1. Basic concepts, definitions and laws of chemistry
1.3. The chemical amount of the substance. Mole. Molar mass
The chemical amount of the substance. Mole. Molar mass
Characterizing a portion of the taken substance, use its mass or volume. However, for the same purpose, it is possible to indicate the number of structural units in the taken portion of the substance. It is extremely important to know this number, since in chemical reactions substances interact in ratios proportional to the number of structural units, and not to masses. For example, the notation 2H 2 + O 2 \u003d 2H 2 O means that the numbers (but not the masses!) Of the reacting molecules H 2 and O 2 are, respectively, 2: 1.
For the convenience of counting the number of structural units, the content of which in any measurable portion of a substance is huge, a new physical quantity - the amount of a substance, which in chemical calculations is also called the chemical amount of a substance.
Chemical amount of a substance - a physical quantity proportional to the number of structural units (atoms, molecules, PU) contained in a given portion of the substance.
The chemical quantity is designated by the letter n (less often ν).
The unit of a chemical quantity of a substance is a mole.
A mole is a portion of a substance containing as many of its elementary structural units as atoms are contained in a portion of the C-12 nuclide weighing 12 g.
The number of atoms in the indicated portion of the C-12 nuclide is approximately equal to 6.02 ⋅ 10 23. The physical quantity equal to 6.02 ⋅ 10 23 mol −1 is called avogadro constant and denoted by N A:
N A \u003d 6.02 ⋅ 10 23 mol - 1 \u003d 6.02 ⋅ 10 23 mol - 1.
The unit of the numerator in the value N A is not indicated, since for different cases it can be different, for example:
N A \u003d 6.02 ⋅ 10 23 atoms mol,
N A \u003d 6.02 ⋅ 10 23 molecules mol,
N A \u003d 6.02 ⋅ 10 23 PU mol.
The physical meaning of Avogadro's constant is that its numerical value (6.02 ⋅ 10 23) shows the number of structural units in 1 mole of substance. For example, 1 mol of sodium (m \u003d 23 g) contains 6.02 ⋅ 10 23 Na atoms; 1 mol of sulfuric acid (m \u003d 98 g) contains 6.02 ⋅ 10 23 molecules of H 2 SO 4; 1 mol of calcium carbonate (m \u003d 100 g) contains 6.02 ⋅ 10 23 CaCO 3 formula units.
A mole is a portion of a substance containing 6.02 ⋅ 10 23 of its structural units
The number of structural units of a substance N (B) and the chemical amount of a substance n (B) are related by the ratio
n (B) \u003d N (B) N A, (1.8)
N (B) \u003d n (B) N A. (1.9)
Knowing the chemical amount of any substance, you can use it chemical formula calculate the chemical amount of individual atoms included in its composition.
One mole of any substance numerically contains the same chemical number of atoms as there are (atoms) in one molecule (formula unit) of a substance
For instance:
- the P 4 molecule contains 4 P atoms, and the composition of 1 mol of P 4 contains 4 moles of P atoms;
- the formula unit Na 3 PO 4 contains 3 Na atoms, 1 P atom and 4 O atoms, and 1 mol of Na 3 PO 4 contains 3 mol of Na atoms, 1 mol of P atoms and 4 mol of O atoms.
With an increase (decrease) in the chemical amount of a substance, the chemical amount of its constituent atoms proportionally increases (decreases). For example: 0.5 mol Na 3 PO 4 contains 3 · 0.5 \u003d 1.5 (mol) Na atoms; 5 mol of Р 4 contains 5 4 \u003d \u003d 20 (mol) R.
For such calculations, you can use the so-called stoichiometric circuits... The principles of drawing up stoichiometric schemes and performing calculations are shown using the example of K 2 SO 4 with a chemical amount of 0.3 mol:
x \u003d n (K) \u003d 0.3 ⋅ 2 1 \u003d 0.6 (mol);
y \u003d n (S) \u003d 0.3 ⋅ 1 1 \u003d 0.3 (mol);
z \u003d n (O) \u003d 0.3 ⋅ 4 1 \u003d 1.2 (mol).
The concept of a mole is applicable to all substances, and the concept of a molecule is not applicable to all, but only to substances of a molecular structure. For example, both concepts are applicable to water (water has molecular structure), but in the case of calcium carbonate (non-molecular structure), only the term "mole" is applicable.
The term "mole" is also used in the case of ions, electrons, protons, neutrons and chemical bonds... For example, if N (PO 4 3 -) \u003d 3.01 ⋅ 10 23, then
n (PO 4 3 -) \u003d 3.01 10 23 / 6.02 ⋅ 10 23 \u003d 0.5 (mol);
N (e) \u003d 1.505 ⋅ 10 22,
n (e) \u003d N (e) / N A \u003d 1.505 10 22 / 6.02 ⋅ 10 23 \u003d 0.025 (mol);
2 mol of Н 2 (Н – Н) molecules contain 2 mol of hydrogen - hydrogen bonds, and 3 mol of Н 2 О (Н – О – Н) molecules - 6 mol of Н – О bonds (each molecule contains two Н – О bonds) ...
Molar mass M (B) is a physical quantity, equal ratio mass of a substance to its chemical amount:
M (B) \u003d m (B) n (B). (1.10)
Formulas for calculating the mass of a substance follow from expression (1.10):
m (B) \u003d n (B) ⋅ M (B) (1.11)
and its chemical amount:
n (B) \u003d m (B) M (B). (1.12)
Since at n (B) \u003d 1 mol the numerical values \u200b\u200bof n (B) and M (B) are the same, it is often said that the molar mass is the mass of 1 mol of a substance. This, of course, is not true, since only the numerical values \u200b\u200bof these quantities coincide, and their physical meaning and units of measurement are different.
With the help of molar mass, you can easily calculate the mass of a molecule or formula unit of a substance:
m mol, FE \u003d M (B) N A. (1.13)
In addition, the molar mass can be found by the formula
M (B) \u003d m mol, FE ⋅ N A. (1.14)
It is easy to show that when using a unit of molar mass, gram per mole, its numerical value coincides:
- with A r for simple substances of atomic structure:
A r (O) \u003d 16, M (O) \u003d 16 g mol;
- M r complex substances of molecular and non-molecular structure:
M r (H 2 O) \u003d 18, M (H 2 O) \u003d 18 g mol;
M r (KOH) \u003d 56, M (KOH) \u003d 56 g mol.
Indeed:
M (B) \u003d m mol (B) ⋅ N A \u003d M r (B) ⋅ u ⋅ N A \u003d M r (B) ⋅ 1 N A ⋅ N A \u003d M r (B)
M (B) \u003d m aт ⋅ N A \u003d A r (B) ⋅ u ⋅ N A \u003d A r (B) ⋅ 1 N A ⋅ N A \u003d A r (B).
Example 1.5. The mass of a molecule of a substance is 7.31 ⋅ 10 −23 g. Calculate the molar mass of a substance.
Decision. First way. From formula (1.14) it follows:
M (B) \u003d m mol (B) ⋅ N A
M (B) \u003d 7.31 ⋅ 10 - 23 g ⋅ 6.02 ⋅ 10 23 1 mol \u003d 44 g / mol.
Second way. We use formula (1.5):
M r (B) \u003d m mol (B) u \u003d 7.31 ⋅ 10 - 23 g 1.66 ⋅ 10 - 24 g \u003d 44;
M (B) \u003d 44 g / mol.
Answer: 44 g / mol.
Gas laws. Gas mixtures
Substances can be in three states of aggregation: gaseous, liquid and solid. Liquid and solid states are called condensed... For most substances, the states of aggregation are mutually transient: when heated, the solid first melts, then evaporates; when cooling, the gas first condenses - turns into liquid state, then the liquid freezes (crystallizes). An increase in pressure and a decrease in temperature contribute to the transition of a substance into a condensed state with a smaller volume (and vice versa - a decrease in pressure and an increase in temperature contribute to the transition of a substance into a gaseous state).
The pressure of a gas in a closed vessel is directly proportional to the number of its molecules (or chemical quantity)
During the transition of a substance from a solid to a liquid state, and then to a gaseous state, the distance between particles gradually increases, and in the case of a gas, this distance is hundreds of times larger than the dimensions of the molecules themselves. It follows from this that the volume of a portion of a gas is determined not by the nature of the gas (the size of its molecules), but by the distance between the molecules (in fact, the volume that the gas occupies is the volume of free space between the molecules).
The distance between gas molecules depends on temperature and pressure, which means that under the same external conditions, the distance between molecules of different gases is the same.
Hence follows the position known as Avogadro's law (1811): equal volumes of different gases under the same conditions contain the same number of molecules
Three consequences follow from Avogadro's law.
1. Same number molecules of different gases at the same pressure and temperature occupy the same volume.
2. Under normal conditions (n.o .: T \u003d 273 K or 0 ° C, p \u003d 101.3 kPa), the volume of a portion of any gas with a chemical amount of 1 mol, or the molar volume V m,
V m \u003d 22.4 dm 3 / mol.
3. Masses of equal volumes of two gases are referred to as their molar (relative molecular) masses. This relationship is called relative density gas A for gas B and is denoted as D B (A):
m (A) m (B) \u003d D B (A) \u003d M (A) M (B) \u003d M r (A) M r (B). (1.15)
Using V m, the volume and chemical amount of the gas are found:
V (B) \u003d n (B) ⋅ V m; (1.16)
n (B) \u003d V (B) / V m. (1.17)
Formula (1.15) allows, knowing the relative density of the unknown gas X from the known gas, to find M (M r) of the unknown gas:
M (X) \u003d D B (X) ⋅ M (B). (1.18)
For example, if the relative density of gas X in air (M air \u003d 29 g / mol) is 1.517, then the molar mass of this gas
M (X) \u003d 29 ⋅ 1.517 \u003d 44 (g / mol).
Relative density is dimensionless and does not depend on temperature and pressure.
Knowing the molar mass of the gas, one can easily calculate the density ρ of the gas (in g / dm 3):
ρ (B) \u003d M (B) V m \u003d M (B) 22.4. (1.19)
For example, for nitrogen
ρ (N 2) \u003d M (N 2) V m \u003d 28 g / mol 22.4 dm 3 / mol \u003d 1.25 g / dm 3.
According to the density of the gas, its molar mass is found:
M (B) \u003d ρ (B) V m. (1.20)
The density of a gas depends on the temperature T and pressure P: with an increase in T and a decrease in P, the density decreases.
If the densities ρ of two gases are equal (ρ 1 \u003d ρ 2), then their molar (relative molecular) masses are also equal, i.e. M 1 \u003d M 2 (and vice versa - if the molar masses of gases are equal, then their densities are also equal)
In the case of gases, it is also true gay-Lussac's law of volumetric relations (1805-1808): in chemical reactions, the volumes of reacting and produced gases are referred to as small whole numbers equal to their stoichiometric coefficients
For example, for the reaction
4NH 3 + 5O 2 \u003d 4NO + 6H 2 O
V (NH 3) V (O 2) \u003d 4 5;
V (O 2) V (NO) \u003d 5 4.
Example 1.6. The relative density (n.o.) of a certain gas X with respect to argon is 1.2. Find the mass of the gas molecule X.
Decision . Using formula (1.18), we find the molar mass of gas X:
M (X) \u003d D Ar (X) ⋅ M (Ar),
M (X) \u003d 1.2 ⋅ 40 \u003d 48 g / mol.
Using formula (1.13), we calculate the mass of a gas molecule X:
m mol (X) \u003d M (X) N A \u003d 48 6.02 10 23 \u003d 7.97 ⋅ 10 - 23 (g).
You can also use formula (1.7):
m mol (X) \u003d M r (X) u \u003d 48 ⋅ 1.66 ⋅ 10 - 24 \u003d 7.97 ⋅ 10 - 23 (g).
Answer: 7.97 ⋅ 10 −23 g.
Methods for collecting gases. Molar gas concentration
Consider laboratory methods for collecting gases. There are two such ways (Fig. 1.1).
Fig. 1.1. Laboratory methods for collecting oxygen by heating KMnO 4:
A - a way of displacing water; b - air displacement method
It's obvious that by displacing water you can collect only those gases that do not dissolve in water and do not interact with it (hydrogen, methane, nitrogen, oxygen). In this way, you cannot collect gases that dissolve well in water or interact with it (HCl, HBr, HI, HF, NH 3). Carbon monoxide (IV) CO 2 dissolves in water relatively poorly, so it can be collected in this way.
When collecting gas air displacement you need to position the tubes correctly:
- neck up if the gas is heavier than air, i.e. M (gas)\u003e M (air). Examples: CO 2, SO 2, HCl;
- neck down if the gas is lighter than air, i.e. M (gas)< M (возд) . Примеры: H 2 , Ne, NH 3 , CH 4 .
For gas characteristics use molar concentration c, equal to the ratio of the chemical amount of gas to the volume of the gas portion:
c (X) \u003d n (X) V (X)
Gas mixtures, like individual gases, are characterized by molar (relative molecular) mass, density ρ, relative density D for another gas, as well as mass w and volumetric φ fractions of individual gases:
M (mixtures) \u003d m (mixtures) n (mixtures), (1.22)
w \u003d m (gas) m (mixture), (1.23)
φ \u003d V (gas) V (mixture), (1.24)
φ \u003d n (gas) n (mixture), (1.25)
D A (mixtures) \u003d M (mixtures) M (A), (1.26)
ρ (mixture) \u003d M (mixture) V m \u003d m (mixture) V (mixture). (1.27)
It is convenient to find the molar mass of a gas mixture by volume fractions and molar masses of individual gases:
M (mixtures) \u003d M 1 φ 1 + M 2 φ 2 + M 3 φ 3 + ... + M n φ n. (1.28)
Obviously:
φ 1 + φ 2 + φ 3 + ... + φ n \u003d 1.
For a mixture of two gases (φ 1 + φ 2 \u003d 1) φ 2 \u003d 1 - φ 1. Then
M (mixtures) \u003d M 1 φ 1 + M 2 φ 2 \u003d M 1 φ 1 + M 2 (1 - φ 1). (1.29)
Example 1.7. Find the molar mass of a gas mixture (n.u.) consisting of nitrogen with a volume (n.a.) 1.12 dm 3 and oxygen with a mass of 5.76 g.
Decision . Using formulas (1.12) and (1.17), we find the chemical amount of gases and mixture:
n (O 2) \u003d m (O 2) M (O 2) \u003d 5.76 32 \u003d 0.18 (mol),
n (N 2) \u003d V (N 2) V m \u003d 1.12 22.4 \u003d 0.05 (mol).
Thus,
n (mixture) \u003d n (O 2) + n (N 2) \u003d 0.05 + 0.18 \u003d 0.23 (mol).
Using the formula (1.25), we find the volume fractions of gases in the mixture:
φ (N 2) \u003d 0.05 0.23 \u003d 0.217,
φ (O 2) \u003d 0.18 0.23 \u003d 0.783
or (since the mixture consists of two gases):
φ (O 2) \u003d 1 - 0.217 \u003d 0.783.
Using the formula (1.29), we find the molar mass of the mixture:
M (mixtures) \u003d M (O 2) φ (O 2) + M (N 2) φ (N 2);
M (mixtures) \u003d 32 ⋅ 0.783 + 28 ⋅ 0.217 \u003d 31.2 (g / mol).
Answer: 31.2 g / mol.
1. The molar mass of a mixture of gases is between the values \u200b\u200bof the molar mass of the lightest and heaviest gas in the mixture. For example, the molar mass of a mixture of NH 3 (M \u003d 17 g / mol) and CO 2 (M \u003d 44 g / mol), depending on the volume fraction of gases, can take values \u200b\u200b17< M (смеси) < 44 (г/моль).
2. If the molar masses of the gases in the mixture are the same, then the molar mass of the mixture does not depend on the volume fractions of individual gases. For example, the molar mass of a mixture of CO, C 2 H 2 and N 2 is always equal to 28 g / mol, regardless of the volume fraction of the components.
3. If a gas is added to a mixture of gases, M of which is greater than M of the heaviest gas in the mixture, then M (mixture) increases. For example, if CO 2 is added to mixtures of N 2 and O 2 of different composition, then M (mixture) will increase.
4. If a gas is added to the mixture of gases, M of which is less than the M of the lightest gas in the mixture, then the M (mixture) of the mixture decreases. For example, if He is added to mixtures of Ne and Ar of different composition, then M (mixtures) will decrease.
5. If the volume fractions of gases in the mixture are equal, the molar mass of the mixture is equal to the arithmetic mean of the molar masses of individual gases. For example, for a mixture of equal volumes of CO 2 and O 2:
M (mixtures) \u003d M (O 2) + M (CO 2) 2 \u003d 32 + 44 2 \u003d 38 g / mol.
Amount of substance in chemistry (moles):
Formulas in chemistry determine what a substance is made of. Now we will learn how to determine in what quantities these substances are present in compounds.
Amount of substance is, in fact, the number of smallest particles (or structural units) that make up the substance. The smallest particles are either atoms (Fe) (they have only one element) or molecules (H 2 O) (from different elements).
Amount of substance in chemistry expressed through (this is the Greek letter "nu", which is similar to the English "v", only with rounded tops).
Even in a grain of matter, there are billions of molecules, so everyone does not count them, but uses special units of measurement - moths.
1 mol is an amount of a substance equal to 6.02 * 10 23 structural units of a substance... That is exactly how many (6.02 * 10 23) molecules, for example, in one mole of water or sugar or something else.
As you can see, this is very, very much - a billion multiplied by a billion, by another 100,000 and by 6 !!! If you take so many one-kopeck coins and lay them out the entire surface of the Earth (as well as all the seas and oceans), you get a layer 1 km thick!
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The amount of substance contained in a body is determined by the number of molecules (or atoms) in that body. Since the number of molecules in macroscopic bodies is very large, to determine the amount of matter in the body, the number of molecules in it is compared with the number of atoms in 0.012 kg of the carbon isotope \\ (~ ^ (12) _6C \\).
Amount of substance ν is a quantity equal to the ratio of the number of molecules (atoms) N in a given body to the number of atoms N A in 0.012 kg of carbon isotope \\ (~ ^ (12) _6C \\):
\\ (~ \\ nu \u003d \\ frac (N) (N_A). \\ qquad (2) \\)
In SI, the unit of the amount of a substance is the mole. 1 mole - the amount of a substance that contains the same structural elements (atoms, molecules, ions), how many atoms are in 0.012 kg of the carbon isotope \\ (~ ^ (12) _6C \\).
The number of particles in one mole of a substance is called avogadro constant.
\\ (~ N_A \u003d \\ frac (0.012) (m_ (0C)) \u003d \\ frac (0.012) (1.995 \\ cdot 10 ^ (- 26)) \\) \u003d 6.02 10 23 mol -1. (3)
Thus, 1 mole of any substance contains the same number of particles - N A particles. Since the mass m 0 particles are different for different substances, then the mass N The A of particles is different for different substances.
The mass of a substance taken in an amount of 1 mol is called molar mass M:
\\ (~ M \u003d m_0 N_A. \\ Qquad (4) \\)
In SI, the unit of molar mass is kilogram per mole (kg / mol).
Between molar mass Μ and relative molecular weight M r the following relationship exists:
\\ (~ M \u003d M_r \\ cdot 10 ^ (- 3). \\)
So, the molecular weight of carbon dioxide is 44, the molar weight is 44 · 10 -3 kg / mol.
Knowing the mass of a substance and its molar mass M, you can find the number of moles (amount of substance) in the body \\ [~ \\ nu \u003d \\ frac (m) (M) \\].
Then from formula (2) the number of particles in the body
\\ (~ N \u003d \\ nu N_A \u003d \\ frac (m) (M) N_A. \\)
Knowing the molar mass and Avogadro's constant, you can calculate the mass of one molecule:
\\ (~ m_0 \u003d \\ frac (M) (N_A) \u003d \\ frac (m) (N). \\)
Literature
Aksenovich L.A. Physics in high school: Theory. Tasks. Tests: Textbook. allowance for institutions providing the receipt of obs. environments, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Minsk: Adukatsya i vyhavanne, 2004. - P. 124-125.
Lesson objectives:
- Introduce the concept of the amount of a substance and its units of measurement: mol, mmol, kmol.
- Give an idea of \u200b\u200bthe Avogadro constant.
- Show the relationship between mass, amount of matter and number of particles.
Lesson Objectives:
- 1. To contribute to the formation of students' worldview ideas about the relationship between different properties of the phenomena of the surrounding world.
- 2. Develop the ability of students to establish causal relationships, as well as observe, generalize and draw conclusions.
Basic terms:
- Nonmetals - chemical elements that form simple substances in free form that do not have physical properties metals.
- a mole is an amount of any substance that contains as many structural elements as atoms contained in 12g. nuclide carbon-12
DURING THE CLASSES
Amount of substance
In chemistry (as well as in physics and other natural sciences), you have to deal with large quantities the smallest particles - with the so-called structural elements of matter (molecules, atoms , ions, electrons, etc.).
In order to express the number of such particles, a unit of quantity was introduced - the mole. 1 mole is the amount of any substance that contains the same number of structural elements as atoms are contained in 12g. nuclide carbon-12. It was found experimentally that the number of structural elements corresponding to 1 mol is 6.02 ∙ 1023 (the constant 6.02 ∙ 1023 mol-1 is called Avogadro's constant. Cylinders with substances in 1 mol).
Fig. 1. Avogadro's constant
Illustration of the consequence of Avogadro's law
Fig. 2. - a unit of the amount of a substance
Mole is a unit of amount of a substance 
Fig. 3. Amount of substance
This portion of a substance has a mass called molar mass. Designated by M, which is found by the formula M \u003d m / n. Guess what units the molar mass will be measured in?
The molar mass coincides in value with the relative atomic or molecular mass, but differ in units of measurement (M - g / mol; Mr, Ar - dimensionless quantities). 
Fig. 4. Amount of substance in moles 
Fig. 5. Molar mass
Control unit
№1.
The mass of 3 mol H2O is ____ g
The mass of 20 mol H2O is ____ g
№2.
36 g of H2O is ______ mol
180g H2O is _______ mol
Homework
How many molecules are there in 180 g of water?
Find the mass of 24x1023 ozone molecules?
Oxygen is the most abundant chemical element in the earth's crust. Oxygen is a part of almost all substances around us. So, for example, water, sand, many rocks and minerals that make up earth crustcontain oxygen. Oxygen is also an important part of many organic compounds, for example, proteins, fats and carbohydrates, which are of exceptional importance in the life of plants, animals and humans.
In 1772 the Swedish chemist K.V. Scheele established that air consists of oxygen and nitrogen. In 1774, D. Priestley obtained oxygen by decomposition of mercury oxide (2). Oxygen is a colorless gas, odorless and tasteless, it is relatively slightly soluble in water, slightly heavier than air: 1 liter of oxygen under normal conditions weighs 1.43 g, and 1 liter of air weighs 1.29 g (Normal conditions - abbreviated: n. - temperature 0 ° C and pressure 760 mm Hg, or 1 atm). At a pressure of 760 mm Hg. Art. and at a temperature of - 183 ° C, oxygen liquefies, and when the temperature drops to - 218.8 ° C, it solidifies.
The chemical element oxygen O, in addition to ordinary oxygen O2, exists in the form of another simple substance - ozone O3. Oxygen O2 is converted to ozone in a device called an ozonizer.
It is a gas with a pungent characteristic odor (the name “ozone” in Greek means “smelling”). You probably smelled ozone more than once during a thunderstorm. Ozone is made up of three atoms of the element oxygen. Pure ozone is a blue gas, one and a half times heavier than oxygen, it dissolves better in water.
An ozone layer exists in the air above the Earth at an altitude of 25 km. There, ozone is formed from oxygen under the influence of ultraviolet radiation from the sun. In turn, the ozone layer traps this radiation, which is dangerous for all living beings, which provides normal life on the ground.
Ozone is used for disinfection drinking water, since ozone oxidizes harmful impurities in natural water. In medicine, ozone is used as a disinfectant.
List of references
1. Lesson on the topic "Amount of substance", teacher of biology and chemistry Yakovleva Larisa Alexandrovna, Kurgan region, Petukhovsky district, MOU "Novogeorgievskaya secondary school"
2. F. A. Derkach "Chemistry" - scientific and methodological manual. - Kiev, 2008.
3. LB Tsvetkova "Inorganic Chemistry" - 2nd edition, revised and enlarged. - Lviv, 2006.
4. V. V. Malinovsky, P. G. Nagorny "Inorganic chemistry" - Kiev, 2009.
4. Glinka N.L. General chemistry. - 27th ed. / Under. ed. V.A. Rabinovich. - L .: Chemistry, 2008 .-- 704 pp.
Edited and sent by Borisenko I.N.