Amount of substance - the number of structural elements (molecules, atoms, ions, etc.) contained in a body or system. The amount of a substance is expressed in moles. A mole is equal to the amount of substance of a system containing as many structural elements as there are atoms in 0.012 kg of the carbon isotope 12 C. The amount of substance of a body (system)
where N - the number of structural elements (molecules, atoms, ions, etc.) that make up the body (system). Avogadro constant N BUT =6,02 10 23 mol -1 .
molar mass of a substance,
where m- mass of a homogeneous body (system); -amount of substance (number of moles) of this body (system). Expressed in units of g/mol (or kg/mol).
A unit of mass equal to 1/12 of the mass of a 12 C carbon atom is called an atomic mass unit (a.m.u.). The masses of atoms or molecules expressed in atomic mass units are called, respectively, the relative atomic or relative molecular mass of the substance. The relative molecular mass of a substance consists of the relative atomic masses of the chemical elements that make up the molecule of the substance. The relative atomic masses of chemical elements are given in the table of D. I. Mendeleev (see also table 8 of the appendix of this manual).
The molar mass of a substance is numerically equal to the relative atomic or molecular weight of a given substance, if the dimension of a.m.u. be replaced by the unit g/mol.
Amount of substance in a mixture of n gases
or 
,
where v i , N i , m i , i - respectively the amount of substance, number of molecules, mass and molar mass i-th component of the mixture ( i=1,2,…,n).
Mendeleev - Clapeyron equation (ideal gas equation of state)
,
where t - mass of gas, - molar mass of gas, R - universal gas constant, ν - amount of substance, T - thermodynamic temperature.
Experimental gas laws, which are special cases of the Mendeleev-Clapeyron equation for isoprocesses:
a) Boyle-Mariotte law (isothermal process: T= const, m= const)
or for the two states of the gas, denoted by the numbers 1 and 2,
,
b) Gay-Lussac's law (isobaric process: R= const, m= const)
or for two states 
,
c) Charles' law (isochoric process: V= const, m= const)
or for two states 
,
d) the combined gas law ( m= const)
or for two states 
.
Normal conditions mean pressure p o \u003d 1 atm (1.013 10 5 Pa), temperature 0 o C ( T=273 K).
Dalton's law, which determines the pressure of the mixture n gases.
,
where p i - partial pressures of the mixture components ( i=1,2,…,n). Partial pressure is the pressure of a gas that this gas would produce if it alone were in the vessel occupied by the mixture.
Molar mass of a mixture of n gases
.
Mass fraction i-th component of the gas mixture (in fractions of a unit or percentage)
,
where t - weight of the mixture.
Molecule concentration
,
where N - the number of molecules contained in a given system; is the density of matter in the system; V- system volume. The formula is valid not only for gases, but also for any state of aggregation of matter.
Van der Waals equation for real gas
,
where a and b- van der Waals coefficients
For an ideal gas, the van der Waals equation turns into the Mendeleev-Clapeyron equation.
Basic molecular equation - kinetic theory gases
,
where p is the average kinetic energy of the translational motion of the molecule.
2.10.1. Calculation of relative and absolute masses of atoms and moleculesThe relative masses of atoms and molecules are determined using the D.I. Mendeleev values of atomic masses. At the same time, when carrying out calculations for educational purposes, the values of the atomic masses of elements are usually rounded to integers (with the exception of chlorine, atomic mass which is taken equal to 35.5).
Example 1 Relative atomic mass of calcium And r (Ca)=40; relative atomic mass of platinum And r (Pt)=195.
The relative mass of a molecule is calculated as the sum of the relative atomic masses of the atoms that make up this molecule, taking into account the amount of their substance.
Example 2. Relative molar mass of sulfuric acid:
M r (H 2 SO 4) \u003d 2A r (H) + A r (S) + 4A r (O) \u003d 2 · 1 + 32 + 4· 16 = 98.
The absolute masses of atoms and molecules are found by dividing the mass of 1 mole of a substance by the Avogadro number.
Example 3. Determine the mass of one atom of calcium.
Solution. The atomic mass of calcium is And r (Ca)=40 g/mol. The mass of one calcium atom will be equal to:
m (Ca) \u003d A r (Ca) : N A \u003d 40: 6.02 · 10 23 = 6,64· 10 -23 years
Example 4 Determine the mass of one molecule of sulfuric acid.
Solution. The molar mass of sulfuric acid is M r (H 2 SO 4) = 98. The mass of one molecule m (H 2 SO 4) is:
m (H 2 SO 4) \u003d M r (H 2 SO 4) : N A \u003d 98: 6.02 · 10 23 = 16,28· 10 -23 years
2.10.2. Calculation of the amount of matter and calculation of the number of atomic and molecular particles from known values of mass and volume
The amount of a substance is determined by dividing its mass, expressed in grams, by its atomic (molar) mass. The amount of a substance in the gaseous state at n.o. is found by dividing its volume by the volume of 1 mol of gas (22.4 l).
Example 5 Determine the amount of sodium substance n(Na) in 57.5 g of metallic sodium.
Solution. The relative atomic mass of sodium is And r (Na)=23. The amount of a substance is found by dividing the mass of metallic sodium by its atomic mass:
n(Na)=57.5:23=2.5 mol.
Example 6 . Determine the amount of nitrogen substance, if its volume at n.o. is 5.6 liters.
Solution. The amount of nitrogen substance n(N 2) we find by dividing its volume by the volume of 1 mol of gas (22.4 l):
n(N 2) \u003d 5.6: 22.4 \u003d 0.25 mol.
The number of atoms and molecules in a substance is determined by multiplying the number of atoms and molecules in the substance by Avogadro's number.
Example 7. Determine the number of molecules contained in 1 kg of water.
Solution. The amount of water substance is found by dividing its mass (1000 g) by the molar mass (18 g / mol):
n (H 2 O) \u003d 1000: 18 \u003d 55.5 mol.
The number of molecules in 1000 g of water will be:
N (H 2 O) \u003d 55.5 · 6,02· 10 23 = 3,34· 10 24 .
Example 8. Determine the number of atoms contained in 1 liter (n.o.) of oxygen.
Solution. The amount of oxygen substance, the volume of which under normal conditions is 1 liter is equal to:
n(O 2) \u003d 1: 22.4 \u003d 4.46 · 10 -2 mol.
The number of oxygen molecules in 1 liter (N.O.) will be:
N (O 2) \u003d 4.46 · 10 -2 · 6,02· 10 23 = 2,69· 10 22 .
It should be noted that 26.9 · 10 22 molecules will be contained in 1 liter of any gas at n.o. Since the oxygen molecule is diatomic, the number of oxygen atoms in 1 liter will be 2 times greater, i.e. 5.38 · 10 22 .
2.10.3. Calculation of the average molar mass of the gas mixture and volume fraction
the gases it contains
The average molar mass of a gas mixture is calculated from the molar masses of the constituent gases of this mixture and their volume fractions.
Example 9 Assuming that the content (in volume percent) of nitrogen, oxygen and argon in the air is 78, 21 and 1, respectively, calculate the average molar mass of air.
Solution.
M air = 0.78 · M r (N 2)+0.21 · M r (O 2)+0.01 · M r (Ar)= 0.78 · 28+0,21· 32+0,01· 40 = 21,84+6,72+0,40=28,96
Or approximately 29 g/mol.
Example 10. The gas mixture contains 12 l of NH 3 , 5 l of N 2 and 3 l of H 2 measured at n.o. Calculate the volume fractions of gases in this mixture and its average molar mass.
Solution. The total volume of the mixture of gases is V=12+5+3=20 l. Volume fractions j of gases will be equal:
φ(NH 3)= 12:20=0.6; φ(N 2)=5:20=0.25; φ(H 2)=3:20=0.15.
The average molar mass is calculated on the basis of the volume fractions of the constituent gases of this mixture and their molecular masses:
M=0.6 · M (NH 3) + 0.25 · M(N2)+0.15 · M (H 2) \u003d 0.6 · 17+0,25· 28+0,15· 2 = 17,5.
2.10.4. Calculation of the mass fraction of a chemical element in a chemical compound
The mass fraction ω of a chemical element is defined as the ratio of the mass of an atom of a given element X contained in a given mass of a substance to the mass of this substance m. Mass fraction is a dimensionless quantity. It is expressed in fractions of a unit:
ω(X) = m(X)/m (0<ω< 1);
or in percentage
ω(X),%= 100 m(X)/m (0%<ω<100%),
where ω(X) is the mass fraction of the chemical element X; m(X) is the mass of the chemical element X; m is the mass of the substance.
Example 11 Calculate the mass fraction of manganese in manganese (VII) oxide.
Solution. The molar masses of substances are equal: M (Mn) \u003d 55 g / mol, M (O) \u003d 16 g / mol, M (Mn 2 O 7) \u003d 2M (Mn) + 7M (O) \u003d 222 g / mol. Therefore, the mass of Mn 2 O 7 with the amount of substance 1 mol is:
m(Mn 2 O 7) = M(Mn 2 O 7) · n(Mn 2 O 7) = 222 · 1= 222
From the formula Mn 2 O 7 it follows that the amount of substance of manganese atoms is twice the amount of substance of manganese oxide (VII). Means,
n(Mn) \u003d 2n (Mn 2 O 7) \u003d 2 mol,
m(Mn)= n(Mn) · M(Mn) = 2 · 55 = 110 g.
Thus, the mass fraction of manganese in manganese(VII) oxide is:
ω(X)=m(Mn) : m(Mn 2 O 7) = 110:222 = 0.495 or 49.5%.
2.10.5. Establishing the formula of a chemical compound by its elemental composition
The simplest chemical formula of a substance is determined on the basis of the known values of the mass fractions of the elements that make up this substance.
Suppose there is a sample of a substance Na x P y O z with a mass m o g. Consider how its chemical formula is determined if the quantities of the substance of the atoms of the elements, their masses or mass fractions in the known mass of the substance are known. The formula of a substance is determined by the ratio:
x: y: z = N(Na) : N(P) : N(O).
This ratio does not change if each of its terms is divided by Avogadro's number:
x: y: z = N(Na)/N A: N(P)/N A: N(O)/N A = ν(Na) : ν(P) : ν(O).
Thus, to find the formula of a substance, it is necessary to know the ratio between the amounts of substances of atoms in the same mass of a substance:
x: y: z = m(Na)/M r (Na) : m(P)/M r (P) : m(O)/M r (O).
If we divide each term of the last equation by the mass of the sample m o , then we get an expression that allows us to determine the composition of the substance:
x: y: z = ω(Na)/M r (Na) : ω(P)/M r (P) : ω(O)/M r (O).
Example 12. The substance contains 85.71 wt. % carbon and 14.29 wt. % hydrogen. Its molar mass is 28 g/mol. Determine the simplest and true chemical formulas of this substance.
Solution. The ratio between the number of atoms in a C x H y molecule is determined by dividing the mass fractions of each element by its atomic mass:
x: y \u003d 85.71 / 12: 14.29 / 1 \u003d 7.14: 14.29 \u003d 1: 2.
Thus, the simplest formula of a substance is CH 2. The simplest formula of a substance does not always coincide with its true formula. In this case, the formula CH 2 does not correspond to the valency of the hydrogen atom. To find the true chemical formula, you need to know the molar mass of a given substance. In this example, the molar mass of the substance is 28 g/mol. Dividing 28 by 14 (the sum of atomic masses corresponding to the formula unit CH 2), we obtain the true ratio between the number of atoms in a molecule:
We get the true formula of the substance: C 2 H 4 - ethylene.
Instead of the molar mass for gaseous substances and vapors, the density for any gas or air can be indicated in the condition of the problem.
In the case under consideration, the gas density in air is 0.9655. Based on this value, the molar mass of the gas can be found:
M = M air · D air = 29 · 0,9655 = 28.
In this expression, M is the molar mass of gas C x H y, M air is the average molar mass of air, D air is the density of gas C x H y in air. The resulting value of the molar mass is used to determine the true formula of the substance.
The condition of the problem may not indicate the mass fraction of one of the elements. It is found by subtracting from unity (100%) the mass fractions of all other elements.
Example 13 An organic compound contains 38.71 wt. % carbon, 51.61 wt. % oxygen and 9.68 wt. % hydrogen. Determine the true formula of this substance if its oxygen vapor density is 1.9375.
Solution. We calculate the ratio between the number of atoms in the molecule C x H y O z:
x: y: z = 38.71/12: 9.68/1: 51.61/16 = 3.226: 9.68: 3.226= 1:3:1.
The molar mass M of a substance is:
M \u003d M (O 2) · D(O2) = 32 · 1,9375 = 62.
The simplest formula of a substance is CH 3 O. The sum of atomic masses for this formula unit will be 12+3+16=31. Divide 62 by 31 and get the true ratio between the number of atoms in the molecule:
x:y:z = 2:6:2.
Thus, the true formula of the substance is C 2 H 6 O 2. This formula corresponds to the composition of dihydric alcohol - ethylene glycol: CH 2 (OH) -CH 2 (OH).
2.10.6. Determination of the molar mass of a substance
The molar mass of a substance can be determined on the basis of its gas vapor density with a known molar mass.
Example 14 . The vapor density of some organic compound in terms of oxygen is 1.8125. Determine the molar mass of this compound.
Solution. The molar mass of an unknown substance M x is equal to the product of the relative density of this substance D and the molar mass of the substance M, by which the value of the relative density is determined:
M x = D · M = 1.8125 · 32 = 58,0.
Substances with the found value of the molar mass can be acetone, propionaldehyde and allyl alcohol.
The molar mass of a gas can be calculated using the value of its molar volume at n.c.
Example 15. Mass of 5.6 liters of gas at n.o. is 5.046 g. Calculate the molar mass of this gas.
Solution. The molar volume of gas at n.s. is 22.4 liters. Therefore, the molar mass of the desired gas is
M = 5.046 · 22,4/5,6 = 20,18.
The desired gas is neon Ne.
The Clapeyron–Mendeleev equation is used to calculate the molar mass of a gas whose volume is given under non-normal conditions.
Example 16 At a temperature of 40 ° C and a pressure of 200 kPa, the mass of 3.0 liters of gas is 6.0 g. Determine the molar mass of this gas.
Solution. Substituting the known quantities into the Clapeyron–Mendeleev equation, we obtain:
M = mRT/PV = 6.0 · 8,31· 313/(200· 3,0)= 26,0.
The gas under consideration is acetylene C 2 H 2.
Example 17 Combustion of 5.6 l (N.O.) of hydrocarbon produced 44.0 g of carbon dioxide and 22.5 g of water. The relative density of the hydrocarbon with respect to oxygen is 1.8125. Determine the true chemical formula of the hydrocarbon.
Solution. The reaction equation for the combustion of hydrocarbons can be represented as follows:
C x H y + 0.5 (2x + 0.5y) O 2 \u003d x CO 2 + 0.5 y H 2 O.
The amount of hydrocarbon is 5.6:22.4=0.25 mol. As a result of the reaction, 1 mol of carbon dioxide and 1.25 mol of water are formed, which contains 2.5 mol of hydrogen atoms. When a hydrocarbon is burned with a quantity of a substance of 1 mole, 4 moles of carbon dioxide and 5 moles of water are obtained. Thus, 1 mol of hydrocarbon contains 4 mol of carbon atoms and 10 mol of hydrogen atoms, i.e. chemical formula of hydrocarbon C 4 H 10 . The molar mass of this hydrocarbon is M=4 · 12+10=58. Its relative oxygen density D=58:32=1.8125 corresponds to the value given in the condition of the problem, which confirms the correctness of the found chemical formula.
The average molecular weight is a conditional value and refers to such a homogeneous gas, in which the number of molecules and the total mass are equal to the number of molecules and the mass of the mixture of gases.
If the value of the gas mixture constant is known, then
Replacing the gas constants R 1 , R 2 , ..., R n with their values from the Clapeyron equation, we obtain an expression for the average molecular weight if the mixture is given by mass fractions:
(3-8)
If the mixture is given by volume fractions, then, as follows from equation (3-6),
Because the 
then
The average molecular weight of a mixture of gases is equal to the sum of the products of volume fractions and the molecular weights of the individual gases that make up the mixture.
Partial pressures
The partial pressure of a gas can be determined in terms of mass fractions from the Clapeyron equation if the main parameters of the gas are known:
(3-10)
To find the partial pressure of each gas when the mixture is specified by volume fractions, one can use the Boyle-Mariotte law, from which it follows that at a constant temperature
(3-11)
The partial pressure of each gas is equal to the product of the total pressure of the mixture of gases and its volume fraction.
Equation (3-11) is usually used in technical calculations and when testing thermal installations. Volume fractions of gases are determined by special devices - gas analyzers.
Specific enthalpy, i.e., enthalpy per 1 kg, is denoted by the letter i and is, by definition, a complex function of the form
The enthalpy differential di is the elementary amount of heat involved in the process at constant pressure. All the heat in the process at constant pressure is spent on the change in enthalpy:
(5-15)
From equation (5-12) it follows that
(5-16)
The enthalpy is greater than the external heat by the amount of work vdp, which is represented on the pv diagram by the elementary area abed (Fig. 5-11). Obviously, the whole square. ABCD is defined by the expression
, which is called disposable or useful work.
The change in enthalpy is completely determined by the initial and final state of the working fluid and does not depend on intermediate states. The change in the enthalpy of the gas in cycles is zero, i.e.
Since enthalpy is a function of the main parameters of the state, then di is the total differential of this function for any independent variables characterizing the state 
gas;
(5-17)
The change in enthalpy in all processes occurring between two points A and B will be the same (Fig. 5-12).
The physical meaning of enthalpy will be clear from consideration of the following example. A weight of mass t kg is placed on a moving piston in a cylinder with 1 kg of gas (Fig. 5-13). Piston area /; internal energy of the working body and. The potential energy of the weight is equal to the product of the mass of the weight m and the height S. Since the gas pressure p is balanced by the mass of the weight, its potential energy can be 
express otherwise:
The product /S is the specific volume of the gas. From here
The product of pressure and volume is the work that must be expended to introduce a gas of volume v into an external medium with pressure p. Thus, the work pv is the potential energy of the gas, which depends on the forces acting on the piston. The greater these external forces, the greater the pressure p and the greater the potential energy of pressure pv.
If we consider the gas in the cylinder and the piston with the load as one system, which we will call the extended system, then the total energy E of this system is the sum of the internal energy of the gas and and the potential energy of the piston with the load equal to pv: 
This shows that the enthalpy i is equal to the energy of the extended system - the body and the environment. This is the physical meaning of enthalpy.
Enthalpy values for vapors, gases and gas mixtures are given in the technical and reference literature. Using these data, it is possible to determine the amount of heat involved in the process at constant pressure. Enthalpy has received great importance and application in the calculations of thermal and refrigeration installations and, as a parameter of the state of the working fluid, greatly simplifies thermal calculations. It allows [to apply graphical methods in the study of various thermodynamic processes and cycles.
It is especially advisable to use enthalpy when p and T are taken as the main parameters. This can be clearly seen if the enthalpy i is compared with the internal energy u. When v \u003d const, the equation of the first law of thermodynamics dq \u003d du + pdv turns into dq v \u003d du, or q v - u 2 -u 1 and when p \u003d const q p \u003d i 3 - i 1.
The enthalpy of an ideal gas, "as well as the internal energy, is a function of temperature and does not depend on other parameters. Indeed, for an ideal gas
hence (since both terms depend only on temperature), i = f(T).
Then, by analogy with the internal energy, we have 
i.e., in any process of changing the state of an ideal gas, the derivative of the change in enthalpy with respect to temperature will be the total derivative.
The numerical values of the enthalpies of ideal gases are given in the Appendix, Table. XIII.
SECTION I. GENERAL CHEMISTRY
Examples of solving typical problems
v. Determination of the average molar mass of a mixture of gases
Formulas and concepts that are used:
where M(mixture) is the average molar mass of a mixture of gases,
M(A), M(B), M(C) are the molar masses of the mixture components A, B and C,
χ(А), χ(B), χ(В) are the molar fractions of the mixture components A, B and C,
φ(A), φ(B), φ(B) are the volume fractions of the mixture components A, B and C,
M (pov.) - molar mass of air, g / mol,
M r (pov.) - relative molecular weight of air.
Problem 23. Calculate the molar mass of a mixture in which the volume fractions of methane and butane are respectively 85 and 15%.
The molar mass of a mixture is the mass of all its components, taken in the total amount of the substance of the mixture 1 mol (M (CH 4) \u003d 16 g / mol, M (C 4 H 10) \u003d 58 g / mol). The average molar mass of a mixture can be calculated using the formula:
Answer: M (mixture) \u003d 22, 3 g / mol.
Problem 24. Determine the density of the gas mixture with nitrogen, in which the volume fractions of carbon (And V) oxide, sulfur (And V) oxide and carbon (II) oxide, respectively, are 35.25 and 40%.
1. Calculate the molar mass of the mixture (M (C O 2) \u003d 44 g / mol, M (SO 2) \u003d 64 g / mol, M (CO) \u003d 28 g / mol):
2. Calculate the relative density of the mixture with nitrogen:
Answer: D N2 (mixtures) = 1.52.
Problem 25. The density of a mixture of acetylene and butene over helium is 11. Determine the volume fraction of acetylene in the mixture.
1. According to the formula, we determine the molar mass of the mixture (M (He) \u003d 4 g / mol):
2. Suppose we have 1 mol of mixture. It contains x mol C 2 H 2, then in accordance
3. Let's write down the expression for calculating the average molar mass of the gas mixture:
Substitute all known data: M (C 2 H 2) \u003d 26 g / mol, M (C 4 H 8) \u003d 56 g / mol:
4. Therefore, 1 mol of the mixture contains 0.4 mol of C 2 H 2. Calculate the mole fraction χ(C 2 H 2):
For gases φ(X) = χ(X). Therefore, φ(C 2 H 4) = 40%.
Definitions of the average molar mass of a mixture of gases - Examples of solving typical problems - Basic chemical concepts. Substance - GENERAL CHEMISTRY - CHEMISTRY - Comprehensive preparation for external independent testing According to the current UPE program - designed to prepare for external independent assessment. It contains theoretical material presented in accordance with the current program in chemistry for secondary schools and the UPE program; examples of solving typical problems; thematic tests.
INTRODUCTION TO GENERAL CHEMISTRY
e-tutorial
Moscow 2013
2. Basic concepts and laws of chemistry. Atomic-molecular doctrine
2.10. Examples of problem solving
2.10.1. Calculation of relative and absolute masses of atoms and molecules
The relative masses of atoms and molecules are determined using the D.I. Mendeleev values of atomic masses. At the same time, when carrying out calculations for educational purposes, the values of the atomic masses of elements are usually rounded up to integers (with the exception of chlorine, the atomic mass of which is assumed to be 35.5).
Example 1 Relative atomic mass of calcium And r (Ca)=40; relative atomic mass of platinum And r (Pt)=195.
The relative mass of a molecule is calculated as the sum of the relative atomic masses of the atoms that make up this molecule, taking into account the amount of their substance.
Example 2. Relative molar mass of sulfuric acid:
The absolute masses of atoms and molecules are found by dividing the mass of 1 mole of a substance by the Avogadro number.
Example 3. Determine the mass of one atom of calcium.
Solution. The atomic mass of calcium is And r (Ca)=40 g/mol. The mass of one calcium atom will be equal to:
m (Ca) \u003d A r (Ca) : N A \u003d 40: 6.02 · 10 23 = 6,64· 10 -23 years
Example 4 Determine the mass of one molecule of sulfuric acid.
Solution. The molar mass of sulfuric acid is M r (H 2 SO 4) = 98. The mass of one molecule m (H 2 SO 4) is:
2.10.2. Calculation of the amount of matter and calculation of the number of atomic and molecular particles from known values of mass and volume
The amount of a substance is determined by dividing its mass, expressed in grams, by its atomic (molar) mass. The amount of a substance in the gaseous state at n.o. is found by dividing its volume by the volume of 1 mol of gas (22.4 l).
Example 5 Determine the amount of sodium substance n(Na) in 57.5 g of metallic sodium.
Solution. The relative atomic mass of sodium is And r (Na)=23. The amount of a substance is found by dividing the mass of metallic sodium by its atomic mass:
Example 6 . Determine the amount of nitrogen substance, if its volume at n.o. is 5.6 liters.
Solution. The amount of nitrogen substance n (N 2) is found by dividing its volume by the volume of 1 mol of gas (22.4 l):
The number of atoms and molecules in a substance is determined by multiplying the number of atoms and molecules in the substance by Avogadro's number.
Example 7. Determine the number of molecules contained in 1 kg of water.
Solution. The amount of water substance is found by dividing its mass (1000 g) by the molar mass (18 g / mol):
The number of molecules in 1000 g of water will be:
N (H 2 O) \u003d 55.5 · 6,02· 10 23 = 3,34· 10 24 .
Example 8. Determine the number of atoms contained in 1 liter (n.o.) of oxygen.
Solution. The amount of oxygen substance, the volume of which under normal conditions is 1 liter is equal to:
n(O 2) \u003d 1: 22.4 \u003d 4.46 · 10 -2 mol.
The number of oxygen molecules in 1 liter (N.O.) will be:
N (O 2) \u003d 4.46 · 10 -2 · 6,02· 10 23 = 2,69· 10 22 .
It should be noted that 26.9 · 10 22 molecules will be contained in 1 liter of any gas at n.o. Since the oxygen molecule is diatomic, the number of oxygen atoms in 1 liter will be 2 times greater, i.e. 5.38 · 10 22 .
2.10.3. Calculation of the average molar mass of the gas mixture and volume fraction
the gases it contains
The average molar mass of a gas mixture is calculated from the molar masses of the constituent gases of this mixture and their volume fractions.
Example 9 Assuming that the content (in volume percent) of nitrogen, oxygen and argon in the air is 78, 21 and 1, respectively, calculate the average molar mass of air.
Solution.
M air = 0.78 · M r (N 2)+0.21 · M r (O 2)+0.01 · M r (Ar)= 0.78 · 28+0,21· 32+0,01· 40 = 21,84+6,72+0,40=28,96
or approximately 29 g/mol.
Example 10. The gas mixture contains 12 l of NH 3 , 5 l of N 2 and 3 l of H 2 measured at n.o. Calculate the volume fractions of gases in this mixture and its average molar mass.
Solution. The total volume of the mixture of gases is V=12+5+3=20 l. Volume fractions j of gases will be equal:
The average molar mass is calculated on the basis of the volume fractions of the constituent gases of this mixture and their molecular masses:
M=0.6 · M (NH 3) + 0.25 · M(N2)+0.15 · M (H 2) \u003d 0.6 · 17+0,25· 28+0,15· 2 = 17,5.
2.10.4. Calculation of the mass fraction of a chemical element in a chemical compound
The mass fraction ω of a chemical element is defined as the ratio of the mass of an atom of a given element X contained in a given mass of a substance to the mass of this substance m. Mass fraction is a dimensionless quantity. It is expressed in fractions of a unit:
ω (X) \u003d m (X) / m (0 ° C and a pressure of 200 kPa, the mass of 3.0 liters of gas is 6.0 g. Determine the molar mass of this gas.
Solution. Substituting the known quantities into the Clapeyron–Mendeleev equation, we obtain:
M = mRT/PV = 6.0 · 8,31· 313/(200· 3,0)= 26,0.
The gas under consideration is acetylene C 2 H 2.
Example 17 Combustion of 5.6 l (N.O.) of hydrocarbon produced 44.0 g of carbon dioxide and 22.5 g of water. The relative density of the hydrocarbon with respect to oxygen is 1.8125. Determine the true chemical formula of the hydrocarbon.
Solution. The reaction equation for the combustion of hydrocarbons can be represented as follows:
The amount of hydrocarbon is 5.6:22.4=0.25 mol. As a result of the reaction, 1 mol of carbon dioxide and 1.25 mol of water are formed, which contains 2.5 mol of hydrogen atoms. When a hydrocarbon is burned with a quantity of a substance of 1 mole, 4 moles of carbon dioxide and 5 moles of water are obtained. Thus, 1 mol of hydrocarbon contains 4 mol of carbon atoms and 10 mol of hydrogen atoms, i.e. chemical formula of hydrocarbon C 4 H 10 . The molar mass of this hydrocarbon is M=4 · 12+10=58. Its relative oxygen density D=58:32=1.8125 corresponds to the value given in the condition of the problem, which confirms the correctness of the found chemical formula.
INTRODUCTION TO GENERAL CHEMISTRY
INTRODUCTION TO GENERAL CHEMISTRY Electronic textbook Moscow 2013 2. Basic concepts and laws of chemistry. Atomic-molecular doctrine 2.10. Examples of problem solving 2.10.1. Calculation of relative
If ideal gases are in communicating cylinders separated by a tap, then when the tap is opened, the gases in the cylinders mix with each other and each of them fills the volume of both cylinders.
For an ideal gas (or two different gases) in communicating cylinders, when the valve is opened, some parameters become the same:
- the pressure of a gas (or a mixture of gases) after opening the valve is equalized:
- gas (or a mixture of gases) after opening the tap occupies the entire volume provided to it, i.e. volume of both vessels:
where V 1 - the volume of the first cylinder; V 2 - the volume of the second cylinder;
- the temperature of the gas (or mixture of gases) after opening the valve is equalized:
- the gas density ρ and its concentration n in both cylinders become the same:
ρ = const, n = const,
If the cylinders have the same volume, then the masses of gas (or a mixture of gases) in each cylinder after opening the valve become the same:
m ′ 1 \u003d m ′ 2 \u003d m ′ \u003d m 1 + m 2 2,
where m ′ 1 is the mass of gas (or a mixture of gases) in the first cylinder after opening the tap; m ′ 2 - mass of gas (or mixture of gases) in the second cylinder after opening the valve; m ′ - the mass of gas (or a mixture of gases) in each cylinder after opening the valve; m 1 - the mass of gas in the first cylinder before opening the valve; m 2 - the mass of gas in the second cylinder before opening the valve.
The mass of gas transferred from one vessel to another as a result of opening a valve is determined by the following expressions:
- change in the mass of gas in the first cylinder
Δm 1 = | m′ 1 − m 1 | = | m 1 + m 2 2 − m 1 | = | m 2 − m 1 | 2;
- change in the mass of gas in the second cylinder
Δm 2 = | m ′ 2 − m 2 | = | m 1 + m 2 2 − m 2 | = | m 1 − m 2 | 2.
Changes in the mass of gas (or mixture of gases) in both cylinders are the same:
Δm 1 = Δm 2 = Δm = | m 2 − m 1 | 2,
those. how much gas left the cylinder with a larger mass of gas - the same amount of gas entered the cylinder with a smaller mass.
If the cylinders have the same volume, then the amount of gas (or mixture of gases) in each cylinder after opening the valve becomes the same:
ν ′ 1 = ν ′ 2 = ν ′ = ν 1 + ν 2 2 ,
where ν ′ 1 - the amount of gas (or a mixture of gases) in the first cylinder after opening the tap; ν ′ 2 - the amount of gas (or a mixture of gases) in the second cylinder after opening the tap; ν' - the amount of gas (or a mixture of gases) in each cylinder after opening the valve; ν 1 - the amount of gas in the first cylinder before opening the valve; ν 2 - the amount of gas in the second cylinder before opening the valve.
The amount of gas transferred from one vessel to another as a result of opening a valve is determined by the following expressions:
- change in the amount of gas in the first cylinder
Δ ν 1 = | ν′ 1 − ν 1 | = | ν 1 + ν 2 2 − ν 1 | = | ν 2 − ν 1 | 2;
- change in the amount of gas in the second cylinder
Δ ν 2 = | ν′ 2 − ν 2 | = | ν 1 + ν 2 2 − ν 2 | = | ν 1 − ν 2 | 2.
Changes in the amount of gas (or mixture of gases) in both cylinders are the same:
Δ ν 1 = Δ ν 2 = Δ ν = | ν 2 − ν 1 | 2,
those. how much gas left the cylinder with a large amount of gas - the same amount of gas entered the cylinder with a smaller amount.
For an ideal gas (or two different gases) in communicating cylinders, when the valve is opened, the pressure becomes the same:
and is determined by Dalton's law (for a mixture of gases) -
where p 1 , p 2 - partial pressures of the components of the mixture.
The partial pressures of the mixture components can be calculated as follows:
- using the Mendeleev-Clapeyron equation; then the pressure is given by
p = (ν 1 + ν 2) R T V 1 + V 2 ,
where ν 1 - the amount of substance of the first component of the mixture; ν 2 - the amount of substance of the second component of the mixture; R is the universal gas constant, R ≈ 8.31 J/(mol ⋅ K); T is the temperature of the mixture; V 1 - the volume of the first cylinder; V 2 - the volume of the second cylinder;
- using the basic equation of molecular kinetic theory; then the pressure is given by
p = (N 1 + N 2) k T V 1 + V 2 ,
where N 1 - the number of molecules of the first component of the mixture; N 2 - the number of molecules of the second component of the mixture; k is the Boltzmann constant, k = 1.38 ⋅ 10 −23 J/K.
Example 26. Determine the average molar mass of a gas mixture consisting of 3.0 kg of hydrogen, 1.0 kg of helium and 8.0 kg of oxygen. The molar masses of hydrogen, helium and oxygen are 2.0, 4.0 and 32 g/mol, respectively.
Solution. The average molar mass of the mixture is given by
where m is the mass of the mixture; ν is the amount of substance in the mixture.
We find the mass of the mixture as the sum of the masses -
where m 1 is the mass of hydrogen; m 2 is the mass of helium; m 3 is the mass of oxygen.
Similarly, we find the amount of substance -
where ν 1 is the amount of hydrogen in the mixture, ν 1 = m 1 / M 1; M 1 - molar mass of hydrogen; ν 2 - the amount of helium in the mixture, ν 2 = m 2 / M 2; M 2 - molar mass of helium; ν 3 - the amount of oxygen in the mixture, ν 3 = m 3 / M 3; M 3 is the molar mass of oxygen.
Substitution of expressions for the mass and amount of substance in the original formula gives
〈 M 〉 = m 1 + m 2 + m 3 ν 1 + ν 2 + ν 3 = m 1 + m 2 + m 3 m 1 M 1 + m 2 M 2 + m 3 M 3 .
〈 M 〉 = 3.0 + 1.0 + 8.0 3.0 2.0 ⋅ 10 − 3 + 1.0 4.0 ⋅ 10 − 3 + 8.0 32 ⋅ 10 − 3 =
6.0 ⋅ 10 − 3 kg/mol = 6.0 g/mol.
Example 27. The density of a gas mixture consisting of helium and hydrogen at a pressure of 3.50 MPa and a temperature of 300 K is 4.50 kg/m 3 . Determine the mass of helium in 4.00 m 3 of the mixture. The molar masses of hydrogen and helium are 0.002 and 0.004 kg/mol, respectively.
Solution. To find the mass of helium m 2 in the indicated volume, it is necessary to determine the density of helium in the mixture:
where ρ 2 is the density of helium; V is the volume of the mixture of gases.
The density of the mixture is defined as the sum of the densities of hydrogen and helium:
where ρ 1 is the density of hydrogen.
However, the written formula contains two unknown quantities - the densities of hydrogen and helium. To determine these quantities, one more equation is required, which includes the densities of hydrogen and helium.
We write Dalton's law for the pressure of a mixture of gases:
where p 1 - hydrogen pressure; p 2 - helium pressure.
To determine the gas pressures, we write the equation of state in the following form:
p 1 = ρ 1 R T M 1 ,
p 2 = ρ 2 R T M 2 ,
where R is the universal gas constant, R ≈ 8.31 J/(mol ⋅ K); T is the temperature of the mixture; M 1 - molar mass of hydrogen; M 2 is the molar mass of helium.
Substituting the expressions for hydrogen and helium pressures into Dalton's law gives
p = ρ 1 R T M 1 + ρ 2 R T M 2 .
Another equation is obtained with two unknown quantities - the density of hydrogen and the density of helium.
Formulas for calculating the density and pressure of the mixture form a system of equations:
ρ = ρ 1 + ρ 2 , p = ρ 1 R T M 1 + ρ 2 R T M 2 , >
which needs to be solved with respect to the density of helium.
To do this, we express the hydrogen densities from the first and second equations
ρ 1 = ρ − ρ 2 , ρ 1 = M 1 R T (p − ρ 2 R T M 2) >
and equate their right-hand sides:
ρ − ρ 2 = M 1 R T (p − ρ 2 R T M 2) .
ρ 2 = M 2 M 2 − M 1 (ρ − p M 1 R T) .
We substitute the resulting expression into the formula for calculating the mass of helium
m 2 = M 2 V M 2 − M 1 (ρ − p M 1 R T)
and do the calculation:
m 2 \u003d 0.004 ⋅ 4.00 0.004 - 0.002 (4.50 - 3.50 ⋅ 10 6 0.002 8.31 ⋅ 300) ≈ 13.6 kg.
The mass of helium in the indicated volume of the mixture is 13.6 kg.
How to find the average molar mass of a mixture of gases
If ideal gases are in communicating cylinders separated by a tap, then when the tap is opened, the gases in the cylinders mix with each other and each of them fills the volume of both cylinders. For