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All formulas in chemistry for the school course. Dictionary of chemical formulas

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 exactly the same size piece from another metal.


mole
(number of moles)

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

NA= 6.02 1023

NA is Avogadro's number. Also "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 values. Therefore, chemists and physicists around the world agreed - let's denote 6.02 1023 particles (atoms, molecules, ions) as 1 mol substances.

1 mol = 6.02 1023 particles

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

Molar mass of a substance

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

Referred to as Mr. It is located according to the periodic table - this 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

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 system.

The law of conservation of mass is The mass of substances that enter into a chemical reaction is always equal to the mass of the formed substances.

If we know the mass (masses) of substances that have entered into a reaction, we can find the mass (masses) of the products of this reaction. And vice versa.

The third formula for solving problems in chemistry is

volume of matter:

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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.o.) has the same volume Vm\u003d 22.413 996 (39) l

That is, if we are given normal conditions in the problem, 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 numberNA

6.02 1023 particles

Amount of substance n (mol)

n=V\22.4 (l\mol)

Mass of matter m (g)

Volume of matter V(l)

V=n 22.4 (l\mol)

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

Key words: Chemistry 8th grade. All formulas and definitions, symbols of 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 With
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 of a 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 weight 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 of the chemical elements that form the compound, taking into account the number of atoms of this 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 a molecular and non-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.
In physical phenomena, molecules are preserved, in chemical phenomena, as a rule, they are destroyed. Atoms in chemical phenomena rearrange, forming molecules of new substances.

THE LAW OF CONSTANT COMPOSITION OF A SUBSTANCE
Each chemically pure substance of 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.

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 of a gaseous substance X by hydrogen

Density of a 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 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= With C + d D

u = kWith a(A) With 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.: Higher 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. 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., 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 reference book for a complete course of 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 take On-line testing on various topics, as well as demo versions of the Unified State Exam

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

Collection of basic formulas for a school course in chemistry

Collection of basic formulas for a school course in chemistry

G. P. Loginova

Elena Savinkina

E. V. Savinkina G. P. Loginova

Collection of basic formulas in chemistry

Student pocket guide

general chemistry

The most important chemical concepts and laws

Chemical element A certain type of atom with the same nuclear charge.

Relative atomic mass(A r) shows how many times the mass of an atom of a given chemical element is greater than the mass of a carbon-12 atom (12 C).

Chemical substance- a collection of any chemical particles.

chemical particles
formula unit- conditional particle, the composition of which corresponds to the given chemical formula, for example:

Ar - substance argon (consists of Ar atoms),

H 2 O - water substance (consists of H 2 O molecules),

KNO 3 - substance potassium nitrate (consists of K + cations and NO 3 ¯ anions).

Relations between physical quantities
Atomic mass (relative) of an element B, Ar(B):

Where *t(atom B) is the mass of an atom of element B;

*t and is the atomic mass unit;

*t and = 1/12 t(atom 12 C) \u003d 1.6610 24 g.

Amount of substance B, n(B), mol:

Where N(B) is the number of particles B;

N A is the Avogadro constant (NA = 6.0210 23 mol -1).

Molar mass of a substance V, M(V), g/mol:

Where t(B)- weight B.

Molar volume of gas AT, V M , l/mol:

Where V M = 22.4 l/mol (a consequence of Avogadro's law), under normal conditions (n.o. - atmospheric pressure p = 101 325 Pa (1 atm); thermodynamic temperature T = 273.15 K or Celsius temperature t = 0°C).

B for hydrogen, D(gas B to H 2):

* Density of a gaseous substance AT by air, D(gas B by air): Mass fraction of the element E in matter B, w(E):

Where x is the number of atoms E in the formula of substance B

The structure of the atom and the Periodic Law D.I. Mendeleev

Mass number (A) - the total number of protons and neutrons in the atomic nucleus:

A = N(p 0) + N(p +).
The charge of the nucleus of an atom (Z) equals the number of protons in the nucleus and the number of electrons in the atom:
Z = N(p+) = N(e¯).
isotopes- atoms of the same element, differing in the number of neutrons in the nucleus, for example: potassium-39: 39 K (19 p + , 20n 0 , 19); potassium-40: 40 K (19 p+, 21n 0 , 19e¯).
*Energy levels and sublevels
*Atomic Orbital(AO) characterizes the region of space in which the probability of an electron having a certain energy to stay is the greatest.
*Shapes of s- and p-orbitals
Periodic Law and Periodic System D.I. Mendeleev
The properties of elements and their compounds are periodically repeated with increasing serial number, which is equal to the charge of the nucleus of the element's atom.

Period number corresponds the number of energy levels filled with electrons, and means last energy level(EU).

Group number A shows and etc.

Group number B shows number of valence electrons ns and (n – 1)d.

s-element section- the energy sublevel (EPL) is filled with electrons ns-epu- IA- and IIA-groups, H and He.

p-elements section- filled with electrons np-epu– IIIA-VIIIA-groups.

d-element section- filled with electrons (P- 1) d-EPU - IB-VIIIB2-groups.

f-element section- filled with electrons (P-2) f-EPU - lanthanides and actinides.

Changes in the composition and properties of hydrogen compounds of elements of the 3rd period of the Periodic system
Non-volatile, decomposed by water: NaH, MgH 2 , AlH 3 .

Volatile: SiH 4 , PH 3 , H 2 S, HCl.

Changes in the composition and properties of higher oxides and hydroxides of elements of the 3rd period of the Periodic system
Basic: Na 2 O - NaOH, MgO - Mg (OH) 2.

Amphoteric: Al 2 O 3 - Al (OH) 3.

Acid: SiO 2 - H 4 SiO 4, P 2 O 5 - H 3 PO 4, SO 3 - H 2 SO 4, Cl 2 O 7 - HClO 4.

chemical bond

Electronegativity(χ) is a value that characterizes the ability of an atom in a molecule to acquire a negative charge.
Mechanisms for the formation of a covalent bond
exchange mechanism- the overlap of two orbitals of neighboring atoms, each of which had one electron.

Donor-acceptor mechanism- overlapping of the free orbital of one atom with the orbital of another atom, which has a pair of electrons.

Orbital overlap during bond formation
*Type of hybridization - geometric shape of the particle - angle between bonds
Hybridization of orbitals of the central atom– alignment of their energy and form.

sp– linear – 180°

sp 2– triangular – 120°

sp 3– tetrahedral – 109.5°

sp 3 d– trigonal-bipyramidal – 90°; 120°

sp 3 d 2– octahedral – 90°

Mixtures and solutions

Solution- a homogeneous system consisting of two or more substances, the content of which can be changed within certain limits.

Solution: solvent (eg water) + solute.

True Solutions contain particles smaller than 1 nanometer.

Colloidal solutions contain particles 1-100 nanometers in size.

Mechanical mixtures(suspensions) contain particles larger than 100 nanometers.

Suspension=> solid + liquid

Emulsion=> liquid + liquid

Foam, fog=> gas + liquid

Heterogeneous mixtures are separated settling and filtering.

Homogeneous mixtures are separated evaporation, distillation, chromatography.

saturated solution is or can be in equilibrium with the solute (if the solute is a solid, then its excess is in the sediment).

Solubility is the content of a solute in a saturated solution at a given temperature.

unsaturated solution less,

Supersaturated solution contains a solute more, than its solubility at a given temperature.

Relationships between physicochemical quantities in solution
Mass fraction of solute AT, w(B); fraction of a unit or %:

Where t(B)- mass B,

t(p) is the mass of the solution.

The mass of the solution m(p), r:

m(p) = m(B) + m(H 2 O) = V(p) ρ(p),
where F(p) is the volume of the solution;

ρ(p) is the density of the solution.

Solution volume, V(p), l:

molar concentration, s(B), mol/l:

Where n(B) is the amount of substance B;

M(B) is the molar mass of substance B.

Changing the composition of the solution
Diluting the solution with water:

> t "(B)= t(B);

> the mass of the solution increases by the mass of the added water: m "(p) \u003d m (p) + m (H 2 O).

Evaporation of water from solution:

> the mass of the solute does not change: t "(B) \u003d t (B).

> the mass of the solution is reduced by the mass of evaporated water: m "(p) \u003d m (p) - m (H 2 O).

Merging two solutions: the masses of the solutions, as well as the masses of the solute, add up:

t "(B) \u003d t (B) + t" (B);

t"(p) = t(p) + t"(p).

Drop of crystals: the mass of the solute and the mass of the solution are reduced by the mass of the precipitated crystals:

m "(B) \u003d m (B) - m (draft); m" (p) \u003d m (p) - m (draft).

The mass of water does not change.

Thermal effect of a chemical reaction

*Enthalpy of formation of matter ΔH° (B), kJ / mol, is the enthalpy of the reaction of formation of 1 mol of a substance from simple substances in their standard states, that is, at a constant pressure (1 atm for each gas in the system or at a total pressure of 1 atm in the absence of gaseous participants in the reaction) and constant temperature (usually 298 K , or 25°C).
*Heat effect of a chemical reaction (Hess' law)
Q = ΣQ(products) - ΣQ(reagents).
ΔН° = ΣΔН°(products) – Σ ΔH°(reagents).
For reaction aA + bB +… = dD + eE +…
ΔH° = (dΔH°(D) + eΔH°(E) +…) – (aΔH°(A) + bΔH°(B) +…),
where a, b, d, e are the stoichiometric quantities of substances corresponding to the coefficients in the reaction equation.

The rate of a chemical reaction

If during the time τ in the volume V amount of reactant or product changed by Δ n, speed reaction:

For a monomolecular reaction А → …:

v=k c(A).
For a bimolecular reaction A + B → ...:
v=k c(A) c(B).
For the trimolecular reaction A + B + C → ...:
v=k c(A) c(B) c(C).
Change in the rate of a chemical reaction
Speed ​​reaction increase:

1) chemically active reagents;

2) promotion reagent concentrations;

3) increase

4) promotion temperature;

5) catalysts. Speed ​​reaction reduce:

1) chemically inactive reagents;

2) downgrade reagent concentrations;

3) decrease surfaces of solid and liquid reagents;

4) downgrade temperature;

5) inhibitors.

*Temperature coefficient of speed(γ) is equal to a number that shows how many times the reaction rate increases when the temperature rises by ten degrees:

Chemical equilibrium

*Law of mass action for chemical equilibrium: in a state of equilibrium, the ratio of the product of molar concentrations of products in powers equal to

Their stoichiometric coefficients, to the product of molar concentrations of reactants in powers equal to their stoichiometric coefficients, at a constant temperature is a constant value (concentration equilibrium constant).

In a state of chemical equilibrium for a reversible reaction:

aA + bB + … ↔ dD + fF + …
K c = [D] d [F] f …/ [A] a [B] b …
*Shift of chemical equilibrium towards the formation of products
1) Increasing the concentration of reagents;

2) decrease in the concentration of products;

3) increase in temperature (for an endothermic reaction);

4) decrease in temperature (for an exothermic reaction);

5) increase in pressure (for a reaction proceeding with a decrease in volume);

6) decrease in pressure (for a reaction proceeding with an increase in volume).

Exchange reactions in solution

Electrolytic dissociation- the process of formation of ions (cations and anions) when certain substances are dissolved in water.

acids formed hydrogen cations and acid anions, for example:

HNO 3 \u003d H + + NO 3 ¯
With electrolytic dissociation grounds formed metal cations and hydroxide ions, for example:
NaOH = Na + + OH¯
With electrolytic dissociation salts(medium, double, mixed) are formed metal cations and acid anions, for example:
NaNO 3 \u003d Na + + NO 3 ¯
KAl (SO 4) 2 \u003d K + + Al 3+ + 2SO 4 2-
With electrolytic dissociation acid salts formed metal cations and acid hydroanions, for example:
NaHCO 3 \u003d Na + + HCO 3 ‾
Some strong acids
HBr, HCl, HClO 4 , H 2 Cr 2 O 7 , HI, HMnO 4 , H 2 SO 4 , H 2 SeO 4 , HNO 3 , H 2 CrO 4
Some strong foundations
RbOH, CsOH, KOH, NaOH, LiOH, Ba(OH) 2 , Sr(OH) 2 , Ca(OH) 2

Degree of dissociation α is the ratio of the number of dissociated particles to the number of initial particles.

At constant volume:

Classification of substances according to the degree of dissociation
Berthollet's rule
Exchange reactions in solution proceed irreversibly if a precipitate, gas, or weak electrolyte is formed as a result.
Examples of molecular and ionic reaction equations
1. Molecular equation: CuCl 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaCl

The "complete" ionic equation: Cu 2+ + 2Cl¯ + 2Na + + 2OH¯ = Cu(OH) 2 ↓ + 2Na + + 2Cl¯

"Short" ionic equation: Сu 2+ + 2OH¯ \u003d Cu (OH) 2 ↓

2. Molecular equation: FeS (T) + 2HCl = FeCl 2 + H 2 S

"Full" ionic equation: FeS + 2H + + 2Cl¯ = Fe 2+ + 2Cl¯ + H 2 S

"Short" ionic equation: FeS (T) + 2H + = Fe 2+ + H 2 S

3. Molecular equation: 3HNO 3 + K 3 PO 4 = H 3 RO 4 + 3KNO 3

"Full" ionic equation: 3H + + 3NO 3 ¯ + ZK + + PO 4 3- \u003d H 3 RO 4 + 3K + + 3NO 3 ¯

"Short" ionic equation: 3H + + PO 4 3- \u003d H 3 PO 4

*Hydrogen index
(pH) pH = – lg = 14 + lg
*PH range for dilute aqueous solutions
pH 7 (neutral medium)
Examples of exchange reactions
Neutralization reaction- an exchange reaction that occurs when an acid and a base interact.

1. Alkali + strong acid: Ba (OH) 2 + 2HCl \u003d BaCl 2 + 2H 2 O

Ba 2+ + 2OH¯ + 2H + + 2Cl¯ = Ba 2+ + 2Cl¯ + 2H 2 O

H + + OH¯ \u003d H 2 O

2. Slightly soluble base + strong acid: Сu (OH) 2 (t) + 2НCl = СuСl 2 + 2Н 2 O

Cu (OH) 2 + 2H + + 2Cl¯ \u003d Cu 2+ + 2Cl¯ + 2H 2 O

Cu (OH) 2 + 2H + \u003d Cu 2+ + 2H 2 O

*Hydrolysis- an exchange reaction between a substance and water without changing the oxidation states of atoms.

1. Irreversible hydrolysis of binary compounds:

Mg 3 N 2 + 6H 2 O \u003d 3Mg (OH) 2 + 2NH 3

2. Reversible hydrolysis of salts:

A) salt is formed strong base cation and strong acid anion:

NaCl = Na + + Сl¯

Na + + H 2 O ≠ ;

Cl¯ + H 2 O ≠

Hydrolysis is absent; the medium is neutral, pH = 7.

B) Salt is formed strong base cation and weak acid anion:

Na 2 S \u003d 2Na + + S 2-

Na + + H 2 O ≠

S 2- + H 2 O ↔ HS¯ + OH¯

Anion hydrolysis; alkaline environment, pH>7.

B) Salt is formed a cation of a weak or sparingly soluble base and an anion of a strong acid:

End of introductory segment.

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Modern symbols of chemical elements were introduced into science in 1813 by J. Berzelius. At his suggestion, the elements are denoted by the initial letters of their Latin names. For example, oxygen (Oxygenium) is denoted 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 of the following is added to the first letter. So, carbon (Carboneum) has the symbol C, calcium (Calcium) - Ca, copper (Cuprum) - Cu.

Chemical symbols are not only abbreviated names of elements: they also express their certain quantities (or masses), i.e. each symbol denotes either one atom of an element, or one mole of its atoms, or the 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.

Formulas of chemicals

The 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 the mass of a substance equal to (or proportional to) its molar mass. For example, H 2 O denotes either one molecule of water, or one mole of water, or 18 mass units (usually (18 g) of water.

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

Formulas in chemistry for solving problems

The formula of the substance is established based on the results of the analysis. For example, according to the analysis, glucose contains 40% (wt.) carbon, 6.72% (wt.) hydrogen and 53.28% (wt.) oxygen. Therefore, the masses of carbon, hydrogen and oxygen are related to each other as 40:6.72:53.28. Let's designate the required glucose formula as C x H y O z , where x, y and z are the numbers of carbon, hydrogen and oxygen atoms in the molecule. The atomic masses of these elements are respectively equal to 12.01; 1.01 and 16.00 amu Therefore, the glucose molecule contains 12.01x a.m.u. carbon, 1.01u a.m.u. hydrogen and 16.00za.u.m. oxygen. The ratio of these masses is 12.01x: 1.01y: 16.00z. But we have already found this ratio, based on the data of glucose analysis. Consequently:

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

According to proportion properties:

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 corresponds to a molecular weight of 30.02. In order to find out the true or molecular formula, it is necessary to know the molecular weight of a given substance. When heated, glucose is destroyed without turning into a 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 the symbols of chemical elements, numerical indices, and some other signs. There are the following types of formulas:

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

molecular , which can be obtained by knowing the simplest formula of a substance and its molecular weight (see the 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 mutual arrangement of atoms in a molecule (it can be two-dimensional (in a plane) or three-dimensional (in space));

electronic, which displays the distribution of electrons in orbits (written only for chemical elements, not for molecules).

Let's take a closer look at the example of an ethanol 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 the complete combustion of oxygen-containing organic matter 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 its relative hydrogen vapor density is 23.
Solution Let's draw up a scheme for the combustion reaction of an organic compound, denoting 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. The values ​​of relative atomic masses taken from the Periodic Table of D.I. Mendeleev, rounded up to integers: Ar(C) = 12 a.m.u., Ar(H) = 1 a.m.u., Ar(O) = 16 a.m.u.

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);

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) \u003d Ar (C) + 2 × Ar (O) \u003d 12+ 2 × 16 \u003d 12 + 32 \u003d 44 g / mol;

M(H 2 O) \u003d 2 × Ar (H) + Ar (O) \u003d 2 × 1 + 16 \u003d 2 + 16 \u003d 18 g / mol.

m(C)=×12=7.2 g;

m(H) \u003d 2 × 16.2 / 18 × 1 \u003d 1.8 g.

m(O) \u003d m (C x H y O z) - m (C) - m (H) \u003d 13.8 - 7.2 - 1.8 \u003d 4.8 g.

Let's define 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 value of the molar mass of an organic substance can be determined using its hydrogen density:

M substance = M(H 2) × D(H 2) ;

M substance \u003d 2 × 23 \u003d 46 g / mol.

M substance / M(C 2 H 6 O) = 46 / 46 = 1.

So the formula of an organic compound will look like 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. Set the molecular formula of oxide.
Solution The mass fraction of the element X in the molecule of the HX composition is calculated by the following formula:

ω (X) = n × Ar (X) / M (HX) × 100%.

Calculate the mass fraction of oxygen in the compound:

ω (O) \u003d 100% - ω (P) \u003d 100% - 56.4% \u003d 43.6%.

Let us denote the number of moles of elements that make up the compound as "x" (phosphorus), "y" (oxygen). Then, the molar ratio will look like this (the values ​​​​of relative atomic masses taken from the Periodic Table of D.I. Mendeleev will be 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 the combination of phosphorus with oxygen will have the form P 2 O 3 and a molar mass of 94 g / mol.

The value of the molar mass of an organic substance can be determined using its density in air:

M substance = M air × D air;

M substance \u003d 29 × 7.59 \u003d 220 g / mol.

To find the true formula of an organic compound, we find the ratio of the obtained 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 look like P 4 O 6.

Answer P 4 O 6

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