annotation

Relevance: Every year, high school students take the Unified State Exam in chemistry. The most problematic topic in the exam is organic chemistry, which includes not only theory, but also solving problems to derive formulas for organic compounds. Having thought about the problem, I want to create an algorithm for solving these problems for successfully completing the Unified State Exam.

Hypothesis: Is it possible to create an algorithm for solving problems of finding the molecular formula of a substance?

Target: Creation of booklets with an algorithm for solving part C problems.

Tasks:

1. Explore several problems in chemistry to derive formulas for organic matter.
2. Determine the types of these tasks.
3. Identify the essence of tasks.
4. Create an algorithm for solving them by variety.
5. Create a solution key and booklets with an algorithm for completing tasks.

Stages of work on the project:

1. Study of information about the general formulas of substances of different classes.
2. Solving problems to find the molecular formula of a substance.
3. Distribution of tasks by type.
4. Identify the essence of performing these tasks.
5. Determination of the algorithm and key for solving problems for deriving formulas of an organic compound.
6. Creation of project products - booklets.
7. Reflection.

View: single-subject, informational.

Type: short.

Project customer: MBOU Secondary School, Druzhba village

Main article

Every year, almost all school graduates take the Unified State Exam in chemistry. When evaluating the exam tests, I realized that the most difficult tasks are C5, the topic of which is the subject of organic chemistry. This requires not only theory, but also solving problems of finding the molecular formula of a substance.

In order to make it easier to complete tasks on the Unified State Exam, I decided to create an algorithm for solving problems to derive the formula of an organic compound. But first, I came up with a hypothesis and set the goal of the project:

Hypothesis: Is it possible to create an algorithm for solving problems of finding the molecular formula of a substance?

Target: creating booklets with an algorithm for solving part C problems.

I had several tasks ahead of me:

1. Explore several problems in chemistry to derive formulas for organic matter.
2. Determine the types of these tasks.
3. Identify the essence of tasks.
4. Create an algorithm for solving them by variety.
5. Create a solution key and booklets with an algorithm for completing tasks.

Stage I. "Informational"

So, to achieve my goal, I studied several problems to find the molecular formula of an organic compound.

To begin with, I researched the general formulas of substances of different classes:

Organic class	General molecular formula
Alkanes	C n H 2n+2
Alkenes	CnH2n
Alkynes	CnH2n-2
Dienes	CnH2n-2
Benzene homologues	CnH2n-6
Saturated monohydric alcohols	C n H 2n+2 O
Polyhydric alcohols	C n H 2n+2 O x
Saturated aldehydes	CnH2nO
Ketones	CnH2nO
Phenols	CnH2n-6O
Saturated carboxylic acids	CnH2nO2
Esters	CnH2nO2
Amines	C n H 2n+3 N
Amino acids	C n H 2n+1 NO 2

Stage II: “Processing information on this problem”

Example 1.

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

Example 1 solution.

Let the mass of the substance be 100g.

Then the mass of C will be equal to 84.21 g, and the mass of H will be 15.79 g.

Let's find the amount of substance of each atom:

V(C) = m / M = 84.21 /12 = 7.0175 mol,

V(H) = 15.79 / 1 = 15.79 mol.

We determine the molar ratio of C and H atoms:

C: H = 7.0175: 15.79 (reduce both numbers by the smaller number) = 1: 2.25 (multiply by 4) = 4: 9.

Thus, the simplest formula is C 4 H 9.

Using relative density, we calculate the molar mass:

M = D(air) * 29 = 114 g/mol.

The molar mass corresponding to the simplest formula C 4 H 9 is 57 g/mol, which is 2 times less than the true molar mass.

So the true formula is C 8 H 18

Answer: C 8 H 18

Example 2.

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

Solution to example 2.

The general formula of alkyne is C n H 2n-2.

Given the density of a gaseous alkyne, how can you find its molar mass? Density p is the mass of 1 liter of gas under normal conditions.

Since 1 mole of a substance occupies a volume of 22.4 liters, it is necessary to find out how much 22.4 liters of such gas weigh:

M = (density p) * (molar volume V m) = 2.41 g/l * 22.4 l/mol = 54 g/mol.

14 * n - 2 = 54, n = 4.

This means that the alkyne has the formula C 4 H 6

Answer: C 4 H 6

Example 3.

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

Example 3 solution.

In this problem, the number of molecules and the corresponding mass are given. Based on these data, we need to again find the molar mass of the substance.

To do this, you need to remember how many molecules are contained in 1 mole of a substance.

This is Avogadro's number: N a = 6.02*10 23 (molecules).

This means that you can find the amount of aldehyde substance: ‘

V = N / N a = 3 * 10 22 / 6.02 * 10 23 = 0.05 mol, and molar mass:

M = m / n = 4.3 / 0.05 = 86 g/mol.

The general formula of saturated aldehyde is C n H 2 n O, that is, M = 14n + 16 = 86, n = 5.

Answer: C 5 H 10 O, pentanal.

Example 4.

448 ml (n.s.) of gaseous saturated non-cyclic hydrocarbon was burned, and

The reaction products were passed through an excess of lime water, resulting in the formation of 8 g of precipitate. What hydrocarbon was taken?

Solution to example 4.

The general formula of a gaseous saturated non-cyclic hydrocarbon (alkane) is C n H 2n+2.

Then the combustion reaction diagram looks like this:

C n H 2n+2 + O2 - CO2+ H2O

It is easy to see that upon combustion of 1 mole of alkane, n moles of carbon dioxide will be released.

We find the amount of an alkane substance by its volume (don’t forget to convert milliliters to liters!):

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

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

CO 2 + Ca(OH) 2 = CaCO 3 + H 2 O

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

This means that its amount of substance y (CaCO 3) = 8 / 100 = 0.08 mol.

The amount of carbon dioxide substance is also 0.08 mol.

The amount of carbon dioxide is 4 times greater than the alkane, which means the formula of the alkane is C 4 H 10.

Answer: C 4 H 10.

Example5.

The relative vapor density of an organic compound with respect to nitrogen is 2. When 9.8 g of this compound is burned, 15.68 liters of carbon dioxide (NO) and 12.6 g of water are formed. Derive the molecular formula of an organic compound.

Example solution5.

Since a substance upon combustion turns into carbon dioxide and water, it means that it consists of atoms C, H and, possibly, O. Therefore, its general formula can be written as CxHyOz.

We can write the combustion reaction diagram (without arranging the coefficients):

CxHyOz + O 2 - CO 2 + H 2 O

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

We find the amounts of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain:

V (CO 2) = V / Vm = 15.68 / 22.4 = 0.7 mol.

There is one C atom per CO 2 molecule, which means there is the same mole of carbon as CO 2.

V(C) = 0.7 mol

V(H 2 O) = m / M = 12.6 /18 = 0.7 mol.

One molecule of water contains two H atoms, which means the amount of hydrogen is twice that of water.

V(H) = 0.7 * 2 = 1.4 mol.

We check the presence of oxygen in the substance. To do this, the masses of C and H must be subtracted from the mass of the entire starting substance. t(C) = 0.7 * 12 = 8.4 g, m(H) = 1.4 * 1 = 1.4 g The mass of the entire substance is 9.8 g .

m(O) = 9.8 - 8.4 - 1.4 = 0, i.e. there are no oxygen atoms in this substance.

If oxygen were present in a given substance, then by its mass it would be possible to find the amount of the substance and calculate the simplest formula based on the presence of three different atoms.

The next steps are already familiar to you: searching for the simplest and true formulas.

S: H = 0.7: 1.4 = 1: 2

The simplest formula is CH 2.

We look for the true molar mass by the relative density of the gas relative to nitrogen (don’t forget that nitrogen consists of diatomic N2 molecules and its molar mass is 28 g/mol):

M ist. = D by N2 * M (N2) = 2 * 28 = 56 g/mol.

The true formula is CH2, its molar mass is 14.

The true formula is C 4 H 8.

Answer: C 4 H 8.

Example6.

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

Example solution6.

The substance contains C, H and N atoms. Since the mass of nitrogen in the combustion products is not given, it will have to be calculated based on the mass of all organic matter. Combustion reaction scheme: CxHyNz + 02 - CO2 + H20 + N2

We find the amounts of substances C02 and H20, and determine how many moles of C and H atoms they contain:

V(CO 2) = m / M = 17.6 / 44 = 0.4 mol. V(C) = 0.4 mol.

V(H 2 O) = m / M = 12.6 /18 = 0.7 mol. V(H) = 0.7 * 2 = 1.4 mol.

Find the mass of nitrogen in the starting substance.

To do this, the masses of C and H must be subtracted from the mass of the entire starting substance.

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

The mass of the total substance is 9.8 g.

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

C: H: N = 0.4: 1.4: 0.2 = 2: 7: 1 The simplest formula is C 2 H 7 N.

True molar mass

M = Dn0 H2 * M(H2) = 22.5 2 = 45 g/mol.

It coincides with the molar mass calculated for the simplest formula. That is, this is the true formula of the substance.

Answer: C 2 H 7 N.

Example7. Determine the formula of alkadiene if 80 g of 2% bromine solution can decolorize it.

Example solution7.

The general formula of alkadienes is CnH2n-2.

Let's write the equation for the reaction of bromine adding to alkadiene, not forgetting that there are two double bonds in the diene molecule and, accordingly, 2 moles of bromine will react with 1 mole of diene:

C n H 2 n-2 + 2Br 2 - C n H 2 n-2 Br 4

Since the problem gives the mass and percentage concentration of the bromine solution that reacted with the diene, we can calculate the amount of the reacted bromine substance:

m(Br 2) = m solution * ω = 80 * 0.02 = 1.6g

V(Br 2) = m/ M = 1.6/160 = 0.01 mol.

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

C n H 2n-2 + 2 Br 2 - C n H 2n-2 Br 4

M diene = m / v = 3.4 / 0.05 = 68 g/mol.

We find the formula of alkadiene using its general formulas, expressing the molar mass in terms of n:

This is pentadiene C5H8.

Answer: C 5 H 8.

Example8.

When 0.74 g of saturated monohydric alcohol interacted with sodium metal, hydrogen was released in an amount sufficient for the hydrogenation of 112 ml of propene (n.o.). What kind of alcohol is this?

Solution to example 8.

The formula of saturated monohydric alcohol is C n H 2n+1 OH. Here it is convenient to write the alcohol in a form in which it is easy to construct the reaction equation - i.e. with a separate OH group.

Let's create reaction equations (we must not forget about the need to equalize reactions):

2C n H 2 n+1 OH + 2Na - 2C n H 2n+1 ONa + H 2

C 3 H 6 + H 2 - C 3 H 8

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

V(C 3 H 6) = V / Vm = 0.112 / 22.4 = 0.005 mol => v(H2) = 0.005 mol,

Uspirta = 0.005 * 2 = 0.01 mol.

Find the molar mass of alcohol and n:

M alcohol = m / v = 0.74 / 0.01 = 74 g/mol,

Alcohol - butanol C 4 H 7 OH.

Answer: C 4 H 7 OH.

Example 9.

Determine the formula of the ester, upon hydrolysis of 2.64 g of which 1.38 g of alcohol and 1.8 g of monobasic carboxylic acid are released.

Solution to Example 9.

The general formula of an ester consisting of an alcohol and an acid with a different number of carbon atoms can be represented as follows:

C n H 2 n+1 COOC m H 2m+1

Accordingly, the alcohol will have the formula

C m H 2 m+1 OH, and acid

C n H 2 n+1 COOH

Ester hydrolysis equation:

C n H 2 n+1 COOC m H 2m+1 + H 2 O - C m H 2 m+1 OH + C n H 2 n+1 COOH

According to the law of conservation of mass of substances, the sum of the masses of the starting substances and the sum of the masses of the reaction products are equal.

Therefore, from the data of the problem you can find the mass of water:

m H 2 O = (mass of acid) + (mass of alcohol) - (mass of ether) = 1.38 + 1.8 - 2.64 = 0.54g

V H2 O = m / M = 0.54 /18 = 0.03 mol

Accordingly, the amounts of acid and alcohol substances are also equal to moles.

You can find their molar masses:

M acid = m / v = 1.8 / 0.03 = 60 g/mol,

M alcohol = 1.38 / 0.03 = 46 g/mol.

We get two equations from which we find the type:

M C nH2 n+1 COO H = 14n + 46 = 60, n = 1 - acetic acid

M C mH2 m+1OH = 14m + 18 = 46, m = 2 - ethanol.

Thus, the ester we are looking for is the ethyl ester of acetic acid, ethyl acetate.

Answer: CH 3 SOOS 2 H 5.

Conclusion: From the analysis of problem solving it is clear that they can be divided into several types.

Stage III. "Typology of tasks"

Looking at these tasks, it is clear that they are divided into three types:

— by mass fractions of chemical elements ( examples No. 1,2,3);

— by combustion products ( examples No. 4,5,6);

- according to the chemical equation ( examples No. 7,8,9).

Stage IV. “Identification of the essence of tasks”

Based on this, the essence of each type of task is visible.

Type I: instead of the class of substance, the mass fractions of elements are indicated;

Type II: the mass of the substance, masses and volumes of its combustion products are indicated;

III type: the class of the substance being sought, the masses and volumes of the two participants in the reaction are indicated.

Stage V “Creating an algorithm for solving problems”

In order to make it easier to complete chemistry tasks to find the molecular formula of a substance, I created an algorithm for solving them:

Algorithm for solving type I problems (by mass fractions of elements):

1. Find the mole ratio of atoms in a substance

(the ratio of the indices is the ratio of the quotients of the mass fraction of an element divided by its relative atomic mass);

1. Using the molar mass of the substance, determine the formula.

Algorithm for solving type II problems (by combustion products):

1. Find the amount of substance of elements in combustion products

(C,H,O,N,S and others);

1. Their relation is the relation of indices.

Algorithm for solving problems of type III (by chemical equation):

1. Draw up general formulas of substances;
2. Express molar masses in terms of n;
3. Equate the amounts of substances taking into account the coefficients.

Stage VI "Key creation"

In addition, in order to better remember the rules, you also need a key for solving problems to derive the formula of an organic compound:

I-th (finding the formula of an organic compound based on the mass fractions of chemical elements):

For A x B y C z:

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

II (finding the formula of an organic compound from combustion products):

For substance C x H y N z:

x:y:z = v (CO 2):2v(H 2 O):2v(N 2)

III (finding the formula of an organic compound using a chemical equation):

For the process C n H 2 n - C n H 2 n+1 OH:

m(alkene)/ 14n = m(alcohol)/ (14n+18)

VII stage. “Creation of a project product - booklet”

The final stage was the creation of booklets. These are the booklets I distributed to my classmates ( application):

VIII stage. "Reflection"

At an open lesson-game on generalizing oxygen-containing organic compounds, I proposed an algorithm for solving problems of finding the molecular formula of a substance in booklets. The guys were happy to receive the booklets. Now they won’t have any problems with C5 assignments on the Unified State Exam!

Bibliography:

1. O.S. Gabrielyan. Chemistry. Grade 10. Basic level: textbook. for general education institutions / O.S. Gabrielyan. – 5th ed., stereotype. – M.: Bustard, 2009.
2. http://infobusiness2.ru/node/16412
3. http://www.liveedu.ru/2013/03/

Chemistry. Thematic tests for preparing for the Unified State Exam. Tasks of a high level of complexity (C1-C5). Ed. Doronkina V.N.

3rd ed. - R.n / D: 2012. - 234 p. R. n/d: 2011. - 128 p.

The proposed manual is compiled in accordance with the requirements of the new Unified State Examination specification and is intended to prepare for the Unified State Exam in Chemistry. The book includes tasks of a high level of complexity (C1-C5). Each section contains the necessary theoretical information, analyzed (demonstration) examples of completing tasks, which allow you to master the methodology for completing tasks in Part C, and groups of training tasks by topic. The book is addressed to students in grades 10-11 of general education institutions who are preparing for the Unified State Exam and planning to get a high result in the exam, as well as teachers and methodologists who organize the process of preparing for the chemistry exam. The manual is part of the educational and methodological complex “Chemistry. Preparation for the Unified State Exam”, which includes such manuals as “Chemistry.

Preparation for the Unified State Examination 2013", "Chemistry. 10-11 grades. Thematic tests for preparing for the Unified State Exam. Basic and advanced levels”, etc. Format: (2012 pdf

, 3rd ed., rev. and additional, 234 pp.) Size:

2.9 MB 14 Watch, download:

.12.2018, links removed at the request of the Legion publishing house (see note)
CONTENT
Introduction 3
Question C1. Redox reactions. Metal corrosion and methods of protection against it 4
Asking question C1 12
Question C2. Reactions confirming the relationship between various classes of inorganic substances 17
Asking question C2 28
SZ question. Reactions confirming the relationship between hydrocarbons and oxygen-containing organic compounds 54
Asking question SZ 55
Question C4. Calculations: masses (volume, amount of substance) of reaction products, if one of the substances is given in excess (has impurities), if one of the substances is given in the form of a solution with a certain mass fraction of the dissolved substance 68
Asking question C4 73
Question C5. Finding the molecular formula of a substance 83
Asking question C5 85
Answers 97
Application. Interrelation of various classes of inorganic substances.
Additional tasks 207
Tasks 209

Solving problems 218
This book is intended to prepare you for completing tasks of a high level of complexity in general, inorganic and organic chemistry (part C tasks).
For each of the questions C1 - C5, a large number of tasks are given (more than 500 in total), which will allow graduates to test their knowledge, improve existing skills, and, if necessary, learn the factual material included in the test tasks of Part C.
The content of the manual reflects the features of the Unified State Exam variants offered in recent years and corresponds to the current specifications. The questions and answers correspond to the wording of the Unified State Examination tests.
Part C tasks have varying degrees of difficulty. The maximum score for a correctly completed task is from 3 to 5 points (depending on the degree of complexity of the task). Testing of tasks in this part is carried out on the basis of comparing the graduate’s answer with an element-by-element analysis of the given sample answer; each correctly completed element is scored 1 point. For example, in the SZ task you need to create 5 equations for reactions between organic substances, describing the sequential transformation of substances, but you can only create 2 (let’s say the second and fifth equations). Be sure to write them down in the answer form, you will receive 2 points for the SZ task and will significantly improve your result in the exam.
We hope that this book will help you successfully pass the Unified State Exam.

In my practice, I often encounter problems when teaching how to solve chemistry problems. One of the difficult tasks in the Unified State Examination tasks was task C 5.

Let me give you a few examples:

Example 1.

Determine the formula of a substance if it contains 84.21% carbon and 15.79% hydrogen, and has a relative density in air equal to 3.93.

Solution:

1. Let the mass of the substance be 100 g. Then the mass of C will be equal to 84.21 g, and the mass of H will be 15.79 g.

2. Find the amount of substance of each atom:

n(C) = m / M = 84.21 / 12 = 7.0175 mol,

n(H) = 15.79 / 1 = 15.79 mol.

3. Determine the molar ratio of C and H atoms:

C: H = 7.0175: 15.79 (reduce both numbers by the smaller number) = 1: 2.25 (multiply by 4) = 4: 9.

Thus, the simplest formula is C 4 H 9.

4. Using relative density, calculate the molar mass:

M = D(air) 29 = 114 g/mol.

5. The molar mass corresponding to the simplest formula C 4 H 9 is 57 g/mol, which is 2 times less than the true molar mass.

This means that the true formula is C 8 H 18.

Example 2.

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

Solution:

1. General formula of alkyne C n H 2n−2

2. Density ρ is the mass of 1 liter of gas under normal conditions. Since 1 mole of a substance occupies a volume of 22.4 liters, you need to find out how much 22.4 liters of such gas weigh:

M = (density ρ) (molar volume V m) = 2.41 g/l 22.4 l/mol = 54 g/mol.

14 n − 2 = 54, n = 4.

This means that the alkyne has the formula C 4 H 6.

Answer: C 4 H 6.

Example 3.

The relative vapor density of an organic compound with respect to nitrogen is 2. When 9.8 g of this compound is burned, 15.68 liters of carbon dioxide (NO) and 12.6 g of water are formed. Derive the molecular formula of an organic compound.

Solution:

1. Since a substance upon combustion turns into carbon dioxide and water, it means that it consists of atoms C, H and, possibly, O. Therefore, its general formula can be written as CxHyOz.

2. We can write the combustion reaction diagram (without setting the coefficients):

CxHyOz + O 2 → CO 2 + H 2 O

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

We find the amounts of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain:

a) n(CO 2) = V / V m = 15.68 / 22.4 = 0.7 mol.

(There is one C atom per CO 2 molecule, which means there is the same mole of carbon as CO 2. n(C) = 0.7 mol)

b) n(H 2 O) = m / M = 12.6 / 18 = 0.7 mol.

(One water molecule contains two H atoms, which means the amount of hydrogen is twice as much as water. n(H) = 0.7 2 = 1.4 mol)

4. Check for the presence of oxygen in the substance. To do this, the masses of C and H must be subtracted from the mass of the entire starting substance.

m(C) = 0.7 12 = 8.4 g, m(H) = 1.4 1 = 1.4 g

The mass of the total substance is 9.8 g.

m(O) = 9.8 − 8.4 − 1.4 = 0, i.e. There are no oxygen atoms in this substance.

5. Search for the simplest and true formulas.

C: H = 0.7: 1.4 = 1: 2. The simplest formula is CH 2.

6. We look for the true molar mass by the relative density of the gas compared to nitrogen (do not forget that nitrogen consists of diatomic molecules N2 and its molar mass is 28 g/mol):

M source = D(N 2) M(N 2) = 2 28 = 56 g/mol.

The true formula is CH 2, its molar mass is 14. 56 / 14 = 4. The true formula is: (CH 2) 4 = C 4 H 8.

Answer: C 4 H 8.

Example 4.

When 25.5 g of saturated monobasic acid reacted with an excess of sodium bicarbonate solution, 5.6 l (n.s.) of gas was released. Determine the molecular formula of the acid.

Solution:

1. C n H 2n+1 COOH + NaHCO 3 à C n H 2n+1 COONa + H 2 O + CO 2

2. Find the amount of substance CO 2

n(CO 2) = V/Vm = 5.6 l: 22.4 l/mol = 0.25 mol

3. n(CO 2) = n(acids) = 0.25 mol (from the equation you can see this 1:1 ratio)

Then the molar mass of the acid is:

M(k-ty) = m/n = 25.5g: 0.25 mol = 102g/mol

4. M(k-ty) = 12n+2n+1+12+16+16 (from the general formula, M = Ar(C)*n + Ar(H)*n + Ar(O)*n = 12* n + 1*(2n+1)+ 12+16+16+1)

M(k-ty) = 12n +2n +46 = 102; n = 4; The acid formula is C 4 H 9 COOH.

Tasks for self-solving C5:

1. The mass fraction of oxygen in a monobasic amino acid is 42.67%. Determine the molecular formula of the acid.

2. Establish the molecular formula of a tertiary amine if it is known that its combustion produced 0.896 l (n.s.) of carbon dioxide, 0.99 g of water and 0.112 l (n.s.) nitrogen.

3. To completely burn 2 liters of hydrocarbon gas, 13 liters of oxygen were required, and 8 liters of carbon dioxide were formed. Find the molecular formula of hydrocarbon.

4. When 3 liters of hydrocarbon gas are burned, 6 liters of carbon dioxide and a certain amount of water are obtained. Determine the molecular formula of a hydrocarbon if it is known that 10.5 liters of oxygen were required for complete combustion.

5. The dichloro derivative of an alkane contains 5.31% hydrogen by weight. Determine the molecular formula of dichloroalkane. Give the structural formula of one of the possible isomers and name it

6. During the combustion of a gaseous organic substance that does not contain oxygen, 4.48 liters of carbon dioxide (n.o.), 3.6 g of water and 2 g of hydrogen fluoride were released. Determine the molecular formula of the compound.

In 2-3 months it is impossible to learn (repeat, improve) such a complex discipline as chemistry.

There are no changes to the 2020 Unified State Exam KIM in chemistry.

Don't put off preparing for later.

1. When starting to analyze tasks, first study theory. The theory on the site is presented for each task in the form of recommendations on what you need to know when completing the task. will guide you in the study of basic topics and determine what knowledge and skills will be required when completing Unified State Examination tasks in chemistry. To successfully pass the Unified State Exam in chemistry, theory is most important.
2. The theory needs to be supported practice, constantly solving problems. Since most of the mistakes are due to the fact that I read the exercise incorrectly and did not understand what is required in the task. The more often you solve thematic tests, the faster you will understand the structure of the exam. Training tasks developed based on demo versions from FIPI give such an opportunity to decide and find out the answers. But don't rush to peek. First, decide for yourself and see how many points you get.

Points for each chemistry task

• 1 point - for tasks 1-6, 11-15, 19-21, 26-28.
• 2 points - 7-10, 16-18, 22-25, 30, 31.
• 3 points - 35.
• 4 points - 32, 34.
• 5 points - 33.

Total: 60 points.

Structure of the examination paper consists of two blocks:

1. Questions requiring a short answer (in the form of a number or a word) - tasks 1-29.
2. Problems with detailed answers – tasks 30-35.

The exam time in chemistry is 3.5 hours (210 minutes).

There will be three cheat sheets on the exam. And you need to understand them

This is 70% of the information that will help you pass the chemistry exam successfully. The remaining 30% is the ability to use the provided cheat sheets.

• If you want to get more than 90 points, you need to spend a lot of time on chemistry.
• To successfully pass the Unified State Exam in chemistry, you need to solve a lot: training tasks, even if they seem easy and of the same type.
• Distribute your strength correctly and do not forget about rest.

Dare, try and you will succeed!

We discussed the general algorithm for solving problem No. 35 (C5). It's time to look at specific examples and offer you a selection of problems to solve on your own.

Example 2. The complete hydrogenation of 5.4 g of some alkyne requires 4.48 liters of hydrogen (n.s.). Determine the molecular formula of this alkyne.

Solution. We will act in accordance with the general plan. Let a molecule of an unknown alkyne contain n carbon atoms. General formula of the homologous series C n H 2n-2. The hydrogenation of alkynes proceeds according to the equation:

C n H 2n-2 + 2H 2 = C n H 2n+2.

The amount of hydrogen that reacted can be found using the formula n = V/Vm. In this case, n = 4.48/22.4 = 0.2 mol.

The equation shows that 1 mole of alkyne adds 2 moles of hydrogen (recall that in the problem statement we are talking about complete hydrogenation), therefore, n(C n H 2n-2) = 0.1 mol.

Based on the mass and amount of the alkyne, we find its molar mass: M(C n H 2n-2) = m(mass)/n(amount) = 5.4/0.1 = 54 (g/mol).

The relative molecular weight of an alkyne is the sum of n atomic masses of carbon and 2n-2 atomic masses of hydrogen. We get the equation:

12n + 2n - 2 = 54.

We solve the linear equation, we get: n = 4. Alkyne formula: C 4 H 6.

Answer: C 4 H 6 .

I would like to draw attention to one significant point: the molecular formula C 4 H 6 corresponds to several isomers, including two alkynes (butyn-1 and butyn-2). Based on these problems, we will not be able to unambiguously establish the structural formula of the substance under study. However, in this case this is not required!

Example 3. When 112 liters (n.a.) of an unknown cycloalkane are burned in excess oxygen, 336 liters of CO 2 are formed. Establish the structural formula of the cycloalkane.

Solution. The general formula of the homologous series of cycloalkanes: C n H 2n. With complete combustion of cycloalkanes, as with the combustion of any hydrocarbons, carbon dioxide and water are formed:

C n H 2n + 1.5n O 2 = n CO 2 + n H 2 O.

Please note: the coefficients in the reaction equation in this case depend on n!

During the reaction, 336/22.4 = 15 moles of carbon dioxide were formed. 112/22.4 = 5 moles of hydrocarbon entered the reaction.

Further reasoning is obvious: if 15 moles of CO 2 are formed per 5 moles of cycloalkane, then 15 molecules of carbon dioxide are formed per 5 molecules of hydrocarbon, i.e., one cycloalkane molecule produces 3 CO 2 molecules. Since each molecule of carbon monoxide (IV) contains one carbon atom, we can conclude: one cycloalkane molecule contains 3 carbon atoms.

Conclusion: n = 3, cycloalkane formula - C 3 H 6.

As you can see, the solution to this problem does not “fit” into the general algorithm. We did not look for the molar mass of the compound here, nor did we create any equation. According to formal criteria, this example is not similar to the standard problem C5. But I already emphasized above that it is important not to memorize the algorithm, but to understand the MEANING of the actions being performed. If you understand the meaning, you yourself will be able to make changes to the general scheme at the Unified State Exam and choose the most rational solution.

There is one more “oddity” in this example: it is necessary to find not only the molecular, but also the structural formula of the compound. In the previous task we were not able to do this, but in this example - please! The fact is that the formula C 3 H 6 corresponds to only one isomer - cyclopropane.

Answer: cyclopropane.

Example 4. 116 g of some saturated aldehyde were heated for a long time with an ammonia solution of silver oxide. The reaction produced 432 g of metallic silver. Determine the molecular formula of the aldehyde.

Solution. The general formula of the homologous series of saturated aldehydes is: C n H 2n+1 COH. Aldehydes are easily oxidized to carboxylic acids, in particular, under the action of an ammonia solution of silver oxide:

C n H 2n+1 COH + Ag 2 O = C n H 2n+1 COOH + 2 Ag.

Note. In reality, the reaction is described by a more complex equation. When Ag 2 O is added to an aqueous ammonia solution, a complex compound OH is formed - diammine silver hydroxide. It is this compound that acts as an oxidizing agent. During the reaction, an ammonium salt of a carboxylic acid is formed:

C n H 2n+1 COH + 2OH = C n H 2n+1 COONH 4 + 2Ag + 3NH 3 + H 2 O.

Another important point! The oxidation of formaldehyde (HCOH) is not described by the given equation. When HCOH reacts with an ammonia solution of silver oxide, 4 moles of Ag per 1 mole of aldehyde are released:

НCOH + 2Ag2O = CO2 + H2O + 4Ag.

Be careful when solving problems involving the oxidation of carbonyl compounds!

Let's return to our example. Based on the mass of released silver, you can find the amount of this metal: n(Ag) = m/M = 432/108 = 4 (mol). According to the equation, 2 moles of silver are formed per 1 mole of aldehyde, therefore, n(aldehyde) = 0.5n(Ag) = 0.5*4 = 2 moles.

Molar mass of aldehyde = 116/2 = 58 g/mol. Try to do the next steps yourself: you need to create an equation, solve it and draw conclusions.

Answer: C 2 H 5 COH.

Example 5. When 3.1 g of a certain primary amine reacts with a sufficient amount of HBr, 11.2 g of salt is formed. Determine the formula of the amine.

Solution. Primary amines (C n H 2n + 1 NH 2) when reacting with acids form alkylammonium salts:

С n H 2n+1 NH 2 + HBr = [С n H 2n+1 NH 3 ] + Br - .

Unfortunately, based on the mass of the amine and the salt formed, we will not be able to find their quantities (since the molar masses are unknown). Let's take a different path. Let us remember the law of conservation of mass: m(amine) + m(HBr) = m(salt), therefore, m(HBr) = m(salt) - m(amine) = 11.2 - 3.1 = 8.1.

Pay attention to this technique, which is very often used when solving C 5. Even if the mass of the reagent is not given explicitly in the problem statement, you can try to find it from the masses of other compounds.

So, we are back on track with the standard algorithm. Based on the mass of hydrogen bromide, we find the amount, n(HBr) = n(amine), M(amine) = 31 g/mol.

Answer: CH 3 NH 2 .

Example 6. A certain amount of alkene X, when reacting with an excess of chlorine, forms 11.3 g of dichloride, and when reacting with an excess of bromine, 20.2 g of dibromide. Determine the molecular formula of X.

Solution. Alkenes add chlorine and bromine to form dihalogen derivatives:

C n H 2n + Cl 2 = C n H 2n Cl 2,

C n H 2n + Br 2 = C n H 2n Br 2.

In this problem it is pointless to try to find the amount of dichloride or dibromide (their molar masses are unknown) or the amount of chlorine or bromine (their masses are unknown).

We use one non-standard technique. The molar mass of C n H 2n Cl 2 is 12n + 2n + 71 = 14n + 71. M(C n H 2n Br 2) = 14n + 160.

The masses of dihalides are also known. You can find the amounts of substances obtained: n(C n H 2n Cl 2) = m/M = 11.3/(14n + 71). n(C n H 2n Br 2) = 20.2/(14n + 160).

By convention, the amount of dichloride is equal to the amount of dibromide. This fact allows us to create the equation: 11.3/(14n + 71) = 20.2/(14n + 160).

This equation has a unique solution: n = 3.

Answer: C 3 H 6

In the final part, I offer you a selection of C5 type problems of varying difficulty. Try to solve them yourself - it will be excellent training before taking the Unified State Exam in Chemistry!