The Law of the Vanguard in Chemistry. Where is the Avogadro number used? Moth. Molar mass. Molar volume
Avogadro's law was formulated by the Italian chemist Amadeo Avogadro in 1811 and was of great importance for the development of chemistry at that time. However, even today it has not lost its relevance and significance. Let's try to formulate Avogadro's law, it will sound something like this.
Formulation of Avogadro's law
So, Avogadro's law states that at the same temperatures and pressures, equal volumes of gases will contain the same number of molecules, regardless of both their chemical nature and physical properties. This number is a certain physical constant, equal to the number of molecules, ions contained in one mole.
Initially, Avogadro's law was only a scientist's hypothesis, but later this hypothesis was confirmed by a large number of experiments, after which it entered science under the name "Avogadro's law", which was destined to become the basic law for ideal gases.
Avogadro's law formula
The discoverer of the law himself believed that the physical constant is a large quantity, but he did not know which one. Already after his death, in the course of numerous experiments, the exact number of atoms contained in 12 g of carbon (namely, 12 g is the atomic mass unit of carbon) or in the molar volume of gas equal to 22.41 liters was established. This constant, in honor of the scientist, was called the "Avogadro number", it is designated as NA, less often L and it is equal to 6.022 * 1023. In other words, the number of molecules of any gas in a volume of 22.41 liters will be the same for both light and heavy gases.
The mathematical formula for Avogadro's law can be written as follows:
Where, V is the volume of gas; n is the amount of a substance, which is the ratio of the mass of a substance to its molar mass; VM is a constant of proportionality or molar volume.
Application of Avogadro's Law
Further practical application of Avogadro's law greatly helped chemists to determine the chemical formulas of many compounds.
Mole and Avogadro's number, video
And finally, an educational video on the topic of our article.
Avogadro's law, discovered in 1811, played a major role in the development of chemistry. First of all, he contributed to the recognition of the atomic-molecular doctrine, formulated for the first time in the middle of the 18th century. M.V. Lomonosov. So, for example, using the Avogadro number:
it turned out to be possible to calculate not only the absolute masses of atoms and molecules, but also the actual linear dimensions of these particles. According to Avogadro's law:
"Equal volumes of different gases at constant pressure and temperature contain the same number of molecules, equal to"
A number of important consequences follow from Avogadro's law regarding the molar volume and density of gases. So, it directly follows from Avogadro's law that the same number of molecules of different gases will occupy the same volume, equal to 22.4 liters. This volume of gases is called molar volume. The reverse is also true - the molar volume of various gases is the same and equal to 22.4 liters:
Indeed, since 1 mole of any substance contains the same number of molecules, equal to , it is obvious that their volumes in the gaseous state under the same conditions will be the same. Thus, under normal conditions (n.o.), i.e. at pressure 
and temperature, the molar volume of various gases will be 
. The amount of substance, volume and molar volume of gases can be related to each other in the general case by a relation of the form:
from where respectively:
In the general case, normal conditions (n.s.) are distinguished:
standard conditions include:
To convert a Celsius temperature to a Kelvin temperature, use the following relationship:
The mass of the gas itself can be calculated from the value of its density, i.e.
Because as shown above:
then obviously:
from where respectively:
From the above relations of the form:
after substitution in the expression:
it also follows that:
from where respectively:
and thus we have:
Since under normal conditions, 1 mole of any occupies a volume equal to:
then respectively:
The ratio obtained in this way is quite important for understanding the 2nd consequence of Avogadro's law, which in turn is directly related to such a concept as the relative density of gases. In the general case, the relative density of gases is a value showing how many times one gas is heavier or lighter than another, i.e. how many times the density of one gas is greater or less than the density of another, i.e. we have a relation of the form:
So, for the first gas we have:
respectively for the second gas:
then obviously:
and thus:
In other words, the relative density of a gas is the ratio of the molecular weight of the gas under investigation to the molecular weight of the gas with which the comparison is made. The relative density of a gas is a dimensionless quantity. Thus, in order to calculate the relative density of one gas from another, it is sufficient to know the molecular relative molecular weights of these gases. In order to make it clear with which gas the comparison is made, an index is put. For example, it means that the comparison is made with hydrogen and then they talk about the density of the gas with respect to hydrogen, without using the word “relative” already, taking it as if by default. Similarly, measurements are carried out, taking air as the reference gas. In this case, it is indicated that the comparison of the test gas is carried out with air. In this case, the average molecular weight of air is assumed to be 29, and since the relative molecular weight and the molar mass are numerically the same, then:
The chemical formula of the gas under study is put next to it in brackets, for example:
and is read as - the density of chlorine by hydrogen. Knowing the relative density of one gas with respect to another, one can calculate the molecular as well as the molar mass of the gas, even if the formula of the substance is unknown. All the above relations refer to the so-called normal conditions.
Story
The first quantitative studies of reactions between gases belong to the French scientist Gay-Lussac. He is the author of the laws on thermal expansion of gases and the law of volumetric ratios. These laws were explained in 1811 by the Italian physicist Amedeo Avogadro.
Consequences of the law
First consequence from Avogadro's law: one mole of any gas under the same conditions occupies the same volume.
In particular, under normal conditions, i.e. at 0 °C (273K) and 101.3 kPa, the volume of 1 mole of gas is 22.4 liters. This volume is called the molar volume of gas V m . You can recalculate this value to other temperatures and pressures using the Mendeleev-Clapeyron equation:
.Second consequence from Avogadro's law: the molar mass of the first gas is equal to the product of the molar mass of the second gas and the relative density of the first gas according to the second.
This position was of great importance for the development of chemistry, since it makes it possible to determine the partial weight of bodies capable of passing into a gaseous or vaporous state. If through m we denote the partial weight of the body, and through d is its specific gravity in the vapor state, then the ratio m / d should be constant for all bodies. Experience has shown that for all the bodies studied, passing into steam without decomposition, this constant is equal to 28.9, if, when determining the partial weight, we proceed from the specific gravity of air, taken as a unit, but this constant will be equal to 2, if we take the specific gravity as a unit hydrogen. Denoting this constant, or, what is the same, the partial volume common to all vapors and gases through FROM, we have from the formula on the other hand m = dC. Since the specific gravity of steam is easily determined, substituting the value d in the formula, the unknown partial weight of the given body is also displayed.
Elemental analysis, for example, of one of the polybutylenes indicates that the share ratio of carbon to hydrogen in it is 1 to 2, and therefore its partial weight can be expressed by the formula CH 2 or C 2 H 4, C 4 H 8 and in general (CH 2) n. The partial weight of this hydrocarbon is immediately determined, following Avogadro's law, since we know the specific gravity, i.e., the density of its vapor; it was determined by Butlerov and turned out to be 5.85 (in relation to air); i.e., its partial weight will be 5.85 28.9 = 169.06. Formula C 11 H 22 corresponds to a partial weight of 154, formula C 12 H 24 - 168, and C 13 H 26 - 182. The formula C 12 H 24 closely corresponds to the observed value, and therefore it should express the size of our hydrocarbon particle CH 2 .
Notes
Links
- // Encyclopedic Dictionary of Brockhaus and Efron: In 86 volumes (82 volumes and 4 additional). - St. Petersburg. , 1890-1907.
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See what "Avogadro's Law" is in other dictionaries:
LAW OF AVOGADRO- equal volumes of any ideal gases under the same conditions (temperature, pressure) contain the same number of particles (molecules, atoms). Equivalent formulation: at the same pressure and temperature, the same amounts of a substance of different ... ... Great Polytechnic Encyclopedia
Avogadro's law- - the law according to which equal volumes of ideal gases at the same temperature and pressure contain the same number of molecules. Dictionary of Analytical Chemistry ... Chemical terms
Avogadro's law- Avogadro dėsnis statusas T sritis Standartizacija ir metrologija apibrėžtis Apibrėžtį žr. priede. priedas(ai) Grafinis formatas atitikmenys: engl. Avogadro's hypothesis; Avogadro's law; Avogadro's principle vok. Avogadrosche Regel, f;… … Penkiakalbis aiskinamasis metrologijos terminų žodynas
Avogadro's law- Avogadro dėsnis statusas T sritis fizika atitikmenys: engl. Avogadro's hypothesis; Avogadro's law vok. Avogadrosche Regel, f; Avogadrosches Gesetz, n; Satz des Avogadro, m rus. Avogadro's law, m pranc. hypothese d'Avogadro, f; loi d'Avogadro, f … Fizikos terminų žodynas
Avogadro's law- Avogadro dėsnis statusas T sritis Energetika apibrėžtis Apibrėžtį žr. priede. priedas(ai) MS Word formatas atitikmenys: engl. Avogadro's law vok. Avogadrosches Gesetz, n rus. Avogadro's law, m pranc. loi d'Avogadro, f ... Aiškinamasis šiluminės ir branduolinės technikos terminų žodynas
See Chemistry and Gases. Z. the eternity of matter, or the conservation of the mass of matter, see Substance, Lavoisier, Chemistry. Z. Henry Dalton, see Solutions. Z. Gibs Le Chatelier, see Reversibility of chemical reactions. Z. (heat capacities) of Dulong and Petit, see Heat and Chemistry. Z.… … Encyclopedic Dictionary F.A. Brockhaus and I.A. Efron
Necessary, essential, stable, recurring relationship between phenomena. 3. expresses the connection between objects, the constituent elements of a given object, between the properties of things, as well as between the properties within a thing. There are 3.… … Philosophical Encyclopedia
AVOGADRO LAW- (Avogadro), based on the hypothesis expressed in 1811 by the Italian physicist Avogadro, which states that "under the same conditions of t ° and pressure, equal volumes of all gases contain the same number of molecules." From this hypothesis... Big Medical Encyclopedia
- (Avogadro) Amedeo, Count di Quaregna (1776-1856), Italian physicist and chemist. In 1811, he put forward a hypothesis (now known as Avogadro's law) that equal volumes of gases at the same pressure and the same temperature contain the same number ... ... Scientific and technical encyclopedic dictionary
- (Avogadro) Amedeo (1776-1856), Italian physicist and chemist. Founder of the molecular theory of the structure of matter (1811). He established one of the gas laws (1811; Avogadro's law), according to which in equal volumes of ideal gases with the same ... ... Modern Encyclopedia
The study of the properties of gases allowed the Italian physicist A. Avogadro in 1811. to make a hypothesis, which was later confirmed by experimental data, and became known as Avogadro's law: equal volumes of different gases under the same conditions (temperature and pressure) contain the same number of molecules.
An important consequence follows from Avogadro's law: a mole of any gas under normal conditions (0C (273 K) and a pressure of 101.3 kPa ) occupies a volume equal to 22.4 liters. This volume contains 6.02 10 23 gas molecules (Avogadro's number).
It also follows from Avogadro's law that the masses of equal volumes of different gases at the same temperature and pressure are related to each other as the molar masses of these gases:
where m 1 and m 2 are masses,
M 1 and M 2 are the molecular weights of the first and second gases.
Since the mass of a substance is determined by the formula
where ρ is the gas density,
V is the volume of gas,
then the densities of various gases under the same conditions are proportional to their molar masses. The simplest method for determining the molar mass of substances in the gaseous state is based on this consequence of Avogadro's law.
.
From this equation, you can determine the molar mass of the gas:
.
2.4 Law of volume ratios
The first quantitative studies of reactions between gases belong to the French scientist Gay-Lussac, the author of the well-known law on the thermal expansion of gases. By measuring the volumes of gases that have entered into a reaction and formed as a result of reactions, Gay-Lussac came to a generalization known as the law of simple volumetric ratios: the volumes of reacting gases are related to each other and the volumes of gaseous reaction products formed as small integers equal to their stoichiometric coefficients .
For example, 2H 2 + O 2 \u003d 2H 2 O when two volumes of hydrogen and one volume of oxygen interact, two volumes of water vapor are formed. The law is valid when the measurements of volumes are carried out at the same pressure and the same temperature.
2.5 Law of equivalents
The introduction of the concepts of "equivalent" and "molar mass of equivalents" into chemistry made it possible to formulate a law called the law of equivalents: the masses (volumes) of substances reacting with each other are proportional to the molar masses (volumes) of their equivalents .
We should dwell on the concept of the volume of mole equivalents of gas. As follows from Avogadro's law, a mole of any gas under normal conditions occupies a volume equal to 22,4 l. Accordingly, to calculate the volume of mole equivalents of a gas, it is necessary to know the number of mole equivalents in one mole. Since one mole of hydrogen contains 2 moles of hydrogen equivalents, then 1 mole of hydrogen equivalents occupies a volume under normal conditions:
3 Solving typical problems
3.1 Mol. Molar mass. Molar volume
Task 1. How many moles of iron (II) sulfide are contained in 8.8 g of FeS?
Solution Determine the molar mass (M) of iron (II) sulfide.
M(FeS)= 56 +32 = 8 8 g/mol
Let's calculate how many moles are contained in 8.8 g of FeS:
n = 8.8 ∕ 88 = 0.1 mol.
Task 2. How many molecules are there in 54 g of water? What is the mass of one molecule of water?
Solution Determine the molar mass of water.
M (H 2 O) \u003d 18 g / mol.
Therefore, 54 g of water contains 54/18 = 3 mol H 2 O. One mole of any substance contains 6.02 10 23 molecules. Then 3 moles (54g H 2 O) contain 6.02 10 23 3 = 18.06 10 23 molecules.
Let's determine the mass of one molecule of water:
m H2O \u003d 18 ∕ (6.02 10 23) \u003d 2.99 10 23 g.
Task 3. How many moles and molecules are contained in 1 m 3 of any gas under normal conditions?
Solution 1 mole of any gas under normal conditions occupies a volume of 22.4 liters. Therefore, 1 m 3 (1000 l) will contain 44.6 moles of gas:
n \u003d 1000 / 22.4 \u003d 44.6 mol.
1 mole of any gas contains 6.02 10 23 molecules. It follows from this that 1 m 3 of any gas under normal conditions contains
6.02 10 23 44.6 \u003d 2.68 10 25 molecules.
Task 4. Express in prayers:
a) 6.02 10 22 C 2 H 2 molecules;
b) 1.80 10 24 nitrogen atoms;
c) 3.01 10 23 NH 3 molecules.
What is the molar mass of these substances?
Solution A mole is the amount of a substance that contains the number of particles of any particular kind, equal to the Avogadro constant. From here
a) n C2H2 \u003d 6.02 10 22 / 6.02 10 23 \u003d 0.1 mol;
b) n N \u003d 1.8 10 24 / 6.02 10 23 \u003d 3 mol;
c) n NH3 \u003d 3.01 10 23 / 6.02 10 23 \u003d 0.5 mol.
The molar mass of a substance in grams is numerically equal to its relative molecular (atomic) mass.
Therefore, the molar masses of these substances are equal:
a) M (C 2 H 2) \u003d 26 g / mol;
b) М(N) = 14 g/mol;
c) M (NH 3) \u003d 17 g / mol.
Task 5. Determine the molar mass of a gas if, under normal conditions, 0.824 g of it occupies a volume of 0.260 liters.
Solution Under normal conditions, 1 mole of any gas occupies a volume of 22.4 liters. By calculating the mass of 22.4 liters of a given gas, we find out its molar mass.
0.824 g of gas occupy a volume of 0.260 liters
X g of gas occupy a volume of 22.4 liters
X \u003d 22.4 0.824 ∕ 0.260 \u003d 71 g.
Therefore, the molar mass of the gas is 71 g/mol.
3.2 Equivalent. Equivalence factor. Molar mass equivalents
Task 1. Calculate the equivalent, the equivalence factor and the molar mass of the equivalents of H 3 PO 4 in exchange reactions that form acidic and normal salts.
Solution Let us write down the reaction equations for the interaction of phosphoric acid with alkali:
H 3 PO 4 + NaOH = NaH 2 PO 4 + H 2 O; (one)
H 3 PO 4 + 2NaOH \u003d Na 2 HPO 4 + 2H 2 O; (2)
H 3 PO 4 + 3NaOH \u003d Na 3 PO 4 + 3H 2 O. (3)
Since phosphoric acid is a tribasic acid, it forms two acid salts (NaH 2 PO 4 - sodium dihydrogen phosphate and Na 2 HPO 4 - sodium hydrogen phosphate) and one middle salt (Na 3 PO 4 - sodium phosphate).
In reaction (1), phosphoric acid exchanges one hydrogen atom for a metal, i.e. behaves like a monobasic acid, so f e (H 3 PO 4) in reaction (1) is 1; E (H 3 RO 4) \u003d H 3 RO 4; M e (H 3 RO 4) \u003d 1 M (H 3 RO 4) \u003d 98 g / mol.
In reaction (2), phosphoric acid exchanges two hydrogen atoms for a metal, i.e. behaves like a dibasic acid, so f e (H 3 PO 4) in reaction (2) is 1/2; E (H 3 RO 4) \u003d 1/2H 3 RO 4; M e (H 3 RO 4) \u003d 1/2 M (H 3 RO 4) \u003d 49 g / mol.
In reaction (3), phosphoric acid behaves like a tribasic acid, so f e (H 3 PO 4) in this reaction is 1/3; E (H 3 RO 4) \u003d 1/3H 3 RO 4; M e (H 3 RO 4) \u003d 1/3 M (H 3 RO 4) \u003d 32.67 g / mol.
Task 2. An excess of potassium hydroxide acted on solutions of: a) potassium dihydrogen phosphate; b) dihydroxovismuth (III) nitrate. Write the equations for the reactions of these substances with KOH and determine their equivalents, equivalence factors and molar mass equivalents.
Solution We write down the equations of the occurring reactions:
KN 2 RO 4 + 2KOH \u003d K 3 RO 4 + 2 H 2 O;
Bi (OH) 2 NO 3 + KOH \u003d Bi (OH) 3 + KNO 3.
Various approaches can be used to determine the equivalent, the equivalence factor, and the molar mass of the equivalent.
The first is based on the fact that substances react in equivalent quantities.
Potassium dihydrogen phosphate reacts with two equivalents of potassium hydroxide, since E (KOH) \u003d KOH. 1/2 KH 2 PO 4 interacts with one equivalent of KOH, therefore, E (KH 2 PO 4) \u003d 1 / 2KH 2 PO 4; f e (KH 2 PO 4) = 1/2; Me (KH 2 PO 4) \u003d 1/2 M (KH 2 PO 4) \u003d 68 g / mol.
Dihydroxovismuth (III) nitrate interacts with one equivalent of potassium hydroxide, therefore, E (Bi (OH) 2 NO 3) \u003d Bi (OH) 2 NO 3; f e (Bi (OH) 2 NO 3) = 1; M e (Bi (OH) 2 NO 3) \u003d 1 M (Bi (OH) 2 NO 3) \u003d 305 g / mol.
The second approach is based on the fact that the equivalence factor of a complex substance is equal to one divided by the equivalence number, i.e. the number of formed or rearranged bonds.
Potassium dihydrogen phosphate, when interacting with KOH, exchanges two hydrogen atoms for a metal, therefore, f e (KH 2 RO 4) \u003d 1/2; E (KN 2 RO 4) \u003d 1/2 KN 2 RO 4; M e (1/2 KH 2 RO 4) \u003d 1/2 M (KH 2 RO 4) \u003d 68 g / mol.
Dihydroxovismuth (III) nitrate, when reacting with potassium hydroxide, exchanges one NO 3 - group, therefore, (Bi (OH) 2 NO 3) \u003d 1; E (Bi (OH) 2 NO 3) \u003d Bi (OH) 2 NO 3; M e (Bi (OH) 2 NO 3) \u003d 1 M e (Bi (OH) 2 NO 3) \u003d 305 g / mol.
Task 3. When 16.74 g of divalent metal was oxidized, 21.54 g of oxide was formed. Calculate the molar mass equivalents of a metal and its oxide. What is the molar and atomic mass of a metal?
Rsolution According to the law of conservation of mass of substances, the mass of the metal oxide formed during the oxidation of the metal with oxygen is equal to the sum of the masses of the metal and oxygen.
Therefore, the mass of oxygen required to form 21.5 g of oxide during the oxidation of 16.74 g of metal will be:
21.54 - 16.74 \u003d 4.8 g.
According to the law of equivalents
m Me ∕ M e (Me) = mO 2 ∕ M e (O 2); 16.74 ∕ M e (Me) = 4.8 ∕ 8.
Therefore, M e (Me) \u003d (16.74 8) ∕ 4.8 \u003d 28 g / mol.
The molar mass of the oxide equivalent can be calculated as the sum of the molar masses of the metal and oxygen equivalents:
Me (MeO) \u003d M e (Me) + M e (O 2) \u003d 28 + 8 + 36 g / mol.
The molar mass of a divalent metal is:
M (Me) \u003d Me (Me) ∕ fe (Me) \u003d 28 ∕ 1 ∕ 2 \u003d 56 g / mol.
The atomic mass of the metal (Ar (Me)), expressed in amu, is numerically equal to the molar mass Ar (Me) = 56 amu.
A physical quantity equal to the number of structural elements (which are molecules, atoms, etc.) per one mole of a substance is called Avogadro's number. Its currently officially accepted value is NA = 6.02214084(18)×1023 mol −1, it was approved in 2010. In 2011, the results of new studies were published, they are considered more accurate, but at the moment they are not officially approved.
Avogadro's law is of great importance in the development of chemistry, he allowed to calculate the weight of bodies that can change state, becoming gaseous or vaporous. It was on the basis of Avogadro's law that the atomic-molecular theory, which follows from the kinetic theory of gases, began its development.
Moreover, with the help of Avogadro's law, a method has been developed to obtain the molecular weight of solutes. To do this, the laws of ideal gases were extended to dilute solutions, based on the idea that the solute will be distributed over the volume of the solvent, as a gas is distributed in a vessel. Also, Avogadro's law made it possible to determine the true atomic masses of a number of chemical elements.
Practical use of Avogadro's number
The constant is used in the calculation of chemical formulas and in the process of compiling equations of chemical reactions. With the help of it, the relative molecular masses of gases and the number of molecules in one mole of any substance are determined.
Through the Avogadro number, the universal gas constant is calculated, it is obtained by multiplying this constant by the Boltzmann constant. In addition, by multiplying the Avogadro number and the elementary electric charge, one can obtain the Faraday constant.
Using the consequences of Avogadro's law
The first consequence of the law says: "One mole of gas (any) under equal conditions will occupy one volume." Thus, under normal conditions, the volume of one mole of any gas is 22.4 liters (this value is called the molar volume of gas), and using the Mendeleev-Clapeyron equation, you can determine the volume of gas at any pressure and temperature.
The second consequence of the law: "The molar mass of the first gas is equal to the product of the molar mass of the second gas by the relative density of the first gas to the second." In other words, under the same conditions, knowing the ratio of the density of two gases, one can determine their molar masses.
At the time of Avogadro, his hypothesis was theoretically unprovable, but it made it easy to experimentally establish the composition of gas molecules and determine their mass. Over time, a theoretical basis was brought under his experiments, and now Avogadro's number is used