Finding the molecular formula of the hydrocarbon. Solving problems to determine the formulas of organic substances (preparation for the exam in chemistry)

With tasks for withdrawal chemical formula Substances students encounter during the chemistry program from grades 8 to 11. In addition, this type of problem is quite often found in olympiad tasks, control and measuring materials of the exam (parts B and C). The range of complexity of these tasks is wide enough. Experience shows that schoolchildren often have difficulties already at the first stages of the solution when deriving the molar mass of a substance.

In this development, problems are proposed for finding the formula of a substance, based on different parameters in conditions. The presented tasks contain different ways finding the molar mass of a substance. The tasks are designed in such a way that students can master best practices and different options solutions. The most common decision techniques are clearly demonstrated. For students, solved problems are offered on the principle of increasing complexity and problems for independent solution.

An example of solving the problem for the application of the Mendeleev - Cliperon equation is given.

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

The relative density of the hydrocarbon for hydrogen, having the composition: w (C) = 85.7%; w (Н) = 14.3%, equal to 21. Print the molecular formula of the hydrocarbon.

Determine the molecular formula of an alkane if it is known that its vapor is 2.5 times heavier than argon.

The mass fraction of carbon in the compound is 39.97%, hydrogen 6, 73%, oxygen 53.30%. Weight 300 ml. (n.o.) of this compound is 2.41 g. Derive the molecular formula of this substance.

In some problems, the elemental composition of the required substance is not obvious from the text of the condition. Most often this applies to combustion reactions organic matter... Uncertainty in the composition is usually associated with the possibility of the presence of oxygen in the burnt substance. In the first step of solving such problems, it is necessary to reveal the elemental composition of the desired substance by calculation.

Task 2.11.
As a result of combustion of 1.74 g of an organic compound, 5.58 g of a mixture of CO 2 and H 2 O was obtained. The amounts of CO 2 and H 2 O in this mixture turned out to be equal. Determine the molecular formula of an organic compound if the relative density of its vapor for oxygen is 1.8125.
Given:
the mass of the organic compound: m org v.wa = 1.74 g;
the total mass of the products of the district: m (CO 2) + m (H 2 O) = 5.58 g;
the ratio of the quantities of substances of the products of the district: n(CO 2) = n(H 2 O);
the relative density of the vapor of the starting material for oxygen: D (O 2) = 1.8125.
Find: the molecular formula of the burnt compound.
Solution:
Step 1. The class of the burned organic compound is not specified, therefore, the elemental composition can be judged only by the reaction products. Carbon and hydrogen were uniquely part of the burned substance, since these elements are present in the combustion products, and only oxygen from the air took part in the reaction. Moreover, all of the carbon and all of the hydrogen were completely transferred from the initial substance to CO 2 and H 2 O. Perhaps, the composition of the desired compound also included oxygen.
The situation with the presence or absence of oxygen can be clarified according to the data from the problem statement. We know the mass of the burned organic compound and the quantitative data,
related to products. Obviously, if the total mass of carbon from CO 2 and hydrogen from H 2 O turns out to be equal to the mass of the original organic matter, then there was no oxygen in its composition. Otherwise, if

m [(C) (in CO 2)] + m [(H) (in H 2 O)]> m org. in-va

oxygen was part of the original substance, and its mass is determined by the difference:

m org. in-va - m (C) (in CO 2) - m (H) (in H 2 O) = m (O) (in out. in-ve).

Let us determine the mass of carbon and hydrogen in the reaction products and compare it with the mass of the initial substance.
1. The condition contains information on the total mass of the reaction products, and therefore, first of all, we need to identify the masses of each of the products separately. To do this, we denote the amount of substance formed carbon dioxide value " a". Then, according to the condition:

n (CO 2) = n (H 2 O) = a mol.

Using the value "a" as known, we find the mass of СО 2 and Н 2 О:

m (CO 2) = M (CO 2). n(CO 2) = (44. a) d,
m (H 2 O) = M (H 2 O). n(H 2 O) = (18. A) g.

We summarize the obtained expressions and equate to the value of the total mass of the reaction products from the condition:

(44 . a) + (18 . a) = 5,58.

We got a mathematical equation with one unknown. Solving it, we find the value of the unknown quantity: a = 0,09.

With this value, we denoted the amount of substance of each of the products:

n(CO 2) = n(H 2 O) = 0.09 mol.

2. Let's find the mass of carbon in СО2 according to the algorithm:

n (CO 2) ---> n (C) (in CO 2) ---> m (C) (in CO 2)
n (C) (in CO 2) = n (CO 2) = 0.09 mol (according to the indices in the formula).
m (C) (in CO 2) = n (C) (in CO 2). M (C) = 0.09. 12 = 1.08 g = m (C) (in original in-ve)

3. Let's find the mass of hydrogen in the formed water according to the algorithm:

n (H 2 O) ---> n (H) (in H 2 O) ---> m (H) (in H 2 O)
n (H) (in H 2 O)> n (H 2 O) 2 times (by indices in the formula)
n (H) (in H 2 O) = 2. n (H 2 O) = 2. 0.09 = 0.18 mol
m (H) (in H2O) = n (H) (in H2O). M (H) = 0.18. 1 \ u003d 0.18 g \ u003d m (H) (in original in-ve)

4. Compare the total mass of carbon and hydrogen with the mass of the starting material:

m (C) (in CO 2) + m (H) (in H 2 O) = 1.08 + 0.18 = 1.26 g;
m org. in-va = 1.74 g.
m (C) (in CO 2) + m (H) (in H 2 O)> m org. c.v.-a,

therefore, oxygen is part of the starting material.

m (O) (in out. in-ve) = m org. in-va - m (C) (in CO 2) - m (H) (in H 2 O) = 1.74 -1.26 = 0.48 g.

5. So, the original substance contains: carbon, hydrogen and oxygen.
Further actions are no different from the examples of the previously considered tasks. Let's designate the required substance as C x H y O z.

Step 2. Let's draw up a combustion reaction scheme:

C x H y O z. + О 2 ---> СО 2 + Н 2 О

Step 3. Let us determine the ratio of the amounts of substance ( n) carbon, hydrogen and oxygen in the original sample of organic matter. We have already determined the amount of carbon and hydrogen matter in the first step.
The amount of substance ( n) oxygen is found according to the data on its mass:

Step 4. We find the simplest formula:

N (C): N (H): N (O) = 0.09: 0.18: 0.03

Choose the smallest value (in this case "0.03") and divide all three numbers by it:

Got a set of smallest integers:

N (C): N (H): N (O) = 3: 6: 1

This makes it possible to write down the simplest formula: C 3 H 6 O 1

Step 5. Revealing the true formula.
According to the data on the relative vapor density of the desired substance for oxygen, we determine the true molar mass:

M truths. = D (O 2). M (O 2) = 1.8125. 32 = 58 g / mol.

Let's determine the value of the molar mass for the simplest formula:

M is simple. = 3 .12 + 6. 1 +1. 16 = 58 g / mol.

M is simple. = M true. therefore, the simplest formula is true.

C 3 H 6 O - the molecular formula of the burnt substance.

Answer: C 3 H 6 O.

Chemistry, part C. Problem C5. Determination of formulas of organic substances.

Types of tasks in task C5.

Determination of the formula of a substance by mass fractions chemical elements or according to the general formula of the substance;

Determination of the formula of a substance by combustion products;

Determination of the formula of a substance by its chemical properties.

Necessary theoretical information.

Mass fraction of an element in a substance. The mass fraction of an element is its content in a substance as a percentage by mass. For example, a substance with the composition C 2 H 4 contains 2 carbon atoms and 4 hydrogen atoms. If we take 1 molecule of such a substance, then its molecular weight will be: Мr (С 2 Н 4) = 2 12 + 4 1 = 28 amu. and it contains 2 12 amu. carbon. To find the mass fraction of carbon in this substance, it is necessary to divide its mass by the mass of the whole substance: ω (C) = 12 2/28 = 0.857 or 85.7%. If a substance has the general formula C x H y O z, then the mass fractions of each of their atoms are also equal to the ratio of their mass to the mass of the whole substance. The mass x of the C atoms is - 12x, the mass of the H atoms is y, the mass z of the oxygen atoms is 16z. Then ω (C) = 12x / (12x + y + 16z) If we write this formula in general view, you get the following expression:

Molecular and simplest formula of a substance. Examples.

Relative density of gas X for gas Y - D poU (NS). The relative density D is a value that shows how many times gas X is heavier than gas Y. It is calculated as the ratio of the molar masses of gases X and Y: D in Y (X) = M (X) / M (Y) Often used for calculations relative densities of gases by hydrogen and by air... The relative density of gas X in terms of hydrogen: D in H2 = M (gas X) / M (H2) = M (gas X) / 2 Air is a mixture of gases, so only the average molar mass can be calculated for it. Its value is taken as 29 g / mol (based on the approximate average composition). Therefore: D by air. = M (gas X) / 29

The absolute density of the gas under normal conditions. The absolute density of a gas is the mass of 1 liter of gas under normal conditions. It is usually measured in g / l for gases. ρ = m (gas) / V (gas) If we take 1 mol of gas, then: ρ = M / V m, and the molar mass of the gas can be found by multiplying the density by the molar volume.

General formulas of substances of different classes.

Determination of formulas of substances by mass fractions of atoms included in its composition.

The solution to such problems consists of two parts:

first, the molar ratio of atoms in a substance is found - it corresponds to its simplest formula. For example, for a substance of composition A x B y, the ratio of the amounts of substances A and B corresponds to the ratio of the number of their atoms in the molecule: x: y = n (A): n (B);

then using molar mass of matter determine its true formula.

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 of 3.93.

Solution of example 1.

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

Let's find the amount of matter of each atom: ν (C) = m / M = 84.21 / 12 = 7.0175 mol, ν (H) = 15.79 / 1 = 15.79 mol.

We determine the molar ratio of atoms C and H: C: H = 7.0175: 15.79 (we will reduce both numbers to a smaller one) = 1: 2.25 (multiply by 4) = 4: 9. Thus, the simplest formula is C 4 H 9.

Based on the 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. Hence, the true formula is C 8 H 18.

There is a much simpler method for solving such a problem, but, unfortunately, they will not give a full score for it... But it is suitable for checking the true formula, i.e. with it you can check your solution. Method 2: We find the true molar mass (114 g / mol), and then we find the masses of carbon and hydrogen atoms in this substance by their mass fractions. m (C) = 114 0.8421 = 96; those. number of C atoms 96/12 = 8 m (H) = 114 0.1579 = 18; that is, the number of atoms H 18/1 = 18. The formula of the substance is C 8 H 18.

Answer: C 8 H 18.

Example 2. Determine the formula for alkyne with a density of 2.41 g / l under normal conditions.

Solution of example 2. General formula of alkyne С n H 2n − 2 How, having the density of gaseous alkyne, to find its molar mass? The density ρ is the mass of 1 liter of gas under normal conditions. Since 1 mole of a substance occupies a volume of 22.4 liters, it is necessary to find out how much 22.4 liters of such a gas weigh: M = (density ρ) (molar volume V m) = 2.41 g / l 22.4 l / mol = 54 g / mol. Next, let's make an equation connecting the molar mass and n: 14 n - 2 = 54, n = 4. So, alkyne has the formula C 4 H 6.

Answer: C 4 H 6.

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

Solution of example 3. In this problem, the number of molecules and the corresponding mass are given. Based on these data, we need to find the value of the molar mass of the substance again. 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 the aldehyde substance: ν = N / Na = 3 10 22 / 6.02 10 23 = 0.05 mol, and the molar mass: M = m / n = 4.3 / 0.05 = 86 g / mole. Next, as in the previous example, we compose an equation and find n. The general formula for the limiting aldehyde is C n H 2n O, that is, M = 14n + 16 = 86, n = 5.

Answer: C 5 H 10 O, pentanal.

Example 4. Determine the formula of dichloroalkane containing 31.86% carbon.

Solution example 4. The general formula of dichloroalkane: C n H 2n Cl 2, there are 2 chlorine atoms and n carbon atoms. Then the mass fraction of carbon is equal to: ω (C) = (number of C atoms in the molecule) (atomic weight C) / (molecular weight of dichloroalkane) 0.3186 = n 12 / (14n + 71) n = 3, the substance is dichloropropane.

Answer: C 3 H 6 Cl 2, dichloropropane.

Determination of formulas of substances by combustion products.

In combustion problems, the amounts of substances of the elements included in the test substance are determined by the volumes and masses of combustion products - carbon dioxide, water, nitrogen and others. The rest of the solution is the same as in the first type of problems.

Example 5. 448 ml (n.a.) gaseous saturated non-cyclic hydrocarbon was burned, and the reaction products were passed through an excess of lime water, with the formation of 8 g of sediment. What hydrocarbon was taken?

Solution example 5.

The general formula of the gaseous limiting non-cyclic hydrocarbon (alkane) is C n H 2n + 2 Then the combustion reaction scheme looks like this: C n H 2n + 2 + O 2 → CO 2 + H 2 O It is easy to see that when 1 mol of alkane is burned, n mol of carbon dioxide will be released. We find the amount of alkane substance by its volume (do not forget to convert milliliters to liters!): Ν (C n H 2n + 2) = 0.488 / 22.4 = 0.02 mol.

When carbon dioxide is passed through Ca (OH) 2 lime water, a calcium carbonate precipitate is formed: CO 2 + Ca (OH) 2 = CaCO 3 + H 2 O The mass of the calcium carbonate precipitate is 8 g, the molar mass of calcium carbonate is 100 g / mol. This means that its amount of substance is ν (CaCO 3) = 8/100 = 0.08 mol. The amount of carbon dioxide is also 0.08 mol.

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

Answer: С 4 Н 10.

Example 6. The relative density of the vapors of the organic compound with respect to nitrogen is 2. When 9.8 g of this compound are burned, 15.68 liters of carbon dioxide (n.a.) and 12.6 g of water are formed. Derive the molecular formula of an organic compound.

Solution of example 6. Since the substance during 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 C x H y O z.

We can write down the combustion reaction scheme (without arranging the coefficients): С х Н у О z + О 2 → CO 2 + H 2 O All carbon from the initial substance goes into carbon dioxide, and all hydrogen goes into water.

We find the amount of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain: ν (CO 2) = V / V m = 15.68 / 22.4 = 0.7 mol. One CO 2 molecule accounts for one atom C, which means that carbon is as much a mole as CO 2.

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

One water molecule contains two atom H means the amount of hydrogen twice as much than water. ν (H) = 0.7 2 = 1.4 mol.

We check the presence of oxygen in the substance. To do this, subtract the masses of C and H. from the mass of the entire initial substance m (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 further actions are already familiar to you: the search for the simplest and true formulas. C: 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 in nitrogen (do not forget that nitrogen consists of diatomic molecules of N 2 and its molar mass 28 g / mol): M source. = D by N2 M (N2) = 2 28 = 56 g / mol. The true formula is CH 2, its molar mass is 14. 56/14 = 4. The true formula is C 4 H 8.

Answer: C 4 H 8.

Example 7. Determine the molecular formula of the substance, upon combustion of 9 g of which 17.6 g of CO 2, 12.6 g of water and nitrogen were formed. The relative density of this substance for hydrogen is 22.5. Determine the molecular formula of the substance.

Solution of example 7.

The substance contains atoms C, H and N. 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: С х Н у N z + O 2 → CO 2 + H 2 O + N 2

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

ν (CO 2) = m / M = 17.6 / 44 = 0.4 mol. ν (C) = 0.4 mol. ν (H 2 O) = m / M = 12.6 / 18 = 0.7 mol. ν (H) = 0.7 2 = 1.4 mol.

Find the mass of nitrogen in the starting material. For this, the masses C and H must be subtracted from the mass of the entire initial 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, ν (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. The true molar mass is M = D by H2 M (H 2) = 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: С 2 Н 7 N.

Example 8. The substance contains C, H, O and S. During the combustion of 11 g of it, 8.8 g of CO 2, 5.4 g of H 2 O were released, and sulfur was completely converted into barium sulfate, the mass of which turned out to be 23.3 g. Determine the formula of the substance.

Solution of example 8. The formula for a given substance can be represented as C x H y S z O k. When burned, carbon dioxide, water and sulfur dioxide are obtained, which are then converted into barium sulfate. Accordingly, all of the sulfur from the starting material is converted to barium sulfate.

We find the amounts of substances carbon dioxide, water and barium sulfate and the corresponding chemical elements from the substance under study:

ν (CO 2) = m / M = 8.8 / 44 = 0.2 mol. ν (C) = 0.2 mol. ν (H 2 O) = m / M = 5.4 / 18 = 0.3 mol. ν (H) = 0.6 mol. ν (BaSO 4) = 23.3 / 233 = 0.1 mol. ν (S) = 0.1 mol.

We calculate the estimated mass of oxygen in the starting material:

m (C) = 0.2 12 = 2.4 g m (H) = 0.6 1 = 0.6 g m (S) = 0.1 32 = 3.2 g m (O) = m substance - m (C) - m (H) - m (S) = 11 - 2.4 - 0.6 - 3.2 = 4.8 g, ν (O) = m / M = 4.8 / 16 = 0 , 3 mol

We find the molar ratio of elements in the substance: C: H: S: O = 0.2: 0.6: 0.1: 0.3 = 2: 6: 1: 3 The formula of the substance is C 2 H 6 SO 3. It should be noted that in this way we obtained only the simplest formula. However, the resulting formula is true, since when trying to double this formula (C 4 H 12 S 2 O 6), it turns out that there are 12 H atoms per 4 carbon atoms, in addition to sulfur and oxygen, and this is impossible.

Answer: C 2 H 6 SO 3.

Determination of formulas of substances by chemical properties.

Example 9. Determine the formula of alkadiene if 80 g of a 2% bromine solution can discolor it.

Solution of example 9.

The general formula for alkadienes is C n H 2n − 2. Let us write down the reaction equation for the addition of bromine to alkadiene, not forgetting that in the diene molecule two double bonds and, accordingly, 2 mol of bromine will react with 1 mol of diene: С n H 2n − 2 + 2Br 2 → С n H 2n − 2 Br 4

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

m (Br 2) = m solution ω = 80 0.02 = 1.6 g ν (Br 2) = m / M = 1.6 / 160 = 0.01 mol.

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

		С n H 2n − 2 Br 4

1. M diene = m / ν = 3.4 / 0.05 = 68 g / mol.

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

14n - 2 = 68 n = 5.

This is C 5 H 8 pentadiene.

Answer: C 5 H 8.

Example 10. With the interaction of 0.74 g of the limiting monohydric alcohol with metallic sodium, hydrogen evolved in an amount sufficient for hydrogenation of 112 ml of propene (n.a.). What is this alcohol?

Solution of example 10.

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

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

2C n H 2n + 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, by the reaction we find the amount of alcohol substance:

ν (C 3 H 6) = V / V m = 0.112 / 22.4 = 0.005 mol => ν (H 2) = 0.005 mol, ν alcohol = 0.005 2 = 0.01 mol.

Find the molar mass of alcohol and n:

M alcohol = m / ν = 0.74 / 0.01 = 74 g / mol, 14n + 18 = 74 14n = 56 n = 4.

Alcohol - butanol C 4 H 7 OH.

Answer: C 4 H 7 OH.

Example 11. Define a formula ester, during the hydrolysis of 2.64 g of which 1.38 g of alcohol and 1.8 g of monobasic carboxylic acid are released.

Solution of example 11.

The general formula of an ester consisting of an alcohol and an acid with different numbers of carbon atoms can be represented as follows: C n H 2n + 1 COOC m H 2m + 1 Accordingly, the alcohol will have the formula C m H 2m + 1 OH, and the acid C n H 2n + 1 COOH. Ester hydrolysis equation: C n H 2n + 1 COOC m H 2m + 1 + H 2 O → C m H 2m + 1 OH + C n H 2n + 1 COOH

According to the law of conservation of mass of substances, the sum of the masses of the initial 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 H2O = (mass of acid) + (mass of alcohol) - (mass of ether) = 1.38 + 1.8 - 2.64 = 0.54 g ν H2O = m / M = 0.54 / 18 = 0.03 mole

Accordingly, the amounts of acid and alcohol substances are also equal to moles. You can find their molar masses:

M acid = m / ν = 1.8 / 0.03 = 60 g / mol, M alcohol = 1.38 / 0.03 = 46 g / mol.

We get two equations, from which we find m and n:

M CnH2n + 1COOH = 14n + 46 = 60, n = 1 - acetic acid M CmH2m + 1OH = 14m + 18 = 46, m = 2 - ethanol.

Thus, the desired ester is ethyl acetate, ethyl acetate.

Answer: CH 3 COOC 2 H 5.

Example 12. Determine the formula of an amino acid if, when acting on 8.9 g of its excess sodium hydroxide, you can get 11.1 g sodium salt this acid.

Solution of example 12.

The general formula of an amino acid (if we assume that it does not contain any other functional groups, except for one amino group and one carboxyl group): NH 2 –CH (R) –COOH. It could be written in different ways, but for the convenience of writing the reaction equation, it is better to separate the functional groups in the amino acid formula separately.

You can draw up the equation for the reaction of this amino acid with sodium hydroxide: NH 2 –CH (R) –COOH + NaOH → NH 2 –CH (R) –COONa + H 2 O The amounts of the amino acid substance and its sodium salt are equal. At the same time, we cannot find the mass of any of the substances in the reaction equation. Therefore, in such problems, it is necessary to express the quantities of amino acid substances and its salts in terms of molar masses and equate them:

M (amino acids NH 2 –CH (R) –COOH) = 74 + М RM (salts NH 2 –CH (R) –COONa) = 96 + М R ν amino acids = 8.9 / (74 + М R), ν salts = 11.1 / (96 + M R) 8.9 / (74 + M R) = 11.1 / (96 + M R) M R = 15

It is easy to see that R = CH 3. This can be done mathematically if we assume that R - C n H 2n + 1. 14n + 1 = 15, n = 1. This is alanine - aminopropanoic acid.

Answer: NH 2 –CH (CH 3) –COOH.

Tasks for an independent solution.

Part 1. Determination of the formula of a substance by composition.

1–1. The density of the hydrocarbon under normal conditions is 1.964 g / l. The mass fraction of carbon in it is 81.82%. Derive the molecular formula for this hydrocarbon.

1–2. The mass fraction of carbon in the diamine is 48.65%, the mass fraction of nitrogen is 37.84%. Derive the molecular formula of diamine.

1–3. The relative vapor density of the saturated dibasic carboxylic acid in air is 4.07. Derive the molecular formula of a carboxylic acid.

1–4. 2 l alkadiene at normal conditions has a mass equal to 4.82 g. Derive the molecular formula of alkadiene.

1-5. (Unified State Exam – 2011) Establish the formula for the limiting monobasic carboxylic acid, the calcium salt of which contains 30.77% calcium.

Part 2. Determination of the formula of a substance by combustion products.

2–1. The relative density of the vapors of the organic compound in terms of sulfur dioxide is 2. When 19.2 g of this substance are burned, 52.8 g of carbon dioxide (NU) and 21.6 g of water are formed. Derive the molecular formula of an organic compound.

2–2. When organic matter with a mass of 1.78 g was burned in an excess of oxygen, 0.28 g of nitrogen, 1.344 l (NU) CO 2 and 1.26 g of water were obtained. Determine the molecular formula of the substance, knowing that the indicated sample of the substance contains 1.204 10 22 molecules.

2–3. Carbon dioxide obtained from the combustion of 3.4 g of hydrocarbon was passed through an excess of calcium hydroxide solution to obtain 25 g of precipitate. Print the simplest formula for a hydrocarbon.

2–4. During the combustion of organic matter containing C, H and chlorine, 6.72 liters (n.u.) of carbon dioxide, 5.4 g of water, 3.65 g of hydrogen chloride were released. Establish the molecular formula of the burnt substance.

2-5. (Unified State Exam – 2011) The combustion of amine released 0.448 l (standard) carbon dioxide, 0.495 g of water and 0.056 l of nitrogen. Determine the molecular formula of this amine.

Part 3. Determination of the formula of a substance by chemical properties.

3–1. Determine the formula of an alkene, if it is known that it is 5.6 g of it when water is added to form 7.4 g of alcohol.

3–2. The oxidation of 2.9 g of saturated aldehyde to acid required 9.8 g of copper (II) hydroxide. Determine the formula for the aldehyde.

3–3. A monobasic monoamino acid weighing 3 g with an excess of hydrogen bromide forms 6.24 g of salt. Determine the amino acid formula.

3–4. During the interaction of a saturated dihydric alcohol weighing 2.7 g with an excess of potassium, 0.672 liters of hydrogen were released. Determine the formula for alcohol.

3-5. (Unified State Exam – 2011) In the oxidation of the saturated monohydric alcohol with copper (II) oxide, 9.73 g of aldehyde, 8.65 g of copper and water were obtained. Determine the molecular formula of this alcohol.

Answers and comments to problems for independent solution.

1-1. C 3 H 8

1-2. C 3 H 6 (NH 2) 2

1-3. C 2 H 4 (COOH) 2

1-5. (HCOO) 2 Ca - calcium formate, formic acid salt

2-1. C 8 H 16 O

2-2. C 3 H 7 NO

2-3. С 5 Н 8 (we find the mass of hydrogen by subtracting the mass of carbon from the mass of the hydrocarbon)

2-4. C 3 H 7 Cl (do not forget that hydrogen atoms are contained not only in water, but also in HCl)

2-5. C 4 H 11 N

3-1. C 4 H 8

3-2. C 3 H 6 O

3–3. C 2 H 5 NO 2

3-4. C 4 H 8 (OH) 2

Theory for task 35 from the exam in chemistry

Finding the molecular formula of a substance

Finding the chemical formula of a substance by mass fractions of elements

The mass fraction of an element is the ratio of its mass to the total mass of the substance, of which it is included:

$ W = (m (e-ta)) / (m (in-va)) $

Mass fraction of an element ($ W $) is expressed in fractions of a unit or as a percentage.

Problem 1. The elementary composition of the substance is as follows: mass fraction of iron $ 72.41% $, mass fraction of oxygen $ 27.59% $. Get the chemical formula.

Given:

$ W (Fe) = 72.41% = 0.7241 $

$ W (O) = 27.59% = 0.2759 $

Solution:

1. For calculations, we choose the mass of the oxide $ m $ (oxide) $ = 100 $ g. Then the masses of iron and oxygen will be as follows:

$ m (Fe) = m_ (oxide) W (Fe); m (Fe) = 100 0.7241 = 72.41 $ g.

$ m (O) = m_ (oxide) W (O); m (O) = 100 0.2759 = $ 27.59 g.

2. The amounts of iron and oxygen are equal, respectively:

$ ν (Fe) = (m (Fe)) / (M (Fe)); ν (Fe) = (72.41) / (56) = 1.29. $

$ ν (O) = (m (O)) / (M (O)); ν (O) = (27.59) / (16) = 1.72. $

3. Find the ratio of the amount of iron and oxygen substances:

$ ν (Fe): ν (O) = 1.29: 1.72. $

The smaller number is taken as $ 1 (1.29 = 1) $ and we find:

$ Fe: O = 1: 1.33 $.

4. Since the formula must contain an integer number of atoms, this ratio is reduced to integers:

$ Fe: O = 1: 1.33 = 2: 2.66 = 3 3.99 = 3: 4 $.

5. Substitute the found numbers and get the oxide formula:

$ Fe: O = 3: 4 $, that is, the formula of the substance is $ Fe_3O_4 $.

Answer: $ Fe_3O_4 $.

Finding the chemical formula of a substance by mass fractions of elements, if the density or relative density of a given substance in a gaseous state is indicated

Problem 2. The mass fraction of carbon in the hydrocarbon is $ 80%. The relative density of the hydrocarbon in terms of hydrogen is $ 15 $.

Given:

Solution:

1. Let's designate the formula of substance $ C_ (x) H_ (y) $.

2. Let's find the number of moles of carbon and hydrogen atoms in $ 100 $ g of a given compound:

$ x = n (C); y = ν (H). $

$ ν (C) = (m (C)) / (M (C)) = (80) / (12) = 6.6; ν (H) = (m (H)) / (M (H)) = ( 20) / (1) = 20. $

1 way.

3. The relationship between atoms:

$ x: y = 6.6: 20 = 1: 3 $, or $ 2: 6 $.

The simplest formula for the substance is $ CH_3 $.

4. Determine the molecular weight of the hydrocarbon by the relative density of its vapors.

$ M_r $ (substances) $ = 2D (H_2) = 32D (O_2) = 29D $ (air).

$ M_x = 2D (H_2) = 2 15 = 30 $ g / mol.

5. We calculate the relative molecular weight of the hydrocarbon using the simplest formula:

$ M_r (CH_3) = A_r (C) + 3A_r (H) = 12 + 3 = 15 $.

6. The values ​​of $ M_x $ and $ M_r $ do not match, $ M_r = (1) / (2) M_x $, hence the hydrocarbon formula is $ C_2H_6 $.

Check: $ M_r (C_2H_6) = 2A_r (C) + 6A_r (H) = 2 12 + 6 1 = 30 $.

Answer: the molecular formula of the $ C_2H_6 $ hydrocarbon is ethane.

Method 2.

3. The relationship between atoms:

$ (x) / (y) = (6.6) / (20); (x) / (y) = (1) / (3.03); y = 3.03x. $

5. Molar mass can be represented as:

$ M_r (C_xH_y) = A_r (C) _x + A_r (H) _y; M_r (C_xH_y) = 12x + y $ or $ 30 = 12x + 1y $.

6. We solve the system of two equations with two unknowns:

$ \ (\ table \ y = 3.03x; \ 12x + y = 30; $ $ 12x + 3.03x = 30; x = 2; y = 6. $

Answer: the formula $ C_2H_6 $ is ethane.

Finding the chemical formula of a substance according to the data on the starting substance and on the products of its combustion (according to the chemical reaction equation)

Problem 3. Find the molecular formula of a hydrocarbon with a density of $ 1.97 $ g / l, if the combustion of $ 4.4 $ g of it in oxygen produces $ 6.72 $ l of carbon monoxide (IV) (n.a.) and $ 7.2 $ g of water.

Given:

$ m (C_xH_y) = 4.4 $ g

$ ρ (C_xH_y) = 1.97 $ g / l

$ V (CO_2) = 6.72 $ l

$ m (H_2O) = 7.2 $ g

Solution:

1. Let's write the scheme of the hydrocarbon combustion equation

$ (C_xH_y) ↖ (4.4g) + O_2 → (CO_2) ↖ (6.72L) + (H_2O) ↖ (7.2g) $

2. Calculate the molar mass $ C_xH_y · M = ρ · V_m $,

$ M = 1.97 $ g / l $ 22.4 $ l / mol $ = 44 $ g / mol.

Relative molecular weight $ M_r = 44 $.

3. Determine the amount of substance:

$ ν (C_xH_y) = (m) / (M) $ or $ ν (C_xH_y) = (4.4) / (44) = 0.1 $ mol.

4. Using the value of the molar volume, we find:

$ ν (CO_2) = (m) / (M) $ or $ ν (H_2O) = (7.2) / (18) = 0.4 $ mol.

6. Therefore: $ ν (C_xH_y): ν (CO_2): νH_2O = 0.1 $ mol $: 0.3 $ mol $: 0.4 $ mol or $ 1: 3: 4 $, which should correspond to the coefficients in the equation and allows you to set the number of carbon atoms and hydrogen:

$ C_xH_y + O_2 → 3CO + 4H_2O $.

The final form of the equation:

$ C_3H_8 + 5O_2 → 3CO_2 + 4H_2O $.

Answer: the hydrocarbon formula $ C_3H_8 $ is propane.