I will solve the exam chemistry task 9 35. Using the electronic balance method, create a reaction equation

To solve problems of this type, you need to know the general formulas for classes of organic substances and general formulas for calculating the molar mass of substances of these classes:

Majority decision algorithm molecular formula problems includes the following actions:

— writing reaction equations in general form;

— finding the amount of substance n for which the mass or volume is given, or the mass or volume of which can be calculated according to the conditions of the problem;

— finding the molar mass of a substance M = m/n, the formula of which needs to be established;

— finding the number of carbon atoms in a molecule and drawing up the molecular formula of a substance.

Examples of solving problem 35 of the Unified State Exam in chemistry to find the molecular formula of an organic substance from combustion products with an explanation

The combustion of 11.6 g of organic matter produces 13.44 liters of carbon dioxide and 10.8 g of water. The vapor density of this substance in air is 2. It has been established that this substance interacts with an ammonia solution of silver oxide, is catalytically reduced by hydrogen to form a primary alcohol, and can be oxidized with an acidified solution of potassium permanganate to carboxylic acid. Based on this data:
1) establish the simplest formula of the starting substance,
2) make up its structural formula,
3) give the reaction equation for its interaction with hydrogen.

Solution: general formula of organic matter is CxHyOz.

Let's convert the volume of carbon dioxide and the mass of water into moles using the formulas:

n = m/M And n = V/ Vm,

Molar volume Vm = 22.4 l/mol

n(CO 2) = 13.44/22.4 = 0.6 mol, => the original substance contained n(C) = 0.6 mol,

n(H 2 O) = 10.8/18 = 0.6 mol, => the original substance contained twice as much n(H) = 1.2 mol,

This means that the required compound contains oxygen in the amount of:

n(O)= 3.2/16 = 0.2 mol

Let's look at the ratio of the C, H and O atoms that make up the original organic substance:

n(C) : n(H) : n(O) = x: y: z = 0.6: 1.2: 0.2 = 3: 6: 1

We found the simplest formula: C 3 H 6 O

To find out the true formula, we find the molar mass of an organic compound using the formula:

М(СxHyOz) = Dair(СxHyOz) *M(air)

M source (СxHyOz) = 29*2 = 58 g/mol

Let's check whether the true molar mass corresponds to the molar mass of the simplest formula:

M (C 3 H 6 O) = 12*3 + 6 + 16 = 58 g/mol - corresponds, => the true formula coincides with the simplest one.

Molecular formula: C 3 H 6 O

From the problem data: “this substance interacts with an ammonia solution of silver oxide, is catalytically reduced by hydrogen to form a primary alcohol and can be oxidized with an acidified solution of potassium permanganate to a carboxylic acid,” we conclude that it is an aldehyde.

2) When 18.5 g of saturated monobasic carboxylic 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.

3) A certain saturated carboxylic monobasic acid weighing 6 g requires the same mass of alcohol for complete esterification. This yields 10.2 g of ester. Determine the molecular formula of the acid.

4) Determine the molecular formula of acetylene hydrocarbon if the molar mass of the product of its reaction with excess hydrogen bromide is 4 times greater than the molar mass of the original hydrocarbon

5) When an organic substance weighing 3.9 g was burned, carbon monoxide (IV) weighing 13.2 g and water weighing 2.7 g were formed. Derive the formula of the substance, knowing that the vapor density of this substance with respect to hydrogen is 39.

6) When an organic substance weighing 15 g was burned, carbon monoxide (IV) with a volume of 16.8 liters and water weighing 18 g were formed. Derive the formula of the substance, knowing that the vapor density of this substance for hydrogen fluoride is 3.

7) When 0.45 g of gaseous organic matter was burned, 0.448 l (n.s.) of carbon dioxide, 0.63 g of water and 0.112 l (n.s.) of nitrogen were released. The density of the initial gaseous substance by nitrogen is 1.607. Determine the molecular formula of this substance.

8) The combustion of oxygen-free organic matter produced 4.48 liters (n.s.) of carbon dioxide, 3.6 g of water and 3.65 g of hydrogen chloride. Determine the molecular formula of the burnt compound.

9) When an organic substance weighing 9.2 g was burned, carbon monoxide (IV) with a volume of 6.72 l (n.s.) and water weighing 7.2 g were formed. Establish the molecular formula of the substance.

10) During the combustion of an organic substance weighing 3 g, carbon monoxide (IV) with a volume of 2.24 l (n.s.) and water weighing 1.8 g were formed. It is known that this substance reacts with zinc.
Based on the data of the task conditions:
1) make the calculations necessary to establish the molecular formula of an organic substance;
2) write down the molecular formula of the original organic substance;
3) draw up a structural formula of this substance, which unambiguously reflects the order of bonds of atoms in its molecule;
4) write the equation for the reaction of this substance with zinc.

To solve problems of this type, you need to know the general formulas for classes of organic substances and general formulas for calculating the molar mass of substances of these classes:

Majority decision algorithm molecular formula problems includes the following actions:

— writing reaction equations in general form;

— finding the amount of substance n for which the mass or volume is given, or the mass or volume of which can be calculated according to the conditions of the problem;

— finding the molar mass of a substance M = m/n, the formula of which needs to be established;

— finding the number of carbon atoms in a molecule and drawing up the molecular formula of a substance.

Examples of solving problem 35 of the Unified State Exam in chemistry to find the molecular formula of an organic substance from combustion products with an explanation

Solution: general formula of organic matter is CxHyOz.

Let's convert the volume of carbon dioxide and the mass of water into moles using the formulas:

n = m/M And n = V/ Vm,

Molar volume Vm = 22.4 l/mol

n(CO 2) = 13.44/22.4 = 0.6 mol, => the original substance contained n(C) = 0.6 mol,

n(H 2 O) = 10.8/18 = 0.6 mol, => the original substance contained twice as much n(H) = 1.2 mol,

This means that the required compound contains oxygen in the amount of:

n(O)= 3.2/16 = 0.2 mol

Let's look at the ratio of the C, H and O atoms that make up the original organic substance:

n(C) : n(H) : n(O) = x: y: z = 0.6: 1.2: 0.2 = 3: 6: 1

We found the simplest formula: C 3 H 6 O

To find out the true formula, we find the molar mass of an organic compound using the formula:

М(СxHyOz) = Dair(СxHyOz) *M(air)

M source (СxHyOz) = 29*2 = 58 g/mol

Let's check whether the true molar mass corresponds to the molar mass of the simplest formula:

M (C 3 H 6 O) = 12*3 + 6 + 16 = 58 g/mol - corresponds, => the true formula coincides with the simplest one.

Molecular formula: C 3 H 6 O

2) When 18.5 g of saturated monobasic carboxylic 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.

3) A certain saturated carboxylic monobasic acid weighing 6 g requires the same mass of alcohol for complete esterification. This yields 10.2 g of ester. Determine the molecular formula of the acid.

8) The combustion of oxygen-free organic matter produced 4.48 liters (n.s.) of carbon dioxide, 3.6 g of water and 3.65 g of hydrogen chloride. Determine the molecular formula of the burnt compound.

9) When an organic substance weighing 9.2 g was burned, carbon monoxide (IV) with a volume of 6.72 l (n.s.) and water weighing 7.2 g were formed. Establish the molecular formula of the substance.

Problems No. 35 on the Unified State Exam in chemistry

Algorithm for solving such tasks

1. General formula of the homologous series

The most commonly used formulas are summarized in the table:

Homologous series	General formula





Saturated monohydric alcohols
Saturated aldehydes	C n H 2n+1 SON
Saturated monocarboxylic acids	C n H 2n+1 COOH

2. Reaction equation

1) ALL organic substances burn in oxygen to form carbon dioxide, water, nitrogen (if N is present in the compound) and HCl (if chlorine is present):

C n H m O q N x Cl y + O 2 = CO 2 + H 2 O + N 2 + HCl (without coefficients!)

2) Alkenes, alkynes, dienes are prone to addition reactions (reactions with halogens, hydrogen, hydrogen halides, water):

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

C n H 2n + H 2 = C n H 2n+2

C n H 2n + HBr = C n H 2n+1 Br

C n H 2n + H 2 O = C n H 2n+1 OH

Alkynes and dienes, unlike alkenes, add up to 2 moles of hydrogen, chlorine or hydrogen halide per 1 mole of hydrocarbon:

C n H 2n-2 + 2Cl 2 = C n H 2n-2 Cl 4

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

When water is added to alkynes, carbonyl compounds are formed, not alcohols!

3) Alcohols are characterized by reactions of dehydration (intramolecular and intermolecular), oxidation (to carbonyl compounds and, possibly, further to carboxylic acids). Alcohols (including polyhydric) react with alkali metals to release hydrogen:

C n H 2n+1 OH = C n H 2n + H 2 O

2C n H 2n+1 OH = C n H 2n+1 OC n H 2n+1 + H 2 O

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

4) The chemical properties of aldehydes are very diverse, but here we will only remember redox reactions:

C n H 2n+1 COH + H 2 = C n H 2n+1 CH 2 OH (reduction of carbonyl compounds in the addition of Ni),

C n H 2n+1 COH + [O] = C n H 2n+1 COOH

important point: the oxidation of formaldehyde (HCO) does not stop at the formic acid stage, HCOOH is further oxidized to CO 2 and H 2 O.

5) Carboxylic acids exhibit all the properties of “ordinary” inorganic acids: they interact with bases and basic oxides, react with active metals and salts of weak acids (for example, with carbonates and bicarbonates). The esterification reaction is very important - the formation of esters when interacting with alcohols.

C n H 2n+1 COOH + KOH = C n H 2n+1 COOK + H 2 O

2C n H 2n+1 COOH + CaO = (C n H 2n+1 COO) 2 Ca + H 2 O

2C n H 2n+1 COOH + Mg = (C n H 2n+1 COO) 2 Mg + H 2

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

C n H 2n+1 COOH + C 2 H 5 OH = C n H 2n+1 COOC 2 H 5 + H 2 O

3. Finding the amount of a substance by its mass (volume)

formula connecting the mass of a substance (m), its quantity (n) and molar mass (M):

m = n*M or n = m/M.

For example, 710 g of chlorine (Cl 2) corresponds to 710/71 = 10 mol of this substance, since the molar mass of chlorine = 71 g/mol.

For gaseous substances, it is more convenient to work with volumes rather than masses. Let me remind you that the amount of a substance and its volume are related by the following formula: V = V m *n, where V m is the molar volume of the gas (22.4 l/mol under normal conditions).

4. Calculations using reaction equations

This is probably the main type of calculations in chemistry. If you do not feel confident in solving such problems, you need to practice.

The basic idea is this: the quantities of reactants and products formed are related in the same way as the corresponding coefficients in the reaction equation (which is why it is so important to place them correctly!)

Consider, for example, the following reaction: A + 3B = 2C + 5D. The equation shows that 1 mol A and 3 mol B upon interaction form 2 mol C and 5 mol D. The amount of B is three times greater than the amount of substance A, the amount of D is 2.5 times greater than the amount of C, etc. If in If the reaction is not 1 mol A, but, say, 10, then the amounts of all other participants in the reaction will increase exactly 10 times: 30 mol B, 20 mol C, 50 mol D. If we know that 15 mol D were formed (three times times more than indicated in the equation), then the amounts of all other compounds will be 3 times greater.

5. Calculation of the molar mass of the test substance

The mass X is usually given in the problem statement; we found the quantity X in paragraph 4. It remains to use the formula M = m/n again.

6. Determination of the molecular formula of X.

The final stage. Knowing the molar mass of X and the general formula of the corresponding homologous series, you can find the molecular formula of the unknown substance.

Let, for example, the relative molecular weight of the limiting monohydric alcohol be 46. The general formula of the homologous series: C n H 2n+1 OH. Relative molecular weight consists of the mass of n carbon atoms, 2n+2 hydrogen atoms and one oxygen atom. We get the equation: 12n + 2n + 2 + 16 = 46. Solving the equation, we find that n = 2. The molecular formula of alcohol is: C 2 H 5 OH.

Don't forget to write down your answer!

Example 1 . 10.5 g of some alkene can add 40 g of bromine. Identify the unknown alkene.

Solution. Let a molecule of an unknown alkene contain n carbon atoms. General formula of the homologous series C n H 2n. Alkenes react with bromine according to the equation:

CnH2n + Br2 = CnH2nBr2.

Let's calculate the amount of bromine that entered the reaction: M(Br 2) = 160 g/mol. n(Br 2) = m/M = 40/160 = 0.25 mol.

The equation shows that 1 mol of alkene adds 1 mol of bromine, therefore, n(C n H 2n) = n(Br 2) = 0.25 mol.

Knowing the mass of the reacted alkene and its quantity, we will find its molar mass: M(C n H 2n) = m(mass)/n(amount) = 10.5/0.25 = 42 (g/mol).

Now it is quite easy to identify an alkene: the relative molecular weight (42) is the sum of the mass of n carbon atoms and 2n hydrogen atoms. We get the simplest algebraic equation:

The solution to this equation is n = 3. The alkene formula is: C 3 H 6 .

Answer: C 3 H 6 .

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 mol of alkyne adds 2 mol of hydrogen (remember that the problem statement refers to 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 .

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.

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 interacting 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.

In our last article, we talked about the basic tasks in the Unified State Exam in Chemistry 2018. Now, we have to analyze in more detail the tasks of an increased (in the 2018 Unified State Exam codifier in chemistry - high level of complexity) level of complexity, previously called part C.

Tasks of an increased level of complexity include only five (5) tasks - No. 30, 31, 32, 33, 34 and 35. Let's consider the topics of the tasks, how to prepare for them and how to solve complex tasks in the Unified State Exam in Chemistry 2018.

Example of task 30 in the Unified State Examination in Chemistry 2018

Aimed at testing the student's knowledge about oxidation-reduction reactions (ORR). The assignment always gives an equation for a chemical reaction with substances missing from either side of the reaction (the left side is the reactants, the right side is the products). A maximum of three (3) points may be awarded for this assignment. The first point is given for correctly filling in the gaps in the reaction and correct equalization of the reaction (arrangement of coefficients). The second point can be obtained by correctly describing the ORR balance, and the last point is given for correctly determining who is the oxidizing agent in the reaction and who is the reducing agent. Let's look at the solution to task No. 30 from the demo version of the Unified State Exam in Chemistry 2018:

Using the electron balance method, create an equation for the reaction

Na 2 SO 3 + … + KOH à K 2 MnO 4 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

The first thing you need to do is arrange the charges of the atoms indicated in the equation, it turns out:

Na + 2 S +4 O 3 -2 + … + K + O -2 H + à K + 2 Mn +6 O 4 -2 + … + H + 2 O -2

Often after this action, we immediately see the first pair of elements that changed the oxidation state (CO), that is, from different sides of the reaction, the same atom has a different oxidation state. In this particular task, we do not observe this. Therefore, it is necessary to take advantage of additional knowledge, namely, on the left side of the reaction, we see potassium hydroxide ( CON), the presence of which tells us that the reaction occurs in an alkaline environment. On the right side, we see potassium manganate, and we know that in an alkaline reaction medium, potassium manganate is obtained from potassium permanganate, therefore, the gap on the left side of the reaction is potassium permanganate ( KMnO 4 ). It turns out that on the left we had manganese at CO +7, and on the right at CO +6, which means we can write the first part of the OVR balance:

Mn +7 +1 e — à Mn +6

Now, we can guess what else should happen in the reaction. If manganese receives electrons, then someone must have given them to it (we follow the law of conservation of mass). Let's consider all the elements on the left side of the reaction: hydrogen, sodium and potassium are already in CO +1, which is the maximum for them, oxygen will not give up its electrons to manganese, which means sulfur remains in CO +4. We conclude that sulfur gives up electrons and goes into the sulfur state with CO +6. Now we can write the second part of the balance sheet:

S +4 -2 e — à S +6

Looking at the equation, we see that on the right hand side, there is no sulfur or sodium anywhere, which means they must be in the gap, and the logical compound to fill it is sodium sulfate ( NaSO 4 ).

Now the OVR balance is written (we get the first point) and the equation takes the form:

Na 2 SO 3 + KMnO 4 + KOHà K 2 MnO 4 + NaSO 4 + H 2 O

Mn +7 +1 e — à Mn +6	1	2
S +4 -2e —à S+6	2	1

It is important at this point to immediately write who is the oxidizing agent and who is the reducing agent, since students often concentrate on balancing the equation and simply forget to do this part of the task, thereby losing a point. By definition, an oxidizing agent is the particle that receives electrons (in our case, manganese), and a reducing agent is the particle that gives up electrons (in our case, sulfur), so we get:

Oxidizer: Mn +7 (KMnO 4 )

Reducing agent: S +4 (Na 2 SO 3 )

Here we must remember that we are indicating the state of the particles in which they were when they began to exhibit the properties of an oxidizing or reducing agent, and not the states to which they came as a result of redox reaction.

Now, in order to get the last point, you need to correctly equalize the equation (arrange the coefficients). Using the balance, we see that in order for it to be sulfur +4, to go into the +6 state, two manganese +7 must become manganese +6, and what matters is we put 2 in front of the manganese:

Na 2 SO 3 + 2KMnO 4 + KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

Now we see that we have 4 potassium on the right, and only three on the left, which means we need to put 2 in front of potassium hydroxide:

Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

As a result, the correct answer to task No. 30 looks like this:

Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O

Mn +7 +1e —à Mn +6	1	2
S +4 -2e —à S+6	2	1

Oxidizer: Mn +7 (KMnO 4)

Reducing agent: S +4 (Na 2 SO 3 )

Solution to task 31 in the Unified State Exam in chemistry

This is a chain of inorganic transformations. To successfully complete this task, you must have a good understanding of the reactions characteristic of inorganic compounds. The task consists of four (4) reactions, for each of which you can get one (1) point, for a total of four (4) points for the task. It is important to remember the rules for completing the assignment: all equations must be equalized, even if a student wrote the equation correctly but did not equalize, he will not receive a point; it is not necessary to solve all the reactions, you can do one and get one (1) point, two reactions and get two (2) points, etc., and it is not necessary to complete the equations strictly in order, for example, a student can do reaction 1 and 3, which means you need to do this and get two (2) points, the main thing is to indicate that these are reactions 1 and 3. Let’s look at the solution to task No. 31 from the demo version of the Unified State Exam in Chemistry 2018:

Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess of sodium hydroxide solution. The brown precipitate that formed was filtered and calcined. The resulting substance was heated with iron.
Write equations for the four reactions described.

To make the solution easier, you can draw up the following diagram in a draft:

To complete the task, of course, you need to know all the proposed reactions. However, there are always hidden clues in the condition (concentrated sulfuric acid, excess sodium hydroxide, brown precipitate, calcined, heated with iron). For example, a student does not remember what happens to iron when interacting with conc. sulfuric acid, but he remembers that the brown precipitate of iron after treatment with alkali is most likely iron hydroxide 3 ( Y = Fe(OH) 3 ). Now we have the opportunity, by substituting Y into the written diagram, to try to make equations 2 and 3. The subsequent steps are purely chemical, so we will not describe them in such detail. The student must remember that heating iron hydroxide 3 results in the formation of iron oxide 3 ( Z = Fe 2 O 3 ) and water, and heating iron oxide 3 with pure iron will lead them to the middle state - iron oxide 2 ( FeO). Substance X, which is a salt obtained after reaction with sulfuric acid, yielding iron hydroxide 3 after treatment with alkali, will be iron sulfate 3 ( X = Fe 2 (SO 4 ) 3 ). It is important to remember to balance the equations. As a result, the correct answer to task No. 31 is as follows:

1) 2Fe + 6H 2 SO 4 (k) a Fe2(SO4)3+ 3SO 2 + 6H 2 O

2) Fe2(SO4)3+ 6NaOH (g) à 2 Fe(OH)3+ 3Na2SO4

3) 2Fe(OH) 3à Fe 2 O 3 + 3H 2 O

4) Fe 2 O 3 + Fe à 3FeO

Task 32 Unified State Exam in Chemistry

Very similar to task No. 31, only it contains a chain of organic transformations. The design requirements and solution logic are similar to task No. 31, the only difference is that in task No. 32 five (5) equations are given, which means you can score five (5) points in total. Due to its similarity to task No. 31, we will not consider it in detail.

Solution to task 33 in chemistry 2018

A calculation task, to complete it you need to know the basic calculation formulas, be able to use a calculator and draw logical parallels. Assignment 33 is worth four (4) points. Let's look at part of the solution to task No. 33 from the demo version of the Unified State Exam in Chemistry 2018:

Determine the mass fractions (in %) of iron (II) sulfate and aluminum sulfide in the mixture if, when treating 25 g of this mixture with water, a gas was released that completely reacted with 960 g of a 5% solution of copper sulfate. In your answer, write down the reaction equations that indicated in the problem statement, and provide all the necessary calculations (indicate the units of measurement of the required physical quantities).

We get the first (1) point for writing the reactions that occur in the problem. Obtaining this particular point depends on knowledge of chemistry, the remaining three (3) points can only be obtained through calculations, therefore, if a student has problems with mathematics, he must receive at least one (1) point for completing task No. 33:

Al 2 S 3 + 6H 2 Oà 2Al(OH) 3 + 3H 2 S

CuSO 4 + H 2 Sà CuS + H2SO4

Since further actions are purely mathematical, we will not go into detail here. You can watch a selection of the analysis on our YouTube channel (link to the video analysis of task No. 33).

Formulas that will be required to solve this task:

Chemistry assignment 34 2018

Calculation task, which differs from task No. 33 in the following:

- - If in task No. 33 we know between which substances the interaction occurs, then in task No. 34 we must find what reacted;
  - In task No. 34 organic compounds are given, while in task No. 33 inorganic processes are most often given.

In fact, task No. 34 is the reverse of task No. 33, which means the logic of the task is reverse. For task No. 34 you can get four (4) points, and, as in task No. 33, only one of them (in 90% of cases) is obtained for knowledge of chemistry, the remaining 3 (less often 2) points are obtained for mathematical calculations . To successfully complete task No. 34 you must:

Know the general formulas of all main classes of organic compounds;

Know the basic reactions of organic compounds;

Be able to write an equation in general form.

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Features of solving and evaluating tasks 30-35 of the Unified State Exam in chemistry

Prepared by: Arnautova Natalya Zakharovna,

teacher of chemistry and biology

MBOU "Secondary school No. 4, Shebekino, Belgorod region"

2017

Methodology for assessing tasks with a detailed answer (main approaches to determining criteria and rating scales for completing tasks)

The basis of the methodology for assessing tasks with a detailed answer is a number of general provisions. The most important among them are the following:

Testing and evaluation of tasks with a detailed answer is carried out only through an independent examination based on the method of element-by-element analysis of the examinees' answers.

The use of the element-by-element analysis method makes it necessary to ensure that the wording of the task conditions clearly corresponds to the content elements being checked. The list of content elements tested by any assignment is consistent with the standard requirements for the level of preparation of high school graduates.

The criterion for assessing the completion of a task using the method of element-by-element analysis is to establish the presence in the answers of the examinees of the answer elements given
in the response model. However, another answer model proposed by the examinee may be accepted if it does not distort the essence of the chemical component of the task conditions.

The rating scale for task performance is established depending on the number of content elements included in the response model, and taking into account factors such as:

Level of complexity of the content being tested;

A specific sequence of actions that should be carried out when completing a task;

Unambiguous interpretation of the task conditions and possible options for wording the answer;

Compliance of the assignment conditions with the proposed assessment criteria for individual content elements;

Approximately the same level of difficulty for each of the content elements tested by the task.

When developing assessment criteria, the features of the content elements of all five long-answer tasks included in the examination paper are taken into account. It is also taken into account that the records of the examinees’ answers can be either very general, streamlined and not specific, or too brief
and insufficiently reasoned. Close attention is paid to highlighting the elements of the answer that are worth one point. This takes into account the inevitability of a gradual increase in the difficulty of obtaining each subsequent point
for a correctly formulated element of content.

When drawing up a scale for grading calculation problems (33 and 34), the possibility of different ways of solving them is taken into account, and therefore, the presence in the examinee’s answer of the main stages and results of completing the tasks indicated
in the evaluation criteria. Let us illustrate the methodology for assessing tasks with a detailed answer using specific examples.

2017-2018 academic year

Tasks

Maximum score

Job Level

Task 30

2016-2017

Tasks 30 are aimed at testing the ability to determine the degree of oxidation of chemical elements, determine the oxidizing agent and the reducing agent, predict the products of redox reactions, establish the formulas of substances missed in the reaction scheme, draw up an electronic balance, and on its basis assign coefficients in reaction equations.

The scale for assessing the performance of such tasks includes the following elements:

 an electronic balance has been compiled – 1 point;

 the oxidizing agent and the reducing agent are indicated – 1 point.

 formulas of missing substances are determined and coefficients are assigned
in the equation of redox reaction – 1 point.

Example task:

Using the electron balance method, create an equation for the reaction

Na 2 SO 3 + … + KOH K 2 MnO 4 + … + H 2 O

Identify the oxidizing agent and the reducing agent.

Points

Possible answer

Mn +7 + ē → Mn +6

S +4 – 2ē → S +6

Sulfur in the +4 oxidation state (or sodium sulfite due to sulfur in the +4 oxidation state) is a reducing agent.

Manganese in oxidation state +7 (or potassium permanganate due to manganese
in the oxidation state +7) – oxidizing agent.

Na 2 SO 3 + 2KMnO 4 + 2KOH = Na 2 SO 4 + 2K 2 MnO 4 + H 2 O

The answer is correct and complete:

the degree of oxidation of elements that are, respectively, an oxidizing agent and a reducing agent in the reaction is determined;

oxidation and reduction processes were recorded, and an electronic (electron-ion) balance was compiled on their basis;

the missing substances in the reaction equation are determined, all the coefficients are placed

Maximum score

When assessing the examinee’s answer, it is necessary to take into account that there are no uniform requirements for the formatting of the answer to this task. As a result, the compilation of both electronic and electron-ion balances is accepted as the correct answer, and the indication of the oxidizing agent and reducing agent can be done in any clearly understandable way. However, if the answer contains elements of the answer that are mutually exclusive in meaning, then they cannot be considered correct.

2018 format tasks

1. Task 30 (2 points)

To complete the task, use the following list of substances: potassium permanganate, hydrogen chloride, sodium chloride, sodium carbonate, potassium chloride. It is permissible to use aqueous solutions of substances.

From the proposed list of substances, select substances between which an oxidation-reduction reaction is possible, and write down the equation for this reaction. Make an electronic balance, indicate the oxidizing agent and reducing agent.

Explanation.

Let's write the reaction equation:

Let's create an electronic balance:

Chlorine in oxidation state −1 is a reducing agent. Manganese in the oxidation state +7 is an oxidizing agent.TOTAL 2 points

substances are selected, the equation of the redox reaction is written, and all the coefficients are set.

oxidation and reduction processes were recorded, and an electronic (electron-ion) balance was compiled on their basis; which are respectively the oxidizing agent and the reducing agent in the reaction;

There was an error in only one of the response elements listed above

There were errors in two of the above response elements

All elements of the answer are written incorrectly

Maximum score

2018 format tasks

1. Task 31 (2 points)

To complete the task, use the following list of substances: potassium permanganate, potassium bicarbonate, sodium sulfite, barium sulfate, potassium hydroxide. It is permissible to use aqueous solutions of substances.

Explanation.

Possible answer:

2. Task 31

To complete the task, use the following list of substances: hydrogen chloride, silver(I) nitrate, potassium permanganate, water, nitric acid. It is permissible to use aqueous solutions of substances.

From the proposed list of substances, select substances between which an ion exchange reaction is possible. Write down the molecular, complete and abbreviated ionic equations for this reaction.

Explanation.

Possible answer:

Task 32. 2018 format tasks

In the condition of task 32, testing knowledge of the genetic relationship of various classes of inorganic substances, a description of a specific chemical experiment is proposed, the progress of which the examinees will have to illustrate using the equations of the corresponding chemical reactions. The grading scale for the task remains, as in 2016, equal to 4 points: each correctly written reaction equation is scored 1 point.

Example task:

Write equations for the four reactions described.

Contents of the correct answer and assessment instructions(other wording of the answer is allowed that does not distort its meaning)

Points

Possible answer

Four equations for the described reactions are written:

1) 2Fe + 6H 2 SO 4
Fe 2 (SO 4 ) 3 + 3SO 2 + 6H 2 O

2) Fe 2 (SO 4 ) 3 + 6NaOH = 2Fe(OH) 3 + 3Na 2 SO 4

3) 2Fe(OH) 3
Fe 2 O 3 + 3H 2 O

4) Fe 2 O 3 + Fe = 3FeO

All reaction equations are written incorrectly

Maximum score

It should be noted that the absence of coefficients (at least one) before the formulas of substances in reaction equations is considered an error. No points are awarded for such an equation.

Task 33. 2018 format tasks

Tasks 33 test the assimilation of knowledge about the relationship of organic substances and provide for checking five elements of content: the correctness of writing five reaction equations corresponding to the diagram - a “chain” of transformations. When writing reaction equations, examinees must use the structural formulas of organic substances. The presence of each checked content element in the answer is scored 1 point. The maximum number of points for completing such tasks is 5.

Example task:

Write the reaction equations that can be used to carry out the following transformations:

When writing reaction equations, use the structural formulas of organic substances.

Contents of the correct answer and assessment instructions
Other wording of the answer is allowed that does not distort its meaning)

Points

Possible answer

Five reaction equations have been written corresponding to the transformation scheme:

Five reaction equations written correctly

Four reaction equations written correctly

Three reaction equations written correctly

Two reaction equations written correctly

One reaction equation written correctly

All elements of the answer are written incorrectly

Maximum score

Let us note that in the examinee’s answer it is permissible to use structural formulas of different types (expanded, contracted, skeletal), which unambiguously reflect the bond order of atoms and the relative arrangement of substituents and functional groups
in a molecule of organic matter.

Task 34. 2018 format tasks

Tasks 34 are calculation problems. Their implementation requires knowledge of the chemical properties of substances and involves the implementation of a certain set of actions to ensure that the correct answer is obtained. Among such actions we name the following:

– drawing up equations of chemical reactions (according to the data of the problem conditions) necessary to perform stoichiometric calculations;

– performing calculations necessary to find answers to questions
in the problem statement there are questions;

– formulating a logically substantiated answer to all questions posed in the task conditions (for example, establishing a molecular formula).

However, it should be borne in mind that not all of the named actions must necessarily be present when solving any calculation problem, and in some cases some of them can be used more than once.

The maximum score for completing the task is 4 points. When checking, you should first of all pay attention to the logical validity of the actions performed, since some tasks can be solved in several ways. At the same time, in order to objectively evaluate the proposed method for solving the problem, it is necessary to check the correctness of the intermediate results that were used to obtain the answer.

Example task:

Determine the mass fractions (in%) of iron(II) sulfate and aluminum sulfide
in a mixture if, when 25 g of this mixture is treated with water, a gas is released, which completely reacts with 960 g of a 5% solution of copper sulfate.

In your answer, write down the reaction equations that are indicated in the problem statement,
and provide all the necessary calculations (indicate the units of measurement of the required physical quantities).

Points

Possible answer

The reaction equations were compiled:

The amount of hydrogen sulfide is calculated:

The amount of substance and mass of aluminum sulfide and iron(II) sulfate are calculated:

The mass fractions of iron(II) sulfate and aluminum sulfide in the initial mixture were determined:

ω(FeSO 4 ) = 10 / 25 = 0.4, or 40%

ω(Al 2 S 3 ) = 15 / 25 = 0.6, or 6 0%

The answer is correct and complete:

the answer correctly contains the reaction equations corresponding to the conditions of the task;

calculations have been carried out correctly using the necessary physical quantities specified in the task conditions;

a logically substantiated relationship between the physical quantities on the basis of which calculations are carried out is demonstrated;

in accordance with the task conditions, the required physical quantity is determined

There was an error in only one of the response elements listed above

All elements of the answer are written incorrectly

Maximum score

When checking the answer, the examinee must take into account the fact that if the answer contains an error in calculations in one of the three elements (second, third or fourth), which led to an incorrect answer, the mark for completing the task is reduced by only 1 point.

Task 35. 2018 format tasks

Tasks 35 involve determining the molecular formula of a substance. Completing this task includes the following sequential operations: carrying out the calculations necessary to establish the molecular formula of an organic substance, writing the molecular formula of an organic substance, drawing up a structural formula of a substance that uniquely reflects the order of bonds of atoms in its molecule, writing a reaction equation that meets the conditions of the task.

The grading scale for task 35 in part 2 of the examination paper will be 3 points.

Tasks 35 use a combination of tested content elements - calculations, on the basis of which they come to determine the molecular formula of a substance, compile a general formula of a substance, and then determine the molecular and structural formula of a substance on its basis.

All these actions can be performed in different sequences. In other words, the examinee can come to the answer in any logical way available to him. Therefore, when assessing a task, the main attention is paid to the correctness of the chosen method for determining the molecular formula of a substance.

Example task:

When a sample of some organic compound weighing 14.8 g is burned, 35.2 g of carbon dioxide and 18.0 g of water are obtained.

It is known that the relative vapor density of this substance with respect to hydrogen is 37. During the study of the chemical properties of this substance, it was established that when this substance interacts with copper(II) oxide, a ketone is formed.

Based on the data of the task conditions:

1) make the calculations necessary to establish the molecular formula of an organic substance (indicate the units of measurement of the required physical quantities);

write down the molecular formula of the original organic substance;

2) draw up a structural formula of this substance, which unambiguously reflects the order of bonds of atoms in its molecule;

3) write the equation for the reaction of this substance with copper(II) oxide using the structural formula of the substance.

Contents of the correct answer and assessment instructions

(other wording of the answer is allowed that does not distort its meaning)

Points

Possible answer

The amount of combustion product substance found:

The general formula of the substance is C x H y O z

n(CO2) = 35.2 / 44 = 0.8 mol; n (C) = 0.8 mol

n(H 2 O) = 18.0 / 18 = 1.0 mol; n(H) = 1.0 ∙ 2 = 2.0 mol

m(O) = 14.8 – 0.8 ∙ 12 – 2 = 3.2 g; n(O) = 3.2 ⁄ 16 = 0.2 mol

The molecular formula of the substance was determined:

x:y:z = 0.8:2:0.2 = 4:10:1

The simplest formula is C 4 H 10 O

M simple (C 4 H 10 O) = 74 g/mol

M source (C x H y O z ) = 37 ∙ 2 = 74 g/mol

Molecular formula of the starting substance – C 4 H 10 O

The structural formula of the substance has been compiled:

The equation for the reaction of a substance with copper(II) oxide is written:

The answer is correct and complete:

the calculations necessary to establish the molecular formula of a substance have been correctly performed; the molecular formula of the substance is written down;

the structural formula of an organic substance is written down, which reflects the bond order and relative arrangement of substituents and functional groups in the molecule in accordance with the assignment conditions;

the equation of the reaction, which is indicated in the task conditions, is written using the structural formula of an organic substance

There was an error in only one of the response elements listed above

There were errors in two of the above response elements

There were errors in three of the above response elements

All elements of the answer are written incorrectly

Maximum score

TOTAL part 2

2+2+ 4+5+4 +3=20 points

Bibliography

1. Methodological materials for chairmen and members of subject commissions of the constituent entities of the Russian Federation on checking the completion of tasks with a detailed answer of the 2017 Unified State Examination papers. Article “Methodological recommendations for assessing the completion of Unified State Examination tasks with a detailed question.” Moscow, 2017.

2. FIPI project of control and measuring materials for the Unified State Exam 2018.

3. Demo versions, specifications, codifiers of the Unified State Exam 2018. FIPI website.

4. Information about planned changes to the 2018 CMM. FIPI website.

5.Site “I will solve the Unified State Exam”: chemistry, for an expert.