The amount of matter contained in a body is determined by the number of molecules (or atoms) in that body. Since the number of molecules in macroscopic bodies is very large, to determine the amount of substance in a body, the number of molecules in it is compared with the number of atoms in 0.012 kg of the carbon isotope \(~^(12)_6C\).
Quantity of substanceν is a value equal to the ratio of the number of molecules (atoms) N in a given body to the number of atoms N A in 0.012 kg of carbon isotope \(~^(12)_6C\):
\(~\nu = \frac(N)(N_A) . \qquad (2)\)
The SI unit of quantity of a substance is the mole. 1 mole- the amount of a substance that contains the same number of structural elements (atoms, molecules, ions) as there are atoms in 0.012 kg of the carbon isotope \(~^(12)_6C\).
The number of particles in one mole of a substance is called Avogadro's constant.
\(~N_A = \frac(0.012)(m_(0C))= \frac(0.012)(1.995 \cdot 10^(-26))\) = 6.02·10 23 mol -1. (3)
Thus, 1 mole of any substance contains the same number of particles - N A particles. Since the mass m 0 particles are different for different substances, then so is the mass N The A of particles is different for different substances.
The mass of a substance taken in an amount of 1 mole is called molar mass M:
\(~M = m_0 N_A . \qquad (4)\)
The SI unit of molar mass is kilogram per mole (kg/mol).
Between molar mass Μ and relative molecular weight M r there is the following relationship:
\(~M = M_r \cdot 10^(-3) .\)
Thus, the molecular mass of carbon dioxide is 44, the molar mass is 44·10 -3 kg/mol.
Knowing the mass of a substance and its molar mass M, you can find the number of moles (amount of substance) in the body\[~\nu = \frac(m)(M)\].
Then from formula (2) the number of particles in the body
\(~N = \nu N_A = \frac(m)(M) N_A .\)
Knowing the molar mass and Avogadro's constant, you can calculate the mass of one molecule:
\(~m_0 = \frac(M)(N_A) = \frac(m)(N) .\)
Literature
Aksenovich L. A. Physics in secondary school: Theory. Tasks. Tests: Textbook. allowance for institutions providing general education. environment, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Mn.: Adukatsiya i vyakhavanne, 2004. - P. 124-125.
The algorithm for finding the amount of a substance is quite simple; it can be useful for simplifying the solution. Also become familiar with another concept that you will need to calculate the amount of a substance: molar mass, or the mass of one mole of an individual atom of an element. Already from the definition it is noticeable that it is measured in g/mol. Use a standard table that contains molar mass values for some elements.
What is the amount of a substance and how is it determined?
In this case, the mass of hydrogen participating in the reaction is approximately 8 times less than the mass of oxygen (since the atomic mass of hydrogen is approximately 16 times less than the atomic mass of oxygen). When the heat of reaction is written as it is in this equation, it is assumed that it is expressed in kilojoules per stoichiometric unit ("mole") of the reaction of the written equation. Heats of reactions are always tabulated per mole of compound formed.
In order to understand what an amount of substance is in chemistry, let’s give the term a definition. To understand what the amount of a substance is, we note that this quantity has its own designation. Eighth-graders who do not yet know how to write chemical equations do not know what an amount of a substance is or how to use this quantity in calculations. After becoming acquainted with the law of constancy of the mass of substances, the meaning of this quantity becomes clear. By it we mean the mass that corresponds to one mole of a specific chemical substance. Not a single problem in a school chemistry course related to calculations using an equation is complete without the use of such a term as “amount of substance.”
2.10.5. Establishing the formula
chemical compound by its elemental
composition
We get the true formula of the substance: C2H4 - ethylene. 2.5 mol hydrogen atoms.
Denoted as Mr. It is found according to the periodic table - it is simply the sum of the atomic masses of a substance. The law of conservation of mass - the mass of substances that enter into a chemical reaction is always equal to the mass of the formed substances. That is, if in the problem we are given normal conditions, then, knowing the number of moles (n), we can find the volume of the substance. Basic formulas for solving problems in chemistry These are formulas.
Where in the Periodic Table are the elements corresponding to simple substances and metals? From the sentences below, write down the numbers corresponding to metals in one column, and the numbers corresponding to non-metals in another column. To obtain a certain amount of a product (in a chemical laboratory or in a factory), it is necessary to take strictly defined quantities of starting substances. Chemists, conducting experiments, noticed that the composition of the products of some reactions depends on the proportions in which the reacting substances were taken. How many atoms will there be in this mass?
N is the number of structural links, and NA is Avogadro’s constant. Avogadro's constant is a proportionality coefficient that ensures the transition from molecular to molar relationships. V is the gas volume (l), and Vm is the molar volume (l/mol).
The unit of measurement for the quantity of a substance in the International System of Units (SI) is the mole. Definition. Write down the formula for calculating this energy and the names of the physical quantities included in the formula. This question relates to the section “10-11″ grades.
Lesson 1.
Topic: Amount of substance. Mole
Chemistry is the science of substances. How to measure substances? In what units? In the molecules that make up substances, but this is very difficult to do. In grams, kilograms or milligrams, but this is how mass is measured. What if we combine the mass that is measured on a scale and the number of molecules of a substance, is this possible?
a) H-hydrogen
A n = 1a.u.m.
1a.u.m = 1.66*10 -24 g
Let's take 1g of hydrogen and count the number of hydrogen atoms in this mass (have students do this using a calculator).
N n = 1g / (1.66*10 -24) g = 6.02*10 23
b) O-oxygen
A o = 16 a.u.m = 16 * 1.67 * 10 -24 g
N o = 16 g / (16 * 1.66 * 10 -24) g = 6.02 * 10 23
c) C-carbon
A c = 12a.u.m = 12*1.67*10 -24 g
N c = 12g / (12* 1.66*10 -24) g = 6.02*10 23
Let us conclude: if we take a mass of a substance that is equal to the atomic mass in size, but taken in grams, then there will always be (for any substance) 6.02 * 10 23 atoms of this substance.
H 2 O - water
18 g / (18 * 1.66 * 10 -24) g = 6.02 * 10 23 water molecules, etc.
N a = 6.02*10 23 - Avogadro’s number or constant.
A mole is the amount of a substance that contains 6.02 * 10 23 molecules, atoms or ions, i.e. structural units.
There are moles of molecules, moles of atoms, moles of ions.
n is the number of moles (the number of moles is often denoted),
N is the number of atoms or molecules,
N a = Avogadro's constant.
Kmol = 10 3 mol, mmol = 10 -3 mol.
Display a portrait of Amedeo Avogadro on a multimedia installation and briefly talk about him, or instruct the student to prepare a short report on the life of the scientist.
Lesson 2.
Topic: “Molar mass of a substance”
What is the mass of 1 mole of a substance? (Students can often draw the conclusion themselves.)
The mass of one mole of a substance is equal to its molecular mass, but expressed in grams. The mass of one mole of a substance is called molar mass and is denoted by M.
Formulas:
M - molar mass,
n - number of moles,
m is the mass of the substance.
The mass of a mole is measured in g/mol, the mass of a kmole is measured in kg/kmol, the mass of a mmol is measured in mg/mol.
Fill out the table (tables are distributed).
Substance |
Number of molecules |
Molar mass |
Number of moles |
Mass of substance |
5mol |
||||
H2SO4 |
||||
12 ,0 4*10 26 |
Lesson 3.
Topic: Molar volume of gases
Let's solve the problem. Determine the volume of water, the mass of which under normal conditions is 180 g.
Given:
Those. We calculate the volume of liquid and solid bodies through density.
But, when calculating the volume of gases, it is not necessary to know the density. Why?
The Italian scientist Avogadro determined that equal volumes of different gases under the same conditions (pressure, temperature) contain the same number of molecules - this statement is called Avogadro's law.
Those. if, under equal conditions, V(H 2) =V(O 2), then n(H 2) =n(O 2), and vice versa, if, under equal conditions, n(H 2) =n(O 2), then the volumes of these gases will be the same. And a mole of a substance always contains the same number of molecules 6.02 * 10 23.
We conclude - under the same conditions, moles of gases should occupy the same volume.
Under normal conditions (t=0, P=101.3 kPa. or 760 mm Hg.), moles of any gases occupy the same volume. This volume is called molar.
V m =22.4 l/mol
1 kmol occupies a volume of -22.4 m 3 /kmol, 1 mmol occupies a volume of -22.4 ml/mmol.
Example 1.(To be solved on the board):
Example 2.(You can ask students to solve):
| Given: | Solution: |
m(H 2)=20g |
Have students fill out the table.
Substance |
Number of molecules |
Mass of substance |
Number of moles |
Molar mass |
Volume |
The decision about the need to maintain such a notebook did not come immediately, but gradually, with the accumulation of work experience.
In the beginning, this was a space at the end of the workbook - a few pages for writing down the most important definitions. Then the most important tables were placed there. Then came the realization that most students, in order to learn to solve problems, need strict algorithmic instructions, which they, first of all, must understand and remember.
That’s when the decision came to keep, in addition to the workbook, another mandatory notebook in chemistry - a chemical dictionary. Unlike workbooks, of which there may even be two during one academic year, a dictionary is a single notebook for the entire chemistry course. It is best if this notebook has 48 sheets and a durable cover.
We arrange the material in this notebook as follows: at the beginning - the most important definitions, which the children copy from the textbook or write down under the dictation of the teacher. For example, in the first lesson in 8th grade, this is the definition of the subject “chemistry”, the concept of “chemical reactions”. During the school year in the 8th grade, more than thirty of them accumulate. I conduct surveys on these definitions in some lessons. For example, an oral question in a chain, when one student asks a question to another, if he answered correctly, then he already asks the next question; or, when one student is asked questions by other students, if he cannot answer, then they answer themselves. In organic chemistry, these are mainly definitions of classes of organic substances and main concepts, for example, “homologues”, “isomers”, etc.
At the end of our reference book, material is presented in the form of tables and diagrams. On the last page is the very first table “Chemical elements. Chemical signs". Then the tables “Valence”, “Acids”, “Indicators”, “Electrochemical series of metal voltages”, “Electronegativity series”.
I especially want to dwell on the contents of the table “Correspondence of acids to acid oxides”:
| Correspondence of acids to acid oxides | ||||
| Acid oxide | Acid | |||
| Name | Formula | Name | Formula | Acid residue, valency |
| carbon(II) monoxide | CO2 | coal | H2CO3 | CO3(II) |
| sulfur(IV) oxide | SO 2 | sulfurous | H2SO3 | SO3(II) |
| sulfur(VI) oxide | SO 3 | sulfuric | H2SO4 | SO 4 (II) |
| silicon(IV) oxide | SiO2 | silicon | H2SiO3 | SiO3(II) |
| nitric oxide (V) | N2O5 | nitrogen | HNO3 | NO 3 (I) |
| phosphorus(V) oxide | P2O5 | phosphorus | H3PO4 | PO 4 (III) |
Without understanding and memorizing this table, it is difficult for 8th grade students to compile equations for the reactions of acid oxides with alkalis.
When studying the theory of electrolytic dissociation, we write down diagrams and rules at the end of the notebook.
Rules for composing ionic equations:
1. The formulas of strong electrolytes soluble in water are written in the form of ions.
2. The formulas of simple substances, oxides, weak electrolytes and all insoluble substances are written in molecular form.
3. The formulas of poorly soluble substances on the left side of the equation are written in ionic form, on the right - in molecular form.
When studying organic chemistry, we write into the dictionary general tables on hydrocarbons, classes of oxygen- and nitrogen-containing substances, and diagrams on genetic connections.
| Physical quantities | |||
| Designation | Name | Units | Formulas |
| amount of substance | mole | = N / N A ; = m / M; V / V m (for gases) |
|
| N A | Avogadro's constant | molecules, atoms and other particles | N A = 6.02 10 23 |
| N | number of particles | molecules, atoms and other particles |
N = N A |
| M | molar mass | g/mol, kg/kmol | M = m / ; /M/ = M r |
| m | weight | g, kg | m = M ; m = V |
| Vm | molar volume of gas | l/mol, m 3/kmol | Vm = 22.4 l / mol = 22.4 m 3 / kmol |
| V | volume | l, m 3 | V = V m (for gases); |
| density | g/ml; | =m/V; M / V m (for gases) |
|
Over the 25-year period of teaching chemistry at school, I had to work using different programs and textbooks. At the same time, it was always surprising that practically no textbook teaches how to solve problems. At the beginning of studying chemistry, to systematize and consolidate knowledge in the dictionary, my students and I compile a table “Physical quantities” with new quantities:
When teaching students how to solve calculation problems, I attach great importance to algorithms. I believe that strict instructions for the sequence of actions allow a weak student to understand the solution of problems of a certain type. For strong students, this is an opportunity to reach a creative level in their further chemical education and self-education, since first you need to confidently master a relatively small number of standard techniques. On the basis of this, the ability to correctly apply them at different stages of solving more complex problems will develop. Therefore, I have compiled algorithms for solving calculation problems for all types of school course problems and for elective classes.
I will give examples of some of them.
Algorithm for solving problems using chemical equations.
1. Briefly write down the conditions of the problem and compose a chemical equation.
2. Write the problem data above the formulas in the chemical equation, and write the number of moles under the formulas (determined by the coefficient).
3. Find the amount of substance, the mass or volume of which is given in the problem statement, using the formulas:
M/M; = V / V m (for gases V m = 22.4 l / mol).
Write the resulting number above the formula in the equation.
4. Find the amount of a substance whose mass or volume is unknown. To do this, reason according to the equation: compare the number of moles according to the condition with the number of moles according to the equation. If necessary, make a proportion.
5. Find the mass or volume using the formulas: m = M; V = Vm.
This algorithm is the basis that the student must master so that in the future he will be able to solve problems using equations with various complications.
Problems with excess and deficiency.
If in the problem conditions the quantities, masses or volumes of two reacting substances are known at once, then this is a problem with excess and deficiency.
When solving it:
1. You need to find the quantities of two reacting substances using the formulas:
M/M; = V/V m .
2. Write the resulting mole numbers above the equation. Comparing them with the number of moles according to the equation, draw a conclusion about which substance is given in deficiency.
3. Based on the deficiency, make further calculations.
Problems on the fraction of the yield of the reaction product practically obtained from the theoretically possible.
Using the reaction equations, theoretical calculations are carried out and theoretical data for the reaction product are found: theor. , m theor. or V theory. . When carrying out reactions in the laboratory or in industry, losses occur, so the practical data obtained are practical. ,
m pract. or V practical. always less than theoretically calculated data. The yield share is designated by the letter (eta) and is calculated using the formulas:
(this) = practical. / theory = m pract. / m theor. = V practical / V theor.
It is expressed as a fraction of a unit or as a percentage. Three types of tasks can be distinguished:
If in the problem statement the data for the starting substance and the fraction of the yield of the reaction product are known, then you need to find a practical solution. , m practical or V practical. reaction product.
Solution procedure:
1. Carry out a calculation using the equation based on the data for the starting substance, find the theory. , m theor. or V theory. reaction product;
2. Find the mass or volume of the reaction product practically obtained using the formulas:
m pract. = m theoretical ; V practical = V theor. ; pract. = theoretical .
If in the problem statement the data for the starting substance and practice are known. , m practical or V practical. the resulting product, and you need to find the yield fraction of the reaction product.
Solution procedure:
1. Calculate using the equation based on the data for the starting substance, find
Theor. , m theor. or V theory. reaction product.
2. Find the yield fraction of the reaction product using the formulas:
Pract. / theory = m pract. / m theor. = V practical /V theor.
If the practical conditions are known in the problem conditions. , m practical or V practical. the resulting reaction product and its yield fraction, while you need to find data for the starting substance.
Solution procedure:
1. Find theory, m theory. or V theory. reaction product according to the formulas:
Theor. = practical / ; m theor. = m pract. / ; V theor. = V practical / .
2. Perform calculations using the equation based on the theory. , m theor. or V theory. product of the reaction and find the data for the starting substance.
Of course, we consider these three types of problems gradually, practicing the skills of solving each of them using the example of a number of problems.
Problems on mixtures and impurities.
A pure substance is the one that is more abundant in the mixture, the rest are impurities. Designations: mass of mixture – m cm, mass of pure substance – m p.h., mass of impurities – m approx. , mass fraction of pure substance - p.h.
The mass fraction of a pure substance is found using the formula: p.h. = m h.v. / m cm, it is expressed in fractions of one or as a percentage. Let's distinguish 2 types of tasks.
If the problem statement gives the mass fraction of a pure substance or the mass fraction of impurities, then the mass of the mixture is given. The word “technical” also means the presence of a mixture.
Solution procedure:
1. Find the mass of a pure substance using the formula: m h.v. = h.v. m cm
If the mass fraction of impurities is given, then you first need to find the mass fraction of the pure substance: p.h. = 1 - approx.
2. Based on the mass of the pure substance, make further calculations using the equation.
If the problem statement gives the mass of the initial mixture and n, m or V of the reaction product, then you need to find the mass fraction of the pure substance in the initial mixture or the mass fraction of impurities in it.
Solution procedure:
1. Calculate using the equation based on the data for the reaction product and find n p.v. and m h.v.
2. Find the mass fraction of the pure substance in the mixture using the formula: p.h. = m h.v. / m see and mass fraction of impurities: approx. = 1 - h.v
Law of volumetric relations of gases.
The volumes of gases are related in the same way as their quantities of substances:
V 1 / V 2 = 1 / 2
This law is used when solving problems using equations in which the volume of a gas is given and you need to find the volume of another gas.
Volume fraction of gas in the mixture.
Vg / Vcm, where (phi) is the volume fraction of gas.
Vg – gas volume, Vcm – volume of gas mixture.
If the problem statement gives the volume fraction of the gas and the volume of the mixture, then, first of all, you need to find the volume of the gas: Vg = Vcm.
The volume of the gas mixture is found using the formula: Vcm = Vg /.
The volume of air spent on combustion of a substance is found through the volume of oxygen found by the equation:
Vair = V(O 2) / 0.21
Derivation of formulas of organic substances using general formulas.
Organic substances form homologous series that have common formulas. This allows:
1. Express the relative molecular weight in terms of the number n.
M r (C n H 2n + 2) = 12 n + 1 (2n + 2) = 14n + 2.
2. Equate M r, expressed through n, to the true M r and find n.
3. Draw up reaction equations in general form and make calculations based on them.
Deriving formulas of substances based on combustion products.
1. Analyze the composition of combustion products and draw a conclusion about the qualitative composition of the burned substance: H 2 O -> H, CO 2 -> C, SO 2 -> S, P 2 O 5 -> P, Na 2 CO 3 -> Na, C.
The presence of oxygen in the substance requires verification. Denote the indices in the formula by x, y, z. For example, CxHyOz (?).
2. Find the amount of substances in combustion products using the formulas:
n = m / M and n = V / Vm.
3. Find the amounts of elements contained in the burned substance. For example:
n (C) = n (CO 2), n (H) = 2 ћ n (H 2 O), n (Na) = 2 ћ n (Na 2 CO 3), n (C) = n (Na 2 CO 3) etc.
Vm = g/l 22.4 l/mol; r = m/V.
b) if the relative density is known: M 1 = D 2 M 2, M = D H2 2, M = D O2 32,
M = D air 29, M = D N2 28, etc.
Method 1: find the simplest formula of the substance (see previous algorithm) and the simplest molar mass. Then compare the true molar mass with the simplest one and increase the indices in the formula by the required number of times.
Method 2: find the indices using the formula n = (e) Mr / Ar(e).
If the mass fraction of one of the elements is unknown, then it needs to be found. To do this, subtract the mass fraction of the other element from 100% or from unity.
Gradually, in the course of studying chemistry in the chemical dictionary, algorithms for solving problems of various types occur. And the student always knows where to find the right formula or the necessary information to solve a problem.
Many students like keeping such a notebook; they themselves supplement it with various reference materials.
As for extracurricular activities, my students and I also keep a separate notebook for writing down algorithms for solving problems that go beyond the scope of the school curriculum. In the same notebook, for each type of problem we write down 1-2 examples; they solve the rest of the problems in another notebook. And, if you think about it, among the thousands of different problems that appear on the chemistry exam in all universities, you can identify 25 - 30 different types of problems. Of course, there are many variations among them.
In developing algorithms for solving problems in elective classes, A.A.’s manual helped me a lot. Kushnareva. (Learning to solve problems in chemistry, - M., School - press, 1996).
The ability to solve problems in chemistry is the main criterion for creative mastery of the subject. It is through solving problems of various levels of complexity that a chemistry course can be effectively mastered.
If a student has a clear understanding of all possible types of problems and has solved a large number of problems of each type, then he will be able to cope with the chemistry exam in the form of the Unified State Exam and when entering universities.
Stoichiometry- quantitative relationships between reacting substances.
If reagents enter into a chemical interaction in strictly defined quantities, and as a result of the reaction substances are formed, the amount of which can be calculated, then such reactions are called stoichiometric.
Laws of stoichiometry:
The coefficients in chemical equations before the formulas of chemical compounds are called stoichiometric.
All calculations using chemical equations are based on the use of stoichiometric coefficients and are associated with finding quantities of a substance (number of moles).
The amount of substance in the reaction equation (number of moles) = the coefficient in front of the corresponding molecule.
N A=6.02×10 23 mol -1.
η - ratio of the actual mass of the product m p to a theoretically possible m t, expressed in fractions of a unit or as a percentage.
If the yield of reaction products is not indicated in the condition, then in the calculations it is taken equal to 100% (quantitative yield).
Calculation scheme using chemical reaction equations:
- Write an equation for a chemical reaction.
- Above the chemical formulas of substances write known and unknown quantities with units of measurement.
- Under the chemical formulas of substances with known and unknowns, write down the corresponding values of these quantities found from the reaction equation.
- Compose and solve a proportion.
Example. Calculate the mass and amount of magnesium oxide formed during complete combustion of 24 g of magnesium.
|
Given: m(Mg) = 24 g Find: ν (MgO) m (MgO) |
Solution: 1. Let's create an equation for a chemical reaction: 2Mg + O 2 = 2MgO. 2. Under the formulas of substances we indicate the amount of substance (number of moles) that corresponds to the stoichiometric coefficients: 2Mg + O2 = 2MgO 2 mole 2 mole 3. Determine the molar mass of magnesium: Relative atomic mass of magnesium Ar (Mg) = 24. Because the molar mass value is equal to the relative atomic or molecular mass, then M (Mg)= 24 g/mol. 4. Using the mass of the substance specified in the condition, we calculate the amount of the substance:
5. Above the chemical formula of magnesium oxide MgO, the mass of which is unknown, we set xmole, above the magnesium formula Mg we write its molar mass: 1 mole xmole 2Mg + O2 = 2MgO 2 mole 2 mole
According to the rules for solving proportions:
Amount of magnesium oxide ν (MgO)= 1 mol. 7. Calculate the molar mass of magnesium oxide: M (Mg)=24 g/mol, M(O)=16 g/mol. M(MgO)= 24 + 16 = 40 g/mol. We calculate the mass of magnesium oxide: m (MgO) = ν (MgO) × M (MgO) = 1 mol × 40 g/mol = 40 g. Answer: ν (MgO) = 1 mol; m (MgO) = 40 g. |
