Preparation for the exam in chemistry, calculation tasks for the derivation of the formula of a substance. The amount of substance of the combustion products was found

The task of deriving the molecular formula of a substance

For a job well done, you'll get 3 points. Approximately 10-15 minutes.

To complete task 35 in chemistry, you must:

1. know three basic formulas:
   • mass fraction w\u003d m (in-va) / m (r-va) * 100%
   • mole fraction by mass M=m/v
   • mole fraction by volume Vm=V/v
2. correctly write the structural and molecular formulas of substances.
3. use

Tasks for training

Organic matter A contains 11.97% nitrogen, 9.40% hydrogen and 27.35% oxygen by mass and is formed by the interaction organic matter B with propanol-2. It is known that substance B is of natural origin and is able to interact with both acids and alkalis.
Based on these conditions of the assignment:
1) carry out the necessary calculations (indicate the units of measurement of the desired physical quantities) and establish the molecular formula of the original organic matter;
2) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule;
3) write the reaction equation for obtaining substance A from substance B and propanol-2 (use the structural formulas of organic substances).


Solution

1. Burned 18 g of organic matter, while 17.92 liters (n.o.) were released carbon dioxide, 25.2 g of water and 4.48 liters of nitrogen. The density of the substance with respect to hydrogen is 22.5. During the study chemical properties substances found that when it interacts with nitrous acid, nitrogen is released.
   
   

   
   Solution

2. The hydrocarbon contains 85.71% carbon by mass. The density of matter in air is 1.931. In the course of studying the chemical properties of the substance, it was found that when it interacts with hydrogen chloride, only one monochlorine derivative is formed.
   1) carry out the necessary calculations (indicate the units of measurement of the initial values) and establish the molecular formula of the initial organic matter;
   2) make a structural formula of this substance, unambiguously reflecting the order of bonding of atoms in its molecule;
   3) write the reaction equation for the interaction of this substance with hydrogen chloride, when writing the reaction equation, use the structural formulas of organic substances.

   
   Solution

3. With the complete combustion of organic matter that does not contain oxygen, 0.448 l (n.o.) of carbon dioxide, 0.18 g of water and 0.4 g of hydrogen fluoride were released. It is known that the substance is formed as a result of the halogenation reaction of an alkene.
   1) carry out the necessary calculations (indicate the units of measurement of the initial values) and establish the molecular formula of the initial organic matter;
   2) make a structural formula of this substance, unambiguously reflecting the order of bonding of atoms in its molecule;
   3) write the reaction equation for the interaction of this substance with nitrous acid; when writing the reaction equation, use the structural formulas of organic substances.

   
   Solution

4. The organic matter reacted with metallic sodium. As a result of the reaction, a substance weighing 10.25 g was obtained and a gas with a volume of 1.4 liters was released. (well.). During the study of the original substance, it was found that it can be obtained as a result of the hydration of the alkene,
   1) carry out the necessary calculations (indicate the units of measurement of the initial values) and establish the molecular formula of the initial organic matter;
   2) make a structural formula of this substance, unambiguously reflecting the order of bonding of atoms in its molecule;
   3) write the reaction equation for the interaction of this substance with metallic sodium; when writing the reaction equation, use the structural formulas of organic substances.

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

Majority Decision Algorithm tasks to find the molecular formula includes the following steps:

— writing reaction equations in general view;

- 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 condition of the problem;

- finding the molar mass of the substance M = m / n, the formula of which must be established;

- finding the number of carbon atoms in a molecule and compiling the molecular formula of a substance.

Examples of solving problem 35 of the Unified State Examination in chemistry to find the molecular formula of organic matter by 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 is capable of being oxidized by an acidified solution of potassium permanganate to a carboxylic acid. Based on these data:
1) install the simplest formula starting material,
2) make its structural formula,
3) give the reaction equation for its interaction with hydrogen.

Solution: the general formula of organic matter is CxHyOz.

Let's translate 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) \u003d 13.44 / 22.4 \u003d 0.6 mol, => the original substance contained n (C) \u003d 0.6 mol,

n (H 2 O) \u003d 10.8 / 18 \u003d 0.6 mol, => the original substance contained twice as much n (H) \u003d 1.2 mol,

This means that the desired compound contains oxygen in the amount:

n(O)= 3.2/16 = 0.2 mol

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

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, find the molar mass organic compound according to the formula:

M (CxHyOz) = Dair (CxHyOz) * M (air)

M ist (CxHyOz) \u003d 29 * 2 \u003d 58 g / mol

Let's check if the true molar mass molar mass of the simplest formula:

M (C 3 H 6 O) \u003d 12 * 3 + 6 + 16 \u003d 58 g / mol - corresponds, \u003d\u003e the true formula coincides with the simplest.

Molecular formula: C 3 H 6 O

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

2) In the interaction of 18.5 g of saturated monobasic carboxylic acid with an excess of sodium bicarbonate solution, 5.6 l (n.o.) of gas were released. Determine the molecular formula of the acid.

3) Some limiting carboxylic monobasic acid with a mass of 6 g requires the same mass of alcohol for complete esterification. This gives 10.2 g ester. Set the molecular formula of the acid.

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

5) During the combustion of organic matter with a mass of 3.9 g, carbon monoxide (IV) with a mass of 13.2 g and water with a mass of 2.7 g were formed. Derive the formula of the substance, knowing that the hydrogen vapor density of this substance is 39.

6) During the combustion of organic matter with a mass of 15 g, carbon monoxide (IV) with a volume of 16.8 l and water with a mass of 18 g were formed. Derive the formula of the substance, knowing that the vapor density of this substance in terms of hydrogen fluoride is 3.

7) During the combustion of 0.45 g of gaseous organic matter, 0.448 l (n.o.) of carbon dioxide, 0.63 g of water and 0.112 l (n.o.) of nitrogen were released. Source Density gaseous substance nitrogen 1.607. Find the molecular formula of this substance.

8) The combustion of oxygen-free organic matter produced 4.48 l (N.O.) of carbon dioxide, 3.6 g of water and 3.65 g of hydrogen chloride. Determine the molecular formula of the burned compound.

9) During the combustion of organic matter weighing 9.2 g, carbon monoxide (IV) was formed with a volume of 6.72 l (n.o.) and water with a mass of 7.2 g. Set the molecular formula of the substance.

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

USE in chemistry

34 task

Bratyakova S.B.

Bratyakova S.B.

For me, a chemistry problem looks something like this: Two camels flew - one red, the other to the left. How much does 1kg of asphalt weigh if the hedgehog is 12 years old?

Bratyakova S.B.

1 . Organic substance A contains 13.58% nitrogen, 8.80% hydrogen and 31.03% oxygen by mass and is formed by the interaction of organic substance B with ethanol in a molar ratio of 1: 1. It is known that substance B is of natural origin and is able to interact with both acids and alkalis.

Bratyakova S.B.

4) write the equation for the reaction of obtaining substance A from substance B and ethanol.

Bratyakova S.B.

Response elements:

ω (C) \u003d 100 - 13.58-8.80-31.03 \u003d 46.59%

x:y:z:g

46,59 ⁄ 12: 8,80 ⁄ 1: 13,58 ⁄ 14: 31,03 ⁄ 16

3,88: 8,80: 0,97: 1,94

FROM 4 H 9 NO 2

Bratyakova S.B.

4) An equation was drawn up for the reaction of obtaining substance A from substance B and ethanol:

Bratyakova S.B.

2. When 4.12 g of organic matter is burned, 3.584 liters of carbon dioxide (n.o.), 448 ml of nitrogen (n.o.) and 3.24 g of water are obtained. When heated with hydrochloric acid, this substance undergoes hydrolysis, the products of which are a compound of composition C 2 H 6 NO 2 Cl and monohydric alcohol.

Bratyakova S.B.

Based on these conditions of the assignment:

1) make the calculations necessary to establish the molecular formula of an organic substance;

2) write down the molecular formula of the original organic matter;

3) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule;

4) write the reaction equation for the hydrolysis of the starting material in the presence of of hydrochloric acid.

Bratyakova S.B.

Response elements:

The general formula of the substance is CxHyNzOg

n(CO 2 ) = 3.584 / 22.4 = 0.16 mol; n(C) = 0.16 mol

n(H 2 O) \u003d 3.24 / 18 \u003d 0.18 mol; n(H) \u003d 0.18 ∙ 2 \u003d 0.36 mol

n(N 2 ) = 0.448 / 22.4 = 0.02 mol; n(N) \u003d 0.02 ∙ 2 \u003d 0.04 mol

2) The mass and amount of substance of oxygen atoms are established,

and defined molecular formula substances:

m(C + H + N) = 0.16 ∙ 12 + 0.36 ∙ 1 + 0.04 ∙ 14 = 2.84 g

m (O) \u003d 4.12 - 2.84 \u003d 1.28 g

n(O) = 1.28 / 16 = 0.08 mol

Bratyakova S.B.

n(C) : n(H) : n(N) : n(O)

0,16: 0,36: 0,04: 0,08 4: 9: 1: 2

Molecular formula - C 4 H 9 NO 2

3) The structural formula of organic matter is given:

Bratyakova S.B.

3. Organic substance A contains 11.97% nitrogen, 9.40% hydrogen and 27.35% oxygen by mass and is formed by the interaction of organic substance B with propanol-2 in a molar ratio of 1: 1. It is known that substance B is of natural origin and able to interact with both acids and alkalis.

Bratyakova S.B.

Based on these conditions of the assignment:

1) make the calculations necessary to establish the molecular formula of organic substance A;

2) write down the molecular formula of substance A;

3) make a structural formula of substance A, which unambiguously reflects the order of bonding of atoms in its molecule;

4) write the reaction equation for obtaining substance A from substance B and propanol-2.

Bratyakova S.B.

Response elements:

The general formula of the substance is CxHyNzOg

1) The ratio of carbon, hydrogen, nitrogen and oxygen atoms in the compound is found:

ω(C) \u003d 100 - 11.97 - 9.40 - 27.35 \u003d 51.31%

x:y:z:g

51,31 ⁄ 12: 9,4 ⁄ 1: 11,97 ⁄ 14: 27,35 ⁄ 16

4,27: 9,4: 0,855: 1,71

5:11:1:2

2) The molecular formula of substance A has been determined.

FROM 5 H 11 NO 2

3) Compiled the structural formula of substance A:

Bratyakova S.B.

4) An equation was drawn up for the reaction of obtaining substance A from substance B and propanol-2:

Bratyakova S.B.

4. When burning a dipeptide sample natural origin with a mass of 6.4 g, 5.376 liters of carbon dioxide (n.s.), 4.32 g of water and 896 ml of nitrogen (n.s.) were obtained. Upon hydrolysis of this dipeptide in the presence of hydrochloric acid, one salt is formed.

Bratyakova S.B.

Based on these conditions of the assignment:

1) make the calculations necessary to establish the molecular formula of the dipeptide;

2) write down the molecular formula of the dipeptide;

3) make a structural formula of this dipeptide, which unambiguously reflects the order of bonding of atoms in its molecule;

4) write the reaction equation for the hydrolysis of the dipeptide in the presence of hydrochloric acid.

Bratyakova S.B.

Response elements:

1) The amount of substance of the combustion products was found:

n(CO 2 ) = 5.376 / 22.4 = 0.24 mol; n(C) = 0.24 mol

n(H 2 O)= 4.32/18= 0.24 mol; n(H)= 0.24∙2= 0.48 mol

n(N 2 ) = 0.896 / 22.4 = 0.04 mol; n(N) \u003d 0.04 ∙ 2 \u003d 0.08 mol

2) The mass and quantity of the substance of oxygen atoms are established, and the molecular formula of the substance is determined:

m(C + H + N) = 0.24 ∙ 12 + 0.48 ∙ 1 + 0.08 ∙ 14 = 4.48 g

m (O) \u003d 6.4 - 4.48 \u003d 1.92 g; n(O) = 1.92/16 = 0.12 mol

n(C) : n(H) : n(N) : n(O) = 0.24: 0.48: 0.08: 0.12 6: 12: 2: 3

Molecular formula - C 6 H 12 N 2 O 3

Bratyakova S.B.

3) The structural formula of the dipeptide is given:

4) An equation for the hydrolysis reaction in the presence of hydrochloric acid was compiled:

Bratyakova S.B.