Chemistry is one of the most difficult subjects in the school curriculum, so schoolchildren often face many problems in achieving the desired result and success in studying. The fact is that everything here is confusing and to some extent incomprehensible, since if you skip just one topic, it will be difficult to solve an example or problem, or conduct an experiment. And schoolchildren sometimes lack perseverance, so this experimental science is very difficult. And besides, school discipline is difficult, so, as a rule, a student needs additional help in order to quickly absorb and learn everything.

As for the need to study an in-depth type of chemistry, as a rule, such a need arises when a student wants to go to specialized higher education institutions related to chemical activity, medicine, the food industry, and so on. Only knowledge of chemistry will help you present yourself from a positive side and create the necessary foundation for further learning.

But the question still remains: how can you study chemistry on your own easily and quickly? The question, at first glance, seems difficult and impossible, but in fact, if you carefully consider certain tips, the result will not be long in coming.

Studying chemistry at home

Individual chemistry training at home– this is an opportunity to master all the rules and key subtleties of this subject, without resorting to the services of teachers and tutors. Do you think there will be no results? You are mistaken, because on the Internet and in thematic books there are a lot of programs aimed at self-study. Here are games to facilitate the learning process, step-by-step tasks, interesting and detailed material on topics

An alternative and very good option would be to study chemistry remotely, for which it will be enough just to have a computer with Internet access. And the programs, presented in electronic form, will be able to provide both theoretical and certain practical skills in the field of chemistry.

But remember that you can count on getting an optimal result only if you have desire and aspiration in the educational process, since this is an excellent opportunity to cope with all the difficulties on the way to fully studying the subject.

E.N.Frenkel

Chemistry tutorial

A manual for those who do not know, but want to learn and understand chemistry

Part I. Elements of general chemistry
(first difficulty level)

I, Frenkel Evgenia Nikolaevna, Honored Worker of Higher Education of the Russian Federation, graduate of the Faculty of Chemistry of Moscow State University in 1972, teaching experience 34 years. In addition, I am the mother of three children and grandmother of four grandchildren, the eldest of whom is in school. I am concerned about the problem of school textbooks. The main problem of many of them is the difficult language, which requires additional “translation” into a language that students can understand to present educational material. Secondary school students often come to me with the following request: “Translate the text of the textbook so that it is clear.” Therefore, I wrote a “Self-Teacher in Chemistry”, in which many complex issues are presented in a completely accessible and at the same time scientific manner. Based on this “Self-Teacher”, which was written in 1991, I developed the program and content of the preparatory courses. Hundreds of schoolchildren studied there. Many of them started from scratch and after 40 lessons understood the subject so much that they passed exams with “4” and “5”. That’s why in our city my self-teaching manuals are selling like hotcakes. Maybe others will find my work useful too?

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Preface

Dear readers! The “Chemistry Self-Teacher” we bring to your attention is not an ordinary textbook. It does not simply state some facts or describe the properties of substances. The “self-teacher” explains and teaches even if, unfortunately, you do not know or understand chemistry, and you cannot or are embarrassed to turn to the teacher for clarification. In manuscript form, this book has been used by schoolchildren since 1991, and there has not been a single student who failed the chemistry exam both at school and in universities. Moreover, many of them did not know chemistry at all.

The “Self-Teacher” is designed for independent work by the student. The main thing is to answer the questions that appear in the text as you read. If you could not answer the question, then read the text carefully again - all the answers are nearby. It is also advisable to perform all the exercises that occur during the explanation of new material. Numerous training algorithms that are practically not found in other textbooks will help with this. With their help you will learn:

Draw up chemical formulas based on valency;

Draw up equations of chemical reactions, arrange coefficients in them, including in equations of redox processes;

Compose electronic formulas (including short electronic formulas) of atoms and determine the properties of the corresponding chemical elements;

Predict the properties of certain compounds and determine whether a given process is possible or not.

The manual has two difficulty levels. Self-instruction manual first difficulty level consists of three parts.

Part I. Elements of general chemistry ( published).

Part II. Elements of inorganic chemistry.

Part III. Elements of organic chemistry.

Books second difficulty level also three.

Theoretical foundations of general chemistry.

Theoretical foundations of inorganic chemistry.

Theoretical foundations of organic chemistry.

Chapter 1. Basic concepts of chemistry. Exercises for Chapter 1. Chapter 2. The most important classes of inorganic compounds. 2.1. Oxides. 2.2. Acids. 2.3. Grounds. Exercises for Chapter 2. Chapter 3. Basic information about the structure of the atom. Periodic law of D.I.Mendeleev. Exercises for Chapter 3. Chapter 4. The concept of chemical bonding. Chapter 5. Solutions. Chapter 6. Electrolytic dissociation. 6.1. The concept of pH (hydrogen value). 6.2. Hydrolysis of salts. Exercises for Chapter 6. Chapter 7. The concept of redox reactions. Chapter 8. Calculations using chemical formulas and equations. 8.1. Basic calculation concepts. 8.2. Problems solved using standard formulas. 8.2.1. Problems on the topic “Gases”. 8.2.2. Problems on the topic “Methods of expressing the concentration of solutions.” 8.2.3. Problems on the topic “Quantitative composition of matter.” 8.3. Problems solved using reaction equations. 8.3.1. Preparation of calculations using reaction equations. 8.3.2. Problems on the topic “Quantitative composition of solutions and mixtures.” 8.3.3. Tasks to establish the formula of a substance. 8.4. Problems for independent solution. Application.

Chapter 1. Basic concepts of chemistry

What is chemistry? Where do we encounter chemical phenomena?

Chemistry is everywhere. Life itself is a countless variety of chemical reactions thanks to which we breathe, see the blue sky, smell the amazing smell of flowers.

What does chemistry study?

Chemistry studies substances, as well as the chemical processes in which these substances participate.

What is a substance?

Matter is what the world around us and ourselves are made of.

What is a chemical process (phenomenon)?

TO chemical phenomena These include processes that result in changes in the composition or structure of the molecules that form a given substance*. The molecules have changed - the substance has changed (it has become different), its properties have changed. For example, fresh milk turned sour, green leaves turned yellow, raw meat changed its smell when fried.

All these changes are a consequence of complex and diverse chemical processes. However, the signs of simple chemical reactions, as a result of which the composition and structure of molecules change, are the same: a change in color, taste or smell, the release of gas, light or heat, the appearance of a precipitate.

What are molecules, the change of which entails such diverse manifestations?

Molecules are the smallest particles of a substance, reflecting its qualitative and quantitative composition and its chemical properties.

By studying the composition and structure of one molecule, it is possible to predict many properties of a given substance as a whole. Such research is one of the main tasks of chemistry.

How are molecules structured? What are they made of?

Molecules are made up of atoms. The atoms in a molecule are connected by chemical bonds. Each atom is designated by symbol(chemical sign). For example, H is a hydrogen atom, O is an oxygen atom.

The number of atoms in a molecule is indicated using index – numbers at the bottom right after the symbol.

For example:

Examples of molecules:

O 2 is an oxygen molecule consisting of two oxygen atoms;

H 2 O is a water molecule consisting of two hydrogen atoms and one oxygen atom.

If the atoms are not connected by a chemical bond, then their number is indicated using coefficient – numbers before the symbol:

The number of molecules is depicted similarly:

2H 2 – two hydrogen molecules;

3H 2 O – three water molecules.

Why do hydrogen and oxygen atoms have different names and different symbols? Because these are atoms of different chemical elements.

A chemical element is a type of atom with the same nuclear charge.

What is the nucleus of an atom? Why is the nuclear charge a sign that an atom belongs to a given chemical element? To answer these questions, it is necessary to clarify: do atoms change in chemical reactions, what does an atom consist of?

A neutral atom has no charge, although it consists of a positively charged nucleus and negatively charged electrons:

During chemical reactions the number of electrons of any atom can change, but the charge of the atomic nucleus does not change. Therefore, the charge of the nucleus of an atom is a kind of “passport” of a chemical element. All atoms with a nuclear charge of +1 belong to the chemical element called hydrogen. Atoms with a nuclear charge of +8 belong to the chemical element oxygen.

Each chemical element is assigned a chemical symbol (sign), a serial number in D.I. Mendeleev’s table (the serial number is equal to the charge of the atomic nucleus), a specific name, and for some chemical elements a special reading of the symbol in the chemical formula (Table 1).

Table 1

Symbols (signs) of chemical elements

No.	No. in the table of D.I. Mendeleev	Symbol	Reading in the formula	Name
1	1	H	ash	Hydrogen
2	6	C	this	Carbon
3	7	N	en	Nitrogen
4	8	O	O	Oxygen
5	9	F	fluorine	Fluorine
6	11	Na	sodium	Sodium
7	12	Mg	magnesium	Magnesium
8	13	Al	aluminum	Aluminum
9	14	Si	silicium	Silicon
10	15	P	pe	Phosphorus
11	16	S	es	Sulfur
12	17	Cl	chlorine	Chlorine
13	19	K	potassium	Potassium
14	20	Ca	calcium	Calcium
15	23	V	vanadium	Vanadium
16	24	Cr	chromium	Chromium
17	25	Mn	manganese	Manganese
18	26	Fe	ferrum	Iron
19	29	Cu	cuprum	Copper
20	30	Zn	zinc	Zinc
21	35	Br	bromine	Bromine
22	47	Ag	argentum	Silver
23	50	Sn	stannum	Tin
24	53	I	iodine	Iodine
25	56	Ba	barium	Barium
26	79	Au	aurum	Gold
27	80	Hg	hydrargyrum	Mercury
28	82	Pb	plumbum	Lead

There are substances simple And complex . If a molecule consists of atoms of one chemical element, it is simple substance. Simple substances - Ca, Cl 2, O 3, S 8, etc.

Molecules complex substances consist of atoms of different chemical elements. Complex substances - H 2 O, NO, H 3 PO 4, C 12 H 22 O 11, etc.

Task 1.1. Indicate the number of atoms in the molecules of complex substances H 2 O, NO, H 3 PO 4, C 12 H 22 O 11, name these atoms.

The question arises: why is the formula H 2 O always written for water, and not HO or HO 2? Experience proves that the composition of water obtained by any method or taken from any source always corresponds to the formula H 2 O (we are talking about pure water).

The fact is that the atoms in a water molecule and in a molecule of any other substance are connected through chemical bonds. A chemical bond connects at least two atoms. Therefore, if a molecule consists of two atoms and one of them forms three chemical bonds, then the other also forms three chemical bonds.

Number of chemical bonds formed by an atom is called valence.

If we designate each chemical bond with a dash, then for a molecule of two AB atoms we obtain AB, where three dashes indicate the three bonds formed by elements A and B between each other.

In this molecule, atoms A and B are trivalent.

It is known that the oxygen atom is divalent, the hydrogen atom is monovalent.

Question. How many hydrogen atoms can attach to one oxygen atom?

ANSWER: Two atoms. The composition of water is described by the formula H–O–H, or H 2 O.

REMEMBER! A stable molecule cannot have “free” or “extra” valences. Therefore, for a two-element molecule, the number of chemical bonds (valences) of the atoms of one element is equal to the total number of chemical bonds of the atoms of the other element.

Valence of atoms of some chemical elements constant(Table 2).

table 2

The value of constant valences of some elements

For other atoms, valence** can be determined (calculated) from the chemical formula of the substance. In this case, it is necessary to take into account the rule stated above about chemical bonds. For example, let's define the valency x manganese Mn according to the formula of the substance MnO 2:

The total number of chemical bonds formed by one and the other element (Mn and O) is the same:
x· 1 = 4; II · 2 = 4. Hence X= 4, i.e. In this chemical formula, manganese is tetravalent.

PRACTICAL CONCLUSIONS

1. If one of the atoms in the molecule is monovalent, then the valency of the second atom is equal to the number of atoms of the first element (see index!):

2. If the number of atoms in a molecule is the same, then the valence of the first atom is equal to the valency of the second atom:

3. If one of the atoms does not have an index, then its valency is equal to the product of the valency of the second atom and its index:

4. In other cases, put the valences “crosswise”, i.e. the valence of one element is equal to the index of another element:

Task 1.2. Determine the valencies of elements in compounds:

CO 2, CO, Mn 2 O 7, Cl 2 O, P 2 O 3, AlP, Na 2 S, NH 3, Mg 3 N 2.

Clue. First, indicate the valence of atoms for which it is constant. The valence of atomic groups OH, PO 4, SO 4, etc. is determined in the same way.

Task 1.3. Determine the valencies of atomic groups (underlined in the formulas):

H 3 P.O. 4 ,Ca( OH) 2 , Ca 3 ( P.O. 4) 2, H 2 SO 4,Cu SO 4 .

(Note! The same groups of atoms have the same valences in all compounds.)

Knowing the valencies of an atom or group of atoms, you can create a formula for a compound. To do this, use the following rules.

If the valencies of the atoms are the same, then the number of atoms is the same, i.e. We don’t put indexes:

If the valences are multiples (both are divided by the same number), then the number of atoms of the element with a lower valency is determined by division:

In other cases, the indices are determined “crosswise”:

Task 1.4. Make up the chemical formulas of the compounds:

Substances whose composition is reflected by chemical formulas can participate in chemical processes (reactions). The graphical notation corresponding to a given chemical reaction is called reaction equation. For example, when coal burns (interacts with oxygen), a chemical reaction occurs:

C + O 2 = CO 2.

The recording shows that one carbon atom C, combining with one molecule of oxygen O 2, forms one molecule of carbon dioxide CO 2. The number of atoms of each chemical element before and after the reaction must be the same. This rule is a consequence of the law of conservation of mass of matter. Law of conservation of mass: the mass of the starting substances is equal to the mass of the reaction products.

The law was discovered in the 18th century. M.V. Lomonosov and, independently of him, A.L. Lavoisier.

In fulfilling this law, it is necessary to arrange the coefficients in the equations of chemical reactions so that the number of atoms of each chemical element does not change as a result of the reaction. For example, the decomposition of Berthollet salt KClO 3 produces salt KCl and oxygen O 2:

KClO 3 KCl + O 2.

The number of potassium and chlorine atoms is the same, but the number of oxygen atoms is different. Let's equalize them:

Now the number of potassium and chlorine atoms before the reaction has changed. Let's equalize them:

Finally, you can put an equal sign between the right and left sides of the equation:

2KClO 3 = 2KСl + 3О 2.

The resulting record shows that the decomposition of the complex substance KClO 3 produces two new substances - the complex KCl and the simple one - oxygen O 2. The numbers in front of the formulas of substances in the equations of chemical reactions are called coefficients.

When selecting coefficients, it is not necessary to count individual atoms. If the composition of some atomic groups has not changed during the reaction, then the number of these groups can be taken into account, considering them as a single whole. Let's create an equation for the reaction of the substances CaCl 2 and Na 3 PO 4:

CaCl 2 + Na 3 PO 4 ……………….

Sequencing

1) Let's determine the valency of the starting atoms and the PO 4 group:

2) Let’s write the right side of the equation (without subscripts for now, the formulas of the substances in brackets need to be clarified):

3) Let’s compile the chemical formulas of the resulting substances based on the valences of their constituent parts:

4) Let's pay attention to the composition of the most complex compound Ca 3 (PO 4) 2 and equalize the number of calcium atoms (there are three of them) and the number of PO 4 groups (there are two):

5) The number of sodium and chlorine atoms before the reaction is now six. Let's put the corresponding coefficient on the right side of the diagram before the NaCl formula:

3CaCl 2 + 2Na 3 PO 4 = Ca 3 (PO 4) 2 + 6NaCl.

Using this sequence, it is possible to equalize the schemes of many chemical reactions (with the exception of more complex redox reactions, see Chapter 7).

Types of chemical reactions. Chemical reactions are of different types. The main four types are connection, decomposition, substitution and exchange.

1. Compound reactions– from two or more substances one substance is formed:

For example:

Ca + Cl 2 = CaCl 2.

2. Decomposition reactions– from one substance two or more substances are obtained:

For example:

Ca(HCO 3) 2 CaCO 3 + CO 2 + H 2 O.

3. Substitution reactions– simple and complex substances react, simple and complex substances are also formed, and the simple substance replaces part of the atoms of the complex substance:

A + BX AX + B.

For example:

Fe + CuSO 4 = Cu + FeSO 4.

4. Exchange reactions– here two complex substances react and two complex substances are obtained. During the reaction, complex substances exchange their constituent parts:

Exercises for Chapter 1

1. Learn the table. 1. Test yourself, write the chemical symbols: sulfur, zinc, tin, magnesium, manganese, potassium, calcium, lead, iron and fluorine.

2. Write the symbols of the chemical elements that are pronounced in formulas as: “ash”, “o”, “cuprum”, “es”, “pe”, “hydrargyrum”, “stannum”, “plumbum”, “en”, “ferrum” , "ce", "argentum". Name these elements.

3. Indicate the number of atoms of each chemical element in the formulas of the compounds:

Al 2 S 3, CaS, MnO 2, NH 3, Mg 3 P 2, SO 3.

4. Determine which substances are simple and which are complex:

Na 2 O, Na, O 2, CaCl 2, Cl 2.

Read the formulas of these substances.

5. Learn the table. 2. Make up chemical formulas of substances based on the known valency of elements and atomic groups:

6. Determine the valence of chemical elements in compounds:

N 2 O, Fe 2 O 3, PbO 2, N 2 O 5, HBr, SiH 4, H 2 S, MnO, Al 2 S 3.

7. Arrange the coefficients and indicate the types of chemical reactions:

a) Mg + O 2 MgO;

b) Al + CuCl 2 AlCl 3 + Cu;

c) NaNO 3 NaNO 2 + O 2;

d) AgNO 3 + BaCl 2 AgCl + Ba(NO 3) 2;

e) Al + HCl AlCl 3 + H 2;

e) KOH + H 3 PO 4 K 3 PO 4 + H 2 O;

g) CH 4 C 2 H 2 + H 2 .

* There are substances that are not built from molecules. But these substances will be discussed later (see Chapter 4).

** Strictly speaking, according to the rules below, it is not the valency that is determined, but the oxidation state (see Chapter 7). However, in many compounds the numerical values ​​of these concepts coincide, so the valence can also be determined using the formula of a substance.

Reprinted with continuation

Chemistry is considered one of the most complex and difficult subjects. Moreover, difficulties arise in mastering this subject for both schoolchildren and students. Why? Students expect tricks, interesting experiments and demonstrations from the lesson. But after the first classes, they are disappointed: there is not much laboratory work with reagents, basically you have to learn new terminology and do extensive homework. Chemical language is completely different from everyday language, so you need to quickly learn terms and names. In addition, you need to be able to think logically and apply mathematical knowledge.

Is it possible to learn chemistry on your own?

Nothing is impossible. Despite the complexity of science, chemistry can be learned from scratch. In some cases, when the topic is particularly complex or requires additional knowledge, you can use the services of an online tutor. The most convenient way to learn is with the help of chemistry tutors on Skype. Distance learning allows you to study a particular topic in detail or clarify complex points. You can contact a qualified teacher via Skype at any time.

In order for the learning process to be effective, several factors are needed:

• Motivation. In any business, you need a goal to strive for. It doesn’t matter why you study chemistry - for admission to a medical institute or the Faculty of Biology, just for self-development. The main thing is to set a goal and determine a way to achieve it. Motivation will be the main driving factor that will force you to continue self-learning.
• The importance of details. It is simply impossible to learn a large amount of information in a short time. To learn chemistry effectively and be able to use knowledge correctly, you need to pay attention to details: formulas, solve a large number of examples, problems. For high-quality assimilation of the material, systematization of information is required: they independently study a new topic, in addition solve problems and examples, learn formulas, etc.
• Check of knowledge . To consolidate the material covered, it is recommended to periodically do verification work. The ability to understand and analyze logically allows you to assimilate knowledge better than cramming. Teachers recommend periodically doing tests and tests for yourself. It would be useful to review the material covered. Workbooks and self-instruction books help you learn chemistry on your own.
• Practice and practice again... It is not enough to have good theoretical knowledge; you need to be able to apply it in practice when solving problems. Practical exercises help to identify weak points in knowledge and consolidate the material covered. In addition, analytical skills and logical construction of a decision chain are developed. While solving examples and problems, you draw conclusions and systematize the acquired knowledge. When the tasks become absolutely clear, you can begin to study the next topic.
• Teach yourself. Not sure about fully mastering chemistry? Try teaching this subject to someone. While explaining the material, weak points in knowledge are identified and consistency is built. It is important to take your time, paying attention to details and practicalities.

You can learn chemistry on your own from scratch if you have strong motivation and time. If the material is complex, professional tutors will help you understand the intricacies of the topic. Whether this will be face-to-face counseling or via Skype is up to you. It is not necessary to take a full course from a tutor; in some cases, you can take a lesson on a separate topic.

If you have entered the university, but by this time have not understood this difficult science, we are ready to reveal a few secrets to you and help you study organic chemistry from scratch (for dummies). All you have to do is read and listen.

Basics of organic chemistry

Organic chemistry is distinguished as a separate subtype due to the fact that the object of its study is everything that contains carbon.

Organic chemistry is a branch of chemistry that deals with the study of carbon compounds, the structure of such compounds, their properties and methods of joining.

As it turned out, carbon most often forms compounds with the following elements - H, N, O, S, P. By the way, these elements are called organogens.

Organic compounds, the number of which today reaches 20 million, are very important for the full existence of all living organisms. However, no one doubted it, otherwise the person would have simply thrown the study of this unknown into the back burner.

The goals, methods and theoretical concepts of organic chemistry are presented as follows:

• Separation of fossil, animal or plant materials into individual substances;
• Purification and synthesis of various compounds;
• Identification of the structure of substances;
• Determination of the mechanics of chemical reactions;
• Finding the relationship between the structure and properties of organic substances.

A little history of organic chemistry

You may not believe it, but back in ancient times, the inhabitants of Rome and Egypt understood something about chemistry.

As we know, they used natural dyes. And often they had to use not a ready-made natural dye, but extract it by isolating it from a whole plant (for example, alizarin and indigo contained in plants).

We can also remember the culture of drinking alcohol. The secrets of producing alcoholic beverages are known in every nation. Moreover, many ancient peoples knew recipes for preparing “hot water” from starch- and sugar-containing products.

This went on for many, many years, and only in the 16th and 17th centuries did some changes and small discoveries begin.

In the 18th century, a certain Scheele learned to isolate malic, tartaric, oxalic, lactic, gallic and citric acid.

Then it became clear to everyone that the products that had been isolated from plant or animal raw materials had many common features. At the same time, they were very different from inorganic compounds. Therefore, the servants of science urgently needed to separate them into a separate class, and this is how the term “organic chemistry” appeared.

Despite the fact that organic chemistry itself as a science appeared only in 1828 (it was then that Mr. Wöhler managed to isolate urea by evaporating ammonium cyanate), in 1807 Berzelius introduced the first term into the nomenclature in organic chemistry for dummies:

The branch of chemistry that studies substances obtained from organisms.

The next important step in the development of organic chemistry is the theory of valence, proposed in 1857 by Kekule and Cooper, and the theory of chemical structure of Mr. Butlerov from 1861. Even then, scientists began to discover that carbon was tetravalent and capable of forming chains.

In general, since then, science has regularly experienced shocks and excitement thanks to new theories, discoveries of chains and compounds, which allowed the active development of organic chemistry.

Science itself emerged due to the fact that scientific and technological progress was unable to stand still. He went on and on, demanding new solutions. And when there was no longer enough coal tar in industry, people simply had to create a new organic synthesis, which over time grew into the discovery of an incredibly important substance, which to this day is more expensive than gold - oil. By the way, it was thanks to organic chemistry that its “daughter” was born - a subscience that was called “petrochemistry”.

But this is a completely different story that you can study for yourself. Next, we invite you to watch a popular science video about organic chemistry for dummies:

Well, if you have no time and urgently need help professionals, you always know where to find them.

I.I want to learn how to findrelative molecular weight.

Wonderful! Let's start learning. suppose we need to find the relative molecular weight of sodium sulfate Na 2 SO 4 ,our actions:

1. Found sodium in the Periodic Table (No. 11)
2. We saw the number 22.9 under the name and rounded it to 23.
3. Since there are two sodium atoms, we multiply 23 by 2 and get 46.
4. Found sulfur in the periodic table (No. 16)
5. We saw the number 32 under the name, we will not multiply, because one sulfur atom.
6. Found oxygen in the Periodic Table (No. 8)
7. Under the name it says 15.9, round it up, we get 16. There are 4 oxygen atoms in the molecule, which means we need to multiply 16 by 4. We get 64.
8 Last action:
46+32+64=142 Hurray! we found the relative molecular weight of sodium sulfate.

Perhaps you should practice on your own.
Try to calculate for:

H2SO4 you should get 98

Ca(OH)2 you should get 74

K3PO4 you should get 212

If you did it, please accept our congratulations. You have taken the first step in solving problems.

We remind you that molar mass is numerically equal to molecular mass, but is measured in grams/per mole (g/mol).

II.I want to learn how to findnumber of moles of a substance.

You will need formulas:

n= m/M use if we are given a mass

n= v /V M we use if we are given a volume

n= N/ N A we use if given to us number of atoms or molecules.

Task: find the amount of sodium sulfate substance weighing 7.1 g.

Given: solution:
m(Na 2 SO 4) = 7.1 g We are given mass, so we will use the formula with mass
_____________ n=m / M, where M- molar mass (if we don’t know how to count it, see paragraph I)

Find: n M((Na 2 SO 4) =46+32+16*4=142 g/mol

n= 7.1g / 142g/mol= 0.05 mol

Answer: n=0.05 mol

Try to find the amount of the substance yourself, if given.
1. 196g H2SO4(answer 2 moles)
2.20.2g KNO 3 (answer 0.2 mol)
3. 16g NaOH (answer 0.4 mol)

We suggest you solve the following problems yourself: ( don't be afraid, you can do it!)

1. Find the amount of substance that makes up 49 g of copper (II) hydroxide.
2. How many molecules do 4.48 liters of hydrogen contain?
3. Find the mass of 5.6 liters of nitrogen.
4. What volume does oxide sulfur (IV) weighing 80 g occupy?

Show the solution to these problems to your chemistry teacher. Ask questions if something is not clear.

III. I want to learn how to solve problems using reaction equations.

Problem: find the mass of magnesium oxide that can dissolve in 12.6 g of nitric acid.

Given:
m(HNO 3)=12.6g
___________
Find:m(MgO)

Solution: 1 . The first action of any such task is find the number of moles of a given substance .
for this we use the formula (see paragraph 2). Since we are given mass, our formula is: n=m / M
n(HNO 3)= 12.6g /M(HNO 3)=12.6 / (1+14+48) =12.6 / 63 = 0.2 mol

2. Second action - write down the reaction equation and arrange the coefficients.

3. Third action - write the number of moles , calculated in the first action, over the substance for which it was calculated, and put over the required one X
X ,,.0.2 mol
MgO +2 HNO 3 = Mg(NO 3) 2 + H 2 O

4. Under these substances write down the amount of substance needed according to the equation - reaction coefficient:
x............0.2 mol
MgO+2 HNO 3 = Mg(NO 3) 2 + H 2 O
1 mole 2 mole

5. Thus we have a recording
x............0.2 mol
MgO +2 HNO 3 = Mg(NO 3) 2 + H 2 O
1 mole 2 mole

In this entry we will see the proportion: x: 1 = 0.2: 2, solve the equation(the product of the extreme terms of a proportion is equal to the product of the middle terms) x= 0.1, that is n(MgO )=0.1mol
5. We found the amount of a substance, but they ask us the mass. We use the formula: m=n*M
m( MgO )=0.1*(24+16) =4g

Answer: the mass of magnesium oxide is 4g.

Try solving the problems yourself :( Be careful and careful!)

1. Calculate the volume of sulfur (IV) oxide that is formed when 80 g of sulfur is burned.
2.Calculate the mass of aluminum oxide that is formed when aluminum burns in 4.48 liters of oxygen .

3. Find the volume of oxygen required to burn 12 g of magnesium.
4. Find the mass of zinc oxide, which is formed when zinc reacts with 11.2 liters of oxygen.
5. Find the mass of sodium oxide, which is formed when sodium reacts with 4.48 liters of oxygen.
6. Find the volume of carbon dioxide that is formed during the combustion of 60 g of coal.
7. Find the volume of oxygen that is consumed during the combustion of 3.1 g of phosphorus.
8. Find the mass of phosphorus that can burn in 4.48 liters of oxygen.
9. Find the volume of chlorine that is required to react with 5.4 g of aluminum.
10.Find the volume of chlorine with which 4.6 g of sodium can react.
11. Find the mass of aluminum chloride, which is formed when 1.12 liters of chlorine reacts with aluminum.
12.Find the mass of iron (III) chloride, which is formed when 11.2 liters of chlorine react with iron.
13.How many liters of oxygen are required to burn 6.2 g of phosphorus?
14. Find the mass of carbon monoxide (IV), which is formed when coal burns in 8.96 liters of oxygen.

IV. I want to learn how to write down the electron configuration diagram of an atom (electronic passport)

To do this, you need to remember that the s-orbital can contain no more than 2 electrons, the p-orbital - no more than six, the d-orbital - no more than 10, and the f-orbital - no more than 14.
So:
S - 2
P-6
d - 10
f - 14
The filling of orbitals with electrons occurs in the following order:
1s2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f5d 6p 7s 5f6d 7p

Please note that the number of the d-orbital is always one less than the number of the s-orbital after which it is located.

The number of electrons in an atom is equal to the number of protons in its nucleus, equal to the number of the element in Periodic system.

So, let's assume that we need to diagram the electronic configuration of the potassium atom.
Its number is 19, which means there are 19 electrons in the atom.
We start in order from 1s, filling the orbitals with the maximum number of electrons possible for them and writing this number as a power above the orbital symbol:

1s 2 2s 2 2p 63s 2 3p 6 4s 1

There is one electron in the last orbital, because we need to “attach” only 19 electrons and having written 1s 2 2s 2 2p 6 3s 2 3p 6 we are left with just one electron. It will be located in the next orbital 4s.
Let's check ourselves:
Potassium is found in fourth period hence external level fourth.
Potassium is in the first group, therefore It has one electron in its outer level.
We wrote the diagram correctly.

We need to practice.
Try writing electronic passports for the atoms of chlorine, sodium, nitrogen, magnesium, and oxygen. And then try tungsten, antimony, iodine, barium, etc. You will succeed, you just need to be careful and persistent.

Test yourself:

Cl 1s 2 2s 2 2p 6 3s 2 3p 5

Na 1s 2 2s 2 2p 6 3s 1

N 1s 2 2s 2 2p 3

Mg 1s 2 2s 2 2p 6 3s 2

O 1s 2 2s 2 2p 4
VI want to learn how to predict the properties of an element by its position in the Periodic Table.

The most energetically favorable state is with a filled outer level (8 electrons). The elements have this configuration neon, argon, krypton, xenon and radon. They are called inert (noble) gases because they do not interact. They are located in the 8th group.
Other elements
or add electrons missing up to 8
for example Cl in the outer level has 7 electrons (write down his passport) , missing 1 will add 1 electron.
for example O in the outer level 6 electrons (write down his passport) , missing 2, will add 2 electrons.
Elements that have 4-7 electrons at the outer level will add electrons. Such elements are called NON-METALS. The more easily an element gains electrons, the more active a nonmetal it is.
Look at the Periodic Table and explain why they exhibit non-metallic properties phosphorus, arsenic, selenium, bromine, nitrogen, fluorine, carbon.

or give up electrons from the external level, as a result of which the filled previous level becomes external.
for example, sodium has 1 electron in its outer level, and 8 in the previous one, (write down his passport) , so sodium donates 1 electron.
for example, aluminum has 3 electrons in its outer level, and 8 in the previous one( write down his passport) therefore aluminum donates 3 electrons.
Elements that have 1-3 electrons in their outer level will donate electrons. These are METALS. The more easily an element gives off electrons, the more active the metal it is.
Find in the Periodic Table magnesium, potassium, indium, rubidium, calcium and explain their properties.

The ability to take or donate electrons also depends on distances between core and outer level, which determines the force of attraction of electrons from the outer level to the nucleus.
for example, nitrogen and bismuth are located in group V, which means they have 5 electrons in the outer level and they both must be non-metals. But bismuth is a metal, since its outer level ( 6th, we found this out by the period number) is located far from the nucleus, the attraction of electrons of the outer level to the nucleus is small and bismuth does not add electrons, but gives up , therefore metal.
Compare the properties of carbon and tin, sulfur and polonium. Using "magic words" - distance and attraction, - explain your conclusion.

I want to learn to recognize the main classes of inorganic substances and know their properties.

There are 4 main classes of substances:
oxides, acids, bases and salts.
You need to learn the definitions:

oxides - complex substances consisting of two elements, one of which is oxygen.
oxide - ExOy for example: Na 2 O - sodium oxide, CuO - copper (II) oxide, P 2 O 5 - phosphorus oxide

acids - complex substances consisting of hydrogen atoms and an acid residue.
acid - HxA, where A is an acidic residue.
for example HCl - hydrochloric acid, H 2 SO 4 - sulfuric acid, HNO 3 - nitric acid

grounds - complex substances consisting of metal atoms and OH hydroxyl groups.

base - Me(OH)x
for example: KOH - potassium hydroxide, Ca(OH) 2 - calcium hydroxide

salt - complex substances consisting of metal atoms and an acid residue.

salt- MechAu

for example: Na 2 SO 4 - sodium sulfate, Cu(NO 3) 2 - copper (II) nitrate.

Let's check how you understand the classification.
Find the extra substance in each line:

1. NaOH HCl Mg(OH) 2 Fe(OH) 3
2. HNO 3 H 2 SO 4 H 2 O HCl
3. Cl 2 O 7 MnO NaOH K 2 O
4. Ca(OH) 2 CuCl 2 Na 2 SO 3 Mn(NO 3) 2
5. CuSO 4 NaCl FeCO 3 H 3 PO 4

check your answers:
1. HCl is an acid, and all other substances are bases
2. H 2 O is an oxide, and all other substances are acids.
3. NaOH is a base, and all other substances are oxides.
4. Ca(OH) 2 is a base, and all other substances are salts.
5. H 3 PO 4 is an acid, and all other substances are salts.

Now it’s time to look at the chemical properties.

The properties of oxides depend on which element forms the oxide.
if the element is a metal, then in most cases it forms a basic oxide,
if an element is a non-metal, then its oxide, in most cases, is acidic.

basic oxide + water = alkali (soluble base) 1
+ acid = salt and water 2
+acid oxide = salt 3

Remember this diagram! It will be very useful to us. How will this diagram help us write reaction equations?
For example:
you need to complete the equation and arrange the coefficients:
CaO + HNO 3 =

Your actions:
1. Find out what kind of substances react:
CaO - oxide, metal oxide, means basic oxide
HNO 3 - acid
2. determine the property number
basic oxide + acid - property No. 2, which means you should get salt and water
3. What is salt? ( This is a complex substance consisting of metal atoms and an acid residue)
who is metal here? ( calcium)
Where is the acid residue?( it is part of the acid, it is NO 3)
4. To correctly write the formula of a salt, you need to take into account the valency (for calcium - II, for acid residue - 1).

We write the equation:

CaO + HNO 3 = Ca(NO 3) 2 + H 2 O

5. Set the odds:

CaO+ 2 HNO 3 = Ca(NO 3) 2 + H 2 O

Now try to complete the reaction equations yourself and arrange the coefficients:
MgO+HCl=
Na 2 O +H 3 PO 4 =
CuO + H2SO4 =
BaO + H2O =
Na 2 O +H 2 O =
CaO + CO 2 =
MgO + P 2 O 5 =

If you want to check the result, show what you got to Alexandra Evgenievna. She will be happy to check, explain the errors, and if there are none, she will put a “5” in the journal.

Let's now look at the properties of acidic oxides.

acid oxide + water = acid 4
+ alkali = salt + water 5
+ basic oxide = salt 6

Let's say we need to complete the following equation and arrange the coefficients:
P 2 O 5 +H 2 O =
Our reasoning:
1. What class of substances does the first substance belong to? ( oxide, non-metal oxide, acid oxide).
2. Determine the property number
(acid oxide + water is property No. 4, an acid should be obtained.)
3. What is an acid?( complex substance consisting of hydrogen atoms and an acid residue)
4. Write the equation:
P 2 O 5 +H 2 O = H3PO4
We set the coefficients:
P 2 O 5 +3H 2 O = 2H3PO4

Now try it yourself:
CO 2 +H 2 O=
SiO 2 + KOH =
P 2 O 5 + LiOH =
CO 2 +Ca(OH) 2 =
SO3 + MgO =

If you want to check your work, show it to Alexandra Evgenievna.

Let's consider the properties of acids.

acid + metal( standing in the voltage series up to hydrogen) = salt + H 2
+ metal oxide = salt + water
+ base = salt + water
+ salt = another salt + another acid ( this reaction should form a precipitate or release a gas)

Let's try to practice choosing substances that can react with acids.

What substance can hydrochloric acid react with?

Copper is in the voltage series after hydrogen, so it does not react with hydrochloric acid solution.

Sulfur oxide is an acidic oxide since sulfur is a non-metal. Acids do not react with acid oxides.

You made the right choice. Magnesium hydroxide is a base. Acids react with bases to form salt and water.

Acids do not interact with non-metals, such as oxygen.

And now we work on our own.
In each line, find a substance that can react with a solution of sulfuric acid.
Ag CuO HNO 3 NaCl
Mg(OH) 2 KCl Hg CO 2
P P 2 O 5 K 3 PO 4 K 2 O
Li 2 SO 4 LiCl LiNO 3 Li 2 CO 3

Answers: CuO. Mg(OH)2. K 2 O. Li 2 CO 3

REMEMBER:
All carbonates react with acids, as the result is the formation of unstable carbonic acid, which immediately decomposes into carbon dioxide and water:
H 2 CO 3 = CO 2 + H 2 O

Complete the reaction equations, arrange the coefficients:
HNO3 + Ca(OH)2 =
HNO3 + MgO =
HNO 3 + K 2 CO 3 =
H3PO4+ KOH =
H3PO4 + BaO =
H3PO4 + Na 2 SiO 3 =

Let's consider the properties of bases.

Soluble and insoluble bases differ in properties.

alkali + acid = salt + water
(soluble base)+Acid oxide = salt +water
+salt= another salt + another base
(a precipitate should form)

Finish those reaction equations, arrange the coefficients :
Ca(OH)2+ CuCl2
Ca(OH)2+ Al(NO3)3
Ca(OH)2 + ZnSO4

insoluble base + acid = salt + water
decomposes when heated = oxide + water

Complete the reaction equations and set the coefficients:

Cu(OH)2+ HCl
Zn(OH)2+ HNO3
Cu(OH)2 =

Let's consider the properties of salts:

Salt + acid = another salt + another acid
+ alkali = another salt + another base
+ salt = other salt + other salt
+ more active metal = different salt + different metal
(a precipitate should form in the first three reactions)

Complete the equations of possible reactions, arrange the coefficients, indicate the substance that precipitates:
ZnSO4 +KOH=
ZnSO4 +K 3 PO 4 =
ZnSO 4 +HNO 3 =
Al(NO3)3+ HCl =
Al(NO 3) 3 + P 2 O 5 =
Al(NO 3) 3 + Ca(OH) 2 =
CuCl2+Mg=
CuCl2+Hg=

Learn chemistry, children!

Wonderful! There is nothing complicated about ionic equations. You will need to be attentive and accurate, and, of course, knowledge of the properties of the main classes of compounds, as already described above.

Remember: Oxides, water and insoluble substances do not dissociate into ions.

Let's get started. Suppose we have the equation

Mg(OH) 2 + 2HCl = MgCl 2 + 2H 2 O

Now we need to consider the possibility of dissociation of each of the recorded substances and create an ionic equation. We look at the solubility table and look for Mg(OH)2. We see that it is insoluble. This means we simply rewrite it

Mg(OH)2+ and move on. We look in the solubility table HCl. We see that this substance is soluble. Wonderful! We write those ions that are written in the table:

Mg(OH)2+H + + Cl - , but in the equation before HCl there was a coefficient of 2, which means we have 2H + and 2 Cl -

Mg(OH)2 +2H + + 2Cl -

But in the formula after chlorine there was an index of 2. This means we have 2 chlorines. So in the equation we put 2 in front of the chlorine ion.

Mg(OH)2 +2Н + + 2Cl - =Мg 2+ +2 Cl -

Now water. But we remember: water does not dissociate, we rewrite it as it is.

Mg(OH)2 +2Н + + 2Cl - =Мg 2+ +2 Cl - + 2H2O

We have written down the complete ionic equation. HOORAY! Now we need to find the same ions on the left and right sides of the equation and cancel them, as in an algebraic equation.

Mg(OH)2 +2H + + 2Cl - =Mg 2+ +2 Cl - + 2H2O

Rewrite without them

Mg(OH)2 +2H + =Mg 2+ +2H2O

We have a shortened ionic equation. As a result, our entry looks like this:

Mg(OH) 2 + 2HCl = MgCl 2 + 2H 2 O

Mg(OH)2 +2Н + + 2Cl - =Мg 2+ +2 Cl - + 2H2O

Mg(OH)2 +2H + =Mg 2+ + 2H 2 O

Congratulations. This is your first ionic equation. I hope not the last. Need to practice. First, use ready-made equations:

2HNO 3 + Ca(OH) 2 = Ca(NO 3) 2 + 2H 2 O
2HNO 3 + MgO = Mg(NO 3) 2 + H 2 O
2HNO 3 + K 2 CO 3 = 2KNO 3 + CO 2 + H 2 O Carbonic acid does not exist!

Test yourself, find mistakes

2HNO 3 + Ca(OH) 2 = Ca(NO 3) 2 + 2H 2 O

2H + +2NO 3 - + Ca 2+ +2OH - =Ca 2+ +2NO 3 - + 2H 2 O

2H + +2OH - = 2H 2 O

2HNO 3 + MgO = Mg(NO 3) 2 + H 2 O

2H + +2NO 3 - + MgO =Mg 2+ +2NO 3 - + H 2 O

2H + + MgO =Mg 2+ + H 2 O

2HNO 3 + K 2 CO 3 = 2KNO 3 + CO 2 + H 2 O

2H + +2NO 3 - + 2 K + +CO 3 2- = 2K + + 2NO 3 - + CO 2 + H 2 O

2H + +CO 3 2- = CO 2 + H 2 O

I hope there were few mistakes.

Practice some more. Complete the reaction equations, arrange the coefficients, write down the ionic equations:

ZnSO4 +

Prediction of element properties by position in the Periodic Table e

main classes of inorganic substances
Ionic reaction equations