Molecular weight is one of the basic concepts in modern chemistry. Its introduction became possible after the scientific substantiation of Avogadro's assertion 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 important concepts and laws.
Mass units of elements
Initially, the hydrogen atom was taken as the base unit of atomic and molecular mass as the lightest element in the Universe. But the atomic masses for the most part were calculated on the basis of their oxygen compounds, so it was decided to choose a new standard for determining atomic masses. The atomic mass of oxygen was taken equal to 15, the atomic mass of the lightest substance on Earth, hydrogen, - 1. In 1961, the oxygen system for determining the weight was generally accepted, but 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 amu) is 1/12 of the mass of this standard. At present, atomic mass is the mass of an atom, which must be expressed in amu.
Molecule mass
The mass of a molecule of any substance is equal to the sum of the masses of all atoms that form a given molecule. The lightest molecular weight of a gas is hydrogen, its compound is written as H 2 and has a value close to two. A water molecule consists of an oxygen atom and two hydrogen atoms. This means that its molecular weight is 15.994 + 2 * 1.0079 = 18.0152 amu. Complex organic compounds - proteins and amino acids - have the largest molecular weights. The molecular weight of a protein structural unit ranges from 600 to 10 6 and more, depending on the number of peptide chains in this macromolecular structure.
Moth
Along with the standard units of mass and volume in chemistry, a completely special system unit is used - 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 applying 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 what structural units are in question - for example, the mole of H + ions, the mole of H 2 molecules, and so on.
Molar and molecular weight
The mass of the amount of a substance in 1 mol is measured in g / mol and is called the molar mass. The relationship between molecular and molar mass can be written as the equation
ν = k × m / M, where k is the coefficient of proportionality.
It is easy to say that for any ratios the proportionality coefficient will be equal to one. Indeed, the isotope of carbon has a relative molecular weight of 12 amu, and, according to the definition, the molar mass of this substance is 12 g / mol. The ratio of molecular weight to molar is 1. Hence, we can conclude that the molar and molecular weight have the same numerical values.
Gas volumes
As you know, all substances around us can be in a solid, liquid or gaseous state of aggregation. For solids, the most common base 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 distance in liquids or solids. For example, under normal conditions one mole of water takes up a volume of 18 ml - approximately the same amount fits in one tablespoon. The volume of one mole of fine-crystalline table salt is 58.5 ml, and the volume of 1 mole of sugar is 20 times more than a mole of water. Even more space is required for gases. Under normal conditions, one mole of nitrogen takes up a volume 1240 times greater than one mole of water.
Thus, the volumes of gaseous substances differ significantly from the volumes of liquid and solid. This is due to the difference in distances between molecules of substances in different states of aggregation.
Normal conditions
The state of any gas is highly dependent on temperature and pressure. For example, nitrogen at a temperature of 20 ° C takes up a volume of 24 liters, and at 100 ° C at the same pressure - 30.6 liters. Chemists took into account this dependence, 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, these are:
- Temperature at 0 ° C.
- Pressure 101.3 kPa.
For normal conditions, a special abbreviation is adopted - n.u. Sometimes in problems this designation is not written, 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 gas
As an example, it is easy to calculate one mole of any gas, such as nitrogen. To do this, you first need to find the value of its relative molecular weight:
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 has been experimentally found that under normal conditions the density of nitrogen is 1.25 g / liter.
Substitute this value into the standard formula known from the school physics course, where:
- V is the gas volume;
- m is the mass of the gas;
- ρ is the density of the gas.
We get 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 takes 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 is its structure and physicochemical characteristics, 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 makes it possible to solve many chemical problems associated with measuring the properties of gases under normal conditions.
Outcomes
The molecular weight of gaseous substances is important for 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 fulfilled:
ν = m / M ν = V / V m.
If we remove the constant ν, we can equalize these two expressions:
So you can calculate the mass of one portion of the substance and its volume, and also the molecular weight of the investigated substance becomes known. Using this formula, you can easily calculate the volume-to-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 investigated gas.
It should be remembered that a strict correspondence of the real molecular weight of a substance to that found by 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 in the third or fourth decimal place in the result found. Therefore, for school problems and experiments, the results found are quite plausible.
DEFINITION
The ratio of the mass (m) of a substance to its amount (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 a 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 coefficient of proportionality, which is the same for all substances. Relative molecular weight is a dimensionless quantity. It is calculated using the relative atomic masses of the chemical elements specified in the Periodic Table of D.I. Mendeleev.
The relative atomic mass of atomic nitrogen is 14.0067 amu. Its relative molecular weight will be equal to 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 the 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 just 28 g / mol.
Nitrogen is a colorless gas that has neither odor nor taste (a 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 in the form of two isotopes 14 N (99.635%) and 15 N (0.365%). These isotopes are characterized by a different content of neutrons in the nucleus of an atom, and hence by a different molar mass. In the first case, it will be equal to 14 g / mol, and in the second - 15 g / mol.
The molecular weight of a substance in a gaseous state can be determined using the concept of its molar volume. To do this, find the volume occupied under normal conditions by 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 (calculating molar mass), it is possible to use the equation of state for an 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 is the absolute temperature (K), R is the 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 36 g of magnesium. |
| Solution | Let us write the equation for the reaction of the chemical interaction of magnesium with nitrogen: 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 tasks are given below. 3.3 Questions for self-control1. 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 on the basis of atomic-molecular theory? 5. How is the law of composition constancy formulated? 6. Formulate the law of simple volumetric relations. 7. How is Avogadro's law formulated? 8. Formulate the consequences of Avogadro's law. 9. What is molar volume? What is it equal to? 10. What is the relative density of gases? 11. Knowing the relative density of a gas, how can you determine its molar mass? 12. What parameters characterize the gas state? 13. What units of measure for mass, volume, pressure and temperature do you know? 14. What is the difference between the Celsius and Kelvin temperature scales? 15. What conditions of the gas state are considered normal? 16. How can the gas volume be brought to normal conditions? 17. What is called a substance equivalent? 18. What is a 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 an equivalent molar 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 11. Under the same conditions, taken equal volumes of O 2 and C1 2. What is the mass ratio of both gases? 1) m(O 2)> m(Cl 2), 2) m(O 2)< m(Cl 2), 3) m(O 2) = m(Cl 2). 2. What is the value of the relative density of oxygen in terms of hydrogen? 1) 32, 2) 8, 3) 16, 4) 64. 3. How many moles of sulfuric acid equivalents are contained in 1 mole of the 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 + 3NаОН = Fe (ОН) 3 + 3NаС1? 1) 1/2, 2) 1, 3) 1/3, 4) 1/4, 5) 1/6. 5. What is the mass of zinc in grams, which must be taken in order to release 5.6 liters of hydrogen during the reaction with an acid? 1) 65, 2) 32,5, 3) 16,25, 4) 3,25. See page 26 for answers. Option 21. Mix equal volumes of hydrogen and chlorine. How will the volume of the mixture change after the reaction? 1) Will increase 2 times 2) decrease 2 times 3) will not change.2. The mass of a gas with a volume of 2.24 liters (under normal conditions) is 2.8 g. What is the value of the relative molecular weight of the gas? 1) 14, 2) 28, 3) 28 G / mol, 4) 42.3. Under what number is the formula of nitrogen oxide, the molar mass of nitrogen equivalent in which is equal to 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. Under what number is the value of the volume of hydrogen in liters at normal conditions, which will be released when 18 g of metal is dissolved in acid, the molar mass of the equivalent of which is 9? 1) 22,4, 2) 11,2, 3) 5,6, 4) 2,24. 5. What is the equivalent of hydroxy iron (III) nitrate 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. See page 26 for answers. Problem 80. Answer: Problem 81.
Answer: Problem 82. Here R is the universal gas constant equal to 8.314 J / (mol. K); T is the gas temperature, K; P is the gas pressure, Pa; V - gas volume, m3; M is the molar mass of the 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 g. Problem 83.
The density of a gas in air is equal to the ratio of the molar mass of this gas to the molar mass of air:
Here is the density of the gas in the 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 for the second, denoted by D. Therefore, by the condition 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 of 22.4 liters. Knowing the mass and volume of 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 for the second, denoted by D. Therefore, by the condition of the problem:
The molar mass of air is 29 g / mol. Then M 1 = D . M 2 = 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 a 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 for the second, denoted by D. Therefore, by the condition of the problem:
The molar mass of nitrogen is 28 g / mol. Then the molar mass of sulfur vapor is: M 1 = D . M 2 = 9.14. 2 = 255.92 g / mol. Knowing that Ar (S) = 32g / mol, we find the number of atoms (n) that make up a sulfur molecule:
Thus, the 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 is the gas temperature, K; P is the gas pressure, Pa; V - gas volume, m 3; M is the molar mass of the gas, g / mol.
Answer: 58 g / mol. Problem 89. Expressing these problems in the SI system of 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 is the gas temperature, K; P is the gas pressure, Pa; V - gas volume, m 3; M is the molar mass of the gas, g / mol.
Answer: 58 g / mol. |
