Carrying out quantitative calculations using equations of chemical reactions. Calculations using chemical equations

When solving settlement chemical problems the ability to perform calculations using the equation of a chemical reaction is required. The lesson is devoted to studying the algorithm for calculating the mass (volume, quantity) of one of the reaction participants from the known mass (volume, quantity) of another reaction participant.

Topic: Substances and their transformations

Lesson:Calculations using the chemical reaction equation

Consider the reaction equation for the formation of water from simple substances:

2H 2 + O 2 = 2H 2 O

We can say that two molecules of water are formed from two molecules of hydrogen and one molecule of oxygen. On the other hand, the same entry says that for the formation of every two moles of water, you need to take two moles of hydrogen and one mole of oxygen.

The molar ratio of reaction participants helps produce important chemical synthesis calculations. Let's look at examples of such calculations.

TASK 1. Let us determine the mass of water formed as a result of the combustion of hydrogen in 3.2 g of oxygen.

To solve this problem, you first need to create an equation for a chemical reaction and write down the given conditions of the problem over it.

If we knew the amount of oxygen that reacted, we could determine the amount of water. And then, they would calculate the mass of water, knowing its amount of substance and . To find the amount of oxygen, you need to divide the mass of oxygen by its molar mass.

Molar mass numerically equal to relative . For oxygen, this value is 32. Let’s substitute it into the formula: the amount of oxygen substance is equal to the ratio of 3.2 g to 32 g/mol. It turned out to be 0.1 mol.

To find the amount of water substance, let’s leave the proportion using the molar ratio of the reaction participants:

For every 0.1 mole of oxygen there is an unknown amount of water, and for every 1 mole of oxygen there are 2 moles of water.

Hence the amount of water substance is 0.2 mol.

To determine the mass of water, you need to multiply the found value of the amount of water by its molar mass, i.e. multiply 0.2 mol by 18 g/mol, we get 3.6 g of water.

Rice. 1. Recording a brief condition and solution to Problem 1

In addition to mass, you can calculate the volume of a gaseous reaction participant (at normal conditions) using a formula known to you, according to which the volume of gas at normal conditions. equal to the product amount of gas substance per molar volume. Let's look at an example of solving a problem.

TASK 2. Let's calculate the volume of oxygen (at normal conditions) released during the decomposition of 27 g of water.

Let us write down the reaction equation and the given conditions of the problem. To find the volume of oxygen released, you first need to find the amount of water substance through the mass, then, using the reaction equation, determine the amount of oxygen substance, after which you can calculate its volume at ground level.

The amount of water substance is equal to the ratio of the mass of water to its molar mass. We get a value of 1.5 mol.

Let's make a proportion: from 1.5 moles of water an unknown amount of oxygen is formed, from 2 moles of water 1 mole of oxygen is formed. Hence the amount of oxygen is 0.75 mol. Let's calculate the volume of oxygen at normal conditions. It is equal to the product of the amount of oxygen and the molar volume. Molar volume of any gaseous substance at no. equal to 22.4 l/mol. Substituting numeric values into the formula, we get a volume of oxygen equal to 16.8 liters.

Rice. 2. Recording a brief condition and solution to Problem 2

Knowing the algorithm for solving such problems, it is possible to calculate the mass, volume or amount of substance of one of the reaction participants from the mass, volume or amount of substance of another reaction participant.

1. Collection of problems and exercises in chemistry: 8th grade: for textbooks. P.A. Orzhekovsky and others. “Chemistry. 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p.40-48)

2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 73-75)

3. Chemistry. 8th grade. Textbook for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013. (§23)

4. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§29)

5. Chemistry: inorganic. chemistry: textbook. for 8th grade general education establishment /G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009. (p.45-47)

6. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.

Additional web resources

2. Single collection of digital educational resources ().

Homework

1) p. 73-75 No. 2, 3, 5 from Workbook in chemistry: 8th grade: to the textbook P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.

2) p. 135 No. 3,4 from the textbook P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova “Chemistry: 8th grade,” 2013

Whatever you study, you
you study for yourself.
Petronius

Lesson objectives:

introduce students to the basic methods of solving problems in chemical equations:
find the quantity, mass and volume of reaction products from the quantity, mass or volume of starting substances,
continue to develop skills in working with the text of a problem, the ability to reasonedly choose a way to solve an educational problem, the ability to write equations chemical reactions.
develop the ability to analyze, compare, highlight the main thing, compose action plan, draw conclusions.
cultivate tolerance towards others, independence in decision-making, and the ability to objectively evaluate the results of one’s work.

Forms of work: frontal, individual, pair, group.

Lesson type: combined with the use of ICT

I Organizational moment.

Hello guys. Today, we will learn how to solve problems using equations of chemical reactions. Slide 1 (see presentation).

Lesson objectives Slide 2.

II.Updating knowledge, skills and abilities.

Chemistry is a very interesting and at the same time complex science. In order to know and understand chemistry, you must not only assimilate the material, but also be able to apply the acquired knowledge. You learned what signs indicate the occurrence of chemical reactions, learned how to write equations for chemical reactions. I hope you have a good understanding of these topics and can answer my questions without difficulty.

Which phenomenon is not a sign of chemical transformations:

a) the appearance of sediment; c) change in volume;

b) gas release; d) the appearance of an odor. Slide 3

4Al + 3O 2 = 2Al 2 O 3

MgCO 3 = MgO + CO 2

2HgO= 2Hg + O2

2Na + S=Na 2 S

Zn + Br 2 = ZnBr2

Zn + 2HCl = ZnCl 2 + H 2

Fe + CuSO 4 = FeSO 4 + Cu

Please indicate in numbers:

a) equations of compound reactions

b) equations of substitution reactions

c) equations of decomposition reactions Slide 4

New topic.

In order to learn how to solve problems, it is necessary to create an algorithm of actions, i.e. determine the sequence of actions.

Algorithm for calculations using chemical equations (on each student’s desk)

5. Write down the answer.

Let's start solving problems using an algorithm

Calculating the mass of a substance from the known mass of another substance participating in the reaction

Calculate the mass of oxygen released as a result of decomposition

portions of water weighing 9 g.

Let's find the molar mass of water and oxygen:

M(H 2 O) = 18 g/mol

M(O 2) = 32 g/mol Slide 6

Let's write the equation of the chemical reaction:

2H 2 O = 2H 2 + O 2

Above the formula in the reaction equation we write what we found

the value of the amount of a substance, and under the formulas of substances -

stoichiometric ratios displayed

chemical equation

0.5mol x mol

2H 2 O = 2H 2 + O 2

2mol 1mol

Let's calculate the amount of substance whose mass we want to find.

To do this, we create a proportion

0.5mol = hopmol

2mol 1mol

where x = 0.25 mol Slide 7

Therefore, n(O 2) = 0.25 mol

Find the mass of the substance that needs to be calculated

m(O 2)= n(O 2)*M(O 2)

m(O 2) = 0.25 mol 32 g/mol = 8 g

Let's write down the answer

Answer: m(O 2) = 8 g Slide 8

Calculating the volume of a substance from the known mass of another substance participating in the reaction

Calculate the volume of oxygen (no.) released as a result of the decomposition of a portion of water weighing 9 g.

V(0 2)=?l(n.s.)

M(H 2 O) = 18 g/mol

Vm=22.4l/mol Slide 9

Let's write down the reaction equation. Let's arrange the coefficients

2H 2 O = 2H 2 + O 2

Above the formula in the reaction equation we write the found value of the amount of the substance, and under the formulas of the substances - the stoichiometric ratios displayed by the chemical equation

0.5 mol - x mol

2H 2 O = 2H 2 + O 2 Slide10

2mol - 1mol

Let's calculate the amount of substance whose mass we want to find. To do this, let's create a proportion

where x = 0.25 mol

Let's find the volume of the substance that needs to be calculated

V(0 2)=n(0 2) Vm

V(O 2) = 0.25 mol 22.4 l/mol = 5.6 l (no.)

Answer: 5.6 l Slide 11

III. Consolidation of the studied material.

Tasks for independent solution:

1. When reducing the oxides Fe 2 O 3 and SnO 2 with coal, 20 g of Fe and Sn were obtained. How many grams of each oxide were taken?

2.In which case is more water formed:

a) when reducing 10 g of copper (I) oxide (Cu 2 O) with hydrogen or

b) when reducing 10 g of copper(II) oxide (CuO) with hydrogen? Slide 12

Let's check the solution to problem 1

M(Fe 2 O 3)=160g/mol

M(Fe)=56g/mol,

m(Fe 2 O 3)=, m(Fe 2 O 3)= 0.18*160=28.6g

Answer: 28.6g

Slide 13

Let's check the solution to problem 2

M(CuO) = 80 g/mol

x mol = 0.07 mol,

n(H 2 O)=0.07 mol

m(H 2 O) = 0.07mol*18g/mol=1.26g

Slide 14

CuO + H 2 = Cu + H 2 O

n(CuO) = m/ M(CuO)

n(CuO) = 10g/ 80g/mol = 0.125 mol

0.125mol hops

CuO + H 2 = Cu + H 2 O

1mol 1mol

x mol = 0.125 mol, n(H 2 O) = 0.125 mol

m (H 2 O) = n * M (H 2 O);

m(H 2 O) = 0.125mol*18g/mol=2.25g

Answer: 2.25g Slide 15

Homework: study the textbook material p. 45-47, solve the problem

What is the mass of calcium oxide and what volume? carbon dioxide(Well.)

can be obtained by decomposing calcium carbonate weighing 250 g?

CaCO 3 = CaO + CO Slide 16.

Literature

1. Gabrielyan O.S. Chemistry course program for grades 8-11 in general education institutions. M. Bustard 2006

2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for general education institutions. Bustard. M. 2005

3. Gorbuntsova S.V. Tests on the main sections of the school course. 8th - 9th grades. VAKO, Moscow, 2006.

4. Gorkovenko M.Yu. Lesson developments in chemistry. To the textbooks of O.S. Gabrielyan, L.S. Guzey, V.V. Sorokin, R.P. Surovtseva and G.E. Rudzitis, F.G. Feldman. 8th grade. VAKO, Moscow, 2004.

5. Gabrielyan O.S. Chemistry. Grade 8: Tests and tests. – M.: Bustard, 2003.

6. Radetsky A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for teachers. – M.: Education, 2000

Application.

Calculations using chemical equations

Algorithm of actions.

In order to solve a calculation problem in chemistry, you can use the following algorithm - take five steps:

1. Write an equation for a chemical reaction.

2. Above the formulas of substances, write known and unknown quantities with the corresponding units of measurement (only for pure substances, without impurities). If, according to the conditions of the problem, substances containing impurities enter into a reaction, then you first need to determine the content of the pure substance.

3. Under the formulas of substances with known and unknowns, write down the corresponding values of these quantities found from the reaction equation.

4. Compose and solve a proportion.

5. Write down the answer.

The relationship between some physical and chemical quantities and their units

Mass (m) : g; kg; mg

Quantity of substances (n): mole; kmol; mmol

Molar mass (M): g/mol; kg/kmol; mg/mmol

Volume (V) : l; m 3 /kmol; ml

Molar volume (Vm) : l/mol; m 3 /kmol; ml/mmol

Number of particles (N): 6 1023 (Avagadro number – N A); 6 1026 ; 6 1020

Calculations using chemical equations (stoichiometric calculations) are based on the law of conservation of mass of substances. In real chemical processes, due to incomplete reactions and losses, the mass of products is usually less than theoretically calculated. Reaction output (ŋ) is the ratio of the actual mass of the product (m practical) to the theoretically possible (m theoretical), expressed in fractions of a unit or as a percentage:

ŋ= (m practical / m theoretical) 100%.

If the yield of reaction products is not specified in the problem conditions, it is taken as 100% in the calculations (quantitative yield).

Example 1. How many g of copper are formed when 8 g of copper oxide is reduced with hydrogen if the reaction yield is 82% of theoretical?

Solution: 1. Calculate the theoretical yield of copper using the reaction equation:

CuO + H2 = Cu + H2O

80 g (1 mol) CuO upon reduction can form 64 g (1 mol) Cu; 8 g CuO upon reduction can form X g Cu

2. Let’s determine how many grams of copper are formed at 82% product yield:

6.4 g – 100% yield (theoretical)

X g –– 82%

X = (8 82) / 100 = 5.25 g

Example 2. Determine the yield of the reaction for producing tungsten using the aluminothermy method if 12.72 g of metal was obtained from 33.14 g of ore concentrate containing WO 3 and non-reducing impurities (mass fraction of impurities 0.3).

Solution 1) Determine the mass (g) of WO 3 in 33.14 g of ore concentrate:

ω(WO 3)= 1.0 - 0.3 = 0.7

m(WO 3) = ω(WO 3) m ore = 0.7 33.14 = 23.2 g

2) Let us determine the theoretical yield of tungsten as a result of the reduction of 23.2 g of WO 3 with aluminum powder:

WO 3 + 2Al = Al 2 O 3 + W.

When 232 g (1 g-mol) WO 3 is reduced, 187 g (1 g-mol) W is formed, and from 23.2 g WO 3 – X g W

X = (23.2 187) / 232 = 18.7 g W

3) Let's calculate the practical yield of tungsten:

18.7 g W –– 100%

12.72 g W –– Y%

Y = (12.72 100) / 18.7 = 68%.

Example 3. How many grams of barium sulfate precipitate are formed when solutions containing 20.8 g of barium chloride and 8.0 g of sodium sulfate are combined?

Solution. Reaction equation:

BaCl 2 + Na 2 SO 4 = BaSO 4 + 2NaCl.

The amount of the reaction product is calculated using the original substance taken in deficiency.

1). It is first determined which of the two starting substances is in short supply.

Let us denote the amount of g Na 2 SO 4 –– X.

208 g (1 mol) BaCl 2 reacts with 132 g (1 mol) Na 2 SO 4; 20.8 g –– with X g

X = (20.8 132) / 208 = 13.2 g Na 2 SO 4.

We have established that the reaction with 20.8 g of BaCl 2 will require 13.2 g of Na 2 SO 4, and 18.0 g is given. Thus, sodium sulfate is taken into the reaction in excess, and further calculations should be carried out using BaCl 2 taken in short supply.

2). We determine the number of grams of precipitate BaSO 4. 208 g (1 mol) BaCl 2 forms 233 g (1 mol) BaSO 4; 20.8 g –– Y g

Y = (233 20.8) / 208 = 23.3 g.

Law of Constancy of Composition

It was first formulated by J. Proust (1808).

All individual chemicals molecular structure have a constant qualitative and quantitative composition and a certain chemical structure, regardless of the method of receipt.

From the law of constancy of composition it follows that chemical elements are combined in certain quantitative ratios.

For example, carbon and oxygen form compounds with different mass ratios of the elements carbon and oxygen. CO C: O = 3: 4 CO2 C: O = 3: 8 In no other way do carbon and oxygen combine. This means that CO and CO2 compounds have a constant composition, which is determined by the oxidation states of the carbon valency in the compounds. The valence of each element has certain values (there may be several of them, variable valence), therefore the composition of the compounds is certain.

All of the above applies to substances of molecular structure. Since molecules have a certain chemical formula (composition), the substance they form has a constant composition (obviously coinciding with the composition of each molecule). The exception is polymers (consisting of molecules of different lengths).

The situation is more complicated with substances of non-molecular structure. It's about about substances in condensed (solid and liquid) states. Because NaCl - an ionic compound in the solid state (alternating Na+ and Cl-) in the gaseous state - represents individual NaCl molecules. It is impossible to isolate individual molecules in a drop of liquid or in a crystal. For example FeO

Fe 2+ O 2– Fe 2+ O 2–, etc. perfect crystal

The law of constant composition requires that the number of Fe2+ ions be exactly equal to the number of O2– ions. And these numbers are huge even for very small crystals (a cube, an edge of 0.001 mm is 5 × 1011). This is impossible for a real crystal. In a real crystal, regularity violations are inevitable. Iron(II) oxide may contain a variable amount of oxygen depending on the production conditions. The actual composition of the oxide is expressed by the formula Fe1 – xO, where 0.16 ³ x ³ 0.04. This is berthollide, a compound of variable composition, in contrast to daltonides with x = 0. With a non-stoichiometric composition of the ionic compound, electrical neutrality is ensured. Fe 3+ is present instead of the missing Fe 2+ ion

In an atomic (non-ionic) substance, some atoms may be absent, and some may replace each other. Such compounds are also classified as daltonides. The formula of the intermetallic compound of copper and zinc, which is integral part brass existing in the composition range 40 – 55 at% Zn can be written as follows: (Cu0.9 – 1.0Zn0.1 – 0)(Cu0 –.0.2Zn0 – 0.8) copper atoms can be replaced by zinc atoms and vice versa .

The law of constancy of composition, therefore, is strictly observed for substances of molecular structure (exceptions are high molecular weight) and has limited application for non-molecular substances.

Mass fraction of element ω(E)– is the fraction of one element in total mass substances. Calculated as a percentage or in shares. Designate Greek letterω (omega). ω shows what part the mass is of this element from the entire mass of the substance:

ω(E) = (n Ar(E)) / Mr

where n is the number of atoms; Ar(E) - relative atomic mass element; Mr is the relative molecular mass of the substance.

Knowing the quantitative elemental composition of a compound, it is possible to establish its simplest molecular formula. To establish the simplest molecular formula:

1) Designate the formula of the compound A x B y C z

2) Calculate the ratio X: Y: Z through mass fractions elements:

ω (A) = (x Ar(A)) / Mr(A x B y C z)

ω (B) = (y Ar(B)) / Mr(A x B y C z)

ω (C) = (z Ar(C)) / Mr(A x B y C z)

X = (ω (A) Mr) / Ar(A)

Y = (ω (B) Mr) / Ar(B)

Z = (ω (C) Mr) / Ar(C)

x: y: z = (ω (A) / Ar(A)) : (ω (B) / Ar(B)) : (ω (C) / Ar(C))

3) The resulting numbers are divided by the smallest to obtain the integers X, Y, Z.

4) Write down the formula of the compound.

Law of Multiples

(D. Dalton, 1803)

If two chemical elements give several compounds, then the weight fractions of the same element in these compounds that fall on the same weight fraction of the second element are related to each other as small integers.

N 2 O N 2 O 3 NO 2 (N 2 O 4) N 2 O 5

The number of oxygen atoms in the molecules of these compounds per two nitrogen atoms is in the ratio 1: 3: 4: 5.

Law of volumetric relations

(Gay-Lussac, 1808)

“The volumes of gases entering into chemical reactions and the volumes of gases formed as a result of the reaction are related to each other as small whole numbers.”

Consequence. Stoichiometric coefficients in the equations of chemical reactions for molecules of gaseous substances show in what volume ratios gaseous substances react or are obtained.

Examples.

a) 2CO + O 2 = 2CO 2

When two volumes of carbon (II) oxide are oxidized by one volume of oxygen, 2 volumes of carbon dioxide are formed, i.e. the volume of the initial reaction mixture is reduced by 1 volume.

b) During the synthesis of ammonia from the elements:

N2 + 3H2 = 2NH3

One volume of nitrogen reacts with three volumes of hydrogen; In this case, 2 volumes of ammonia are formed - the volume of the initial gaseous reaction mass will decrease by 2 times.

“A mole is equal to the amount of substance in a system containing the same number of structural elements as there are atoms in carbon - 12 (12 C) weighing 0.012 kg (exactly). When using a mole, the structural elements must be specified and may be atoms, molecules, ions, electrons and other particles or specified groups of particles." We are not talking about carbon in general, but its isotope 12 C, as with the introduction of the atomic mass unit. Since 12 g of carbon 12 C contains 6.02 × 10 23 atoms, we can say that a mole is the amount of substance containing 6.02 × 10 23 of its structural elements (atoms or groups of atoms, molecules, groups of ions (Na 2 SO 4), complex groups etc.). The number N A = 6.02 × 10 23 is named Avogadro's constant. The molar mass of a substance is the mass of one mole. Its usual unit is g/mol, symbol M.

Recall that relative molecular mass (M r) is the ratio of the mass of one molecule to the mass of an atomic mass unit, which is equal to 1/N A g.

Let the relative molecular mass of a substance be equal to M r. Let's calculate its molecular weight M.

Mass of one molecule: m = M r a.m.u. = M r × g

Mass of one mole (N A molecules): M = m N A = M r × = M r. We see that the numerical molar mass in grams coincides with the relative molecular weight. This is a consequence of the choice of a certain atomic mass unit (1/12 of the mass of the carbon isotope 12 C).

When solving computational chemical problems, it is necessary to be able to perform calculations using the equation of a chemical reaction. The lesson is devoted to studying the algorithm for calculating the mass (volume, quantity) of one of the reaction participants from the known mass (volume, quantity) of another reaction participant.

Topic: Substances and their transformations

Lesson:Calculations using the chemical reaction equation

Consider the reaction equation for the formation of water from simple substances:

2H 2 + O 2 = 2H 2 O

The molar ratio of reaction participants helps to make calculations important for chemical synthesis. Let's look at examples of such calculations.

TASK 1. Let us determine the mass of water formed as a result of the combustion of hydrogen in 3.2 g of oxygen.

To solve this problem, you first need to create an equation for a chemical reaction and write down the given conditions of the problem over it.

Molar mass is numerically equal to relative mass. For oxygen, this value is 32. Let’s substitute it into the formula: the amount of oxygen substance is equal to the ratio of 3.2 g to 32 g/mol. It turned out to be 0.1 mol.

To find the amount of water substance, let’s leave the proportion using the molar ratio of the reaction participants:

For every 0.1 mole of oxygen there is an unknown amount of water, and for every 1 mole of oxygen there are 2 moles of water.

Hence the amount of water substance is 0.2 mol.

To determine the mass of water, you need to multiply the found value of the amount of water by its molar mass, i.e. multiply 0.2 mol by 18 g/mol, we get 3.6 g of water.

Rice. 1. Recording a brief condition and solution to Problem 1

In addition to the mass, you can calculate the volume of the gaseous participant in the reaction (at standard conditions) using a formula known to you, according to which the volume of gas at normal conditions. equal to the product of the amount of gas substance and the molar volume. Let's look at an example of solving a problem.

TASK 2. Let's calculate the volume of oxygen (at normal conditions) released during the decomposition of 27 g of water.

The amount of water substance is equal to the ratio of the mass of water to its molar mass. We get a value of 1.5 mol.

Let's make a proportion: from 1.5 moles of water an unknown amount of oxygen is formed, from 2 moles of water 1 mole of oxygen is formed. Hence the amount of oxygen is 0.75 mol. Let's calculate the volume of oxygen at normal conditions. It is equal to the product of the amount of oxygen and the molar volume. The molar volume of any gaseous substance at ambient conditions. equal to 22.4 l/mol. Substituting the numerical values into the formula, we obtain a volume of oxygen equal to 16.8 liters.

Rice. 2. Recording a brief condition and solution to Problem 2

3. Chemistry. 8th grade. Textbook for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013. (§23)

4. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§29)

5. Chemistry: inorganic. chemistry: textbook. for 8th grade general education establishment /G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009. (p.45-47)

6. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.

Additional web resources

2. Unified collection of digital educational resources ().

Homework

1) p. 73-75 No. 2, 3, 5 from the Workbook in Chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.

2) p. 135 No. 3,4 from the textbook P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova “Chemistry: 8th grade,” 2013

Atomic-molecular science.

Basic concepts of chemistry:

Atom- system of interacting elementary particles, consisting of a nucleus and electrons. The type of an atom is determined by the composition of its nucleus. The nucleus consists of protons and neutrons = nucleons.

Element- a collection of atoms with the same nuclear charge, i.e. number of protons.

Electron(from Greek - amber) - an elementary particle carrying a negative charge.

Isotope- nuclides that contain the same number of protons, but a different number of neutrons (differ in mass numbers)

Molecule- smallest particle a substance determined by its properties.

Ions- electrically charged particles are formed when an electron is lost or gained.

Radicals-particles with unpaired elements, if you divide the pairs in half, then it is a radical.

Simple substance- consists of 1 chemical element.

Allotropy- ability chemical elements exist in the form of several bodies.

Polymorphism(manifold) exists in 2 or more structures and properties, forming different crystal lattice. Oxygen => ozone; carbon =>, graphite, diamond.

Isomorphism- ability to gather according to the composition of substances, form mixed crystals.

The atomic mass unit is taken to be 1/12 carbon 12

Relative molecular weight- attitude average weight an atom with its natural isotopic composition to 1/12 the mass of an atom of the carbon isotope 12. The mass of an atom or molecule of any substance is equal to the product of the relative mass and the atomic mass unit.

Moll- a unit of measurement of the amount of a substance that contains such a number of structural atoms, ions, radicals, in 12 g. Carbon.

Law of Conservation of Mass-The mass of all substances participating in a chemical reaction is equal to the mass of all reaction products.

Law of Constancy of Composition-Modern formulation of the law: every chemically pure substance with a molecular structure, regardless of location and method of production, has the same constant qualitative and quantitative composition.

Chemical equation (chemical reaction equation) call the conventional notation of a chemical reaction using chemical formulas, numerical coefficients and mathematical symbols.

Compilation rules

On the left side of the equation, write down the formulas of the substances that reacted, connecting them with a plus sign. On the right side of the equation, write down the formulas of the resulting substances, also connected by a plus sign. An arrow is placed between the parts of the equation. Then they find odds- numbers placed before formulas of substances so that the number of atoms of identical elements on the left and right sides of the equation is equal.

To draw up equations of chemical reactions, in addition to knowing the formulas of the reagents and reaction products, it is necessary to select the correct coefficients. This can be done using simple rules:

1. Before the formula of a simple substance, you can write a fractional coefficient, which shows the amount of substance of the reacting and resulting substances.

2. If the reaction scheme contains a salt formula, then first the number of ions forming the salt is equalized.

3. If the substances involved in the reaction contain hydrogen and oxygen, then the hydrogen atoms are equalized in the penultimate order, and the oxygen atoms in the last place.

4. If there are several salt formulas in the reaction scheme, then it is necessary to begin the equation with the ions that are part of the salt containing a larger number of them.

Calculations using chemical equations

Cheat sheet for calculating chemical equations
In order to solve a calculation problem in chemistry, you can use the following algorithm - take five steps:
1. Write an equation for a chemical reaction.
2. Above the formulas of substances, write known and unknown quantities with the corresponding units of measurement (only for pure substances, without impurities). If, according to the conditions of the problem, substances containing impurities enter into a reaction, then you first need to determine the content of the pure substance.
3. Under the formulas of substances with known and unknowns, write down the corresponding values of these quantities found from the reaction equation.
4. Compose and solve a proportion.
5. Write down the answer.

The relationship between some physical and chemical quantities and their units

Mass (m) : g; kg; mg
Quantity of substances (n): mole; kmol; mmol
Molar mass (M): g/mol; kg/kmol; mg/mmol
Volume (V) : l; m 3 /kmol; ml
Molar volume (V m) : l/mol; m 3 /kmol; ml/mmol
Number of particles (N): 6 10 23 (Avagadro number – N A); 6 10 26 ; 6 10 20

Carrying out quantitative calculations using equations of chemical reactions. Calculations using chemical equations

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