Tasks for determining the formula of organic matter are of several types. Usually, the solution of these problems is not particularly difficult, but often graduates lose points on this problem. There are several reasons:
- Incorrect design;
- The solution is not mathematical, but by enumeration;
- Incorrectly compiled general formula of a substance;
- Errors in the reaction equation involving a substance written in general view.
Types of tasks in task C5.
- Determination of the formula of a substance by mass fractions chemical elements or according to the general formula of the substance;
- Determination of the formula of a substance by combustion products;
- Determination of the formula of a substance by chemical properties.
Necessary theoretical information.
- Mass fraction element in matter.
The mass fraction of an element is its content in a substance as a percentage by mass.
For example, a substance of composition C 2 H 4 contains 2 carbon atoms and 4 hydrogen atoms. If we take 1 molecule of such a substance, then its molecular weight will be equal to:
Mr (C 2 H 4) \u003d 2 12 + 4 1 \u003d 28 a.m.u. and it contains 2 12 a.m.u. carbon.To find the mass fraction of carbon in this substance, it is necessary to divide its mass by the mass of the entire substance:
ω(C) = 12 2 / 28 = 0.857 or 85.7%.
If a substance has the general formula C x H y O z, then the mass fractions of each of their atoms are also equal to the ratio of their mass to the mass of the entire substance. The mass x of C atoms is -12x, the mass y of H atoms is y, the mass z of oxygen atoms is 16z.
Then
ω(C) = 12 x / (12x + y + 16z)If we write this formula in general form, we get the following expression:
- Molecular and simplest formula of a substance.
Molecular (true) formula - a formula that reflects the real number of atoms of each type included in the molecule of a substance.
For example, C 6 H 6 is the true formula of benzene.
The simplest (empirical) formula - shows the ratio of atoms in a substance.
For example, for benzene, the ratio C:H = 1:1, i.e. The simplest formula for benzene is CH.
The molecular formula may coincide with the simplest or be a multiple of it.Examples.
If only the mass fractions of elements are given in the problem, then in the process of solving the problem, only the simplest formula of a substance can be calculated. To obtain the true formula in the problem, additional data is usually given - the molar mass, the relative or absolute density of the substance, or other data that can be used to determine the molar mass of the substance.
- Relative density of gas X by gas Y - D by Y (X).
Relative density D is a value that shows how many times gas X is heavier than gas Y. It is calculated as the ratio of the molar masses of gases X and Y:
D by Y (X) \u003d M (X) / M (Y)
Often used for calculations relative densities of gases for hydrogen and for air.
Relative gas density X for hydrogen:
D by H 2 \u003d M (gas X) / M (H 2) \u003d M (gas X) / 2
Air is a mixture of gases, so only the average molar mass can be calculated for it. Its value is taken as 29 g/mol (based on the approximate average composition).
So:
D by air. = M (gas X) / 29 - The absolute density of a gas under normal conditions.
The absolute density of a gas is the mass of 1 liter of gas under normal conditions. Usually for gases it is measured in g / l.
ρ = m (gas) / V (gas)
If we take 1 mole of gas, then:
ρ \u003d M / V m,
and the molar mass of a gas can be found by multiplying the density by the molar volume. - General formulas of substances of different classes.
Often, to solve problems with chemical reactions, it is convenient to use not the usual general formula, but a formula in which a multiple bond or a functional group is singled out separately.Class of organic substances General molecular formula Formula with highlighted multiple bond and functional group Alkanes C n H 2n+2 — Alkenes C n H 2n C n H 2n+1 -CH=CH 2 Alkynes C n H 2n−2 C n H 2n+1 -C≡CH dienes C n H 2n−2 — Benzene homologues C n H 2n−6 C 6 H 5 -C n H 2n+1 Limit monohydric alcohols C n H 2n+2 O C n H 2n+1 -OH Polyhydric alcohols C n H 2n+2 O x C n H 2n+2−x (OH) x Limit aldehydes C n H 2n O Esters C n H 2n O 2
Determination of formulas of substances by mass fractions of atoms that make up its composition.
The solution to these problems consists of two parts:
- first, the molar ratio of atoms in a substance is found - it corresponds to its simplest formula. For example, for a substance of composition A x B y, the ratio of the amounts of substances A and B corresponds to the ratio of the number of their atoms in the molecule:
x: y = n(A) : n(B); - then, using the molar mass of the substance, determine its true formula.
Example 1
Determine the formula of a substance if it contains 84.21% C and 15.79% H and has a relative density in air of 3.93.
Example 1 solution.
- Let the mass of the substance be 100 g. Then the mass C will be 84.21 g, and the mass H will be 15.79 g.
- Find the amount of matter of each atom:
ν(C) \u003d m / M \u003d 84.21 / 12 \u003d 7.0175 mol,
ν(H) = 15.79 / 1 = 15.79 mol. - We determine the molar ratio of C and H atoms:
C: H \u003d 7.0175: 15.79 (we reduce both numbers by a smaller one) \u003d 1: 2.25 (we multiply by 4) \u003d 4: 9.
Thus, the simplest formula is C 4 H 9. - Calculate the molar mass from the relative density:
M \u003d D (air.) 29 \u003d 114 g / mol.
The molar mass corresponding to the simplest formula C 4 H 9 is 57 g / mol, which is 2 times less than the true molar mass.
Hence, the true formula is C 8 H 18.
There is a much simpler way to solve this problem, but, unfortunately, they will not give a full score for it. But it is suitable for checking the true formula, i.e. with it you can check your solution.
Method 2: We find the true molar mass (114 g / mol), and then we find the masses of carbon and hydrogen atoms in this substance by their mass fractions.
m(C) = 114 0.8421 = 96; those. number of C atoms 96/12 = 8
m(H) = 114 0.1579 = 18; i.e., the number of H atoms 18/1 = 18.
The formula of the substance is C 8 H 18.
Answer: C 8 H 18.
Example 2
Determine the formula of alkyne with a density of 2.41 g/l under normal conditions.
Example 2 solution.
General formula of alkyne С n H 2n−2
How, given the density of a gaseous alkyne, to find its molar mass? Density ρ is the mass of 1 liter of gas under normal conditions.
Since 1 mole of a substance occupies a volume of 22.4 liters, it is necessary to find out how much 22.4 liters of such a gas weigh:
M \u003d (density ρ) (molar volume V m) \u003d 2.41 g / l 22.4 l / mol \u003d 54 g / mol.
Next, we write an equation relating the molar mass and n:
14n − 2 = 54, n = 4.
Hence, alkyne has the formula C 4 H 6.
Answer: C 4 H 6.
Example 3
Determine the formula of the limiting aldehyde if it is known that 3 10 22 molecules of this aldehyde weigh 4.3 g.
Example 3 solution.
In this problem, the number of molecules and the corresponding mass are given. Based on these data, we need to find again the value of the molar mass of the substance.
To do this, you need to remember how many molecules are contained in 1 mol of a substance.
This is Avogadro's number: N a = 6.02 10 23 (molecules).
So, you can find the amount of aldehyde substance:
ν \u003d N / Na \u003d 3 10 22 / 6.02 10 23 \u003d 0.05 mol,
and molar mass:
M \u003d m / n \u003d 4.3 / 0.05 \u003d 86 g / mol.
Further, as in the previous example, we make an equation and find n.
The general formula of the limiting aldehyde is C n H 2n O, that is, M \u003d 14n + 16 \u003d 86, n \u003d 5.
Answer: C 5 H 10 O, pentanal.
Example 4
Determine the formula of dichloroalkane containing 31.86% carbon.
Example 4 solution.
The general formula of dichloroalkane is: C n H 2n Cl 2, there are 2 chlorine atoms and n carbon atoms.
Then the mass fraction of carbon is equal to:
ω(C) = (number of C atoms in a molecule) ( atomic mass C) / (molecular weight of dichloroalkane)
0.3186 = n 12 / (14n + 71)
n = 3, the substance is dichloropropane.
Answer: C 3 H 6 Cl 2, dichloropropane.
Determination of formulas of substances by combustion products.
In tasks for combustion, the amount of substances of elements included in the substance under study is determined by the volumes and masses of combustion products - carbon dioxide, water, nitrogen and others. The rest of the solution is the same as in the first type of problems.
Example 5
448 ml (n.a.) gaseous saturated non-cyclic hydrocarbon was burned, and the reaction products were passed through an excess of lime water, while forming 8 g of a precipitate. What hydrocarbon was taken?
Example 5 solution.
- The general formula of a gaseous saturated non-cyclic hydrocarbon (alkane) is C n H 2n + 2
Then the combustion reaction scheme looks like this:C n H 2n+2 + O 2 → CO 2 + H 2 O
It is easy to see that the combustion of 1 mole of alkane will release n moles of carbon dioxide.The amount of alkane substance is found by its volume (do not forget to convert milliliters to liters!):
ν(C n H 2n+2) = 0.488 / 22.4 = 0.02 mol.
- When carbon dioxide is passed through Ca (OH) 2 lime water, calcium carbonate precipitates:
CO 2 + Ca (OH) 2 \u003d CaCO 3 + H 2 O
The mass of calcium carbonate precipitate is 8 g, the molar mass of calcium carbonate is 100 g/mol.
So its amount of matter
ν (CaCO 3) \u003d 8/100 \u003d 0.08 mol.
The amount of carbon dioxide substance is also 0.08 mol. - The amount of carbon dioxide is 4 times more than alkane, so the formula of alkane is C 4 H 10.
Answer: C 4 H 10.
Example 6
Relative vapor density organic compound nitrogen is 2. When burning 9.8 g of this compound, 15.68 liters of carbon dioxide (n.a.) and 12.6 g of water are formed. Derive the molecular formula of the organic compound.
Example 6 solution.
Since the substance turns into carbon dioxide and water during combustion, it means that it consists of atoms C, H and, possibly, O. Therefore, its general formula can be written as C x H y O z.
- We can write the combustion reaction scheme (without placing the coefficients):
C x H y O z + O 2 → CO 2 + H 2 O
All the carbon from the original substance goes into carbon dioxide, and all the hydrogen goes into water.
- We find the amounts of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain:
ν (CO 2) \u003d V / V m \u003d 15.68 / 22.4 \u003d 0.7 mol.
One molecule of CO 2 accounts for one atom C, which means that there are as many moles of carbon as CO 2.ν(C) = 0.7 mol
One molecule of water contains two atom H, means the amount of hydrogen twice as much than water.
ν(H) \u003d 0.7 2 \u003d 1.4 mol. - We check the presence of oxygen in the substance. To do this, the masses C and H must be subtracted from the mass of the entire initial substance.
m(C) = 0.7 12 = 8.4 g, m(H) = 1.4 1 = 1.4 g
The mass of the entire substance is 9.8 g.
m(O) = 9.8 - 8.4 - 1.4 = 0, i.e. there are no oxygen atoms in this substance.
If oxygen were present in a given substance, then by its mass it would be possible to find the amount of the substance and calculate the simplest formula based on the presence of three different atoms. - The next steps are already familiar to you: the search for the simplest and true formulas.
C: H = 0.7: 1.4 = 1: 2
The simplest formula of CH 2. - We are looking for the true molar mass by the relative density of the gas with respect to nitrogen (do not forget that nitrogen consists of diatomic N 2 molecules and its molar mass is 28 g / mol):
M ist. \u003d D by N 2 M (N 2) \u003d 2 28 \u003d 56 g / mol.
The true formula is CH 2, its molar mass is 14.
56 / 14 = 4.
The true formula is C 4 H 8.
Answer: C 4 H 8.
Example 7
Determine the molecular formula of the substance, during the combustion of 9 g of which 17.6 g of CO 2, 12.6 g of water and nitrogen were formed. The relative density of this substance with respect to hydrogen is 22.5. Determine the molecular formula of the substance.
Example 7 solution.
- Substance contains C,H atoms and N. Since the mass of nitrogen in the combustion products is not given, it will have to be calculated based on the mass of all organic matter.
Combustion reaction scheme:
C x H y N z + O 2 → CO 2 + H 2 O + N 2 - We find the amounts of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain:
ν (CO 2) \u003d m / M \u003d 17.6 / 44 \u003d 0.4 mol.
ν(C) = 0.4 mol.
ν (H 2 O) \u003d m / M \u003d 12.6 / 18 \u003d 0.7 mol.
ν(H) \u003d 0.7 2 \u003d 1.4 mol. - Find the mass of nitrogen in the original substance.
To do this, the masses C and H must be subtracted from the mass of the entire initial substance.M(C) = 0.4 12 = 4.8 g,
m(H) = 1.4 1 = 1.4 gThe mass of the entire substance is 9.8 g.
M(N) = 9 - 4.8 - 1.4 = 2.8 g,
ν(N) \u003d m / M \u003d 2.8 / 14 \u003d 0.2 mol. - C:H:N=0.4:1.4:0.2=2:7:1
The simplest formula is C 2 H 7 N.
True molar mass
M \u003d D according to H 2 M (H 2) \u003d 22.5 2 \u003d 45 g / mol.
It coincides with the molar mass calculated for the simplest formula. That is, this is the true formula of the substance.
Answer: C 2 H 7 N.
Example 8
The substance contains C, H, O and S. When 11 g was burned, 8.8 g CO 2, 5.4 g H 2 O were released, and sulfur was completely converted into barium sulfate, the mass of which turned out to be 23.3 g. Determine substance formula.
Example 8 solution.
The formula of a given substance can be represented as C x H y S z O k . When it is burned, carbon dioxide, water and sulfur dioxide are produced, which are then converted into barium sulfate. Accordingly, all sulfur from the original substance is converted to barium sulfate.
- We find the amounts of substances of carbon dioxide, water and barium sulfate and the corresponding chemical elements from the substance under study:
ν (CO 2) \u003d m / M \u003d 8.8 / 44 \u003d 0.2 mol.
ν(C) = 0.2 mol.
ν (H 2 O) \u003d m / M \u003d 5.4 / 18 \u003d 0.3 mol.
ν(H) = 0.6 mol.
ν (BaSO 4) \u003d 23.3 / 233 \u003d 0.1 mol.
ν(S) = 0.1 mol. - We calculate the estimated mass of oxygen in the initial substance:
M(C) = 0.2 12 = 2.4 g
m(H) = 0.6 1 = 0.6 g
m(S) = 0.1 32 = 3.2 g
m(O) = m substances − m(C) − m(H) − m(S) = 11 − 2.4 − 0.6 − 3.2 = 4.8 g,
ν(O) = m / M = 4.8 / 16 = 0.3 mol - We find the molar ratio of elements in a substance:
C:H:S:O=0.2:0.6:0.1:0.3=2:6:1:3
Substance formula C 2 H 6 SO 3.
It should be noted that in this way we obtained only the simplest formula.
However, the resulting formula is true, because when you try to double this formula (C 4 H 12 S 2 O 6), it turns out that 4 carbon atoms, in addition to sulfur and oxygen, have 12 H atoms, and this is impossible.
Answer: C 2 H 6 SO 3.
Determination of formulas of substances by chemical properties.
Example 9
Determine the formula of alkadiene if 80 g of a 2% bromine solution can decolorize it.
Example 9 solution.
- The general formula for alkadienes is С n H 2n−2.
Let us write the reaction equation for the addition of bromine to alkadiene, not forgetting that in the diene molecule two double bonds and, accordingly, 2 moles of bromine will react with 1 mole of the diene:
С n H 2n−2 + 2Br 2 → С n H 2n−2 Br 4 - Since the mass and percentage concentration of the bromine solution that reacted with the diene are given in the problem, it is possible to calculate the amount of substance of the reacted bromine:
M (Br 2) \u003d m solution ω \u003d 80 0.02 \u003d 1.6 g
ν (Br 2) \u003d m / M \u003d 1.6 / 160 \u003d 0.01 mol. - Since the amount of bromine that reacted is 2 times more than the alkadiene, you can find the amount of the diene and (since its mass is known) its molar mass:
0,005 0,01 C n H 2n−2 + 2Br2 → C n H 2n−2 Br 4 M diene \u003d m / ν \u003d 3.4 / 0.05 \u003d 68 g / mol.
- We find the formula of alkadiene according to its general formulas, expressing the molar mass in terms of n:
14n - 2 = 68
n = 5.This is C 5 H 8 pentadiene.
Answer: C 5 H 8.
Example 10
In the interaction of 0.74 g of saturated monohydric alcohol with metallic sodium, hydrogen was released in an amount sufficient to hydrogenate 112 ml of propene (n.a.). What is this alcohol?
Example 10 solution.
- The formula for limiting monohydric alcohol is C n H 2n + 1 OH. Here it is convenient to write the alcohol in a form in which it is easy to formulate the reaction equation - i.e. with a separate OH group.
- Let's compose the reaction equations (we must not forget about the need to equalize the reactions):
2C n H 2n+1 OH + 2Na → 2C n H 2n+1 ONa + H 2
C 3 H 6 + H 2 → C 3 H 8 - You can find the amount of propene, and from it - the amount of hydrogen. Knowing the amount of hydrogen, by the reaction we find the amount of alcohol substance:
ν (C 3 H 6) \u003d V / V m \u003d 0.112 / 22.4 \u003d 0.005 mol => ν (H 2) \u003d 0.005 mol,
ν alcohol \u003d 0.005 2 \u003d 0.01 mol. - Find the molar mass of alcohol and n:
M alcohol \u003d m / ν \u003d 0.74 / 0.01 \u003d 74 g / mol,
14n + 18 = 74
14n = 56
n = 4.Alcohol - butanol C 4 H 7 OH.
Answer: C 4 H 7 OH.
Example 11.
Define Formula ester, upon hydrolysis of 2.64 g of which 1.38 g of alcohol and 1.8 g of monobasic carboxylic acid.
Example 11 solution.
- The general formula of an ester, consisting of an alcohol and an acid with a different number of carbon atoms, can be represented as follows:
C n H 2n+1 COOC m H 2m+1
Accordingly, alcohol will have the formula
C m H 2m+1 OH,
and acid
C n H 2n+1 COOH .
Ester hydrolysis equation:
C n H 2n+1 COOC m H 2m+1 + H 2 O → C m H 2m+1 OH + C n H 2n+1 COOH - According to the law of conservation of mass of substances, the sum of the masses of the initial substances and the sum of the masses of the reaction products are equal.
Therefore, from the data of the problem, you can find the mass of water:M H 2 O \u003d (mass of acid) + (mass of alcohol) - (mass of ether) \u003d 1.38 + 1.8 - 2.64 \u003d 0.54 g
ν H 2 O \u003d m / M \u003d 0.54 / 18 \u003d 0.03 molAccordingly, the amounts of acid and alcohol substances are also equal to a mole.
You can find their molar masses:M acid \u003d m / ν \u003d 1.8 / 0.03 \u003d 60 g / mol,
M alcohol \u003d 1.38 / 0.03 \u003d 46 g / mol.We get two equations, from which we find m and n:
M C n H 2n + 1 COOH \u003d 14n + 46 \u003d 60, n \u003d 1 - acetic acid
M C m H 2m + 1 OH = 14m + 18 = 46, m = 2 - ethanol.Thus, the desired ester is the ethyl ester of acetic acid, ethyl acetate.
Answer: CH 3 COOC 2 H 5 .
Example 12.
Determine the formula of an amino acid if, by treating 8.9 g of it with an excess of sodium hydroxide, 11.1 g of the sodium salt of this acid can be obtained.
Example 12 solution.
- The general formula of an amino acid (assuming that it does not contain any other functional groups, except for one amino group and one carboxyl group):
NH2-CH(R)-COOH.
It could be written in different ways, but for the convenience of writing the reaction equation, it is better to isolate the functional groups separately in the amino acid formula. - You can write an equation for the reaction of this amino acid with sodium hydroxide:
NH 2 -CH(R)-COOH + NaOH → NH 2 -CH(R)-COONa + H 2 O
The amounts of the substance of the amino acid and its sodium salt are equal. However, we cannot find the mass of any of the substances in the reaction equation. Therefore, in such problems it is necessary to express the amounts of substances of an amino acid and its salt in terms of molar masses and equate them:M (amino acids NH 2 -CH (R) -COOH) \u003d 74 + M R
M(salts NH 2 -CH(R)-COONa) \u003d 96 + M R
ν amino acids = 8.9 / (74 + M R),
ν salt = 11.1 / (96 + M R)
8.9 / (74 + M R) = 11.1 / (96 + M R)
M R = 15It is easy to see that R = CH 3 .
This can be done mathematically if we assume that R - C n H 2n+1 .
14n + 1 = 15, n = 1 . Set the formula of the limiting monobasic carboxylic acid, the calcium salt of which contains 30.77% calcium.Part 2. Determination of the formula of a substance by combustion products.
2-1. The relative vapor density of an organic compound in terms of sulfur dioxide is 2. When burning 19.2 g of this substance, 52.8 g of carbon dioxide (N.O.) and 21.6 g of water are formed. Derive the molecular formula of the organic compound.
2-2. When burning organic matter weighing 1.78 g in excess oxygen, 0.28 g of nitrogen, 1.344 l (n.o.) CO 2 and 1.26 g of water were obtained. Determine the molecular formula of the substance, knowing that the indicated sample of the substance contains 1.204 10 22 molecules.
2-3. Carbon dioxide obtained from the combustion of 3.4 g of hydrocarbon was passed through an excess of calcium hydroxide solution and 25 g of precipitate was obtained. Derive the simplest formula for a hydrocarbon.
2-4. During the combustion of organic matter containing C, H and chlorine, 6.72 l (N.O.) of carbon dioxide, 5.4 g of water, 3.65 g of hydrogen chloride were released. Set the molecular formula of the burned substance.
2-5. (USE-2011) During the combustion of the amine, 0.448 l (n.o.) of carbon dioxide, 0.495 g of water and 0.056 l of nitrogen were released. Determine the molecular formula of this amine.
Part 3. Determination of the formula of a substance by chemical properties.
3-1. Determine the formula of an alkene if it is known that 5.6 g of it, when added to water, form 7.4 g of alcohol.
3-2. For the oxidation of 2.9 g of saturated aldehyde to acid, 9.8 g of copper (II) hydroxide was required. Determine the formula of the aldehyde.
3-3. Monobasic monoamino acid weighing 3 g with an excess of hydrogen bromide forms 6.24 g of salt. Determine the amino acid formula.
3-4. In the interaction of the limiting dihydric alcohol weighing 2.7 g with an excess of potassium, 0.672 liters of hydrogen were released. Determine the formula of alcohol.
3-5. (USE-2011) When saturated monohydric alcohol was oxidized with copper (II) oxide, 9.73 g of aldehyde, 8.65 g of copper and water were obtained. Determine the molecular formula of this alcohol.
Answers and comments to tasks for independent solution.
1-2. C 3 H 6 (NH 2) 2
1-3. C 2 H 4 (COOH) 2
1-5. (HCOO) 2 Ca - calcium formate, salt of formic acid
2-1. C 8 H 16 O
2-2. C 3 H 7 NO
2-3. C 5 H 8 (we find the mass of hydrogen by subtracting the mass of carbon from the mass of hydrocarbon)
2-4. C 3 H 7 Cl (do not forget that hydrogen atoms are found not only in water, but also in HCl)
3-2. C 3 H 6 O
3-3. C 2 H 5 NO 2
Chemical formula is an image with symbols .
Signs of chemical elements
chemical sign or element chemical symbol is the first or two first letters of the Latin name of this element.
For example: Ferrum-Fe , cuprum-Cu , oxygenium-O etc.
Table 1: Information provided by the chemical mark
Intelligence On the example of Cl Element name Chlorine Non-metal, halogen One item 1 chlorine atom (ar) given element Ar(Cl) = 35.5 Absolute atomic mass of a chemical element m = Ar 1.66 10 -24 g = Ar 1.66 10 -27 kg
M (Cl) \u003d 35.5 1.66 10 -24 \u003d 58.9 10 -24 g The name of a chemical sign in most cases is read as the name of a chemical element. For example, K - potassium, Ca - calcium, Mg - magnesium, Mn - manganese.
Cases where the name of the chemical mark is read differently are given in Table 2:
Name of the chemical element chemical sign The name of the chemical symbol (pronunciation)
Nitrogen N En Hydrogen H Ash Iron Fe Ferrum Gold Au Aurum Oxygen O O Silicon Si Silicium Copper Cu Cuprum Tin sn Stanum Mercury hg hydrargium Lead Pb Plumbum Sulfur S Es Silver Ag Argentum Carbon C Tse Phosphorus P Pe Chemical formulas of simple substances
The chemical formulas of most simple substances(of all metals and many non-metals) are signs of the corresponding chemical elements.
So substance iron and chemical element iron are labeled the same Fe .
If it has a molecular structure (exists in the form , then its formula is the chemical sign of the element with index bottom right, indicating number of atoms in a molecule: H2, O2, O 3, N 2, F2, Cl2, Br2, P4, S8.
Table 3: Information provided by the chemical mark
Intelligence For example C Substance name Carbon (diamond, graphite, graphene, carbine) Belonging of an element to a given class of chemical elements Non-metal One element atom 1 carbon atom Relative atomic mass (ar) the element that makes up the substance Ar(C)=12 Absolute atomic mass M (C) \u003d 12 1.66 10-24 \u003d 19.93 10 -24 g One substance 1 mole of carbon, i.e. 6.02 10 23 carbon atoms M(C) = Ar(C) = 12 g/mol Chemical formulas of complex substances
The formula of a complex substance is compiled by writing the signs of the chemical elements of which this substance consists, indicating the number of atoms of each element in the molecule. In this case, as a rule, chemical elements are written in order of increasing electronegativity according to the following practice series:
Me , Si , B , Te , H , P , As , I , Se , C , S , Br , Cl , N , O , F
For example, H2O , CaSO4 , Al2O3 , CS2 , OF 2 , NaH.
The exception is:
- some compounds of nitrogen with hydrogen (for example, ammonia NH3 , hydrazine N 2H4 );
- salts of organic acids (for example, sodium formate HCOONa , calcium acetate (CH 3COO) 2Ca) ;
- hydrocarbons ( CH 4 , C 2 H 4 , C 2 H 2 ).
Chemical formulas substances that exist in the form dimers (NO 2 , P2O 3 , P2O5, monovalent mercury salts, for example: HgCl , HgNO3 etc.), is written in the form N 2 O 4 ,P4 O 6 ,P4 O 10 ,Hg 2 Cl2,Hg 2 ( NO 3) 2 .
The number of atoms of a chemical element in a molecule and a complex ion is determined based on the concept valency or oxidation states and recorded index bottom right from the sign of each element (index 1 is omitted). This is based on the rule:
algebraic sum the oxidation states of all atoms in a molecule should be zero (the molecules are electrically neutral), and in a complex ion, the charge of the ion.
For example:
2Al 3 + + 3SO 4 2- \u003d Al 2 (SO 4) 3
The same rule is used when determining the degree of oxidation of a chemical element according to the formula of a substance or complex. Usually it is an element that has several oxidation states. The oxidation states of the remaining elements forming the molecule or ion must be known.
The charge of a complex ion is the algebraic sum of the oxidation states of all the atoms that form the ion. Therefore, when determining the oxidation state of a chemical element in a complex ion, the ion itself is enclosed in brackets, and its charge is taken out of brackets.
When compiling formulas for valence the substance is represented as a compound consisting of two particles of different types, the valences of which are known. Further enjoy rule:
in a molecule, the product of valence and the number of particles of one type must be equal to the product of valence and the number of particles of another type.
For example:
The number in front of a formula in a reaction equation is called coefficient. She indicates either number of molecules, or number of moles of a substance.
Factor before chemical sign , indicates the number of atoms of a given chemical element, and in the case when the sign is a formula of a simple substance, the coefficient indicates either number of atoms, or the number of moles of this substance.
For example:
- 3 Fe- three iron atoms, 3 moles of iron atoms,
- 2 H- two hydrogen atoms, 2 mol hydrogen atoms,
- H2- one molecule of hydrogen, 1 mole of hydrogen.
The chemical formulas of many substances have been determined empirically, which is why they are called "empirical".
Table 4: Information provided by the chemical formula of a complex substance
Intelligence For example C aCO3 Substance name Calcium carbonate Belonging of an element to a certain class of substances Medium (normal) salt One molecule of a substance 1 molecule of calcium carbonate One mole of a substance 6.02 10 23 molecules CaCO3 Relative molecular weight of the substance (Mr) Mr (CaCO3) \u003d Ar (Ca) + Ar (C) + 3Ar (O) \u003d 100 Molar mass of a substance (M) M (CaCO3) = 100 g/mol Absolute molecular weight of a substance (m) M (CaCO3) = Mr (CaCO3) 1.66 10 -24 g = 1.66 10 -22 g Qualitative composition (what chemical elements form a substance) calcium, carbon, oxygen The quantitative composition of the substance: The number of atoms of each element in one molecule of a substance: The calcium carbonate molecule is made up of 1 atom calcium, 1 atom carbon and 3 atoms oxygen. The number of moles of each element in 1 mole of a substance: In 1 mol CaCO 3(6.02 10 23 molecules) contains 1 mol(6.02 10 23 atoms) calcium, 1 mol(6.02 10 23 atoms) carbon and 3 mol(3 6.02 10 23 atoms) of the chemical element oxygen) Mass composition of the substance: The mass of each element in 1 mole of a substance: 1 mole of calcium carbonate (100g) contains chemical elements: 40g calcium, 12g carbon, 48g oxygen. Mass fractions of chemical elements in a substance (composition of a substance in percent by weight): Composition of calcium carbonate by mass: W (Ca) \u003d (n (Ca) Ar (Ca)) / Mr (CaCO3) \u003d (1 40) / 100 \u003d 0.4 (40%)
W (C) \u003d (n (Ca) Ar (Ca)) / Mr (CaCO3) \u003d (1 12) / 100 \u003d 0.12 (12%)
W (O) \u003d (n (Ca) Ar (Ca)) / Mr (CaCO3) \u003d (3 16) / 100 \u003d 0.48 (48%)
For a substance with an ionic structure (salts, acids, bases) - the formula of a substance gives information about the number of ions of each type in a molecule, their number and mass of ions in 1 mol of a substance: Molecule CaCO 3 is made up of an ion Ca 2+ and ion CO 3 2- 1 mol ( 6.02 10 23 molecules) CaCO 3 contains 1 mol of Ca 2+ ions and 1 mole of ions CO 3 2-;
1 mole (100g) of calcium carbonate contains 40g ions Ca 2+ and 60g ions CO 3 2-
Molar volume of a substance under normal conditions (only for gases) Graphic formulas
For more information about a substance use graphic formulas , which indicate the order in which atoms are connected in a molecule and valency of each element.
Graphic formulas of substances consisting of molecules, sometimes, to one degree or another, reflect the structure (structure) of these molecules, in these cases they can be called structural .
To draw up a graphical (structural) formula of a substance, you must:
- Determine the valence of all chemical elements that form a substance.
- Write down the signs of all chemical elements that form a substance, each in an amount equal to the number of atoms of a given element in a molecule.
- Connect the signs of chemical elements with dashes. Each line denotes a pair that makes a connection between chemical elements and therefore equally belongs to both elements.
- The number of dashes surrounding the sign of a chemical element must correspond to the valence of this chemical element.
- When formulating oxygen-containing acids and their salts, hydrogen atoms and metal atoms are bound to the acid-forming element through an oxygen atom.
- Oxygen atoms are connected to each other only when formulating peroxides.
Examples of graphic formulas:
Chemistry- the science of the composition, structure, properties and transformations of substances.
Atomic-molecular doctrine. Substances are made up of chemical particles (molecules, atoms, ions) that have complex structure and consist of elementary particles(protons, neutrons, electrons).
Atom- a neutral particle consisting of a positive nucleus and electrons.
Molecule- a stable group of atoms linked by chemical bonds.
Chemical element A type of atom with the same nuclear charge. Element designate
where X is the symbol of the element, Z- the serial number of the element in the Periodic system of elements of D.I. Mendeleev, A- mass number. Serial number Z equal to the charge of the atomic nucleus, the number of protons in the atomic nucleus and the number of electrons in the atom. Mass number A is equal to the sum of the numbers of protons and neutrons in an atom. The number of neutrons is equal to the difference A-Z
isotopes Atoms of the same element with different mass numbers.
Relative atomic mass(A r) is the ratio of the average mass of an atom of an element of natural isotopic composition to 1/12 of the mass of an atom of the carbon isotope 12 C.
Relative molecular weight(M r) - the ratio of the average mass of a molecule of a substance of natural isotopic composition to 1/12 of the mass of an atom of the carbon isotope 12 C.
Atomic mass unit(a.u.m) - 1/12 part of the mass of an atom of the carbon isotope 12 C. 1 a.u. m = 1.66? 10 -24 years
mole- the amount of a substance containing so much structural units(atoms, molecules, ions), how many atoms are contained in 0.012 kg of the carbon isotope 12 C. mole- the amount of a substance containing 6.02 10 23 structural units (atoms, molecules, ions).
n = N/N A, where n- amount of substance (mol), N is the number of particles, a N A is the Avogadro constant. The amount of a substance can also be denoted by the symbol v.
Avogadro constant N A = 6.02 10 23 particles/mol.
Molar massM(g / mol) - the ratio of the mass of a substance m(d) to the amount of substance n(mol):
M = m/n, where: m = M n and n = m/M.
Molar volume of gasV M(l/mol) – ratio of gas volume V(l) to the amount of substance of this gas n(mol). Under normal conditions V M = 22.4 l/mol.
Normal conditions: temperature t = 0°C or T = 273 K, pressure p = 1 atm = 760 mm. rt. Art. = 101 325 Pa = 101.325 kPa.
V M = V/n, where: V = V M n and n = V/V M .
The result is a general formula:
n = m/M = V/V M = N/N A .
Equivalent- a real or conditional particle interacting with one hydrogen atom, or replacing it, or equivalent to it in some other way.
Molar mass equivalents M e- the ratio of the mass of a substance to the number of equivalents of this substance: M e = m/n (eq) .
In charge exchange reactions, the molar mass of substance equivalents
with molar mass M equal to: M e = М/(n ? m).
In redox reactions, the molar mass equivalents of a substance with a molar mass M equal to: M e = M/n(e), where n(e) is the number of electrons transferred.
Law of Equivalents– the masses of reactants 1 and 2 are proportional to the molar masses of their equivalents. m1/m2= M E1 / M E2, or m 1 / M E1 \u003d m 2 / M E2, or n 1 \u003d n 2, where m 1 and m2 are the masses of two substances, M E1 and M E2 are the molar masses of equivalents, n 1 and n 2- the number of equivalents of these substances.
For solutions, the law of equivalents can be written in the following form:
c E1 V 1 = c E2 V 2, where with E1, with E2, V 1 and V 2- molar concentrations of equivalents and volumes of solutions of these two substances.
Combined gas law: pV = nRT, where p– pressure (Pa, kPa), V- volume (m 3, l), n- the amount of gas substance (mol), T- temperature (K), T(K) = t(°C) + 273, R- constant, R= 8.314 J / (K? mol), while J \u003d Pa m 3 \u003d kPa l.
2. The structure of the atom and the Periodic Law
Wave-particle duality matter - the idea that each object can have both wave and corpuscular properties. Louis de Broglie proposed a formula linking the wave and particle properties of objects: ? = h/(mV), where h is Planck's constant, ? is the wavelength that corresponds to each body with a mass m and speed v. Although wave properties exist for all objects, but they can be observed only for micro-objects having masses of the order of the mass of an atom and an electron.
Heisenberg Uncertainty Principle: ?(mV x) ?x > h/2n or ?V x ?x > h/(2?m), where m is the mass of the particle, x is its coordinate Vx- speed in direction x, ?– uncertainty, determination error. The uncertainty principle means that it is impossible to simultaneously specify the position (coordinate) of x) and speed (Vx) particles.
Particles with small masses (atoms, nuclei, electrons, molecules) are not particles in the understanding of this by Newtonian mechanics and cannot be studied by classical physics. They are being studied quantum physics.
Principal quantum numbern takes the values 1, 2, 3, 4, 5, 6 and 7 corresponding to the electronic levels (layers) K, L, M, N, O, P and Q.
Level- space where electrons with the same number are located n. Electrons of different levels are spatially and energetically separated from each other, since the number n determines the energy of electrons E(the more n, the more E) and distance R between electrons and the nucleus (the more n, the more R).
Orbital (side, azimuthal) quantum numberl takes values depending on the number n:l= 0, 1,…(n- one). For example, if n= 2, then l = 0.1; if n= 3, then l = 0, 1, 2. Number l characterizes the sublevel (sublayer).
sublevel- the space where the electrons are located with certain n and l. Sublevels of this level are designated depending on the number l:s- if l = 0, p- if l = 1, d- if l = 2, f- if l = 3. The sublevels of a given atom are designated depending on the numbers n and l, ex: 2s (n = 2, l = 0), 3d(n= 3, l = 2), etc. The sublevels of a given level have different energies (the more l, the more E): E s< E < Е А < … and different shape orbitals that make up these sublevels: the s-orbital has the shape of a ball, p-orbital has the shape of a dumbbell, etc.
Magnetic quantum numberm 1 characterizes the orientation of the orbital magnetic moment equal to l, in space relative to the external magnetic field and takes the values: – l,…-1, 0, 1,…l, i.e. total (2l + 1) value. For example, if l = 2, then m 1 =-2, -1, 0, 1, 2.
Orbital(part of a sublevel) - the space where electrons are located (no more than two) with certain n, l, m 1 . Sublevel contains 2l+1 orbital. For example, d– the sublevel contains five d-orbitals. Orbitals of the same sublevel, having different numbers m 1 , have the same energy.
Magnetic spin numberm s characterizes the orientation of the intrinsic magnetic moment of the electron s, equal to?, relative to the external magnetic field and takes two values: +? and _ ?.
Electrons in an atom occupy levels, sublevels, and orbitals according to the following rules.
Pauli's rule: Two electrons in one atom cannot have four identical quantum numbers. They must differ by at least one quantum number.
It follows from the Pauli rule that an orbital can contain no more than two electrons, a sublevel can contain no more than 2(2l + 1) electrons, a level can contain no more than 2n 2 electrons.
Klechkovsky's rule: the filling of electronic sublevels is carried out in ascending order of the amount (n+l), and in the case of the same amount (n+l)- in ascending order of number n.
Graphic form of the Klechkovsky rule.
According to the Klechkovsky rule, the filling of sublevels is carried out in the following order: 1s, 2s, 2p, 3s, Zp, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s,…
Although the filling of sublevels occurs according to the Klechkovsky rule, in the electronic formula, sublevels are written sequentially by levels: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f etc. Thus, the electronic formula of the bromine atom is written as follows: Br (35e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 .
The electronic configurations of a number of atoms differ from those predicted by the Klechkovsky rule. So, for Cr and Cu:
Cr(24e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 and Cu(29e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1.
Hund's (Gund's) rule: the filling of the orbitals of a given sublevel is carried out so that the total spin is maximum. The orbitals of a given sublevel are first filled by one electron.
Electronic configurations of atoms can be written down by levels, sublevels, orbitals. For example, the electronic formula P(15e) can be written:
a) by levels)2)8)5;
b) by sublevels 1s 2 2s 2 2p 6 3s 2 3p 3;
c) by orbitals
Examples of electronic formulas of some atoms and ions:
V(23e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2;
V 3+ (20e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 0.
3. Chemical bond
3.1. Valence bond method
According to the method of valence bonds, the bond between atoms A and B is formed using a common pair of electrons.
covalent bond. Donor-acceptor connection.Valency characterizes the ability of atoms to form chemical bonds and is equal to the number chemical bonds formed by an atom. According to the method of valence bonds, valence is equal to the number of common pairs of electrons, and in the case covalent bond valency is equal to the number of unpaired electrons in the outer level of an atom in its ground or excited states.
Valence of atomsFor example, for carbon and sulfur:
Saturability covalent bond: atoms form a limited number of bonds equal to their valency.
Hybridization of atomic orbitals– mixing of atomic orbitals (AO) of different sublevels of the atom, the electrons of which are involved in the formation of equivalent?-bonds. The equivalence of hybrid orbitals (HO) explains the equivalence of the formed chemical bonds. For example, in the case of a tetravalent carbon atom, there is one 2s– and three 2p-electron. To explain the equivalence of the four?-bonds formed by carbon in the CH 4, CF 4, etc. molecules, the atomic one s- and three R- orbitals are replaced by four equivalent hybrid sp 3-orbitals:
Orientation covalent bond is that it is formed in the direction of maximum overlap of the orbitals that form a common pair of electrons.
Depending on the type of hybridization, hybrid orbitals have a certain spatial arrangement:
sp– linear, the angle between the axes of the orbitals is 180°;
sp 2– triangular, the angles between the axes of the orbitals are 120°;
sp 3– tetrahedral, the angles between the axes of the orbitals are 109°;
sp 3 d 1– trigonal-bipyramidal, angles 90° and 120°;
sp2d1– square, the angles between the axes of the orbitals are 90°;
sp 3 d 2– octahedral, the angles between the axes of the orbitals are 90°.
3.2. Theory of molecular orbitals
According to the theory of molecular orbitals, a molecule consists of nuclei and electrons. In molecules, electrons are in molecular orbitals (MOs). MO of outer electrons have a complex structure and are considered as a linear combination of outer orbitals of the atoms that make up the molecule. The number of formed MOs is equal to the number of AOs participating in their formation. The energies of MOs can be lower (bonding MOs), equal (non-bonding MOs), or higher (loosening, anti-bonding MOs) than the energies of the AOs that form them.
JSC interaction conditions
1. AO interact if they have similar energies.
2. AOs interact if they overlap.
3. AO interact if they have the appropriate symmetry.
For a diatomic AB molecule (or any linear molecule), the MO symmetry can be:
If a given MO has an axis of symmetry,
If a given MO has a plane of symmetry,
If MO has two perpendicular planes of symmetry.
The presence of electrons on bonding MOs stabilizes the system, since it reduces the energy of the molecule compared to the energy of atoms. The stability of a molecule is characterized connection order n, equal to: n \u003d (n sv - n res) / 2, where n sv and n res - the number of electrons in bonding and loosening orbitals.
The filling of an MO with electrons occurs according to the same rules as the filling of an AO in an atom, namely: the Pauli rule (there cannot be more than two electrons on an MO), the Hund rule (the total spin must be maximum), etc.
The interaction of 1s-AO atoms of the first period (H and He) leads to the formation of a bonding?-MO and a loosening?*-MO:
Electronic formulas of molecules, bond orders n, experimental bond energies E and intermolecular distances R for diatomic molecules from atoms of the first period are given in the following table:
Other atoms of the second period contain, in addition to 2s-AO, also 2p x -, 2p y - and 2p z -AO, which can form ?- and ?-MO upon interaction. For O, F, and Ne atoms, the energies of 2s– and 2p-AO are significantly different, and the interaction between the 2s-AO of one atom and the 2p-AO of another atom can be neglected, considering the interaction between the 2s-AO of two atoms separately from the interaction of their 2p-AO. The MO scheme for O 2 , F 2 , Ne 2 molecules has the following form:
For B, C, N atoms, the energies of 2s– and 2p-AO are close in their energies, and the 2s-AO of one atom interacts with the 2p z-AO of another atom. Therefore, the order of MO in B 2 , C 2 and N 2 molecules differs from the order of MO in O 2 , F 2 and Ne 2 molecules. Below is the MO scheme for B 2 , C 2 and N 2 molecules:
Based on the above schemes of MO, one can, for example, write down the electronic formulas of the molecules O 2 , O 2 + and O 2 ?:
O 2 + (11e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *1 ? y *0)
n = 2 R = 0.121 nm;
O 2 (12e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *1 ? y *1)
n = 2.5 R = 0.112 nm;
O2?(13e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *2 ? y *1)
n = 1.5 R = 0.126 nm.
In the case of the O 2 molecule, the MO theory makes it possible to foresee the greater strength of this molecule, since n = 2, the nature of the change in binding energies and internuclear distances in the O 2 + – O 2 – O 2 ? series, as well as the paramagnetism of the O 2 molecule, on the upper MOs of which there are two unpaired electrons.
3.3. Some types of connections
Ionic bond– electrostatic bond between ions of opposite charges. An ionic bond can be considered as an extreme case of a covalent polar bond. An ionic bond is formed if the difference in the electronegativity of atoms? X is greater than 1.5–2.0.
Ionic bond is non-directional non-saturable connection. In a NaCl crystal, the Na + ion is attracted by all Cl ions? and is repelled by all other Na + ions, regardless of the direction of interaction and the number of ions. This predetermines the greater stability of ionic crystals in comparison with ionic molecules.
hydrogen bond- the bond between the hydrogen atom of one molecule and the electronegative atom (F, CI, N) of another molecule.
The existence of a hydrogen bond explains the anomalous properties of water: the boiling point of water is much higher than that of its chemical counterparts: t bale (H 2 O) = 100 ° C, and t bale (H 2 S) = -61 ° C. Hydrogen bonds do not form between H 2 S molecules.
4. Patterns of the course of chemical processes
4.1. Thermochemistry
Energy(E)- the ability to do work. mechanical work(A) is performed, for example, by a gas during its expansion: A \u003d p? V.
Reactions that go with the absorption of energy - endothermic.
Reactions that take place with the release of energy exothermic.
Types of energy: heat, light, electrical, chemical, nuclear energy, etc.
Energy types: kinetic and potential.
Kinetic energy- the energy of a moving body, this is the work that a body can do before it reaches rest.
Heat (Q)- a type of kinetic energy - associated with the movement of atoms and molecules. When imparting a mass to the body (m) and specific heat capacity (c) of heat? Q its temperature rises by an amount? t: ?Q = m with ?t, where? t = ?Q/(c t).
Potential energy- the energy acquired by the body as a result of a change in its position in space or its components. The energy of chemical bonds is a type of potential energy.
First law of thermodynamics: energy can pass from one form to another, but cannot disappear or arise.
Internal energy (U) - the sum of the kinetic and potential energies of the particles that make up the body. The heat absorbed in the reaction is equal to the difference between the internal energy of the reaction products and reactants (Q \u003d? U \u003d U 2 - U 1), provided that the system has not done work on environment. If the reaction proceeds at constant pressure, then the released gases do work against the forces of external pressure, and the heat absorbed during the reaction is equal to the sum of the changes in internal energy ?U and work A \u003d p? V. This heat absorbed at constant pressure is called the enthalpy change: H = ?U + p?V, defining enthalpy as H \u003d U + pV. Reactions of liquid and solids flow without a significant change in volume (?V= 0), so what is for these reactions? H close to ?U (?H = ?U). For reactions with a change in volume, we have ?H > ?U if expansion is in progress, and ?H< ?U if compression is in progress.
The change in enthalpy is usually attributed to the standard state of matter: i.e., for a pure substance in a certain (solid, liquid or gaseous) state, at a pressure of 1 atm = 101 325 Pa, a temperature of 298 K and a concentration of substances 1 mol / l.
Standard enthalpy of formation? H arr- the heat released or absorbed during the formation of 1 mol of a substance from the simple substances that make it up under standard conditions. For example, ?N arr(NaCl) = -411 kJ/mol. This means that in the reaction Na(tv) + ?Cl 2 (g) = NaCl(tv), 411 kJ of energy is released during the formation of 1 mol of NaCl.
Standard enthalpy of reaction?- enthalpy change during a chemical reaction, is determined by the formula: ?H = ?N arr(products) - ?N arr(reagents).
So for the reaction NH 3 (g) + HCl (g) \u003d NH 4 Cl (tv), knowing? H o 6 p (NH 3) \u003d -46 kJ / mol,? H o 6 p (HCl) \u003d -92 kJ / mol and? H o 6 p (NH 4 Cl) = -315 kJ / mol we have:
H \u003d? H o 6 p (NH 4 Cl) -? H o 6 p (NH 3) -? H o 6 p (HCl) \u003d -315 - (-46) - (-92) \u003d -177 kJ.
If a? H< 0, the reaction is exothermic. If a? H > 0, the reaction is endothermic.
Law Hess: the standard enthalpy of reaction depends on the standard enthalpies of the reactants and products and does not depend on the reaction path.
Spontaneous processes can be not only exothermic, i.e., processes with a decrease in energy (?H< 0), but can also be endothermic processes, i.e. processes with an increase in energy (?H > 0). In all these processes, the "disorder" of the system increases.
EntropyS – physical quantity characterizing the degree of disorder of the system. S is the standard entropy, ?S is the change in the standard entropy. If?S > 0, disorder grows if AS< 0, то беспорядок системы уменьшается. Для процессов в которых растет число частиц, ?S >0. For processes in which the number of particles decreases, ?S< 0. Например, энтропия меняется в ходе реакций:
CaO (tv) + H 2 O (l) \u003d Ca (OH) 2 (tv),? S< 0;
CaCO 3 (tv) \u003d CaO (tv) + CO 2 (g), ?S\u003e 0.
Processes proceed spontaneously with the release of energy, i.e. for which? H< 0, and with an increase in entropy, i.e., for which?S > 0. Accounting for both factors leads to an expression for Gibbs energy: G = H - TS or? G \u003d? H - T? S. Reactions in which the Gibbs energy decreases, i.e. ?G< 0, могут идти самопроизвольно. Реакции, в ходе которых энергия Гиббса увеличивается, т. е. ?G >0, spontaneously do not go. The condition? G = 0 means that an equilibrium has been established between the products and the reactants.
At low temperature, when the value T is close to zero, only exothermic reactions take place, since T?S– few and? G = ? H< 0. At high temperatures, the values T?S large, and, neglecting the magnitude? H, we have? G = – T?S, i.e., processes with an increase in entropy will spontaneously occur, for which? S > 0, and ?G< 0. При этом чем больше по absolute value value? G, the more complete this process is.
The value of AG for a particular reaction can be determined by the formula:
G = ?С arr (products) – ?G o b p (reagents).
In this case, the values? G o br, as well as? H arr and?s o br for a large number substances are given in special tables.
4.2. Chemical kinetics
The rate of a chemical reaction(v) is determined by the change in the molar concentration of the reactants per unit time:
where v is the reaction rate, s is the molar concentration of the reagent, t- time.
The rate of a chemical reaction depends on the nature of the reactants and the reaction conditions (temperature, concentration, presence of a catalyst, etc.)
Influence of concentration. AT In the case of simple reactions, the reaction rate is proportional to the product of the concentrations of the reactants, taken in powers equal to their stoichiometric coefficients.
For reaction
where 1 and 2 are respectively the direction of the forward and backward reactions:
v 1 \u003d k 1? [A]m? [B]n and
v 2 \u003d k 2? [C]p? [D] q
where v- speed reaction, k is the rate constant, [A] is the molar concentration of substance A.
Reaction molecularity is the number of molecules involved in the elementary act of the reaction. For simple reactions, for example: mA + nB> pC + qD, molecularity is equal to the sum of the coefficients (m + n). Reactions can be one-molecular, two-molecular and rarely three-molecular. Higher molecular reactions do not occur.
Reaction order is equal to the sum of the indicators of the degrees of concentration in the experimental expression of the rate of a chemical reaction. Yes, for complex reaction
mA + nB > рС + qD the experimental expression for the reaction rate has the form
v 1 = k1? [BUT] ? ? [AT] ? and the reaction order is (? + ?). Wherein? and? are experimental and may not coincide with m and n respectively, since the equation of a complex reaction is the result of several simple reactions.
The effect of temperature. The reaction rate depends on the number of effective collisions of molecules. An increase in temperature increases the number of active molecules, giving them the necessary for the reaction to proceed. activation energy E act and increases the rate of a chemical reaction.
Van't Hoff's rule. With an increase in temperature by 10°, the reaction rate increases by a factor of 2–4. Mathematically, this is written as:
v2 = v1? ?(t 2 - t 1) / 10
where v 1 and v 2 are the reaction rates at the initial (t 1) and final (t 2) temperatures, ? - the temperature coefficient of the reaction rate, which shows how many times the reaction rate increases with an increase in temperature by 10 °.
More precisely, the dependence of the reaction rate on temperature is expressed as Arrhenius equation:
k = A? e - E/(RT) ,
where k is the rate constant, BUT- constant, independent of temperature, e = 2.71828, E is the activation energy, R= 8.314 J/(K? mol) – gas constant; T– temperature (K). It can be seen that the rate constant increases with increasing temperature and decreasing activation energy.
4.3. Chemical equilibrium
A system is in equilibrium if its state does not change with time. The equality of the rates of the direct and reverse reactions is a condition for maintaining the equilibrium of the system.
An example reversible reaction is the reaction
N 2 + 3H 2 - 2NH 3.
Mass action law: the ratio of the product of the concentrations of the reaction products to the product of the concentrations of the starting substances (all concentrations are indicated in powers equal to their stoichiometric coefficients) is a constant called equilibrium constant.
The equilibrium constant is a measure of the progress of a direct reaction.
K = O - no direct reaction;
K =? - the direct reaction goes to the end;
K > 1 - the balance is shifted to the right;
To< 1 - the balance is shifted to the left.
Reaction equilibrium constant To is related to the change in standard Gibbs energy?G for the same reaction:
G= – RT ln K, or ?g= -2.3RT lg K, or K= 10 -0.435?G/RT
If a K > 1, then lg K> 0 and?G< 0, т. е. если равновесие сдвинуто вправо, то реакция – переход от исходного состояния к равновесному – идет самопроизвольно.
If a To< 1, then lg K < 0 и?G >0, i.e. if the equilibrium is shifted to the left, then the reaction does not spontaneously go to the right.
Equilibrium displacement law: If an external influence is exerted on a system in equilibrium, a process arises in the system that counteracts the external influence.
5. Redox reactions
Redox reactions- reactions that go with a change in the oxidation states of elements.
Oxidation is the process of giving up electrons.
Recovery is the process of adding electrons.
Oxidizing agent An atom, molecule, or ion that accepts electrons.
Reducing agent An atom, molecule, or ion that donates electrons.
Oxidizing agents, accepting electrons, go into the reduced form:
F2 [ca. ] + 2e > 2F? [rest.].
Reducing agents, donating electrons, pass into the oxidized form:
Na 0 [restore ] – 1e > Na + [approx.].
The equilibrium between the oxidized and reduced forms is characterized by Nernst equations for redox potential:
where E 0 is the standard value of the redox potential; n is the number of transferred electrons; [rest. ] and [ca. ] are the molar concentrations of the compound in the reduced and oxidized forms, respectively.
Values of standard electrode potentials E 0 are given in tables and characterize the oxidizing and reducing properties of the compounds: the more positive the value E 0, the stronger the oxidizing properties, and the more negative the value E 0, the stronger the restorative properties.
For example, for F 2 + 2e - 2F? E 0 = 2.87 volts, and for Na + + 1e - Na 0 E 0 =-2.71 volts (the process is always recorded for reduction reactions).
The redox reaction is a combination of two half-reactions, oxidation and reduction, and is characterized by an electromotive force (emf)? E 0:?E 0= ?E 0 ok – ?E 0 restore, where E 0 ok and? E 0 restore are the standard potentials of the oxidizing agent and reducing agent for the given reaction.
emf reactions? E 0 is related to the change in the Gibbs free energy?G and the equilibrium constant of the reaction TO:
?G = –nF?E 0 or? E = (RT/nF) ln K.
emf reactions at non-standard concentrations? E is equal to: ? E =?E 0 - (RT / nF)? Ig K or? E =?E 0 -(0,059/n)lg K.
In the case of equilibrium? G \u003d 0 and? E \u003d 0, where? E =(0.059/n)lg K and K = 10n?E/0.059.
For the spontaneous occurrence of the reaction, the following relations must be satisfied: ?G< 0 или K >> 1 that the condition matches? E 0> 0. Therefore, to determine the possibility of a given redox reaction, it is necessary to calculate the value? E 0 . If a? E 0 > 0, the reaction is on. If a? E 0< 0, there is no reaction.
Chemical current sourcesGalvanic cells Devices that convert the energy of a chemical reaction into electrical energy.
Daniel's galvanic cell consists of zinc and copper electrodes immersed in ZnSO 4 and CuSO 4 solutions, respectively. Electrolyte solutions communicate through a porous partition. At the same time, oxidation occurs on the zinc electrode: Zn > Zn 2+ + 2e, and reduction occurs on the copper electrode: Cu 2+ + 2e > Cu. In general, the reaction is going on: Zn + CuSO 4 = ZnSO 4 + Cu.
Anode- the electrode at which oxidation takes place. Cathode- the electrode on which the reduction is taking place. In galvanic cells, the anode is negatively charged and the cathode is positively charged. In the element diagrams, the metal and solution are separated by a vertical line, and two solutions by a double vertical line.
So, for the reaction Zn + CuSO 4 \u003d ZnSO 4 + Cu, the galvanic cell circuit is written: (-) Zn | ZnSO 4 || CuSO4 | Cu(+).
The electromotive force (emf) of the reaction is? E 0 \u003d E 0 ok - E 0 restore= E 0(Cu 2+ /Cu) - E 0(Zn 2+ / Zn) \u003d 0.34 - (-0.76) \u003d 1.10 V. Due to losses, the voltage created by the element will be somewhat less than? E 0 . If the concentrations of solutions differ from the standard ones, equal to 1 mol/l, then E 0 ok and E 0 restore are calculated according to the Nernst equation, and then the emf is calculated. corresponding galvanic cell.
dry element consists of a zinc body, NH 4 Cl paste with starch or flour, a mixture of MnO 2 with graphite and a graphite electrode. In the course of its work, the following reaction takes place: Zn + 2NH 4 Cl + 2MnO 2 = Cl + 2MnOOH.
Element diagram: (-)Zn | NH4Cl | MnO 2 , C(+). emf element - 1.5 V.
Batteries. A lead battery consists of two lead plates immersed in a 30% sulfuric acid solution and covered with a layer of insoluble PbSO 4 . When the battery is charged, the following processes take place on the electrodes:
PbSO 4 (tv) + 2e > Pb (tv) + SO 4 2-
PbSO 4 (tv) + 2H 2 O > РbO 2 (tv) + 4H + + SO 4 2- + 2e
When the battery is discharged, the following processes take place on the electrodes:
Pb(tv) + SO 4 2-> PbSO 4 (tv) + 2e
РbO 2 (tv) + 4H + + SO 4 2- + 2e> PbSO 4 (tv) + 2Н 2 O
The overall reaction can be written as:
To work, the battery needs regular charging and control of the concentration of sulfuric acid, which may decrease slightly during battery operation.
6. Solutions
6.1. Solution concentration
Mass fraction of a substance in solution w is equal to the ratio of the mass of the solute to the mass of the solution: w \u003d m in-va / m solution or w = m in-va / (V ? ?), as m p-ra \u003d V p-pa? ?r-ra.
Molar concentration with is equal to the ratio of the number of moles of the solute to the volume of the solution: c = n(mol)/ V(l) or c = m/(M? V( l )).
Molar concentration of equivalents (normal or equivalent concentration) with e is equal to the ratio of the number of equivalents of the solute to the volume of the solution: with e = n(mol equiv.)/ V(l) or with e \u003d m / (M e? V (l)).
6.2. Electrolytic dissociation
Electrolytic dissociation – decomposition of the electrolyte into cations and anions under the action of polar solvent molecules.
Degree of dissociation? is the ratio of the concentration of dissociated molecules (c diss) to the total concentration of dissolved molecules (c vol): ? = s diss / s rev.
Electrolytes can be divided into strong(?~1) and weak.
Strong electrolytes(for them? ~ 1) - salts and bases soluble in water, as well as some acids: HNO 3, HCl, H 2 SO 4, HI, HBr, HClO 4 and others.
Weak electrolytes(for them?<< 1) – Н 2 O, NH 4 OH, малорастворимые основания и соли и многие кислоты: HF, H 2 SO 3 , H 2 CO 3 , H 2 S, CH 3 COOH и другие.
Ionic reaction equations. AT In ionic reaction equations, strong electrolytes are written as ions, and weak electrolytes, poorly soluble substances and gases are written as molecules. For example:
CaCO 3 v + 2HCl \u003d CaCl 2 + H 2 O + CO 2 ^
CaCO 3 v + 2H + + 2Cl? \u003d Ca 2+ + 2Cl? + H 2 O + CO 2 ^
CaCO 3 v + 2H + = Ca 2+ + H 2 O + CO 2 ^
Reactions between ions go in the direction of the formation of a substance that gives fewer ions, i.e., in the direction of a weaker electrolyte or less soluble substance.
6.3. Dissociation of weak electrolytes
Let us apply the law of mass action to the equilibrium between ions and molecules in a solution of a weak electrolyte, such as acetic acid:
CH 3 COOH - CH 3 COО? + H +
The equilibrium constants of dissociation reactions are called dissociation constants. Dissociation constants characterize the dissociation of weak electrolytes: the smaller the constant, the less the weak electrolyte dissociates, the weaker it is.
Polybasic acids dissociate in steps:
H 3 PO 4 - H + + H 2 PO 4?
The equilibrium constant of the total dissociation reaction is equal to the product of the constants of the individual stages of dissociation:
H 3 PO 4 - ZN + + PO 4 3-
Ostwald's dilution law: the degree of dissociation of a weak electrolyte (a) increases with a decrease in its concentration, i.e., upon dilution:
Effect of a common ion on the dissociation of a weak electrolyte: the addition of a common ion reduces the dissociation of a weak electrolyte. So, when adding a weak electrolyte solution CH 3 COOH
CH 3 COOH - CH 3 COО? + H + ?<< 1
a strong electrolyte containing an ion common with CH 3 COOH, i.e. an acetate ion, for example CH 3 COONa
CH 3 COONa - CH 3 COO? +Na+? = 1
the concentration of the acetate ion increases, and the equilibrium of the dissociation of CH 3 COOH shifts to the left, i.e., the dissociation of the acid decreases.
6.4. Dissociation of strong electrolytes
Ion activity a is the concentration of an ion, which manifests itself in its properties.
Activity factorf is the ratio of ion activity a to concentration with: f= a/c or a = f.c.
If f = 1, then the ions are free and do not interact with each other. This occurs in very dilute solutions, in solutions of weak electrolytes, etc.
If f< 1, то ионы взаимодействуют между собой. Чем меньше f, тем больше взаимодействие между ионами.
The activity coefficient depends on the ionic strength of solution I: the greater the ionic strength, the lower the activity coefficient.
Ionic strength of solution I depends on charges z and concentrations from ions:
I= 0.52?s z2.
The activity coefficient depends on the charge of the ion: the greater the charge of the ion, the lower the activity coefficient. Mathematically, the dependence of the activity coefficient f from ionic strength I and ion charge z is written using the Debye-Hückel formula:
Ion activity coefficients can be determined using the following table:
6.5 Ionic product of water. Hydrogen indicator
Water, a weak electrolyte, dissociates to form H+ and OH? ions. These ions are hydrated, i.e., connected to several water molecules, but for simplicity they are written in non-hydrated form
H 2 O - H + + OH?.
Based on the law of mass action, for this equilibrium:
The concentration of water molecules [H 2 O], i.e., the number of moles in 1 liter of water, can be considered constant and equal to [H 2 O] \u003d 1000 g / l: 18 g / mol \u003d 55.6 mol / l. From here:
To[H 2 O] = To(H 2 O ) = [H + ] = 10 -14 (22°C).
Ionic product of water– the product of concentrations [H + ] and – is a constant value at a constant temperature and equal to 10 -14 at 22°C.
The ionic product of water increases with increasing temperature.
pH value is the negative logarithm of the concentration of hydrogen ions: pH = – lg. Similarly: pOH = – lg.
The logarithm of the ionic product of water gives: pH + pOH = 14.
The pH value characterizes the reaction of the medium.
If pH = 7, then [H + ] = is a neutral medium.
If pH< 7, то [Н + ] >- acid environment.
If pH > 7, then [H + ]< – щелочная среда.
6.6. buffer solutions
Buffer solutions are solutions that have a certain concentration of hydrogen ions. The pH of these solutions does not change when diluted and changes little when small amounts of acids and alkalis are added.
I. A solution of a weak acid HA, concentration - from acid, and its salts with a strong base BA, concentration - from salt. For example, an acetate buffer is a solution of acetic acid and sodium acetate: CH 3 COOH + CHgCOONa.
pH \u003d pK acidic + lg (salt /s acidic).
II. A solution of a weak base BOH, concentration - with basic, and its salts with a strong acid BA, concentration - with salt. For example, an ammonia buffer is a solution of ammonium hydroxide and ammonium chloride NH 4 OH + NH 4 Cl.
pH = 14 - рК basic - lg (from salt / from basic).
6.7. Salt hydrolysis
Salt hydrolysis- the interaction of salt ions with water with the formation of a weak electrolyte.
Examples of hydrolysis reaction equations.
I. Salt is formed by a strong base and a weak acid:
Na 2 CO 3 + H 2 O - NaHCO 3 + NaOH
2Na + + CO 3 2- + H 2 O - 2Na + + HCO 3? +OH?
CO 3 2- + H 2 O - HCO 3? + OH?, pH > 7, alkaline.
In the second stage, hydrolysis practically does not occur.
II. A salt is formed from a weak base and a strong acid:
AlCl 3 + H 2 O - (AlOH)Cl 2 + HCl
Al 3+ + 3Cl? + H 2 O - AlOH 2+ + 2Cl? + H + + Cl?
Al 3+ + H 2 O - AlOH 2+ + H +, pH< 7.
In the second stage, hydrolysis occurs less, and in the third stage it practically does not occur.
III. Salt is formed by a strong base and a strong acid:
K + + NO 3 ? + H 2 O? no hydrolysis, pH? 7.
IV. A salt is formed from a weak base and a weak acid:
CH 3 COONH 4 + H 2 O - CH 3 COOH + NH 4 OH
CH 3 COO? + NH 4 + + H 2 O - CH 3 COOH + NH 4 OH, pH = 7.
In some cases, when the salt is formed by very weak bases and acids, complete hydrolysis occurs. In the solubility table for such salts, the symbol is “decomposed by water”:
Al 2 S 3 + 6H 2 O \u003d 2Al (OH) 3 v + 3H 2 S ^
The possibility of complete hydrolysis should be taken into account in exchange reactions:
Al 2 (SO 4) 3 + 3Na 2 CO 3 + 3H 2 O \u003d 2Al (OH) 3 v + 3Na 2 SO 4 + 3CO 2 ^
Degree of hydrolysish is the ratio of the concentration of hydrolyzed molecules to the total concentration of dissolved molecules.
For salts formed by a strong base and a weak acid:
= ch, pOH = -lg, pH = 14 - pOH.
It follows from the expression that the degree of hydrolysis h(i.e. hydrolysis) increases:
a) with increasing temperature, since K(H 2 O) increases;
b) with a decrease in the dissociation of the acid that forms the salt: the weaker the acid, the greater the hydrolysis;
c) with dilution: the lower c, the greater the hydrolysis.
For salts formed from a weak base and a strong acid
[H + ] = ch, pH = – lg.
For salts formed by a weak base and a weak acid
6.8. Protolytic theory of acids and bases
Protolysis is the proton transfer process.
Protoliths acids and bases that donate and accept protons.
Acid A molecule or ion capable of donating a proton. Each acid has its conjugate base. The strength of acids is characterized by the acid constant To k.
H 2 CO 3 + H 2 O - H 3 O + + HCO 3?
K k = 4 ? 10 -7
3+ + H 2 O - 2+ + H 3 O +
K k = 9 ? 10 -6
Base A molecule or ion that can accept a proton. Each base has its conjugate acid. The strength of the bases is characterized by the base constant K 0 .
NH3? H 2 O (H 2 O) - NH 4 + + OH?
K 0 = 1,8 ?10 -5
Ampholytes- protoliths capable of recoil and proton attachment.
HCO3? + H 2 O - H 3 O + + CO 3 2-
HCO3? - acid.
HCO3? + H 2 O - H 2 CO 3 + OH?
HCO3? - base.
For water: H 2 O + H 2 O - H 3 O + + OH?
K (H 2 O) \u003d [H 3 O +] \u003d 10 -14 and pH \u003d - lg.
Constants K to and K 0 for conjugated acids and bases are linked.
ON + H 2 O - H 3 O + + A ?,
BUT? + H 2 O - ON + OH?,
7. Solubility constant. Solubility
In a system consisting of a solution and a precipitate, two processes take place - the dissolution of the precipitate and precipitation. The equality of the rates of these two processes is the equilibrium condition.
saturated solution A solution that is in equilibrium with the precipitate.
The law of mass action applied to the equilibrium between precipitate and solution gives:
Since = const,
To = K s (AgCl) = .
In general, we have:
BUT m B n(TV) - m A +n+n B -m
K s ( A m B n)= [A +n ] m[AT -m ] n .
Solubility constantKs(or solubility product PR) - the product of ion concentrations in a saturated solution of a sparingly soluble electrolyte - is a constant value and depends only on temperature.
Solubility of an insoluble substance s can be expressed in moles per litre. Depending on the size s substances can be divided into poorly soluble - s< 10 -4 моль/л, среднерастворимые – 10 -4 моль/л? s? 10 -2 mol/l and highly soluble s>10 -2 mol/l.
The solubility of compounds is related to their solubility product.
Precipitation and dissolution condition In the case of AgCl: AgCl - Ag + + Cl?
Ks= :
a) the equilibrium condition between the precipitate and the solution: = K s .
b) settling condition: > K s ; during precipitation, the ion concentrations decrease until equilibrium is established;
c) the condition for the dissolution of the precipitate or the existence of a saturated solution:< K s ; during the dissolution of the precipitate, the concentration of ions increases until equilibrium is established.
8. Coordination compounds
Coordination (complex) compounds are compounds with a donor-acceptor bond.
For K3:
ions of the outer sphere - 3K +,
ion of the inner sphere - 3-,
complexing agent - Fe 3+,
ligands - 6CN?, their denticity - 1,
coordination number - 6.
Examples of complexing agents: Ag +, Cu 2+, Hg 2+, Zn 2+, Ni 2+, Fe 3+, Pt 4+, etc.
Examples of ligands: polar molecules H 2 O, NH 3 , CO and anions CN?, Cl?, OH? and etc.
Coordination numbers: usually 4 or 6, rarely 2, 3, etc.
Nomenclature. They first name the anion (in the nominative case), then the cation (in genitive case). The names of some ligands: NH 3 - ammine, H 2 O - aqua, CN? – cyano, Cl? – chloro, OH? - hydroxo. Names of coordination numbers: 2 - di, 3 - three, 4 - tetra, 5 - penta, 6 - hexa. Indicate the degree of oxidation of the complexing agent:
Cl is diamminesilver(I) chloride;
SO 4 - tetramminecopper(II) sulfate;
K 3 is potassium hexacyanoferrate(III).
Chemical connection.The theory of valence bonds assumes hybridization of the orbitals of the central atom. The location of the resulting hybrid orbitals determines the geometry of the complexes.
Diamagnetic complex ion Fe(CN) 6 4- .
Cyanide ion - donor
Iron ion Fe 2+ - acceptor - has the formula 3d 6 4s 0 4p 0. Taking into account the diamagnetism of the complex (all electrons are paired) and the coordination number (6 free orbitals are needed), we have d2sp3- hybridization:
The complex is diamagnetic, low-spin, intra-orbital, stable (no external electrons are used), octahedral ( d2sp3-hybridization).
Paramagnetic complex ion FeF 6 3- .
Fluoride ion is a donor.
Iron ion Fe 3+ - acceptor - has the formula 3d 5 4s 0 4p 0 . Taking into account the paramagnetism of the complex (electrons are steamed) and the coordination number (6 free orbitals are needed), we have sp 3 d 2- hybridization:
The complex is paramagnetic, high-spin, outer-orbital, unstable (outer 4d-orbitals are used), octahedral ( sp 3 d 2-hybridization).
Dissociation of coordination compounds.Coordination compounds in solution completely dissociate into ions of the inner and outer spheres.
NO 3 > Ag(NH 3) 2 + + NO 3 ?, ? = 1.
Ions of the inner sphere, i.e., complex ions, dissociate into metal ions and ligands, like weak electrolytes, in steps.
where K 1 , To 2 , TO 1 _ 2 are called instability constants and characterize the dissociation of complexes: the smaller the instability constant, the less the complex dissociates, the more stable it is.
several basic concepts and formulas.
All substances have different mass, density and volume. A piece of metal from one element can weigh many times more than exactly the same size piece from another metal.
mole(number of moles)designation: mole, international: mol is a unit of measure for the amount of a substance. Corresponds to the amount of the substance that contains NA particles (molecules, atoms, ions) Therefore, a universal value was introduced - the number of mol. A frequently encountered phrase in tasks is “it was received ... mole of substance"
NA= 6.02 1023
NA is Avogadro's number. Also "number by agreement". How many atoms are there in the tip of a pencil? About a thousand. It is not convenient to operate with such values. Therefore, chemists and physicists around the world agreed - let's denote 6.02 1023 particles (atoms, molecules, ions) as 1 mol substances.
1 mol = 6.02 1023 particles
It was the first of the basic formulas for solving problems.
Molar mass of a substance
Molar mass matter is the mass of one mole of substance.
Referred to as Mr. It is located according to the periodic table - this is simply the sum of the atomic masses of a substance.
For example, we are given sulfuric acid - H2SO4. Let's calculate the molar mass of a substance: atomic mass H = 1, S-32, O-16.
Mr(H2SO4)=1 2+32+16 4=98 g/mol.The second necessary formula for solving problems is
mass formula:
That is, to find the mass of a substance, you need to know the number of moles (n), and we find the molar mass from Periodic system.
The law of conservation of mass is The mass of substances that enter into a chemical reaction is always equal to the mass of the formed substances.
If we know the mass (masses) of substances that have entered into a reaction, we can find the mass (masses) of the products of this reaction. And vice versa.
The third formula for solving problems in chemistry is
volume of matter:
Unfortunately, this image does not comply with our guidelines. To continue posting, please delete the image or upload another one.Where did the number 22.4 come from? From Avogadro's law:
equal volumes of different gases, taken at the same temperature and pressure, contain the same number of molecules.
According to Avogadro's law, 1 mol ideal gas under normal conditions (n.s.) has the same volume Vm\u003d 22.413 996 (39) l
That is, if we are given normal conditions in the problem, then, knowing the number of moles (n), we can find the volume of the substance.
So, basic formulas for solving problems in chemistry
Avogadro's numberNA
6.02 1023 particles
Amount of substance n (mol)
n=V\22.4 (l\mol)
Mass of matter m (g)
Volume of matter V(l)
V=n 22.4 (l\mol)
Unfortunately, this image does not comply with our guidelines. To continue posting, please delete the image or upload another one.These are formulas. Often, to solve problems, you must first write the reaction equation and (required!) Arrange the coefficients - their ratio determines the ratio of moles in the process.
Chemistry is the science of substances, their properties and transformations. .
That is, if nothing happens to the substances around us, then this does not apply to chemistry. But what does "nothing happens" mean? If a thunderstorm suddenly caught us in the field, and we all got wet, as they say, “to the skin”, then is this not a transformation: after all, the clothes were dry, but became wet.If, for example, you take an iron nail, process it with a file, and then assemble iron filings (Fe) , then this is also not a transformation: there was a nail - it became powder. But if after that to assemble the device and hold obtaining oxygen (O 2): heat up potassium permanganate(KMpo 4) and collect oxygen in a test tube, and then place these iron filings red-hot "to red" in it, then they will flare up with a bright flame and after combustion turn into a brown powder. And this is also a transformation. So where is the chemistry? Despite the fact that in these examples the shape (iron nail) and the state of clothing (dry, wet) change, these are not transformations. The fact is that the nail itself, as it was a substance (iron), remained so, despite its different form, and our clothes soaked up the water from the rain, and then it evaporated into the atmosphere. The water itself has not changed. So what are transformations in terms of chemistry?
From the point of view of chemistry, transformations are such phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It does not matter what form it took after being filed, but after being collected from it iron filings placed in an atmosphere of oxygen - it turned into iron oxide(Fe 2 O 3 ) . So, has something really changed? Yes, it has. There was a nail substance, but under the influence of oxygen a new substance was formed - element oxide gland. molecular equation this transformation can be represented by the following chemical symbols:
4Fe + 3O 2 = 2Fe 2 O 3 (1)
For a person uninitiated in chemistry, questions immediately arise. What is the "molecular equation", what is Fe? Why are there numbers "4", "3", "2"? What are the small numbers "2" and "3" in the formula Fe 2 O 3? This means that the time has come to sort things out in order.
Signs of chemical elements.
Despite the fact that they begin to study chemistry in the 8th grade, and some even earlier, many people know the great Russian chemist D. I. Mendeleev. And of course, his famous "Periodic Table of Chemical Elements". Otherwise, more simply, it is called the "Mendeleev's Table".
In this table, in the appropriate order, the elements are located. To date, about 120 of them are known. The names of many elements have been known to us for a long time. These are: iron, aluminum, oxygen, carbon, gold, silicon. Previously, we used these words without hesitation, identifying them with objects: an iron bolt, aluminum wire, oxygen in the atmosphere, a golden ring, etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their respective elements. The whole paradox is that the element cannot be touched, picked up. How so? They are in the periodic table, but you can’t take them! Yes exactly. A chemical element is an abstract (that is, abstract) concept, and is used in chemistry, however, as in other sciences, for calculations, compiling equations, and solving problems. Each element differs from the other in that it is characterized by its own electronic configuration of an atom. The number of protons in the nucleus of an atom is equal to the number of electrons in its orbitals. For example, hydrogen is element #1. Its atom consists of 1 proton and 1 electron. Helium is element number 2. Its atom consists of 2 protons and 2 electrons. Lithium is element number 3. Its atom consists of 3 protons and 3 electrons. Darmstadtium - element number 110. Its atom consists of 110 protons and 110 electrons.
Each element is denoted by a certain symbol, Latin letters, and has a certain reading in translation from Latin. For example, hydrogen has the symbol "N", read as "hydrogenium" or "ash". Silicon has the symbol "Si" read as "silicium". Mercury has a symbol "Hg" and is read as "hydrargyrum". Etc. All these designations can be found in any chemistry textbook for the 8th grade. For us now, the main thing is to understand that when compiling chemical equations, you must operate on the specified element symbols.
Simple and complex substances.
Denoting various substances with single symbols of chemical elements (Hg mercury, Fe iron, Cu copper, Zn zinc, Al aluminum) we essentially denote simple substances, that is, substances consisting of atoms of the same type (containing the same number of protons and neutrons in an atom). For example, if iron and sulfur substances interact, then the equation will take the following form:
Fe + S = FeS (2)
Simple substances include metals (Ba, K, Na, Mg, Ag), as well as non-metals (S, P, Si, Cl 2, N 2, O 2, H 2). And you should pay attention
special attention to the fact that all metals are indicated by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals - either by simple symbols: C, S, P or may have different indices that indicate them molecular structure: H 2 , Cl 2 , O 2 , J 2 , P 4 , S 8 . In the future, this will be of great importance in the formulation of equations. It is not at all difficult to guess that complex substances are substances formed from atoms of different types, for example,one). Oxides:
aluminium oxide Al 2 O 3,
sodium oxide Na 2 O
copper oxide CuO,
zinc oxide ZnO
titanium oxide Ti2O3,
carbon monoxide or carbon monoxide (+2) CO
sulfur oxide (+6) SO 32). Reasons:
iron hydroxide(+3) Fe (OH) 3,
copper hydroxide Cu(OH)2,
potassium hydroxide or potassium alkali KOH,
sodium hydroxide NaOH.3). Acids:
hydrochloric acid HCl
sulfurous acid H2SO3,
Nitric acid HNO34). Salts:
sodium thiosulfate Na 2 S 2 O 3,
sodium sulfate or Glauber's salt Na 2 SO 4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl 25). organic matter:
sodium acetate CH 3 COOHa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6Finally, after we figured out the structure various substances, you can start compiling chemical equations.
Chemical equation.
The word “equation” itself is derived from the word “equalize”, i.e. divide something into equal parts. In mathematics, equations are almost the very essence of this science. For example, you can give such a simple equation in which the left and right sides will be equal to "2":
40: (9 + 11) = (50 x 2): (80 - 30);
And in chemical equations, the same principle: the left and right sides of the equation must correspond to the same number of atoms, the elements participating in them. Or, if an ionic equation is given, then in it number of particles must also meet this requirement. A chemical equation is a conditional record of a chemical reaction using chemical formulas and mathematical signs. A chemical equation inherently reflects a particular chemical reaction, that is, the process of interaction of substances, during which new substances arise. For example, it is necessary write a molecular equation reactions that take part barium chloride BaCl 2 and sulfuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfate BaSO 4 and hydrochloric acid Hcl:
ВаСl 2 + H 2 SO 4 = BaSO 4 + 2НCl (3)
First of all, it is necessary to understand that the large number “2” in front of the HCl substance is called the coefficient, and the small numbers “2”, “4” under the formulas ВаСl 2, H 2 SO 4 , BaSO 4 are called indices. Both the coefficients and indices in chemical equations play the role of factors, not terms. In order to correctly write a chemical equation, it is necessary arrange the coefficients in the reaction equation. Now let's start counting the atoms of the elements on the left and right sides of the equation. On the left side of the equation: the substance BaCl 2 contains 1 barium atom (Ba), 2 chlorine atoms (Cl). In the substance H 2 SO 4: 2 hydrogen atoms (H), 1 sulfur atom (S) and 4 oxygen atoms (O). On the right side of the equation: in the BaSO 4 substance there is 1 barium atom (Ba) 1 sulfur atom (S) and 4 oxygen atoms (O), in the HCl substance: 1 hydrogen atom (H) and 1 chlorine atom (Cl). Whence it follows that on the right side of the equation the number of hydrogen and chlorine atoms is half that on the left side. Therefore, before the HCl formula on the right side of the equation, it is necessary to put the coefficient "2". If we now add the number of atoms of the elements involved in this reaction, both on the left and on the right, we get the following balance:
In both parts of the equation, the number of atoms of the elements participating in the reaction are equal, therefore it is correct.
Chemical equation and chemical reactions
As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are such phenomena in the process of which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:
one). Connection reactions
2). decomposition reactions.The vast majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with a single substance if it is not subjected to external influences (dissolution, heating, light). Nothing characterizes a chemical phenomenon, or reaction, as much as the changes that occur when two or more substances interact. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color change, sedimentation, release of gaseous products, noise.
For clarity, we present several equations that reflect the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl 2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)
Cl 2 + 2Nа = 2NaCl (4)
CuCl 2 + Zn \u003d ZnCl 2 + Cu (5)
AgNO 3 + KCl \u003d AgCl + 2KNO 3 (6)
3HCl + Al(OH) 3 \u003d AlCl 3 + 3H 2 O (7)
Among the reactions of the compound, the following should be especially noted : substitution (5), exchange (6), and how special case exchange reactions - reaction neutralization (7).
Substitution reactions include those in which atoms of a simple substance replace the atoms of one of the elements in a complex substance. In example (5), zinc atoms replace copper atoms from the CuCl 2 solution, while zinc passes into the soluble ZnCl 2 salt, and copper is released from the solution in the metallic state.
Exchange reactions are reactions in which two complex substances exchange their components. In the case of reaction (6), the soluble salts of AgNO 3 and KCl, when both solutions are drained, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are attached to NO 3 anions, and silver cations Ag + - to Cl - anions.
A special, particular case of exchange reactions is the neutralization reaction. Neutralization reactions are reactions in which acids react with bases to form salt and water. In example (7), hydrochloric acid HCl reacts with base Al(OH) 3 to form AlCl 3 salt and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl anions - from the acid. As a result, it happens hydrochloric acid neutralization.
Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex one. As reactions, one can cite those in the process of which 1) decompose. potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate is formed (K 2 MnO 4), manganese oxide(MnO 2) and oxygen (O 2); 3). calcium carbonate or marble; in the process are formed carbonicgas(CO 2) and calcium oxide(Cao)
2KNO 3 \u003d 2KNO 2 + O 2 (8)
2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 \u003d CaO + CO 2 (10)In reaction (8), one complex and one simple substance is formed from a complex substance. In reaction (9) there are two complex and one simple. In reaction (10) there are two complex substances, but simpler in composition
All classes of complex substances undergo decomposition:
one). Oxides: silver oxide 2Ag 2 O = 4Ag + O 2 (11)
2). Hydroxides: iron hydroxide 2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O (12)
3). Acids: sulfuric acid H 2 SO 4 \u003d SO 3 + H 2 O (13)
4). Salts: calcium carbonate CaCO 3 \u003d CaO + CO 2 (14)
5). organic matter: alcoholic fermentation of glucose
C 6 H 12 O 6 \u003d 2C 2 H 5 OH + 2CO 2 (15)
According to another classification, all chemical reactions can be divided into two types: reactions that take place with the release of heat, they are called exothermic, and reactions that go with the absorption of heat - endothermic. The criterion for such processes is thermal effect of the reaction. As a rule, exothermic reactions include oxidation reactions, i.e. interactions with oxygen methane combustion:
CH 4 + 2O 2 \u003d CO 2 + 2H 2 O + Q (16)
and to endothermic reactions - decomposition reactions, already given above (11) - (15). The Q sign at the end of the equation indicates whether heat is released during the reaction (+Q) or absorbed (-Q):
CaCO 3 \u003d CaO + CO 2 - Q (17)
You can also consider all chemical reactions according to the type of change in the degree of oxidation of the elements involved in their transformations. For example, in reaction (17), the elements participating in it do not change their oxidation states:
Ca +2 C +4 O 3 -2 \u003d Ca +2 O -2 + C +4 O 2 -2 (18)
And in reaction (16), the elements change their oxidation states:
2Mg 0 + O 2 0 \u003d 2Mg +2 O -2
These types of reactions are redox . They will be considered separately. To formulate equations for reactions of this type, it is necessary to use half-reaction method and apply electronic balance equation.
After bringing the various types of chemical reactions, you can proceed to the principle of compiling chemical equations, in other words, the selection of coefficients in their left and right parts.
Mechanisms for compiling chemical equations.
Whatever the type of one or the other chemical reaction, its record (chemical equation) must comply with the condition of equality of the number of atoms before the reaction and after the reaction.
There are equations (17) that do not require adjustment, i.e. placement of coefficients. But in most cases, as in examples (3), (7), (15), it is necessary to take actions aimed at equalizing the left and right parts of the equation. What principles should be followed in such cases? Is there any system in the selection of coefficients? There is, and not one. These systems include:
one). Selection of coefficients according to given formulas.
2). Compilation according to the valencies of the reactants.
3). Compilation according to the oxidation states of the reactants.
In the first case, it is assumed that we know the formulas of the reactants both before and after the reaction. For example, given the following equation:
N 2 + O 2 →N 2 O 3 (19)
It is generally accepted that until the equality between the atoms of the elements before and after the reaction is established, the equal sign (=) is not put in the equation, but is replaced by an arrow (→). Now let's get down to the actual balancing. On the left side of the equation there are 2 nitrogen atoms (N 2) and two oxygen atoms (O 2), and on the right side there are two nitrogen atoms (N 2) and three oxygen atoms (O 3). It is not necessary to equalize it by the number of nitrogen atoms, but by oxygen it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were three atoms. Let's make the following diagram:
before reaction after reaction
O 2 O 3Let's define the smallest multiple between the given numbers of atoms, it will be "6".
O 2 O 3
\ 6 /Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved:
N 2 + 3O 2 →N 2 O 3
We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:
N 2 + 3O 2 → 2N 2 O 3
The number of oxygen atoms in both the left and right parts of the equation became equal, respectively, 6 atoms:
But the number of nitrogen atoms in both sides of the equation will not match:
On the left side there are two atoms, on the right side there are four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, putting the coefficient "2":
Thus, the equality for nitrogen is observed and, in general, the equation will take the form:
2N 2 + 3O 2 → 2N 2 O 3
Now in the equation, instead of an arrow, you can put an equal sign:
2N 2 + 3O 2 \u003d 2N 2 O 3 (20)
Let's take another example. The following reaction equation is given:
P + Cl 2 → PCl 5
On the left side of the equation there is 1 phosphorus atom (P) and two chlorine atoms (Cl 2), and on the right side there is one phosphorus atom (P) and five oxygen atoms (Cl 5). It is not necessary to equalize it by the number of phosphorus atoms, but for chlorine it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were five atoms. Let's make the following diagram:
before reaction after reaction
Cl 2 Cl 5Let's define the smallest multiple between the given numbers of atoms, it will be "10".
Cl 2 Cl 5
\ 10 /Divide this number on the left side of the equation for chlorine by "2". We get the number "5", put it in the equation to be solved:
Р + 5Cl 2 → РCl 5
We also divide the number "10" for the right side of the equation by "5". We get the number "2", just put it in the equation to be solved:
Р + 5Cl 2 → 2РCl 5
The number of chlorine atoms in both the left and right parts of the equation became equal, respectively, 10 atoms:
But the number of phosphorus atoms in both sides of the equation will not match:
Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation, putting the coefficient "2":
Thus, the equality for phosphorus is observed and, in general, the equation will take the form:
2Р + 5Cl 2 = 2РCl 5 (21)
When writing equations by valency must be given definition of valence and set values for the most famous elements. Valency is one of the previously used concepts, currently in a number of school programs not used. But with its help it is easier to explain the principles of compiling equations of chemical reactions. By valency is meant the number of chemical bonds that an atom can form with another, or other atoms . Valence has no sign (+ or -) and is indicated by Roman numerals, usually above the symbols of chemical elements, for example:
Where do these values come from? How to apply them in the preparation of chemical equations? Numeric values valencies of elements coincide with their group number of the Periodic system of chemical elements of D. I. Mendeleev (Table 1).
For other elements valency values may have other values, but never greater than the number of the group in which they are located. Moreover, for even numbers of groups (IV and VI), the valencies of elements take only even values, and for odd ones, they can have both even and odd values (Table.2).
Of course, there are exceptions to the valency values for some elements, but in each specific case, these points are usually specified. Now consider general principle compiling chemical equations for given valences for certain elements. Most often, this method is acceptable in the case of compiling equations for chemical reactions of combining simple substances, for example, when interacting with oxygen ( oxidation reactions). Suppose you want to display the oxidation reaction aluminum. But recall that metals are denoted by single atoms (Al), and non-metals that are in a gaseous state - with indices "2" - (O 2). First, we write the general scheme of the reaction:
Al + O 2 → AlO
At this stage, it is not yet known which correct writing should be aluminum oxide. And it is precisely at this stage that knowledge of the valencies of the elements will come to our aid. For aluminum and oxygen, we put them above the proposed formula for this oxide:
IIIII
Al OAfter that, "cross"-on-"cross" these symbols of the elements will put the corresponding indices below:
IIIII
Al 2 O 3Composition of a chemical compound Al 2 O 3 determined. The further scheme of the reaction equation will take the form:
Al + O 2 →Al 2 O 3
It remains only to equalize the left and right parts of it. We proceed in the same way as in the case of formulating equation (19). We equalize the number of oxygen atoms, resorting to finding the smallest multiple:
before reaction after reaction
O 2 O 3
\ 6 /Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved. We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:
Al + 3O 2 → 2Al 2 O 3
In order to achieve equality for aluminum, it is necessary to adjust its amount on the left side of the equation by setting the coefficient "4":
4Al + 3O 2 → 2Al 2 O 3
Thus, the equality for aluminum and oxygen is observed and, in general, the equation will take the final form:
4Al + 3O 2 \u003d 2Al 2 O 3 (22)
Using the valency method, it is possible to predict which substance is formed in the course of a chemical reaction, what its formula will look like. Suppose nitrogen and hydrogen with the corresponding valences III and I entered into the reaction of the compound. Let's write the general reaction scheme:
N 2 + H 2 → NH
For nitrogen and hydrogen, we put down the valencies over the proposed formula of this compound:
As before, "cross"-on-"cross" for these element symbols, we put the corresponding indices below:
III I
N H 3The further scheme of the reaction equation will take the form:
N 2 + H 2 → NH 3
Equalizing in the already known way, through the smallest multiple for hydrogen, equal to "6", we obtain the desired coefficients, and the equation as a whole:
N 2 + 3H 2 \u003d 2NH 3 (23)
When compiling equations for oxidation states reacting substances, it must be recalled that the degree of oxidation of an element is the number of electrons received or given away in the process of a chemical reaction. The oxidation state in compounds basically, numerically coincides with the values of the element's valences. But they differ in sign. For example, for hydrogen, the valence is I, and the oxidation state is (+1) or (-1). For oxygen, the valence is II, and the oxidation state is (-2). For nitrogen, the valencies are I, II, III, IV, V, and the oxidation states are (-3), (+1), (+2), (+3), (+4), (+5), etc. . The oxidation states of the elements most commonly used in equations are shown in Table 3.
In the case of compound reactions, the principle of compiling equations in terms of oxidation states is the same as in compiling in terms of valencies. For example, let's give the reaction equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write the proposed equation:
Cl 2 + O 2 → ClO
We put the oxidation states of the corresponding atoms over the proposed ClO compound:
As in the previous cases, we establish that the desired compound formula will take the form:
7 -2
Cl 2 O 7The reaction equation will take the following form:
Cl 2 + O 2 → Cl 2 O 7
Equalizing for oxygen, finding the smallest multiple between two and seven, equal to "14", we finally establish the equality:
2Cl 2 + 7O 2 \u003d 2Cl 2 O 7 (24)
A slightly different method must be used with oxidation states when compiling exchange, neutralization, and substitution reactions. In some cases, it is difficult to find out: what compounds are formed during the interaction of complex substances?
How do you know what happens in a reaction?
Indeed, how do you know: what reaction products can arise in the course of a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?
Ba (NO 3) 2 + K 2 SO 4 →?
Maybe VAC 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction, compounds are formed: BaSO 4 and KNO 3. And how is this known? And how to write formulas of substances? Let's start with what is most often overlooked: the very concept of "exchange reaction". This means that in these reactions, the substances change with each other in constituent parts. Since the exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will change are metal cations (Na +, Mg 2+, Al 3+, Ca 2+, Cr 3+), H + ions or OH -, anions - acid residues, (Cl -, NO 3 2-, SO 3 2-, SO 4 2-, CO 3 2-, PO 4 3-). In general, the exchange reaction can be given in the following notation:
Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)
Where Kt1 and Kt2 are the metal cations (1) and (2), and An1 and An2 are the anions (1) and (2) corresponding to them. In this case, it must be taken into account that in compounds before and after the reaction, cations are always established in the first place, and anions in the second. Therefore, if it reacts potassium chloride and silver nitrate, both in solution
KCl + AgNO 3 →
then in the process of it substances KNO 3 and AgCl are formed and the corresponding equation will take the form:
KCl + AgNO 3 \u003d KNO 3 + AgCl (26)
In neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):
HCl + KOH \u003d KCl + H 2 O (27)
The oxidation states of metal cations and the charges of anions of acid residues are indicated in the table of the solubility of substances (acids, salts and bases in water). Metal cations are shown horizontally, and anions of acid residues are shown vertically.
Based on this, when compiling the exchange reaction equation, it is first necessary to establish the oxidation states of the particles receiving in this chemical process on its left side. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let's draw up the initial scheme for this reaction:
CaCl + NaCO 3 →
Ca 2+ Cl - + Na + CO 3 2- →
Having performed the already known “cross”-to-“cross” action, we determine the real formulas of the starting substances:
CaCl 2 + Na 2 CO 3 →
Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction:
CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl
We put down the corresponding charges over their cations and anions:
Ca 2+ CO 3 2- + Na + Cl -
Substance formulas are written correctly, in accordance with the charges of cations and anions. Let's make a complete equation by equating the left and right parts of it in terms of sodium and chlorine:
CaCl 2 + Na 2 CO 3 \u003d CaCO 3 + 2NaCl (28)
As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:
VaON + NPO 4 →
We put the corresponding charges over cations and anions:
Ba 2+ OH - + H + RO 4 3- →
Let's define the real formulas of the starting substances:
Va (OH) 2 + H 3 RO 4 →
Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction, taking into account that in the exchange reaction, one of the substances must necessarily be water:
Ba (OH) 2 + H 3 RO 4 → Ba 2+ RO 4 3- + H 2 O
Let's determine the correct record of the formula of the salt formed during the reaction:
Ba (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O
Equate the left side of the equation for barium:
3VA (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O
Since on the right side of the equation the residue of phosphoric acid is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:
3VA (OH) 2 + 2H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O
It remains to match the number of hydrogen and oxygen atoms on the right side of the water. Since the total number of hydrogen atoms on the left is 12, on the right it must also correspond to twelve, therefore, before the water formula, it is necessary put a coefficient"6" (since there are already 2 hydrogen atoms in the water molecule). For oxygen, equality is also observed: on the left 14 and on the right 14. So, the equation has the correct form of writing:
3Ва (ОН) 2 + 2Н 3 РО 4 → Ва 3 (РО 4) 2 + 6Н 2 O (29)
Possibility of chemical reactions
The world is made up of a great variety of substances. The number of variants of chemical reactions between them is also incalculable. But can we, having written this or that equation on paper, assert that a chemical reaction will correspond to it? There is a misconception that if the right arrange odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and drop into it zinc, then we can observe the process of hydrogen evolution:
Zn + H 2 SO 4 \u003d ZnSO 4 + H 2 (30)
But if copper is lowered into the same solution, then the process of gas evolution will not be observed. The reaction is not feasible.
Cu + H 2 SO 4 ≠
If concentrated sulfuric acid is taken, it will react with copper:
Cu + 2H 2 SO 4 \u003d CuSO 4 + SO 2 + 2H 2 O (31)
In reaction (23) between nitrogen and hydrogen gases, thermodynamic balance, those. how many molecules ammonia NH 3 is formed per unit time, the same number of them will decompose back into nitrogen and hydrogen. Shift in chemical equilibrium can be achieved by increasing the pressure and decreasing the temperature
N 2 + 3H 2 \u003d 2NH 3
If you take potassium hydroxide solution and pour on it sodium sulfate solution, then no changes will be observed, the reaction will not be feasible:
KOH + Na 2 SO 4 ≠
Sodium chloride solution when interacting with bromine, it will not form bromine, despite the fact that this reaction can be attributed to a substitution reaction:
NaCl + Br 2 ≠
What are the reasons for such discrepancies? The fact is that it is not enough just to correctly define compound formulas, you need to know the specifics of the interaction of metals with acids, skillfully use the table of solubility of substances, know the rules of substitution in the series of activity of metals and halogens. This article outlines only the most basic principles of how arrange the coefficients in the reaction equations, as write molecular equations , as determine the composition of a chemical compound.
Chemistry, as a science, is extremely diverse and multifaceted. This article reflects only a small part of the processes occurring in real world. Types, thermochemical equations, electrolysis, processes organic synthesis and many many others. But more on that in future articles.
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