved in the supply and implementation of industrial equipment from Japan or...
• A brief dictionary of biochemical terms, Kunizhev S.M.. The dictionary is intended for students of chemical and biological specialties at universities studying a course in general biochemistry, ecology and fundamentals of biotechnology, and can also be used in ...

Modern symbols for chemical elements were introduced into science in 1813 by J. Berzelius. According to his proposal, elements are designated by the initial letters of their Latin names. For example, oxygen (Oxygenium) is designated by the letter O, sulfur (Sulfur) by the letter S, hydrogen (Hydrogenium) by the letter H. In cases where the names of the elements begin with the same letter, one more letter is added to the first letter. Thus, carbon (Carboneum) has the symbol C, calcium (Calcium) - Ca, copper (Cuprum) - Cu.

Chemical symbols are not only abbreviated names of elements: they also express certain quantities (or masses), i.e. Each symbol represents either one atom of an element, or one mole of its atoms, or a mass of an element equal to (or proportional to) the molar mass of that element. For example, C means either one carbon atom, or one mole of carbon atoms, or 12 mass units (usually 12 g) of carbon.

Chemical formulas

Formulas of substances also indicate not only the composition of the substance, but also its quantity and mass. Each formula represents either one molecule of a substance, or one mole of a substance, or a mass of a substance equal to (or proportional to) its molar mass. For example, H2O represents either one molecule of water, or one mole of water, or 18 mass units (usually (18 g) of water.

Simple substances are also designated by formulas showing how many atoms a molecule of a simple substance consists of: for example, the formula for hydrogen H 2. If the atomic composition of a molecule of a simple substance is not precisely known or the substance consists of molecules containing a different number of atoms, and also if it has an atomic or metallic structure rather than a molecular one, the simple substance is designated by the symbol of the element. For example, the simple substance phosphorus is denoted by the formula P, since, depending on conditions, phosphorus can consist of molecules with a different number of atoms or have a polymer structure.

Chemistry formulas for solving problems

The formula of the substance is determined based on the results of the analysis. For example, according to analysis, glucose contains 40% (wt.) carbon, 6.72% (wt.) hydrogen and 53.28% (wt.) oxygen. Therefore, the masses of carbon, hydrogen and oxygen are in the ratio 40:6.72:53.28. Let us denote the desired formula for glucose C x H y O z, where x, y and z are the numbers of carbon, hydrogen and oxygen atoms in the molecule. The masses of the atoms of these elements are respectively equal to 12.01; 1.01 and 16.00 amu Therefore, the glucose molecule contains 12.01x amu. carbon, 1.01u amu hydrogen and 16.00zа.u.m. oxygen. The ratio of these masses is 12.01x: 1.01y: 16.00z. But we have already found this relationship based on glucose analysis data. Hence:

12.01x: 1.01y: 16.00z = 40:6.72:53.28.

According to the properties of proportion:

x: y: z = 40/12.01:6.72/1.01:53.28/16.00

or x:y:z = 3.33:6.65:3.33 = 1:2:1.

Therefore, in a glucose molecule there are two hydrogen atoms and one oxygen atom per carbon atom. This condition is satisfied by the formulas CH 2 O, C 2 H 4 O 2, C 3 H 6 O 3, etc. The first of these formulas - CH 2 O- is called the simplest or empirical formula; it has a molecular weight of 30.02. In order to find out the true or molecular formula, it is necessary to know the molecular mass of a given substance. When heated, glucose is destroyed without turning into gas. But its molecular weight can be determined by other methods: it is equal to 180. From a comparison of this molecular weight with the molecular weight corresponding to the simplest formula, it is clear that the formula C 6 H 12 O 6 corresponds to glucose.

Thus, a chemical formula is an image of the composition of a substance using symbols of chemical elements, numerical indices and some other signs. The following types of formulas are distinguished:

— simplest , which is obtained experimentally by determining the ratio of chemical elements in a molecule and using the values ​​of their relative atomic masses (see example above);

— molecular , which can be obtained by knowing the simplest formula of a substance and its molecular weight (see example above);

— rational , displaying groups of atoms characteristic of classes of chemical elements (R-OH - alcohols, R - COOH - carboxylic acids, R - NH 2 - primary amines, etc.);

— structural (graphic) , showing the relative arrangement of atoms in a molecule (can be two-dimensional (in a plane) or three-dimensional (in space));

— electronic, displaying the distribution of electrons across orbitals (written only for chemical elements, not for molecules).

Let's take a closer look at the example of the ethyl alcohol molecule:

1. the simplest formula of ethanol is C 2 H 6 O;
2. the molecular formula of ethanol is C 2 H 6 O;
3. the rational formula of ethanol is C 2 H 5 OH;

Examples of problem solving

EXAMPLE 1

Exercise	With complete combustion of an oxygen-containing organic substance weighing 13.8 g, 26.4 g of carbon dioxide and 16.2 g of water were obtained. Find the molecular formula of a substance if the relative density of its vapors with respect to hydrogen is 23.
Solution	Let’s draw up a diagram of the combustion reaction of an organic compound, designating the number of carbon, hydrogen and oxygen atoms as “x”, “y” and “z”, respectively: C x H y O z + O z →CO 2 + H 2 O. Let us determine the masses of the elements that make up this substance. Values ​​of relative atomic masses taken from the Periodic Table of D.I. Mendeleev, round to whole numbers: Ar(C) = 12 amu, Ar(H) = 1 amu, Ar(O) = 16 amu. m(C) = n(C)×M(C) = n(CO 2)×M(C) = ×M(C); m(H) = n(H)×M(H) = 2×n(H 2 O)×M(H) = ×M(H); Let's calculate the molar masses of carbon dioxide and water. As is known, the molar mass of a molecule is equal to the sum of the relative atomic masses of the atoms that make up the molecule (M = Mr): M(CO 2) = Ar(C) + 2×Ar(O) = 12+ 2×16 = 12 + 32 = 44 g/mol; M(H 2 O) = 2×Ar(H) + Ar(O) = 2×1+ 16 = 2 + 16 = 18 g/mol. m(C) = ×12 = 7.2 g; m(H) = 2 × 16.2 / 18 × 1 = 1.8 g. m(O) = m(C x H y O z) - m(C) - m(H) = 13.8 - 7.2 - 1.8 = 4.8 g. Let's determine the chemical formula of the compound: x:y:z = m(C)/Ar(C) : m(H)/Ar(H) : m(O)/Ar(O); x:y:z = 7.2/12:1.8/1:4.8/16; x:y:z = 0.6: 1.8: 0.3 = 2: 6: 1. This means the simplest formula of the compound is C 2 H 6 O and the molar mass is 46 g/mol. The molar mass of an organic substance can be determined using its hydrogen density: M substance = M(H 2) × D(H 2) ; M substance = 2 × 23 = 46 g/mol. M substance / M(C 2 H 6 O) = 46 / 46 = 1. This means the formula of the organic compound will be C 2 H 6 O.
Answer	C2H6O

EXAMPLE 2

Exercise	The mass fraction of phosphorus in one of its oxides is 56.4%. The oxide vapor density in air is 7.59. Determine the molecular formula of the oxide.
Solution	The mass fraction of element X in a molecule of the composition NX is calculated using the following formula: ω (X) = n × Ar (X) / M (HX) × 100%. Let's calculate the mass fraction of oxygen in the compound: ω(O) = 100% - ω(P) = 100% - 56.4% = 43.6%. Let us denote the number of moles of elements included in the compound as “x” (phosphorus), “y” (oxygen). Then, the molar ratio will look like this (the values ​​of relative atomic masses taken from D.I. Mendeleev’s Periodic Table are rounded to whole numbers): x:y = ω(P)/Ar(P) : ω(O)/Ar(O); x:y = 56.4/31: 43.6/16; x:y = 1.82:2.725 = 1:1.5 = 2:3. This means that the simplest formula for combining phosphorus with oxygen will be P 2 O 3 and a molar mass of 94 g/mol. The molar mass of an organic substance can be determined using its air density: M substance = M air × D air; M substance = 29 × 7.59 = 220 g/mol. To find the true formula of an organic compound, we find the ratio of the resulting molar masses: M substance / M(P 2 O 3) = 220 / 94 = 2. This means that the indices of phosphorus and oxygen atoms should be 2 times higher, i.e. the formula of the substance will be P 4 O 6.
Answer	P4O6