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.

Solution to example 1.

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.

Solution to example 3.

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 compared 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!

References:

1. O.S. Gabrielyan. Chemistry. 10th grade. 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/

Methods for solving problems in chemistry

When solving problems, you must be guided by a few simple rules:

1. Read the task conditions carefully;
2. Write down what is given;
3. Convert, if necessary, units of physical quantities into SI units (some non-system units are allowed, for example liters);
4. Write down, if necessary, the reaction equation and arrange the coefficients;
5. Solve a problem using the concept of the amount of a substance, and not the method of drawing up proportions;
6. Write down the answer.

In order to successfully prepare for chemistry, you should carefully consider the solutions to the problems given in the text, and also solve a sufficient number of them yourself. It is in the process of solving problems that the basic theoretical principles of the chemistry course will be reinforced. It is necessary to solve problems throughout the entire time of studying chemistry and preparing for the exam.

You can use the problems on this page, or you can download a good collection of problems and exercises with solutions to standard and complicated problems (M. I. Lebedeva, I. A. Ankudimova): download.

Mole, molar mass

Molar mass is the ratio of the mass of a substance to the amount of substance, i.e.

M(x) = m(x)/ν(x), (1)

where M(x) is the molar mass of substance X, m(x) is the mass of substance X, ν(x) is the amount of substance X. The SI unit of molar mass is kg/mol, but the unit g/mol is usually used. Unit of mass – g, kg. The SI unit for quantity of a substance is the mole.

Any chemistry problem solved through the amount of substance. You need to remember the basic formula:

ν(x) = m(x)/ M(x) = V(x)/V m = N/N A , (2)

where V(x) is the volume of the substance X(l), V m is the molar volume of the gas (l/mol), N is the number of particles, N A is Avogadro’s constant.

1. Determine mass sodium iodide NaI amount of substance 0.6 mol.

Given: ν(NaI)= 0.6 mol.

Find: m(NaI) =?

Solution. The molar mass of sodium iodide is:

M(NaI) = M(Na) + M(I) = 23 + 127 = 150 g/mol

Determine the mass of NaI:

m(NaI) = ν(NaI) M(NaI) = 0.6 150 = 90 g.

2. Determine the amount of substance atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g.

Given: m(Na 2 B 4 O 7) = 40.4 g.

Find: ν(B)=?

Solution. The molar mass of sodium tetraborate is 202 g/mol. Determine the amount of substance Na 2 B 4 O 7:

ν(Na 2 B 4 O 7) = m(Na 2 B 4 O 7)/ M(Na 2 B 4 O 7) = 40.4/202 = 0.2 mol.

Recall that 1 mole of sodium tetraborate molecule contains 2 moles of sodium atoms, 4 moles of boron atoms and 7 moles of oxygen atoms (see sodium tetraborate formula). Then the amount of atomic boron substance is equal to: ν(B) = 4 ν (Na 2 B 4 O 7) = 4 0.2 = 0.8 mol.

Calculations using chemical formulas. Mass fraction.

Mass fraction of a substance is the ratio of the mass of a given substance in a system to the mass of the entire system, i.e. ω(X) =m(X)/m, where ω(X) is the mass fraction of substance X, m(X) is the mass of substance X, m is the mass of the entire system. Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. ω(O)=0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. ω(Cl)=0.607.

3. Determine the mass fraction water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

Solution: The molar mass of BaCl 2 2H 2 O is:

M(BaCl 2 2H 2 O) = 137+ 2 35.5 + 2 18 = 244 g/mol

From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol of H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O:

m(H 2 O) = 2 18 = 36 g.

Find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

ω(H 2 O) = m(H 2 O)/ m(BaCl 2 2H 2 O) = 36/244 = 0.1475 = 14.75%.

4. Silver weighing 5.4 g was isolated from a rock sample weighing 25 g containing the mineral argentite Ag 2 S. Determine the mass fraction argentite in the sample.

Given: m(Ag)=5.4 g; m = 25 g.

Find: ω(Ag 2 S) =?

Solution: we determine the amount of silver substance found in argentite: ν(Ag) =m(Ag)/M(Ag) = 5.4/108 = 0.05 mol.

From the formula Ag 2 S it follows that the amount of argentite substance is half as much as the amount of silver substance. Determine the amount of argentite substance:

ν(Ag 2 S)= 0.5 ν(Ag) = 0.5 0.05 = 0.025 mol

We calculate the mass of argentite:

m(Ag 2 S) = ν(Ag 2 S) М(Ag 2 S) = 0.025 248 = 6.2 g.

Now we determine the mass fraction of argentite in a rock sample weighing 25 g.

ω(Ag 2 S) = m(Ag 2 S)/ m = 6.2/25 = 0.248 = 24.8%.

Deriving compound formulas

5. Determine the simplest formula of the compound potassium with manganese and oxygen, if the mass fractions of elements in this substance are 24.7, 34.8 and 40.5%, respectively.

Given: ω(K) =24.7%; ω(Mn) =34.8%; ω(O) =40.5%.

Find: formula of the compound.

Solution: for calculations we select the mass of the compound equal to 100 g, i.e. m=100 g. The masses of potassium, manganese and oxygen will be:

m (K) = m ω(K); m (K) = 100 0.247 = 24.7 g;

m (Mn) = m ω(Mn); m (Mn) =100 0.348=34.8 g;

m (O) = m ω(O); m(O) = 100 0.405 = 40.5 g.

We determine the amounts of atomic substances potassium, manganese and oxygen:

ν(K)= m(K)/ M(K) = 24.7/39= 0.63 mol

ν(Mn)= m(Mn)/ М(Mn) = 34.8/ 55 = 0.63 mol

ν(O)= m(O)/ M(O) = 40.5/16 = 2.5 mol

We find the ratio of the quantities of substances:

ν(K) : ν(Mn) : ν(O) = 0.63: 0.63: 2.5.

Dividing the right side of the equality by a smaller number (0.63) we get:

ν(K) : ν(Mn) : ν(O) = 1: 1: 4.

Therefore, the simplest formula for the compound is KMnO 4.

6. The combustion of 1.3 g of a substance produced 4.4 g of carbon monoxide (IV) and 0.9 g of water. Find the molecular formula substance if its hydrogen density is 39.

Given: m(in-va) =1.3 g; m(CO 2)=4.4 g; m(H 2 O) = 0.9 g; D H2 =39.

Find: formula of a substance.

Solution: Let's assume that the substance we are looking for contains carbon, hydrogen and oxygen, because during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amounts of CO 2 and H 2 O substances in order to determine the amounts of atomic carbon, hydrogen and oxygen substances.

ν(CO 2) = m(CO 2)/ M(CO 2) = 4.4/44 = 0.1 mol;

ν(H 2 O) = m(H 2 O)/ M(H 2 O) = 0.9/18 = 0.05 mol.

We determine the amounts of atomic carbon and hydrogen substances:

ν(C)= ν(CO 2); ν(C)=0.1 mol;

ν(H)= 2 ν(H 2 O); ν(H) = 2 0.05 = 0.1 mol.

Therefore, the masses of carbon and hydrogen will be equal:

m(C) = ν(C) M(C) = 0.1 12 = 1.2 g;

m(N) = ν(N) M(N) = 0.1 1 =0.1 g.

We determine the qualitative composition of the substance:

m(in-va) = m(C) + m(H) = 1.2 + 0.1 = 1.3 g.

Consequently, the substance consists only of carbon and hydrogen (see the problem statement). Let us now determine its molecular weight based on the given condition tasks hydrogen density of a substance.

M(v-va) = 2 D H2 = 2 39 = 78 g/mol.

ν(С) : ν(Н) = 0.1: 0.1

Dividing the right side of the equality by the number 0.1, we get:

ν(С) : ν(Н) = 1: 1

Let us take the number of carbon (or hydrogen) atoms as “x”, then, multiplying “x” by the atomic masses of carbon and hydrogen and equating this sum to the molecular mass of the substance, we solve the equation:

12x + x = 78. Hence x = 6. Therefore, the formula of the substance is C 6 H 6 – benzene.

Molar volume of gases. Laws of ideal gases. Volume fraction.

The molar volume of a gas is equal to the ratio of the volume of the gas to the amount of substance of this gas, i.e.

V m = V(X)/ ν(x),

where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) – volume of gas X; ν(x) is the amount of gas substance X. The molar volume of gases under normal conditions (normal pressure pH = 101,325 Pa ≈ 101.3 kPa and temperature Tn = 273.15 K ≈ 273 K) is V m = 22.4 l /mol.

In calculations involving gases, it is often necessary to switch from these conditions to normal ones or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

──── = ─── (3)

Where p is pressure; V – volume; T - temperature in Kelvin scale; the index “n” indicates normal conditions.

The composition of gas mixtures is often expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

where φ(X) is the volume fraction of component X; V(X) – volume of component X; V is the volume of the system. Volume fraction is a dimensionless quantity; it is expressed in fractions of a unit or as a percentage.

7. Which one volume will take at a temperature of 20 o C and a pressure of 250 kPa ammonia weighing 51 g?

Given: m(NH 3)=51 g; p=250 kPa; t=20 o C.

Find: V(NH 3) =?

Solution: determine the amount of ammonia substance:

ν(NH 3) = m(NH 3)/ M(NH 3) = 51/17 = 3 mol.

The volume of ammonia under normal conditions is:

V(NH 3) = V m ν(NH 3) = 22.4 3 = 67.2 l.

Using formula (3), we reduce the volume of ammonia to these conditions [temperature T = (273 +20) K = 293 K]:

p n TV n (NH 3) 101.3 293 67.2

V(NH 3) =──────── = ───────── = 29.2 l.

8. Define volume, which will be occupied under normal conditions by a gas mixture containing hydrogen, weighing 1.4 g, and nitrogen, weighing 5.6 g.

Given: m(N 2)=5.6 g; m(H 2)=1.4; Well.

Find: V(mixtures)=?

Solution: find the amounts of hydrogen and nitrogen substances:

ν(N 2) = m(N 2)/ M(N 2) = 5.6/28 = 0.2 mol

ν(H 2) = m(H 2)/ M(H 2) = 1.4/ 2 = 0.7 mol

Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of the gases, i.e.

V(mixtures)=V(N 2) + V(H 2)=V m ν(N 2) + V m ν(H 2) = 22.4 0.2 + 22.4 0.7 = 20.16 l.

Calculations using chemical equations

Calculations using chemical equations (stoichiometric calculations) are based on the law of conservation of mass of substances. However, in real chemical processes, due to incomplete reaction and various losses of substances, the mass of the resulting products is often less than that which should be formed in accordance with the law of conservation of mass of substances. The yield of the reaction product (or mass fraction of yield) is the ratio, expressed as a percentage, of the mass of the actually obtained product to its mass, which should be formed in accordance with the theoretical calculation, i.e.

η = /m(X) (4)

Where η is the product yield, %; m p (X) is the mass of product X obtained in the real process; m(X) – calculated mass of substance X.

In those tasks where the product yield is not specified, it is assumed that it is quantitative (theoretical), i.e. η=100%.

9. How much phosphorus needs to be burned? to receive phosphorus (V) oxide weighing 7.1 g?

Given: m(P 2 O 5) = 7.1 g.

Find: m(P) =?

Solution: we write down the equation for the combustion reaction of phosphorus and arrange the stoichiometric coefficients.

4P+ 5O 2 = 2P 2 O 5

Determine the amount of substance P 2 O 5 resulting in the reaction.

ν(P 2 O 5) = m(P 2 O 5)/ M(P 2 O 5) = 7.1/142 = 0.05 mol.

From the reaction equation it follows that ν(P 2 O 5) = 2 ν(P), therefore, the amount of phosphorus required in the reaction is equal to:

ν(P 2 O 5)= 2 ν(P) = 2 0.05= 0.1 mol.

From here we find the mass of phosphorus:

m(P) = ν(P) M(P) = 0.1 31 = 3.1 g.

10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in excess hydrochloric acid. What volume hydrogen, measured under standard conditions, will stand out at the same time?

Given: m(Mg)=6 g; m(Zn)=6.5 g; Well.

Find: V(H 2) =?

Solution: we write down the reaction equations for the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.

Zn + 2 HCl = ZnCl 2 + H 2

Mg + 2 HCl = MgCl 2 + H 2

We determine the amounts of magnesium and zinc substances that reacted with hydrochloric acid.

ν(Mg) = m(Mg)/ М(Mg) = 6/24 = 0.25 mol

ν(Zn) = m(Zn)/ M(Zn) = 6.5/65 = 0.1 mol.

From the reaction equations it follows that the amounts of metal and hydrogen substances are equal, i.e. ν(Mg) = ν(H 2); ν(Zn) = ν(H 2), we determine the amount of hydrogen resulting from two reactions:

ν(H 2) = ν(Mg) + ν(Zn) = 0.25 + 0.1 = 0.35 mol.

We calculate the volume of hydrogen released as a result of the reaction:

V(H 2) = V m ν(H 2) = 22.4 0.35 = 7.84 l.

11. When a volume of 2.8 liters of hydrogen sulfide (normal conditions) was passed through an excess solution of copper (II) sulfate, a precipitate weighing 11.4 g was formed. Determine the exit reaction product.

Given: V(H 2 S)=2.8 l; m(sediment)= 11.4 g; Well.

Find: η =?

Solution: we write down the equation for the reaction between hydrogen sulfide and copper (II) sulfate.

H 2 S + CuSO 4 = CuS ↓+ H 2 SO 4

We determine the amount of hydrogen sulfide involved in the reaction.

ν(H 2 S) = V(H 2 S) / V m = 2.8/22.4 = 0.125 mol.

From the reaction equation it follows that ν(H 2 S) = ν(СuS) = 0.125 mol. This means we can find the theoretical mass of CuS.

m(СuS) = ν(СuS) М(СuS) = 0.125 96 = 12 g.

Now we determine the product yield using formula (4):

η = /m(X)= 11.4 100/ 12 = 95%.

12. Which one weight ammonium chloride is formed by the interaction of hydrogen chloride weighing 7.3 g with ammonia weighing 5.1 g? Which gas will remain in excess? Determine the mass of the excess.

Given: m(HCl)=7.3 g; m(NH 3)=5.1 g.

Find: m(NH 4 Cl) =? m(excess) =?

Solution: write down the reaction equation.

HCl + NH 3 = NH 4 Cl

This task is about “excess” and “deficiency”. We calculate the amounts of hydrogen chloride and ammonia and determine which gas is in excess.

ν(HCl) = m(HCl)/ M(HCl) = 7.3/36.5 = 0.2 mol;

ν(NH 3) = m(NH 3)/ M(NH 3) = 5.1/ 17 = 0.3 mol.

Ammonia is in excess, so we calculate based on the deficiency, i.e. for hydrogen chloride. From the reaction equation it follows that ν(HCl) = ν(NH 4 Cl) = 0.2 mol. Determine the mass of ammonium chloride.

m(NH 4 Cl) = ν(NH 4 Cl) М(NH 4 Cl) = 0.2 53.5 = 10.7 g.

We have determined that ammonia is in excess (in terms of the amount of substance, the excess is 0.1 mol). Let's calculate the mass of excess ammonia.

m(NH 3) = ν(NH 3) M(NH 3) = 0.1 17 = 1.7 g.

13. Technical calcium carbide weighing 20 g was treated with excess water, obtaining acetylene, which, when passed through excess bromine water, formed 1,1,2,2-tetrabromoethane weighing 86.5 g. Determine mass fraction CaC 2 in technical carbide.

Given: m = 20 g; m(C 2 H 2 Br 4) = 86.5 g.

Find: ω(CaC 2) =?

Solution: we write down the equations for the interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.

CaC 2 +2 H 2 O = Ca(OH) 2 + C 2 H 2

C 2 H 2 +2 Br 2 = C 2 H 2 Br 4

Find the amount of tetrabromoethane.

ν(C 2 H 2 Br 4) = m(C 2 H 2 Br 4)/ M(C 2 H 2 Br 4) = 86.5/ 346 = 0.25 mol.

From the reaction equations it follows that ν(C 2 H 2 Br 4) = ν(C 2 H 2) = ν(CaC 2) = 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities).

m(CaC 2) = ν(CaC 2) M(CaC 2) = 0.25 64 = 16 g.

We determine the mass fraction of CaC 2 in technical carbide.

ω(CaC 2) =m(CaC 2)/m = 16/20 = 0.8 = 80%.

Solutions. Mass fraction of solution component

14. Sulfur weighing 1.8 g was dissolved in benzene with a volume of 170 ml. The density of benzene is 0.88 g/ml. Define mass fraction sulfur in solution.

Given: V(C 6 H 6) = 170 ml; m(S) = 1.8 g; ρ(C 6 C 6) = 0.88 g/ml.

Find: ω(S) =?

Solution: to find the mass fraction of sulfur in a solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene.

m(C 6 C 6) = ρ(C 6 C 6) V(C 6 H 6) = 0.88 170 = 149.6 g.

Find the total mass of the solution.

m(solution) = m(C 6 C 6) + m(S) = 149.6 + 1.8 = 151.4 g.

Let's calculate the mass fraction of sulfur.

ω(S) =m(S)/m=1.8 /151.4 = 0.0119 = 1.19%.

15. Iron sulfate FeSO 4 7H 2 O weighing 3.5 g was dissolved in water weighing 40 g. Determine mass fraction of iron (II) sulfate in the resulting solution.

Given: m(H 2 O)=40 g; m(FeSO 4 7H 2 O) = 3.5 g.

Find: ω(FeSO 4) =?

Solution: find the mass of FeSO 4 contained in FeSO 4 7H 2 O. To do this, calculate the amount of the substance FeSO 4 7H 2 O.

ν(FeSO 4 7H 2 O)=m(FeSO 4 7H 2 O)/M(FeSO 4 7H 2 O)=3.5/278=0.0125 mol

From the formula of iron sulfate it follows that ν(FeSO 4) = ν(FeSO 4 7H 2 O) = 0.0125 mol. Let's calculate the mass of FeSO 4:

m(FeSO 4) = ν(FeSO 4) M(FeSO 4) = 0.0125 152 = 1.91 g.

Considering that the mass of the solution consists of the mass of iron sulfate (3.5 g) and the mass of water (40 g), we calculate the mass fraction of ferrous sulfate in the solution.

ω(FeSO 4) =m(FeSO 4)/m=1.91 /43.5 = 0.044 =4.4%.

Problems to solve independently

1. 50 g of methyl iodide in hexane were exposed to metallic sodium, and 1.12 liters of gas were released, measured under normal conditions. Determine the mass fraction of methyl iodide in the solution. Answer: 28,4%.
2. Some alcohol was oxidized to form a monocarboxylic acid. When 13.2 g of this acid was burned, carbon dioxide was obtained, the complete neutralization of which required 192 ml of KOH solution with a mass fraction of 28%. The density of the KOH solution is 1.25 g/ml. Determine the formula of alcohol. Answer: butanol.
3. The gas obtained by reacting 9.52 g of copper with 50 ml of an 81% nitric acid solution with a density of 1.45 g/ml was passed through 150 ml of a 20% NaOH solution with a density of 1.22 g/ml. Determine the mass fractions of dissolved substances. Answer: 12.5% ​​NaOH; 6.48% NaNO 3 ; 5.26% NaNO2.
4. Determine the volume of gases released during the explosion of 10 g of nitroglycerin. Answer: 7.15 l.
5. A sample of organic matter weighing 4.3 g was burned in oxygen. The reaction products are carbon monoxide (IV) with a volume of 6.72 l (normal conditions) and water with a mass of 6.3 g. The vapor density of the starting substance with respect to hydrogen is 43. Determine the formula of the substance. Answer: C 6 H 14.

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. 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 “strange thing” 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!

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", including 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: pdf (2012 , 3rd ed., rev. and additional, 234 pp.)

Size: 2.9 MB

Watch, download: 14 .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
Literature 234

INTRODUCTION
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.

Attention!!!

Changes in the KIM Unified State Exam 2018 in Chemistry of the Year compared to 2017

The following changes have been made in the 2018 exam paper compared to the 2017 paper.

1. In order to more clearly distribute tasks into individual thematic blocks and content lines, the order of tasks of basic and advanced levels of complexity in part 1 of the examination paper has been slightly changed.

2. In the 2018 examination paper, the total number of tasks was increased from 34 (in 2017) to 35 due to an increase in the number of tasks in part 2 of the examination paper from 5 (in 2017) to 6 tasks. This is achieved through the introduction of tasks with a single context. In particular, this format presents tasks No. 30 and No. 31, which are aimed at testing the assimilation of important content elements: “Redox reactions” and “Ion exchange reactions.”

3. The grading scale for some tasks has been changed due to the clarification of the level of difficulty of these tasks based on the results of their completion in the 2017 examination paper:

Task No. 9 of an increased level of complexity, focused on testing the assimilation of the content element “Characteristic chemical properties of inorganic substances” and presented in a format for establishing correspondence between reacting substances and reaction products between these substances, will be assessed with a maximum of 2 points;

Task No. 21 of a basic level of complexity, aimed at testing the assimilation of the content element “Redox reactions” and presented in a format to establish correspondence between the elements of two sets, will be scored 1 point;

Task No. 26 of a basic level of complexity, aimed at testing the assimilation of the content lines “Experimental Fundamentals of Chemistry” and “General Ideas about Industrial Methods for the Production of Essential Substances” and presented in a format to establish a correspondence between the elements of two sets, will be scored 1 point;

Task No. 30 of a high level of complexity with a detailed answer, aimed at testing the assimilation of the content element “Redox reactions”, will be assessed with a maximum of 2 points;

Task No. 31 of a high level of complexity with a detailed answer, aimed at testing the assimilation of the content element “Ion exchange reactions”, will be assessed with a maximum of 2 points.

In general, the adopted changes in the 2018 examination work are aimed at increasing the objectivity of testing the formation of a number of important general educational skills, primarily such as: applying knowledge in the system, independently assessing the correctness of completing an educational and educational-practical task, as well as combining knowledge about chemical objects with an understanding of the mathematical relationship between various physical quantities.

General changes in KIM Unified State Exam 2017 - The structure of the examination paper has been optimized:

1. The structure of part 1 of CMM has been fundamentally changed: tasks with a choice of one answer have been excluded; The tasks are grouped into separate thematic blocks, each of which contains tasks of both basic and advanced levels of difficulty.

2. The total number of tasks has been reduced from 40 (in 2016) to 34.

3. The rating scale has been changed (from 1 to 2 points) for completing tasks at a basic level of complexity, which test the assimilation of knowledge about the genetic connection of inorganic and organic substances (9 and 17).

4. The maximum initial score for completing the work as a whole will be 60 points (instead of 64 points in 2016)

Dear colleagues and students!

An open bank of assignments in 13 subjects, including chemistry, has appeared on the FIPI website.

Open bank of tasks for the Unified State Exam and State Examination in Chemistry

Open task banks for the Unified State Exam and GIA-9 provide the following opportunities:  get acquainted with the tasks collected according to the thematic rubricator,  download tasks on a user-selected topic, divided into 10 tasks per page and the ability to turn pages,  open a task selected by the user in a separate window. Answers to assignments are not provided.

Selection of materials

Tasks C1 (with solutions)

Tasks C2 (with solutions)

C3 tasks

C4 tasks

C5 tasks

I offer a selection of materials (Sikorskaya O.E.) for preparing students for the Unified State Exam:

Main types of problems in Part B:

Main types of tasks in Part C: