Lesson 1.

Topic: Number of substance. Mole

Chemistry is a science of substances.And how to measure substances? What units? In molecules, of which there are substances, but it is very difficult to do it. In grams, kilograms or milligrams, but they measure the mass. And what if you combine the mass, which is measured on the scales and the number of molecules of the substance, is it possible?

a) h-hydrogen

A N \u003d 1A.E.M.

1A.M \u003d 1.66 * 10 -24 g

Take 1 hydrogen and we calculate the number of hydrogen atoms in this mass (offer it to make students using a calculator).

N \u003d 1G / (1.66 * 10 -24) r \u003d 6.02 * 10 23

b) O-oxygen

A O \u003d 16A.E.M \u003d 16 * 1.67 * 10 -24 g

N O \u003d 16g / (16 * 1.66 * 10 -24) r \u003d 6.02 * 10 23

c) C-carbon

And c \u003d 12a.e.m \u003d 12 * 1.67 * 10 -24 g

N c \u003d 12g / (12 * 1.66 * 10 -24) r \u003d 6.02 * 10 23

We conclude: if we take such a mass of a substance that is equal to the atomic mass in magnitude, but taken in grams, it will always be (for any substance) 6.02 * 10 23 atoms of this substance.

H 2 O - Water

18G / (18 * 1.66 * 10 -24) r \u003d 6.02 * 10 23 water molecules, etc.

N A \u003d 6,02 * 10 23 - Number or Permanent Avogadro.

Mol is the amount of substance that contains 6.02 * 10 23 molecules, atoms or ions, i.e. structural units.

It happens mole molecules, moth atoms, mole ions.

n is the number of moles, (the number of moles is often denoted),
N is the number of atoms or molecules,
N a \u003d Permanent Avogadro.

Kolol \u003d 10 3 mol, mmol \u003d 10 -3 mol.

Show the portrait of Amedeo Avogadro on a multimedia installation and briefly tell about him, or entrust the student to prepare a small report on the life of the scientist.

Lesson 2.

Theme "Molar Mass of Substance"

What is the mass of 1 praying substance? (Conclusion Students can often do themselves.)

The mass of one praying substance is equal to its molecular weight, but expressed in grams. The mass of one praying substance is called the molar mass and is indicated - M.

Formulas:

M - molar mass,
n - the number of moles,
M - Mass of the substance.

The weight of praying is measured in g / mol, the mass of the Kbol is measured in kg / kmol, the mass of the mmol is measured in mg / mol.

Fill out the table (the tables are distributed).

Lesson 3.

Topic: Molar volume of gases

We will solve the task. Determine the volume of water, the mass of which under normal conditions is 180 g.

Given:

Those. The volume of liquid and solids we consider through density.

But, when calculating the volume of gases does not necessarily know the density. Why?

Italian scientist Avogadro determined that in equal volumes of various gases under the same conditions (pressure, temperature) contains the same number of molecules - this statement is called the Avogadro law.

Those. If at equal conditions V (H 2) \u003d V (O 2), then n (H 2) \u003d n (O 2), and on the contrary if at an equal condition N (H 2) \u003d n (O 2), the volumes of these Gas will be the same. And the moth of the substance always contains the same number of molecules 6.02 * 10 23.

We conclude - under the same conditions, moths of gases should occupy the same volume.

Under normal conditions (t \u003d 0, p \u003d 101.3 kPa. Or 760 mm Hg. Art.) Moth of any gases occupy the same volume. This volume is called molar.

V m \u003d 22.4 l / mol

1 kmmol occupies a volume -22.4 m 3 / kmol, 1mmol occupies a volume -22.4 ml / mmol.

Example 1. (Solving on the board):

Example 2. (You can propose to solve students):

Suggest students to fill in the table.

The objectives for determining the formula of the organic matter are several species. Usually, the solution of these tasks does not represent special difficulties, but often graduates lose points on this task. There are several reasons:

1. Incorrect design;
2. The decision is not mathematical, but by the method of extinguishing;
3. Incorrectly composed general formula for substance;
4. Errors in the reaction equation with the participation of a substance recorded in general.

Type types in task C5.

1. Determination of the formula for mass fractions chemical elements or according to the general formula of substance;
2. Determination of the formula of the substance in combustion products;
3. Determination of the formula for the chemical properties.

Necessary theoretical information.

1. Mass fraction of the element in the substance.
   The mass fraction of the element is its content in the substance in percentage by weight.
   For example, in the substance of the composition C 2 H 4, there are 2 carbon atoms and 4 hydrogen atoms. If you take 1 molecule of such a substance, then its molecular weight will be equal to:
   Mr (C 2 H 4) \u003d 2 12 + 4 1 \u003d 28 A.M. And there it contains 2 12 AE.M. Carbon.

   To find a mass fraction of carbon in this substance, it is necessary to divide its mass to the mass of the whole substance:
   ω (c) \u003d 12 2/28 \u003d 0.857 or 85.7%.
   If the substance has a general formula with x n y o z, the mass fractions of each of their atoms are also equal to the ratio of their mass to the mass of the entire substance. The mass X of atoms is equal to - 12x, the mass of atoms H - y, the mass Z of oxygen atoms - 16z.
   Then
   ω (c) \u003d 12 x / (12x + y + 16z)

   If you write this formula in general, the following expression will be obtained:

2. Molecular and simplest formula of substance.

   Molecular (true) formula - formula, which reflects the real number of atoms of each type included in the substance molecule.
   For example, from 6 H 6 - the true formula of benzene.
   The simplest (empirical) formula shows the ratio of atoms in the substance.
   For example, for benzene, the ratio C: H \u003d 1: 1, i.e. The simplest formula of benzene - CH.
   The molecular formula can coincide with the simplest or to be multiple to it.

   Examples.

   If only mass fractions of the elements are given in the problem, then in the process of solving the problem, only the simplest formula of the substance can be calculated. To obtain a true formula, additional data is usually given - molar mass, relative or absolute density of matter or other data, with which you can determine the molar mass of the substance.

3. The relative density of gas x on gas y - D for (x).
   The relative density D is a value that shows how many times the gas x is heavier than gas. It is calculated as the ratio of molar mass of gases x and y:
   D (x) \u003d m (x) / m (y)
   Often for calculations use relative gas densities for hydrogen and by air.
   Relative gas density in hydrogen:
   D by h 2 \u003d m (gas x) / m (H 2) \u003d m (gas x) / 2
   Air is a mixture of gases, so it can only be calculated for it to calculate the middle molar mass. Its value is adopted in 29 g / mol (based on the exemplary averaged composition).
   Therefore:
   D on Wards. \u003d M (Gas X) / 29
4. Absolute gas density under normal conditions.

   The absolute gas density is the mass of 1 l gas under normal conditions. Usually for gases, it is measured in g / l.
   ρ \u003d M (gas) / V (gas)
   If you take 1 mol of gas, then:
   ρ \u003d m / v m,
   And the molar mass of gas can be found, multiplying the density to the molar volume.

5. General formulas of substances of different classes.
   Often, to solve problems with chemical reactions, it is convenient to use a non-usual general formula, but a formula in which a separate communication or a functional group is selected.

   Class of organic substances	Total molecular formula	Formula with highlighted tie and functional group
   Alkana	C n h 2n + 2	—
   Alkenes	C n h 2n	C N H 2N + 1 -CH \u003d CH 2
   Alkina	C N H 2N-2	C N H 2N + 1 -C≡CH
   Devices	C N H 2N-2	—
   Homologists Benzol	C N H 2N-6	C 6 H 5 -C N H 2N + 1
   Limit single alcohols	C n h 2n + 2 o	C N H 2N + 1 -OH
   Multiatomic alcohols	C n h 2n + 2 o x	C n h 2n + 2-x (OH) X
   Limit Aldehydes	C n h 2n o
   Essentials	C n h 2n o 2

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

Solution of such tasks consists of two parts:

• first find the molar ratio of atoms in the substance - it corresponds to its simplest formula. For example, for the substance of the composition A x B, 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 \u003d n (a): n (b);
• then, using the molar mass of the substance, determine its true formula.

Example 1.
Determine the formula of substance, if it contains 84.21% C and 15.79% H and has a relative air density equal to 3.93.

Solution of Example 1.

1. Let the mass of the substance equal to 100 g. Then the mass C will be equal to 84.21 g, and the mass H - 15.79
2. Find the amount of substance of each atom:
   ν (c) \u003d m / m \u003d 84,21 / 12 \u003d 7,0175 mol,
   ν (H) \u003d 15.79 / 1 \u003d 15.79 mol.
3. We determine the molar ratio of atoms C and N:
   C: H \u003d 7.0175: 15.79 (reduce both numbers to smaller) \u003d 1: 2.25 (domestic on 4) \u003d 4: 9.
   Thus, the simplest formula - with 4 H 9.
4. By relative density, we calculate the molar mass:
   M \u003d D (Rev.) 29 \u003d 114 g / mol.
   The molar mass corresponding to the simplest formula with 4 H 9 - 57 g / mol, it is 2 times less than the truly molar mass.
   It means that the true formula is from 8 H 18.

There is a much simpler solution to solve such a task, but, unfortunately, it will not put a full score for it.. But it will suit the verification of the true formula, i.e. With it, you can check your decision.

Method 2:We find the true molar mass (114 g / mol), and then we find the mass of carbon and hydrogen atoms in this substance by their mass fractions.
m (C) \u003d 114 0.8421 \u003d 96; those. Number of atoms from 96/12 \u003d 8
m (h) \u003d 114 0.1579 \u003d 18; That is, the number of atoms H 18/1 \u003d 18.
The formula of substances - from 8 H 18.

Answer: from 8 H 18.

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

Solution of Example 2.

The general formula of alkina with N H 2N-2
How, having a density of alkina gaseous, find it molar mass? The density ρ is the mass of 1 liter of gas under normal conditions.
Since 1 mol of substance occupies 22.4 liters, then it is necessary to know how much 22.4 liters of such gas weigh:
M \u003d (density ρ) (molar volume v m) \u003d 2.41 g / l 22.4 l / mol \u003d 54 g / mol.
Next, we will make an equation connecting the molar mass and n:

14 n - 2 \u003d 54, n \u003d 4.
So, Alquin has a formula with 4 H 6.

Answer: With 4 H 6.

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

Solution of Example 3.

This problem is given the number of molecules and the corresponding mass. Based on these data, we need to reiterate the magnitude of the molar mass of the substance.
To do this, you need to recall how the number of molecules is contained in 1 mol of substance.
This is the number of Avogadro: N a \u003d 6.02 10 23 (molecules).
So, you can find the amount of aldehyde substance:
ν \u003d n / na \u003d 3 10 22/6.02 10 23 \u003d 0.05 mol,
and molar mass:
M \u003d m / n \u003d 4.3 / 0.05 \u003d 86 g / mol.
Further, as in the previous example, we compile equation and find N.
The general formula of the limit aldehyde with n h 2n o, that is, m \u003d 14n + 16 \u003d 86, n \u003d 5.

Answer: C 5 H 10 o, Pentanal.

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

Solution of Example 4.

The general formula of dichloroalkan: with 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) \u003d (number of C atoms in the molecule) (atomic mass C) / (dichloroalkane molecular weight)
0.3186 \u003d N 12 / (14N + 71)
N \u003d 3, the substance is dichloropropane.

Answer: C 3 H 6 Cl 2, dichlorpropane.

Determination of formulas for combustion products.

In the combustion tasks of the amount of substances included in the studied substance, determine the volumes and masses of the combustion products - carbon dioxide, water, nitrogen and others. The rest of the solution is the same as in the first type of tasks.

Example 5.
448 ml (n. Y.) The gaseous limit non-cyclic hydrocarbon burned, and the reaction products were missed through an excess of lime water, while 8 g of sediment was formed. What hydrocarbon was taken?

Solution of Example 5.

1. The total formula of the gaseous limit non-cyclic hydrocarbon (alkane) - 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 combustion 1 mole of alkane is highlighted by n mole of carbon dioxide.

   The amount of alkane substance find on its volume (do not forget to translate milliliters to liters!):

   ν (C n h 2n + 2) \u003d 0.488 / 22.4 \u003d 0.02 mol.

2. When carbon dioxide passage through Ca lime water (OH) 2 drops calcium carbonate:

   CO 2 + SA (OH) 2 \u003d Saso 3 + H 2 O

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

   It means its amount of substance
   ν (sasso 3) \u003d 8/100 \u003d 0.08 mol.
   The amount of carbon dioxide substance is also 0.08 mol.

3. The amount of carbon dioxide is 4 times larger than alkan, it means the alkane formula with 4 H 10.

Answer: C 4 H 10.

Example 6.
The relative density of the vapor of organic compound on nitrogen is 2. When burning 9.8 g of this compound, 15.68 liters of carbon dioxide (H. y) and 12.6 g of water are formed. Output the molecular formula of the 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, N and, possibly, O. Therefore, its general formula can be recorded as with x n o z.

1. We can record the combustion reaction diagram (without coefficient alignment):

   With x n o z + o 2 → CO 2 + H 2 O

   All carbon from the starting material goes into carbon dioxide, and all hydrogen in water.

2. We find the amount of substances of CO 2 and H 2 O, and determine how many mole of atoms C and N contains:
   ν (CO 2) \u003d V / V M \u003d 15.68 / 22.4 \u003d 0.7 mol.
   Per molecule CO 2 has to one Atom C, which means, carbon is as much moth as CO 2.

   ν (C) \u003d 0.7 mol
   

   In one water molecule, contain two Atom n, then the amount of hydrogen twice as muchthan water.
   ν (h) \u003d 0.7 2 \u003d 1.4 mol.

3. Check the presence in the substance of oxygen. To do this, the mass of the whole source substance should be subtracted by mass C and N.
   m (C) \u003d 0.7 12 \u003d 8.4 g, M (H) \u003d 1.4 1 \u003d 1.4 g
   Mass of the whole substance 9.8 g
   m (O) \u003d 9.8 - 8.4 - 1,4 \u003d 0, i.e. in this substance there are no oxygen atoms.
   If oxygen in this substance was present, then it would be possible to find the amount of substance and calculate the simplest formula, based on the presence of three different atoms.
4. Further actions are already familiar to you: the search for the simplest and true formulas.
   C: H \u003d 0.7: 1,4 \u003d 1: 2
   The simplest formula CH 2.
5. True molar mass we are looking for a relative gas density on nitrogen (do not forget that nitrogen consists of diKatomny Molecules N 2 and its molar mass 28 g / mol):
   M East. \u003d D for n 2 m (n 2) \u003d 2 28 \u003d 56 g / mol.
   The truth formula CH 2, its molar mass 14.
   56 / 14 = 4.
   True formula with 4 H 8.

Answer: C 4 H 8.

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

Solution of Example 7.

1. The substance contains atoms C, N and N. Since the mass of nitrogen in combustion products is not given, it will have to be calculated, based on the mass of the entire organic matter.
   Burning reaction scheme:
   With x n in n z + o 2 → CO 2 + H 2 O + N 2
2. We find the amount of substances of CO 2 and H 2 O, and determine how many mole of atoms C and N contains:

   ν (CO 2) \u003d m / m \u003d 17.6 / 44 \u003d 0.4 mol.
   ν (C) \u003d 0.4 mol.
   ν (H 2 O) \u003d M / M \u003d 12.6 / 18 \u003d 0.7 mol.
   ν (h) \u003d 0.7 2 \u003d 1.4 mol.

3. We find a lot of nitrogen in the starting matter.
   To do this, the mass of the whole source substance should be subtracted by mass C and N.

   M (C) \u003d 0.4 12 \u003d 4.8 g,
   M (H) \u003d 1.4 1 \u003d 1.4 g

   Mass of the whole substance 9.8 g

   M (n) \u003d 9 - 4.8 - 1.4 \u003d 2.8 g,
   ν (n) \u003d m / m \u003d 2.8 / 14 \u003d 0.2 mol.

4. C: H: N \u003d 0.4: 1.4: 0.2 \u003d 2: 7: 1
   The simplest formula - C 2 H 7 N.
   True molar mass
   M \u003d D on H 2 m (H 2) \u003d 22.5 2 \u003d 45 g / mol.
   It coincides with the molar mass calculated for the simplest formula. That is, this is the true formula of substances.

Answer: C 2 H 7 N.

Example 8.
The substances contain C, N, O and S. 8.8 g of CO 2, 5.4 g H 2 o, and sulfur was completely translated into the barium sulfate, the mass of which was equal to 23.3 g. The formula of substance.

Solution of Example 8.

The formula of a given substance can be represented as C x H y s z o k. When combining it, carbon dioxide, water and sulfur gas is obtained, which is then converted to barium sulfate. Accordingly, the entire sulfur from the starting material is converted to barium sulfate.

1. We find the amount of carbon dioxide, water and barium sulfate and the corresponding chemical elements from the studied substance:

   ν (CO 2) \u003d M / M \u003d 8.8 / 44 \u003d 0.2 mol.
   ν (C) \u003d 0.2 mol.
   ν (H 2 O) \u003d M / M \u003d 5.4 / 18 \u003d 0.3 mol.
   ν (H) \u003d 0.6 mol.
   ν (Baso 4) \u003d 23.3 / 233 \u003d 0.1 mol.
   ν (s) \u003d 0.1 mol.

2. We calculate the estimated mass of oxygen in the starting matter:

   M (C) \u003d 0.2 12 \u003d 2.4 g
   M (H) \u003d 0.6 1 \u003d 0.6 g
   M (S) \u003d 0.1 32 \u003d 3.2 g
   m (o) \u003d m substance - M (C) - M (H) - M (S) \u003d 11 - 2.4 - 0.6 - 3.2 \u003d 4.8 g,
   ν (O) \u003d m / m \u003d 4.8 / 16 \u003d 0.3 mol

3. We find the molar ratio of elements in the substance:
   C: H: S: O \u003d 0.2: 0.6: 0.1: 0.3 \u003d 2: 6: 1: 3
   The formula of the substance C 2 H 6 SO 3.
   It should be noted that in this way we received only the simplest formula.
   However, the resulting formula is true, since when attempting to doubling this formula (C 4 H 12 S 2 O 6), it turns out that by 4 carbon atoms, in addition to sulfur and oxygen, there are 12 n atoms, and this is impossible.

Answer: C 2 H 6 SO 3.

Determination of formulas for chemical properties.

Example 9.
Determine the formula of alkadien, if it can be discouraged by 80 g of 2% of the bromine solution.

Solution of Example 9.

1. The total formula of alkadiennes - with N H 2N-2.
   We write the equation of the reaction of the joining bromine to alkadien, not forgetting that in the diene molecule two double bonds And, accordingly, 2 mol bromine will enter the reaction from 1 mole:
   With N H 2N-2 + 2Br 2 → C N H 2N-2 Br 4
2. Since the task is given a mass and percentage concentration of a bromine solution that has reacted with a diene, one can calculate the amount of substance of the reacted bromine:

   M (br 2) \u003d m solution ω \u003d 80 0.02 \u003d 1.6 g
   ν (br 2) \u003d m / m \u003d 1.6 / 160 \u003d 0.01 mol.

3. Since the amount of bromine that has joined the reaction is 2 times more than alkadien, you can find the amount of diene and (since its mass is known) its molar mass:

   0,005		0,01
   With n h 2n-2	+ 2Br 2 →	With n h 2n-2 br 4

   M diena \u003d m / ν \u003d 3.4 / 0.05 \u003d 68 g / mol.

4. We find the formula of alkadien in its general formula, expressing the molar mass through n:

   14N - 2 \u003d 68
   n \u003d 5.

   This is pentadiene from 5 H 8.

Answer: C 5 H 8.

Example 10.
When the interaction of 0.74 g of limit monohydric alcohol with a metal sodium, hydrogen was distinguished in an amount sufficient for hydrogenation 112 ml of propon (H. Y.). What is this alcohol?

Solution of Example 10.

1. The formula of the limit monohydric alcohol - C n h 2n + 1 oh. Here it is convenient to record alcohol in such a form in which the reaction equation is easy - i.e. With a separately dedicated group, it.
2. We will make the equation of reactions (it is impossible to forget about the need to equalize the reaction):

   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

3. You can find the amount of propacted, and on it - the amount of hydrogen. Knowing the amount of hydrogen, by reaction we find the amount of alcohol substance:

   ν (C 3 H 6) \u003d V / V M \u003d 0.112 / 22.4 \u003d 0.005 mol \u003d\u003e ν (H 2) \u003d 0.005 mol,
   ν alcohol \u003d 0.005 2 \u003d 0.01 mol.

4. We find the molar mass of alcohol and n:

   M alcohol \u003d m / ν \u003d 0.74 / 0.01 \u003d 74 g / mol
   14N + 18 \u003d 74
   14n \u003d 56.
   n \u003d 4.

   Alcohol - Butanol with 4 H 7.

Answer: C 4 H 7 Oh.

Example 11.
To determine the formula of the ester, with hydrolysis of 2.64 g of which 1.38 g of alcohol and 1.8 g of mono-axis carboxylic acid are distinguished.

Solution of Example 11.

1. The general formula for ester consisting of alcohol and acid with different numbers of carbon atoms can be represented in this form:
   C N H 2N + 1 COOC M H 2M + 1
   Accordingly, the alcohol will have a formula
   C M H 2M + 1 OH,
   A acid
   C N H 2N + 1 COOH.
   Equation of hydrolysis of ester:
   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
2. According to the law of preserving the mass of substances, the amount of mass of the starting materials and the sum of the masses of the reaction products are equal.
   Therefore, from these tasks you can find a lot of water:

   M H 2 O \u003d (acid mass) + (alcohol weight) - (ether mass) \u003d 1.38 + 1.8 - 2.64 \u003d 0.54 g
   ν H 2 O \u003d M / M \u003d 0.54 / 18 \u003d 0.03 mol

   Accordingly, the number of substances of acid and alcohol is also equal to mole.
   You can find their molar masses:

   M acid \u003d m / ν \u003d 1.8 / 0.03 \u003d 60 g / mol,
   M alcohol \u003d 1.38 / 0.03 \u003d 46 g / mol.

   We obtain two equations from which we will find M and N:

   M C n H 2N + 1 COOH \u003d 14N + 46 \u003d 60, N \u003d 1 - acetic acid
   M C M H 2M + 1 OH \u003d 14M + 18 \u003d 46, M \u003d 2 - ethanol.

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

Answer: CH 3 COOC 2 H 5.

Example 12.
Determine the formula of amino acids, if under action by 8.9 g its excess sodium hydroxide can be obtained 11.1 g sodium salt This acid.

Solution of Example 12.

1. The general formula of amino acids (if we assume that it does not contain any other functional groups, except for one amino group and one carboxyl):
   NH 2 -CH (R) -COOH.
   It would be possible to record it different ways, But for the convenience of writing the reaction equation, it is better to allocate functional groups in the amino acid formula separately.
2. You can create an equation of 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 a lot of any of the substances in the reaction equation. Therefore, in such problems it is necessary to express the amounts of amino acid substances and its salts through molar masses and equate them:

   M (Amino acids NH 2 -CH (R) -COOH) \u003d 74 + m R
   M (NH 2 -Ch (R) -Coona Salts) \u003d 96 + M R
   ν amino acids \u003d 8.9 / (74 + m R),
   ν salts \u003d 11.1 / (96 + m R)
   8.9 / (74 + m R) \u003d 11.1 / (96 + m R)
   M R \u003d 15

   It is easy to see that R \u003d CH 3.
   You can make it mathematically, if we assume that R is C n h 2n + 1.
   14N + 1 \u003d 15, n \u003d 1. Install the formula for the limiting mono-block of carboxylic acid, the calcium salt of which contains 30.77% calcium.

   Part 2. Definition of the formula of substance on combustion products.

   2-1. The relative density of the vapor of organic compound on sulfur gas is equal to 2. When burning 19.2 g of this substance, 52.8 g of carbon dioxide (N.U.) and 21.6 g of water are formed. Output the molecular formula of the organic compound.

   2-2. When burning an organic matter weighing 1.78 g in an excess of oxygen, 0.28 g of nitrogen was obtained, 1.344 liters (n.) CO 2 and 1.26 g of water. Determine the molecular formula of substance, knowing that in the specified sample of the substance contains 1.204 10 22 molecules.

   2-3. Carbon dioxide obtained during combustion of 3.4 g of hydrocarbon, passed through an excess solution of calcium hydroxide and obtained 25 g of precipitate. Output the simplest hydrocarbon formula.

   2-4. When combustion of an organic matter containing C, N and chlorine, 6.72 liters (N.O.) of carbon dioxide, 5.4 g of water, 3.65 g of chloride is distinguished. Install the molecular formula of the burnt substance.

   2-5. (EGE-2011) With the combustion of the amine, 0.448 liters (N.O.) of carbon dioxide, 0.495 g of water and 0.056 liters of nitrogen were separated. Determine the molecular formula of this amine.

   Part 3. Definition of the formula for the substance for chemical properties.

   3-1. It is known to determine the formula of alkene if it is known that it is 5.6 g of it when water is attached to form 7.4 g of alcohol.

   3-2. For oxidation, 2.9 g of limit aldehyde to the acid was required by 9.8 g of copper hydroxide (II). Determine the formula of aldehyde.

   3-3. A single-axis monoamin acid mass weigh 3 g with an excess of bromomotorodor forms 6.24 g of salts. Determine the formula of amino acids.

   3-4. When the interaction of the limit dioment alcohol, weighing 2.7 g with an excess of potassium was released 0.672 liters of hydrogen. Determine the formula of alcohol.

   3-5. (EGE-2011) In the oxidation of the maximum monohydric alcohol, 9.73 g of aldehyde, 8.65 g of copper and water obtained oxide (II). Determine the molecular formula of this alcohol.

   Answers and comments to tasks for an independent solution.

   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. From 8 n 16 o

   2-2. 3H 7 NO

   2-3. C 5 H 8 (a mass of hydrogen is found, carbon mass from the mass of hydrocarbon)

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

   3-2. C 3N 6 o

   3-3. C 2N 5 NO 2

   In order to characterize the number of molecules or other structural units, the concept of the amount of substance is introduced. Often, even in the smallest masses, there is a huge number of molecules in the substance. Therefore, for counting, as it were, blocks with a certain number of particles.

   Concept Number of substance

   The amount of substance is a private from dividing the number of molecules in this bodyN to the number of atoms that is concluded in 12 g of carbonN a.

   12 g of carbon contains 6.02 · 10 23 molecules. This number is called number Avogadro. On this basis, it can be said that the amount of substance shows the number of blocks of 6.02 · 10 23 molecules each.

   This value is measured in moles.

   Mol is the amount of substance that contains 6.02 · 10 23 molecules (atoms).

   Examples of solving problems

   Consider the task:

   1. What is the number of molecules of 8 miles of chromium?

   Since in one mole of any substance (no matter oxygen, fluorine, aluminum, chromium) contains 6.02 · 10 23 molecules N a. To calculate the number of N molecules, there are enough 8 moles to multiply on the number of molecules in one mole.

   N \u003d v. * N a;

   N \u003d 8 mol * 6 · 10 23 mol -1 \u003d 48 · 10 23

   Answer: 48 · 10 23 molecules.

   In real life, the counting of the amount of substance is associated with some difficulties, since the number of molecules or atoms, even in several grams, can reach several million. Therefore, to find the number of molecules, the mass of the substance is divided into a lot of one pray.

   Molar mass shows how much weighs one mole of substances.

   Using the periodic table of elements with ease, you can find a molar mass. For example, the molar mass of oxygen - 16, hydrogen - 1, phosphorus - 30, nitrogen - 14, fluorine - 19.

   1. Determine the number of molecules in 504 g of Crypton?

   The condition said that it is necessary to find the amount of substance, therefore, we calculate how much one mole of Krypton (molar mass) weighs. In the Mendeleev table, it is located at number 36.

   M (O 2) \u003d 84 g / mol.

   Since we know how many molecules are contained in one mole of any substance (6.02 · 10 23), the amount of this substance should be calculated, that is, the number of moles.

   To do this, we divide the mass of oxygen on the molar mass.

   V.\u003d m / m (O 2);

   V.\u003d \u003d 6 mole.

   After, as in the previous task, multiply the value obtained to the number of Avogadro.

   N \u003d v. * N a;

   N \u003d 6 · 10 23 mol -1 * 6 mol \u003d 36 · 10 23

   Answer: 36 · 10 23 molecules.

   Class: 8

   Purpose: To acquaint students with the concepts of "Number of substance", "Molar Mass" to give an idea of \u200b\u200bconstant Avogadro. Show the relationship of the amount of substance, the number of particles and the constant Avogadro, as well as the relationship of the molar mass, mass and amount of the substance. Teach to make calculations.

   Type of lesson:lesson studying and primary consolidation of new knowledge.

   During the classes

   I. Organizational moment

   II. Verification of d / s on the topic: "Types of chemical reactions"

   III. Studying a new material

   1. The amount of substance is mole

   Substances react into strictly certain ratios. For example, to obtain a substance of water, you need to take so much hydrogen and oxygen so that every two hydrogen molecules accounted for one oxygen molecule:

   2N 2 + O 2 \u003d 2N 2

   To obtain a substance of iron sulphide, you need to take so much iron and sulfur so that each atom of iron accounted for one sulfur atom.

   To obtain a substance of phosphorus oxide, you need to take so many molecules of phosphorus and oxygen so that four phosphorus molecules accounted for five oxygen molecules.

   It is impossible to determine the number of atoms, molecules and other particles in practice - they are too small and are not visible to the naked eye. To determine the number of structural units (atoms, molecules) in chemistry, a special amount is used - number of substance ( v - NU). Unit amount of substance is mole.

   • Mol is the amount of substance that contains so many structural particles (atoms, molecules), how many atoms are contained in 12 g of carbon.

   It is experimentally established that 12 g of carbon contains 6 · 10 23 atoms. So one moth of any substance, regardless of its aggregate state, contains the same number of particles - 6 · 10 23.

   • 1 mole oxygen (O 2) contains 6 · 10 23 molecules.
   • 1 mol of hydrogen (H 2) contains 6 · 10 23 molecules.
   • 1 mol of water (H 2 O) contains 6 · 10 23 molecules.
   • 1 mole of iron (FE) contains 6 · 10 23 molecules.

   The task: Using the information received, respond to questions:

   a) how many oxygen atoms are contained in 1 oxygen mole?

   - 6 · 10 23 · 2 \u003d 12 · 10 23 atoms.

   b) how many hydrogen and oxygen atoms are contained in 1 mall of water (H 2 O)?

   - 6 · 10 23 · 2 \u003d 12 · 10 23 hydrogen atoms and 6 · 10 23 oxygen atoms.

   Number 6 · 10 23 named constant avogadro In honor of the Italian 19th century scientist and is denoted by NA. Units of measurement atoms / mole or molecules / mole.

   2. Solving tasks to find the amount of substance

   Often you need to know how many particles of the substance are contained in a certain amount of substance. Or find the amount of substance at a well-known number of molecules. These calculations can be made by the formula:

   where n is the number of molecules, NA is constant Avogadro, v. - Number of substance. From this formula, you can express the amount of substance.

   v \u003d. N / NA

   Task 1. How many atoms contain in 2 sulfur prayers?

   N \u003d 2 · 6 · 10 23 \u003d 12 · 10 23 atoms.

   Task 2. How many atoms are contained in 0.5 iron chairs?

   N \u003d 0.5 · 6 · 10 23 \u003d 3 · 10 23 atoms.

   Task 3. How many molecules are contained in 5 moles of carbon dioxide?

   N \u003d 5 · 6 · 10 23 \u003d 30 · 10 23 molecules.

   Task 4. What amount of substance is 12 · 10 23 molecules of this substance?

   v \u003d.12 · 10 23/6 · 10 23 \u003d 2 mol.

   Task 5. What amount of substance is 0.6 · 10 23 molecules of this substance?

   v \u003d.0.6 · 10 23/6 · 10 23 \u003d 0.1 mol.

   Task 6. What amount of substance is 3 · 10 23 molecules of this substance?

   v \u003d.3 · 10 23/6 · 10 23 \u003d 0.5 mol.

   3. Molar Massa

   For chemical reactions, it is necessary to take into account the amount of substance in the moles.

   Q: But how in practice, measure 2, or 2.5 mol substances? Which units are best to measure the mass of substances?

   For convenience in chemistry use molar mass.

   The molar mass is the mass of one praying substance.

   It is indicated - M. is measured in g / mol.

   The molar mass is equal to the ratio of the mass of the substance to the corresponding number of substance.

   Molar weight - the magnitude is constant. The numerical value of the molar mass corresponds to the value of the relative atomic or relative molecular weight.

   Q: How can I find the values \u200b\u200bof a relative atomic or relative molecular weight?

   Mr (s) \u003d 32; M (s) \u003d 32 g / mol - What corresponds to 1 praying sulfur

   Mr (H 2 O) \u003d 18; M (H 2 O) \u003d 18 g / mol - which corresponds to 1 praying water.

   4. Solving tasks to find the mass of matter

   Task 7. Determine the mass of 0.5 mol iron.

   Task 8. Determine the mass of 0.25 mole of copper

   Task 9. Determine Mass 2 Mol Carbon Dioxide (CO 2)

   Task 10.How many moles of copper oxide - Cuo make up 160 g of copper oxide?

   v \u003d.160/80 \u003d 8 mole

   Task 11.How many moles of water correspond to 30 g of water

   v \u003d.30/18 \u003d 1.66 mol

   Task 12.How many moles of magnesium correspond to its 40 grams?

   v \u003d.40/24 \u003d 1.66 mol

   IV. Fixing

   Frontal survey:

   1. What is the amount of substance?
   2. What is 1 mol of any substance?
   3. What is a molar mass?
   4. Is the concepts of "mole molecules" and "mole atoms"?
   5. Explain on the example of the ammonia molecule NN3.
   6. Why do you need to know the formulas when solving problems?

   Tasks:

   1. How many molecules are contained in 180 grams of water?
   2. How many molecules is 80 g of carbon dioxide?

   V. Homework

   Examine the text of the paragraph, make up two tasks: to find the amount of substance; To find the mass of the substance.

   Literature:

   1. Gara N.N. Chemistry. Lessons in grade 8: manual for teacher. _ M.: Enlightenment, 2009.
   2. RUDZITES G.E., Feldman F.G. Chemistry. Grade 8: Textbook for general educational institutions - M.: Enlightenment, 2009.

   The most typical processes carried out in chemistry are chemical reactions, i.e. The interactions between some kind of source substances leading to the formation of new substances. Substances react in certain quantitative relations that need to be taken into account in order to obtain the desired products to spend the minimum amount of starting materials and not to create useless production waste. To calculate the masses of the reactant substances, another physical value is necessary, which characterizes the portion of the substance from the point of view of the number of structural units contained in it. In itself, the ego number is extremely great. This is obvious, in particular, from Example 2.2. Therefore, in practical calculations, the number of structural units is replaced by a special value called number Substances.

   The amount of substance is a measure of the number of structural units determined by the expression

   where N (x) - number of structural units of substance X. In real or mentally taken portion of the substance, N a \u003d. 6.02 10 23 - constant (number) of the Avogadro, widely used in science, one of the fundamental physical constants. If necessary, it is possible to use a more accurate value of the Avogadro constant 6.02214 10 23. Portion of a substance containing N A structural units is a single amount of substance - 1 mol. Thus, the amount of substance is measured in moles, and the Avogadro constant has a unit of measurement 1 / mol, or in another mole -1 record.

   For all sorts of reasoning and calculations associated with the properties of the substance and chemical reactions, the concept number of substances Fully replaces the concept the number of structural units. Thanks to this, it disappears the need to use large numbers. For example, instead of saying "6.02 10 23 structural units (molecules) of water", we will say, "take 1 mol of water".

   Any portion of the substance is characterized by both the mass and the amount of substance.

   The ratio of mass of matterX. The amount of substance is called molar massM (x):

   The molar mass is numerically equal to mass 1 mol of substance. This is an important quantitative characteristic of each substance depending only on the mass of structural units. The Avogadro number is established such that the molar mass of the substance, expressed in g / mol, is numerically coincided with the relative molecular weight Mr. For water molecule Mg \u003d. 18. This means that the molar mass of water M (H 2 0) \u003d 18 g / mol. Using the data of the Mendeleev table, you can calculate more accurate values. Mr. and M (x), But in class tasks in chemistry it is usually not required. Of all this, it is clear how easy it is to calculate the molar mass of the substance - it is enough to add atomic masses in accordance with the formula of the substance and put a unit of measurement of g / mol. Therefore, formula (2.4) is practically used to calculate the amount of substance:

   
   

   Example 2.9.Calculate the molar mass of drinking soda NaHC0 3.

   Decision. According to the formula of substances Mg \u003d. 23 + 1 + 12 + 3 16 \u003d 84. Hence, by definition, M (Naiic0 3) \u003d 84 g / mol.

   Example 2.10.What amount of substance make up 16.8 g of drinking soda? Decision. M (NaHC0 3) \u003d 84 g / mol (see above). By Formula (2.5)

   Example 2.11.How many Tolik (structural units) drinking soda is located in 16.8 g of substance?

   Decision. Converting formula (2.3), we find:

   AT (NaHC0 3) \u003d N A N (NaHC0 3);

   tT (NaHC0 3) \u003d 0.20 mol (see example 2.10);

   N (NaHC0 3) \u003d 6.02 10 23 mol "1 0.20 mol \u003d 1,204 10 23.

   Example 2.12.How many atoms are located in 16.8 g of drinking soda?

   Decision. Drinking soda, NaHC0 3, consists of sodium, hydrogen atoms, carbon and oxygen. In total, in the structural unit of substance 1 + 1 + 1+ 3 \u003d 6 atoms. As it was found in Example 2.11, this mass of drinking soda consists of 1.204 10 23 structural units. Therefore, the total number of atoms in the substance is

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