V eq1 and V eq2 – molar volumes of their equivalents.
Using the considered stoichiometric laws, it is possible to solve a wide range of problems. Examples of solving a number of typical problems are given below.
3.3.Questions for self-control
1. What is stoichiometry?
2. What stoichiometric laws do you know?
3. How is the law of conservation of mass of substances formulated?
4. How to explain the validity of the law of conservation of mass of substances based on the atomic-molecular theory?
5. How is the law of constancy of composition formulated?
6. Formulate the law of simple volumetric relations.
7. How is Avogadro's law formulated?
8. Formulate consequences from Avogadro’s law.
9. What is molar volume? What is it equal to?
10. What is the relative density of gases?
11. How, knowing the relative density of a gas, can one determine its molar mass?
12. What parameters characterize the gas state?
13. What units of mass, volume, pressure and temperature do you know?
14. What is the difference between the Celsius and Kelvin temperature scales?
15. What gas conditions are considered normal?
16. How can the volume of gas be brought to normal conditions?
17. What is called the equivalent of a substance?
18. What is the molar mass equivalent?
19. How is the equivalence factor determined for a) oxide,
b) acids, c) bases, d) salts?
20. What formulas can be used to calculate the equivalent for a) oxide, b) acid, c) base, d) salt?
21. What formulas can be used to calculate the molar masses of equivalents for a) oxide, b) acid, c) base, d) salt?
22. What is the molar equivalent volume?
23. How is the law of equivalents formulated?
24. What formulas can be used to express the law of equivalents?
3.4. Tests for self-control on the topic “Equivalent” Option 1
1. Under the same conditions, equal volumes of O 2 and C1 2 are taken. What is the ratio of the masses of both gases?
1) m(O 2) > m(Cl 2), 2) m(O2)< m(Cl 2), 3) m(O2) = m(Cl 2).
2. What is the relative density of oxygen to hydrogen?
1) 32, 2) 8, 3) 16, 4) 64.
3. How many moles of sulfuric acid equivalents are contained in 1 mole of molecules of this substance participating in the complete neutralization reaction?
1) 2, 2) 1, 3) 1/2, 4) 1/6, 5) 1/4.
4. What is the equivalent of iron (III) chloride in the reaction
FeCl 3 + 3NaOH = Fe(OH) 3 + 3NaC1?
1) 1/2, 2) 1, 3) 1/3, 4) 1/4, 5) 1/6.
5. What is the mass of zinc in grams that must be taken in order for the reaction with acid to release hydrogen with a volume of 5.6 liters?
1) 65, 2) 32,5, 3) 16,25, 4) 3,25.
For answers, see page 26.
Option 2
1. Mix equal volumes of hydrogen and chlorine. How will the volume of the mixture change after the reaction occurs?
1) Increase by 2 times 2) decrease by 2 times 3) will not change.
2. The mass of a gas with a volume of 2.24 liters (under normal conditions) is equal to 2.8 g. What is the value of the relative molecular weight gas?
1) 14, 2) 28, 3) 28 G/mol, 4) 42.
3. What number is the formula of nitric oxide, the molar mass of nitrogen equivalent in which is 7 g/mol?
1) N 2 O, 2) NO, 3) N 2 O 3, 4) N 2 O 4, 5) N 2 O 5.
4. What number indicates the volume of hydrogen in liters at standard conditions, which will be released when 18 g of a metal is dissolved in acid, the molar mass of which is equivalent to 9?
1) 22,4, 2) 11,2, 3) 5,6, 4) 2,24.
5. What is the equivalent of iron hydroxyl nitrate (III) in the reaction:
Fe(NO 3) 3 + NaOH = Fe(OH) 2 NO 3 + NaNO 3?
1) 1/4, 2) 1/6, 3) 1, 4) 1/2, 5) 1/3.
For answers, see page 26.
DEFINITION
The ratio of the mass (m) of a substance to its quantity (n) is called molar mass of the substance:
Molar mass is usually expressed in g/mol, less often in kg/kmol. Since one mole of any substance contains the same number of structural units, the molar mass of the substance is proportional to the mass of the corresponding structural unit, i.e. relative atomic mass of a given substance (M r):
where κ is the proportionality coefficient, the same for all substances. Relative molecular weight is a dimensionless quantity. It is calculated using relative atomic masses chemical elements, specified in Periodic table DI. Mendeleev.
The relative atomic mass of atomic nitrogen is 14.0067 amu. Its relative molecular mass will be 14.0064 and its molar mass:
M(N) = M r (N) × 1 mol = 14.0067 g/mol.
It is known that the nitrogen molecule is diatomic - N 2, then the relative atomic mass of the nitrogen molecule will be equal to:
A r (N 2) = 14.0067 × 2 = 28.0134 amu
The relative molecular mass of a nitrogen molecule will be equal to 28.0134, and the molar mass:
M(N 2) = M r (N 2) × 1 mol = 28.0134 g/mol or simply 28 g/mol.
Nitrogen is a colorless gas that has neither odor nor taste (the diagram of the atomic structure is shown in Fig. 1), poorly soluble in water and other solvents with very low melting points (-210 o C) and boiling points (-195.8 o C).
Rice. 1. The structure of the nitrogen atom.
It is known that in nature nitrogen can be found in the form of two isotopes 14 N (99.635%) and 15 N (0.365%). These isotopes are characterized by different neutron contents in the atomic nucleus, and therefore by molar mass. In the first case it will be equal to 14 g/mol, and in the second - 15 g/mol.
Molecular mass of a substance in gaseous state can be determined using the concept of its molar volume. To do this, find the volume occupied at normal conditions a certain mass of a given substance, and then calculate the mass of 22.4 liters of this substance under the same conditions.
To achieve this goal (calculation of molar mass), it is possible to use the equation of state ideal gas(Mendeleev-Clapeyron equation):
where p is the gas pressure (Pa), V is the gas volume (m 3), m is the mass of the substance (g), M is the molar mass of the substance (g/mol), T - absolute temperature(K), R - universal gas constant equal to 8.314 J/(mol×K).
Examples of problem solving
EXAMPLE 1
EXAMPLE 2
| Exercise | Calculate the volume of nitrogen (normal conditions) that can react with magnesium weighing 36 g. |
| Solution | Let's write the reaction equation chemical interaction magnesium with nitrogen: Problem 80. Answer: Problem 81.
Answer: Problem 82. Here R is the universal gas constant equal to 8.314 J/(mol. K); T – gas temperature, K; P – gas pressure, Pa; V – gas volume, m3; M – molar mass of gas, g/mol.
b) 1 mole of any substance contains 6.02 . 10 23 particles (atoms, molecules), then the mass of one molecule is calculated from the ratio:
Answer: M = 28g/mol; m = 4.65 . 10 -23 years Problem 83.
The density of a gas in air is equal to the ratio of the molar mass of a given gas to the molar mass of air:
Here is the gas density in air; - molar mass of gas; - air (29g/mol). Then
Problem 84.
The molar mass of oxygen is 32 g/mol. Then Answer: Problem 85. where m 1 /m 2 is the relative density of the first gas relative to the second, denoted by D. Therefore, according to the conditions of the problem:
The molar mass of nitrogen is 28 g/mol. Then b) 1 mole of any gas under normal conditions (T = 0 0 C and P = 101.325 kPa) occupies a volume equal to 22.4 liters. Knowing the mass and volume of the gas under normal conditions, we calculate molar mass it, making up the proportion:
Answer: M (Gas) = 34 g/mol. Problem 86. where m 1 /m 2 is the relative density of the first gas relative to the second, denoted by D. Therefore, according to the conditions of the problem:
The molar mass of air is 29 g/mol. Then M 1 = D . M2 = 6.92 . 29 = 200.6 g/mol. Knowing that Ar(Hg) = 200.6 g/mol, we find the number of atoms (n) that make up the mercury molecule:
Thus, a mercury molecule consists of one atom. Answer: from one. Problem 87. where m 1 /m 2 is the relative density of the first gas relative to the second, denoted by D. Therefore, according to the conditions of the problem:
The molar mass of nitrogen is 28 g/mol. Then the molar mass of sulfur vapor is equal to: M 1 = D . M2 = 9.14. 2 = 255.92 g/mol. Knowing that Ar(S) = 32 g/mol, we find the number of atoms (n) that make up the sulfur molecule:
Thus, a sulfur molecule consists of one atom. Answer: out of eight. Problem 88. Here R is the universal gas constant equal to 8.314 J/(mol . TO); T – gas temperature, K; P – gas pressure, Pa; V – gas volume, m3; M – molar mass of gas, g/mol.
Answer: 58 g/mol. Problem 89. Expressing these problems in the system of SI units (P = 10.4...104Pa; V = 6.24...10-4m3; m = 1.56...10-3kg; T = 290K) and substituting them into the Clapeyron-Mendeleev equation (equation state of an ideal gas), we find the molar mass of the gas: Here R is the universal gas constant equal to 8.314 J/(mol. K); T – gas temperature, K; P – gas pressure, Pa; V – gas volume, m3; M – molar mass of gas, g/mol.
Answer: 58 g/mol. Molecular mass is one of the basic concepts in modern chemistry. Its introduction became possible after the scientific substantiation of Avogadro’s statement that many substances consist of tiny particles - molecules, each of which, in turn, consists of atoms. Science owes this judgment largely to the Italian chemist Amadeo Avogadro, who scientifically substantiated the molecular structure of substances and gave chemistry many of the most important concepts and laws. Units of mass of elementsInitially, the hydrogen atom was taken as the basic unit of atomic and molecular mass as the lightest element in the Universe. But atomic masses were mostly calculated based on their oxygen compounds, therefore, it was decided to choose a new standard for determining atomic masses. The atomic mass of oxygen was taken to be 15, the atomic mass of the lightest substance on Earth, hydrogen, was 1. In 1961, the oxygen system for determining weight was generally accepted, but it created certain inconveniences.
In 1961, a new scale of relative atomic masses was adopted, the standard for which was the carbon isotope 12 C. The atomic mass unit (abbreviated as amu) is 1/12 of the mass of this standard. Currently, atomic mass is the mass of an atom, which must be expressed in amu. Mass of moleculesThe mass of a molecule of any substance is equal to the sum of the masses of all the atoms that form this molecule. The lightest molecular weight of a gas is hydrogen; its compound is written as H2 and has a value close to two. A water molecule consists of an oxygen atom and two hydrogen atoms. This means that its molecular mass is 15.994 + 2*1.0079=18.0152 amu. The largest molecular weights have complex organic compounds- proteins and amino acids. The molecular weight of a protein structural unit ranges from 600 to 10 6 and higher, depending on the number of peptide chains in this macromolecular structure.
MoleAlong with the standard units of mass and volume, a completely special system unit is used in chemistry - the mole. A mole is the amount of a substance that contains as many structural units (ions, atoms, molecules, electrons) as is contained in 12 grams of the 12 C isotope. When using a measure of the amount of a substance, it is necessary to indicate which structural units are meant. As follows from the concept of “mole”, in each individual case it is necessary to indicate exactly which structural units we're talking about- for example, moles of H + ions, moles of H 2 molecules, etc. Molar and molecular massThe mass of 1 mole of a substance is measured in g/mol and is called molar mass. The relationship between molecular and molar mass can be written as the equation ν = k × m/M, where k is the proportionality coefficient. It is easy to say that for any ratio the proportionality coefficient will be equal to one. Indeed, the carbon isotope has a relative molecular mass of 12 amu, and, according to definition, the molar mass of this substance is 12 g/mol. The ratio of molecular mass to molar mass is 1. From this we can conclude that molar and molecular mass have the same numerical values. Gas volumesAs you know, all the substances around us can be in a solid, liquid or gaseous state of aggregation. For solids, the most common basic measure is mass, for solids and liquids - volume. This is due to the fact that solids retain their shape and finite dimensions. Liquid and gaseous substances do not have finite dimensions. The peculiarity of any gas is that between its structural units - molecules, atoms, ions - the distance is many times greater than the same distances in liquids or solids. For example, one mole of water under normal conditions occupies a volume of 18 ml - approximately the same amount as one tablespoon. The volume of one mole of finely crystalline table salt is 58.5 ml, and the volume of 1 mole of sugar is 20 times greater than a mole of water. Gases require even more space. One mole of nitrogen under normal conditions occupies a volume 1240 times larger than one mole of water.
Thus, the volumes of gaseous substances differ significantly from the volumes of liquid and solid substances. This is due to the difference in distances between molecules of substances in different states of aggregation. Normal conditionsThe state of any gas depends greatly on temperature and pressure. For example, nitrogen at a temperature of 20 °C occupies a volume of 24 liters, and at 100 °C at the same pressure - 30.6 liters. Chemists took this dependence into account, so it was decided to reduce all operations and measurements with gaseous substances to normal conditions. All over the world the parameters of normal conditions are the same. For gaseous chemicals This:
For normal conditions, a special abbreviation has been adopted - no. Sometimes this designation is not written in problems, then you should carefully re-read the conditions of the problem and bring the given gas parameters to normal conditions. Calculation of the volume of 1 mole of gasAs an example, it is not difficult to calculate one mole of any gas, such as nitrogen. To do this, you first need to find the value of its relative molecular mass: M r (N 2) = 2×14 = 28. Since the relative molecular mass of a substance is numerically equal to the molar mass, then M(N 2)=28 g/mol. It was found experimentally that under normal conditions the density of nitrogen is 1.25 g/liter. Let's substitute this value into the standard formula known from school course physics, where:
We find that the molar volume of nitrogen under normal conditions V(N 2) = 25 g/mol: 1.25 g/liter = 22.4 l/mol. It turns out that one mole of nitrogen occupies 22.4 liters. If you perform such an operation with all existing gaseous substances, you can come to an amazing conclusion: the volume of any gas under normal conditions is 22.4 liters. Regardless of what kind of gas we are talking about, what its structure and physicochemical characteristics are, one mole of this gas will occupy a volume of 22.4 liters.
The molar volume of a gas is one of the most important constants in chemistry. This constant allows us to solve many chemical problems related to measuring the properties of gases under normal conditions. ResultsThe molecular weight of gaseous substances is important in determining the amount of a substance. And if a researcher knows the amount of substance of a particular gas, he can determine the mass or volume of such a gas. For the same portion of a gaseous substance, the following conditions are simultaneously satisfied: ν = m/ M ν= V/ V m. If we remove the constant ν, we can equate these two expressions: This way you can calculate the mass of one portion of a substance and its volume, and the molecular mass of the substance under study also becomes known. Using this formula, you can easily calculate the volume-mass ratio. When this formula is reduced to the form M= m V m /V, the molar mass of the desired compound will become known. In order to calculate this value, it is enough to know the mass and volume of the gas under study. It should be remembered that a strict correspondence between the real molecular weight of a substance and that found using the formula is impossible. Any gas contains a lot of impurities and additives that make certain changes in its structure and affect the determination of its mass. But these fluctuations introduce changes to the third or fourth decimal place in the found result. Therefore for school tasks and experiments, the results found are quite plausible. We recommendBasic types of subordinate clauses“The science of tender passion, which Nazon sang...” On the eve of A.S.’s birthday. Pushkin, certain lines of our brilliant poet involuntarily come to mind....
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