Count moles. Chemical formulas. Calculation of the mass fraction of the chemical element in the chemical connection
The amount of substance in chemistry (moths):
Formulas in chemistry are determined from which the substance consists. Now we will learn to determine in what quantities these substances are present in the compounds.
Number of substances - It is essentially the number of smallest particles (or structural units) From which the substance consists. The smallest particles are either atoms (Fe) (only one element) or molecules (H 2 O) (from different elements).
Number of substances in chemistry express through (This is the Greek letter "NU", which is similar to the English "V", only with rounded vertices).
Even in the grains of molecules of billions, so everyone does not consider them, but use special units of measurement - moths.
1 mol is the amount of substance equal to 6.02 * 10 23 structural units of the substance. It is so much (6.02 * 10 23) molecules, for example, in one water mole or sugar or something else.
As you can see, it is very, a lot - a billion, multiplied by a billion, another 100,000 and 6 !!! If you take such a number of single-handed coins and put them all the surface of the earth (as well as all seas and oceans), then it turns out a layer of 1 km thick!
Instruction
One of the solven volume formulas: v \u003d m / p, where V is the volume of the solution (ml), M is the mass (g), P - density (g / ml). If you need to additionally find a mass, then this can be done, knowing the formula and the amount of the desired substance. With the help of the formula, we will find its molar mass, having completed the atomic masses of all elements included in it. For example, M (AGNO3) \u003d 108 + 14 + 16 * 3 \u003d 170 g / mol. Next, we find a mass by the formula: m \u003d n * m, where m is the mass (g), n - the amount of substance (mol), m is the molar mass of the substance (g / mol). It is understood that the amount of substance is given in the task.
Next to find the volume of the solution is derived from the formula molar: C \u003d N / V, where C is a molar concentration of the solution (mol / l), n is the amount of substance (mol), V is the volume of the solution (L). Take: v \u003d n / c. The amount of substance can be additionally found using the formula: n \u003d m / m, where M is a mass, M - molar mass.
The following are formulas for finding the volume of gas. V \u003d N * Vm, where V is the volume of gas (L), n - the amount of substance (mol), Vm is the molar volume of gas (l / mol). With normal, i.e. The pressure is equal to 101 325 Pa 273 to the molar volume of gas is the value of constant and is equal to 22.4 l / mol.
For a gas system, there is a formula :: q (x) \u003d V (x) / v, where Q (x) (F) is the volume fraction of the component, V (x) - the volume of the component (L), V - the volume of the system (L) . From this formula, you can remove 2 others: V (x) \u003d Q * V, as well as V \u003d V (x) / q.
If the problem of the reaction is present in the problem, the task should be solved using it. From the equation you can find the number of any substance, it is equal to the coefficient. For example, Cuo + 2HCl \u003d CUCL2 + H2O. From here we see that with the interaction of 1 praying copper oxide and 2 mole of hydrochloric acid, 1 mol of copper chloride and 1 mol of water turned out. Knowing under the condition of the problem The amount of substance of just one component of the reaction is possible to easily find the amounts of all substances. Let it be the amount of substance of the copper oxide equal to 0.3 mol, it means N (HCl) \u003d 0.6 mol, n (CuCl2) \u003d 0.3 mol, n (H2O) \u003d 0.3 mol.
note
Do not forget about the units of measure!
Sources:
- "Collection of tasks in chemistry", G.P. Homchenko, I.G. Homchenko, 2002.
- mass formula
The mass of any substance, molecules are equal to the sum of the masses of its formation. If when calculating the use of relative atomic masses, then the relative molecular weight of the substance is obtained. The relative molecular weight shows how many times the absolute mass of the molecule of this substance is greater than 1/12 of the absolute mass of the carbon atom. Typically use the approximate values \u200b\u200bof relative atomic and molecular weights. These values \u200b\u200bare dimensionless.
Instruction
Calculate what is equal to each element in the molecule. To learn the relative mass of one atom, look into the periodic system of elements. The first number is atomic weight. You can also calculate it according to the formula Ar (element) \u003d M (element) /1a.e.m. For ease of calculations, approximate values \u200b\u200bare used.
Ar (h) \u003d 1? 2 \u003d 2; ar (o) \u003d 16? 1 \u003d 16ar (Fe) \u003d 56? 2 \u003d 112; ar (s) \u003d 32? 3 \u003d 96; ar (o) \u003d 16? 12 \u003d 192.
Fold the results. This will be the molecular weight of the substance.
Mr (H2O) \u003d 2AR (H) + AR (O) \u003d 2 + 16 \u003d 18
Mr (Fe2 (SO4) 3) \u003d 2Ar (Fe) + 3Ar (S) + 12Ar (O) \u003d 112 + 96 + 192 \u003d 400
In addition to the relative molecular weight in the calculations, the molar mass is used more often. Its unit of measure - g / mol. It is numerically equal to the relative molecular weight of the substance.
M (H2O) \u003d 18 g / mol
M (Fe2 (SO4) 3 \u003d 400 g / mol
Video on the topic
During the chemical reaction, a variety of substances can be formed: gaseous, soluble, uni-soluble. In the latter case, they fall into the sediment. Often there is a need to know what the exact mass of the precipitate formed. How can this be calculated?
You will need
- - Glass funnel;
- - paper filter;
- - Laboratory scales.
Instruction
You can act an experimental way. That is, conduct a chemical, carefully separate the resulting sediment from the filtrate using a conventional glass funnel and a paper filter, for example. A more complete compartment is achieved by vacuum filtering (on the Buchner funnel).
After that, the precipitate is dried - natural way Or under vacuum, and weighing with possible greater accuracy. Best, on sensitive laboratory scales. This is how the task will be solved. It is resorted to this method when the exact number of source substances that have entered the reaction are unknown.
If you know these quantities, then the task can be solved much easier and faster. Suppose it is necessary to calculate how much chloride was formed 20 grams of chloride - table salt - and 17 grams of nitric acid silver. First of all, write the equation: NaCl + AGNO3 \u003d NanO3 + AgCl.
In the course of this reaction, a very small soluble compound is formed - silver chloride falling in the form of a white sediment.
Calculate the molar masses of the source substances. For sodium chloride, it is approximately 58.5 g / mol, for nitric acid silver - 170 g / mol. That is, initially under the conditions of the problem you had 20 / 58.5 \u003d 0.342 praying sodium chloride and 17/170 \u003d 0.1 mole of nitric acid silver.
Thus, it turns out that the sodium chloride was initially taken in excess, that is, the reaction on the second starting substance will pass to the end (all 0.1 praying of nitric acid silver will react, "linked" the same 0.1 praying salt). How many silver chloride is formed? To answer this question, find the molecular weight of the precipitate formed: 108 + 35.5 \u003d 143.5. Multiplying the initial amount of nitric acid silver (17 grams) on the ratio of the molecular masses of the product and the source substance, get the answer: 17 * 143.5 / 170 \u003d 14.3 grams. This will be the exact weight of the sediment formed during the reaction.
Of course, the resulting answer is not very accurate, since you used in the calculations rounded the values \u200b\u200bof atomic masses of the elements. If high accuracy is required, it is necessary to consider that the atomic weight of silver, for example, is not 108, and 107,868. Accordingly, the atomic mass of chlorine is not 35.5, and 35, 453, etc.
Sources:
- calculate the mass of the precipitate that was formed when interacting
In the tasks in the chemistry of the school course, as a rule, it is required to calculate the volume for the gaseous reaction product. You can do this if the number of people of any participant in chemical interaction is known. Or find this amount from other task data.
The algorithm for finding the amount of substance is quite simple, it can be useful for simplifying the decision of the solution. Get acquainted with another concept that will be required to calculate the amount of substance: this is a molar mass, or a mass of one praying of an individual element atom. Already from the definition it is noticeable that it is measured in g / mol. Use the standard table that contains the molar weight values \u200b\u200bfor some items.
What is the amount of substance and how it is determined
At the same time, the mass of hydrogen participating in the reaction is about 8 times less than the mass of oxygen (since the atomic mass of hydrogen is approximately 16 times less than the atomic mass of oxygen). When the heat of the reaction is written as it is made in this equation, it is understood that it is expressed in kilodzhoules on a stoichiometric unit ("mole") of the reaction according to the recorded equation. The heat of the reactions is always tabulated in the calculation of the mol of the resulting compound.
In order to understand what the amount of substance is in chemistry, give the term definition. To understand what is the amount of substance, we note that this value has its own designation. Eighth graders who still do not know how to write chemical equations do not know that such a matter of substance, how to use this value in the calculations. After acquaintance with the law of constancy of the mass of substances, the meaning of this value becomes clear. Under it implies the mass that corresponds to one pray for a specific chemical substance. No task of the school rate of chemistry associated with calculations by equation is not without the use of such a term as "the amount of substance".
2.10.5. Establishment of formula
chemical compound by its element
FIUM
We obtain the true formula of substance: C2N4- ethylene. 2.5 mol of hydrogen atoms.
Referred to as Mr. Located on the Mendeleev table - it is simply the sum of the atomic masses of the substance. The law of preserving the mass is the mass of substances that have entered the chemical reaction are always equal to the mass of the substances. That is, if there are normal conditions in the task, then, knowing the amount of mol (N), we can find the volume of substance. The basic formulas for solving chemistry problems are formulas.
Where in the periodic system are the elements corresponding to these metals? From the proposals below, write numbers corresponding to metals in one column, and the numbers corresponding to nonmetallam into another column. To obtain a certain amount of product (in chemical laboratory Or at the factory) it is necessary to take strictly certain amounts of starting materials. Chemists, conducting experiments, noticed that the composition of the products of some reactions depends on what relations reacting substances were taken. How many atoms will be in this mass?
N is the number of structural links, and Na is constant Avogadro. Permanent Avogadro is a ratio of proportionality that ensures the transition from molecular relationship to molar. V is the volume of gas (L), and Vm is a molar volume (l / mol).
Unit of measurement of the amount of substance in the international system of units (C) - mole. Determination. Record the formula to calculate this energy and the names of the physical quantities included in the formula. This question refers to the section "10-11" classes.
1. Basic concepts, definitions and laws of chemistry
1.3. Chemical amount of substance. Mole. Molar mass
Chemical amount of substance. Mole. Molar mass
Characterizing the portion of the resulting substance, use its mass or volume. However, for the same purpose, the number of structural units in the consistent portion of the substance can be indicated. To know this number is extremely important, since in chemical reactions of the substance interact in relations proportional to the number of structural units, and not the masses. For example, recording 2H 2 + O 2 \u003d 2H 2 O denotes that the numbers (but not mass!) The reactive molecules H 2 and O 2 are respectively as 2: 1.
For the convenience of counting the number of structural units, the content of which in any measurable portion of the substance was huge, a new physical value was introduced - the amount of substance that was also called the chemical amount of substance in chemical calculations.
Chemical amount of substance - The physical quantity proportional to the number of structural units (atoms, molecules, pheet) contained in this portion of the substance.
Denotes the chemical number of the letter N (less often ν).
The unit of chemical amount of matter is moth.
Mol - portion of a substance containing so much of its elementary structural units, how many atoms are contained in the portions of Nuclide C-12 weighing 12 g.
The number of atoms in the specified portion of the nuclide C-12 is approximately 6.02 ⋅ 10 23. Physical value equal to 6.02 ⋅ 10 23 mol -1, called permanent Avogadro and denotes n a:
N a \u003d 6.02 ⋅ 10 23 mol - 1 \u003d 6.02 ⋅ 10 23 mol - 1.
The number of the numerator in the value of n a does not indicate, since for different cases it can be different, for example:
N a \u003d 6.02 ⋅ 10 23 atoms mole,
N a \u003d 6.02 ⋅ 10 23 molecules mole,
N a \u003d 6.02 ⋅ 10 23 Fe moth.
The physical meaning of constant avogadro is that its numerical value (6.02 ⋅ 10 23) shows the number of structural units in 1 mol of substance. For example, 1 mol sodium (m \u003d 23 g) contains 6.02 ⋅ 10 23 Na atoms; 1 mole of sulfuric acid (m \u003d 98 g) contains 6.02 ⋅ 10 23 molecules H 2 SO 4; 1 mol calcium carbonate (m \u003d 100 g) contains 6.02 ⋅ 10 23 CACO 3 formula units.
Mol is a portion of a substance containing 6.02 ⋅ 10 23 of its structural units
The number of structural units of the substance N (b) and the chemical amount of the substance N (b) are associated with the relation
n (b) \u003d n (b) n a, (1.8)
N (b) \u003d n (b) n a. (1.9)
Knowing the chemical amount of any substance, it is possible by its chemical formula to calculate the chemical number of individual atoms included in its composition.
One mol of any substance numely contains the same chemical number of atoms as their (atoms) is contained in the same molecule (formula unit) substance
For example:
- as part of Molecules P 4 contains 4 atoms P, and in composition 1 mol P 4 - 4 mol of atoms P;
- as part of the formula unit Na 3 PO 4, 3 atoms Na, 1 atom p and 4 atom o, and in 1 mol Na 3 PO 4 - 3 mol of Na atoms, 1 mol of atoms P and 4 mol of atoms O.
With an increase in (decreasing), the chemical amount of the substance increases in proportion to (decreases) the chemical number of atoms included in its composition. For example: 0.5 mol Na 3 PO 4 contains 3 · 0.5 \u003d 1.5 (mol) atoms Na; 5 mol p 4 contains 5 · 4 \u003d 20 (mol) of atoms R.
For such calculations, you can use the so-called stoichiometric schemes. The principles for the preparation of stoichiometric schemes and settlements are shown in the example of K 2 SO 4 with a chemical amount of 0.3 mol:
x \u003d n (k) \u003d 0.3 ⋅ 2 1 \u003d 0.6 (mol);
y \u003d n (s) \u003d 0.3 ⋅ 1 1 \u003d 0.3 (mole);
z \u003d n (O) \u003d 0.3 ⋅ 4 1 \u003d 1,2 (mol).
The concept of mol is applicable to all substances, and the concept of a molecule is not all, but only to the substances of the molecular structure. For example, both concepts are applicable to water (water has a molecular structure), but in the case of calcium carbonate (neomolecular structure), only the concept of "mole" is applicable.
The concept of "mole" is also used in the case of ions, electrons, protons, neutrons and chemical bonds. For example, if n (PO 4 3 -) \u003d 3.01 ⋅ 10 23, then
n (PO 4 3 -) \u003d 3.01 ⋅ 10 23 / 6.02 ⋅ 10 23 \u003d 0.5 (mole);
N (E) \u003d 1.505 ⋅ 10 22,
n (e) \u003d n (E) / n a \u003d 1.505 ⋅ 10 22 / 6.02 ⋅ 10 23 \u003d 0.025 (mol);
2 mole molecules H 2 (H-H) contain 2 mol of hydrogen bonds - hydrogen, and 3 mole molecules H 2 O (N-ON) - 6 mol of n-o links (each molecule contains two connections N-O) .
The molar mass M (B) is a physical value equal to the ratio of the mass of the substance to its chemical quantity:
M (b) \u003d m (b) n (b). (1.10)
From the expression (1.10), the formulas for calculating the mass of the substance are followed:
m (b) \u003d n (b) ⋅ M (b) (1.11)
and its chemical quantity:
n (b) \u003d m (b) m (b). (1.12)
Since with n (b) \u003d 1 mol numerical values \u200b\u200bN (b) and M (b) coincide, often say that the molar mass is a mass of 1 mol of matter. This, of course, is incorrect, since only the numerical values \u200b\u200bof these values \u200b\u200bcoincide, and their physical meaning and units of measurement are different.
With the help of molar mass, it is easy to calculate the mass of the molecule or a formula unit of the substance:
m Mol, Fe \u003d M (c) n a. (1.13)
In addition, the molar mass can be found by the formula
M (B) \u003d m Mol, Fe ⋅ n a. (1.14)
It is easy to show that when using the unit of the molar mass of grams to the mol, its numerical value coincides:
- c a R for simple substances of the atomic structure:
A r (o) \u003d 16, m (O) \u003d 16 g mol;
- M R complex substances of the molecular and non-elastic structure:
M R (H 2 O) \u003d 18, M (H 2 O) \u003d 18 g mol;
M R (KOH) \u003d 56, M (KOH) \u003d 56 g mol.
Indeed:
M (b) \u003d m mole (c) ⋅ n a \u003d m r (b) ⋅ u ⋅ n a \u003d m r (b) ⋅ 1 n a ⋅ n a \u003d m r (b)
M (B) \u003d m AT ⋅ N a \u003d a r (b) ⋅ U ⋅ N a \u003d a r (b) ⋅ 1 n A ⋅ N a \u003d a r (b).
Example 1.5. The mass of the substance molecules is 7.31 ⋅ 10 -23. Calculate the molar mass of the substance.
Decision. The first way. From formula (1.14) follows:
M (b) \u003d m mole (b) ⋅ n a
M (b) \u003d 7.31 ⋅ 10 - 23 g ⋅ 6.02 ⋅ 10 23 1 mol \u003d 44 g / mol.
The second way. We use formula (1.5):
M R (b) \u003d m mole (b) u \u003d 7.31 ⋅ 10 - 23 g 1.66 ⋅ 10 - 24 g \u003d 44;
M (b) \u003d 44 g / mol.
Answer: 44 g / mol.
Gas laws. Mixtures gases
Substances can be in three aggregate states: gaseous, liquid and solid. Liquid and solid state are called condensed. For most substances, the aggregate states are interconnected: when heated, the solid is at first melting, then evaporates; When cooled, the gas is initially condensed - goes into a liquid state, then the liquid freezes (crystallizes). Increased pressure and decrease in temperature contribute to the transition of a substance into a condensed state with a smaller volume (and on the contrary - a decrease in pressure and an increase in temperature contribute to the transition of a substance into a gaseous state).
Gas pressure in a closed vessel is directly proportional to the number of its molecules (or chemical quantity)
When moving a substance from a solid state into a liquid, and then - in the gaseous distance between the particles, it increases sequentially, and in the case of gas, this distance is hundreds of times more sizes Molecules themselves. From this it follows that the amount of gas portion is determined not by the nature of the gas (the size of its molecules), and the distance between molecules (essentially, the volume that occupies gas is the volume of free space between molecules).
The distance between gas molecules depends on temperature and pressure, which means that under the same external conditions, the distance between the molecules of various gases is the same.
From here it follows the position known as the Avogadro law (1811): in equal volumes of different gases under the same conditions, the same number of molecules are contained
From the law, the Avogadro flow three consequences.
1. Same number Molecules of various gases at the same pressure and temperature occupy the same volume.
2. Under normal conditions (N.U.: T \u003d 273 K or 0 ° C, P \u003d 101.3 kPa), the volume of portions of any gas with a chemical amount of 1 mol, or a molar volume V M,
V m \u003d 22.4 DM 3 / mol.
3. The masses of the same volumes of two gases are assessed as their molar (relative molecular weight) masses. This relationship is called relative gas density and gas in and denotes how d b (a):
m (a) m (b) \u003d d b (a) \u003d m (a) m (b) \u003d m r (a) m r (b). (1.15)
Using V M find the volume and chemical amount of gas:
V (b) \u003d n (b) ⋅ V m; (1.16)
n (b) \u003d V (b) / v m. (1.17)
Formula (1.15) allows, knowing the relative density of unknown gas from a known gas, to find M (m R) of unknown gas:
M (x) \u003d d b (x) ⋅ m (b). (1.18)
For example, if the relative density of the gas in the air (m is carried out \u003d 29 g / mol) is 1.517, then the molar mass of this gas
M (x) \u003d 29 ⋅ 1,517 \u003d 44 (g / mol).
Relative density - the value is dimensionless and does not depend on temperature and pressure.
Knowing the molar mass of gas, it is easy to calculate the density ρ of gas (in G / DM 3):
ρ (c) \u003d m (c) v m \u003d m (c) 22.4. (1.19)
For example, for nitrogen
ρ (n 2) \u003d m (n 2) v m \u003d 28 g / mol 22.4 dm 3 / mol \u003d 1.25 g / dm 3.
In terms of gas density, its molar mass is found:
M (c) \u003d ρ (c) v m. (1.20)
The gas density depends on the temperature T and pressure P: with the growth of T and decrease, the density decreases.
If the density ρ of two gases (ρ 1 \u003d ρ 2) is equal and their molar (relative molecular weight) masses, i.e. M 1 \u003d m 2 (and vice versa - if molar masses of gases are equal, then their density are equal)
In the case of gases, also fair law of Volume Relations Gay Loursak (1805-1808): In chemical reactions, the volume of reacting and obtained gases relate as small integers equal to their stoichiometric coefficients
For example, for reaction
4NH 3 + 5O 2 \u003d 4NO + 6H 2 O
V (NH 3) V (O 2) \u003d 4 5;
V (O 2) V (NO) \u003d 5 4.
Example 1.6. The relative density (N.U.) of some gas in argon is 1.2. Find a mass of the gas molecule X.
Decision . Using formula (1.18), we will find the molar mass of gas x:
M (x) \u003d d ar (x) ⋅ m (AR),
M (x) \u003d 1.2 ⋅ 40 \u003d 48 g / mol.
By formula (1.13), we calculate the mass of the gas molecule X:
m Mol (x) \u003d m (x) n a \u003d 48 6.02 ⋅ 10 23 \u003d 7.97 ⋅ 10 - 23 (g).
You can also use formula (1.7):
m Mol (x) \u003d m R (x) U \u003d 48 ⋅ 1.66 ⋅ 10 - 24 \u003d 7.97 ⋅ 10 - 23 (g).
Answer: 7.97 ⋅ 10 -23
Methods of gathering gases. Molar gas concentration
Consider laboratory gases collection methods. There are two such methods (Fig. 1.1).
Fig. 1.1. Laboratory methods for collecting oxygen heating KMNO 4:
A - method of displacing water; b - way of air displacement
It's obvious that way of displacing water Only those gases can be collected, which in water do not dissolve and do not interact with it (hydrogen, methane, nitrogen, oxygen). In this way, it is impossible to collect gases that in water are well dissolved or interact with it (HCl, HBr, Hi, HF, NH 3). Carbon oxide (IV) CO 2 in water dissolves relatively badly, so it can be collected in this method.
When collecting Gaza method of overpaying air You need to correctly place test tubes:
- torch up, if the gas is heavier than air, i.e. M (gas)\u003e M (RED). Examples: CO 2, SO 2, HCl;
- sorry down if gas is easier than air, i.e. M (gas)< M (возд) . Примеры: H 2 , Ne, NH 3 , CH 4 .
For characteristics Gas Used molar concentration c equal to the ratio of the chemical amount of gas to the volume of gas portion:
c (x) \u003d n (x) v (x)
Gas mixtures are similar to individual gases are characterized by a molar (relative molecular) mass, density ρ, relative density D on another gas, as well as mass W and volumetric φ shares of individual gases:
M (mixtures) \u003d M (mixtures) n (mixtures), (1.22)
w \u003d m (gas) M (mixtures), (1.23)
φ \u003d V (gas) V (mixtures), (1.24)
φ \u003d n (gas) n (mixtures), (1.25)
D A (mixtures) \u003d m (mixtures) m (a), (1.26)
ρ (mixtures) \u003d m (mixtures) v m \u003d m (mixtures) V (mixtures). (1.27)
The molar weight of the mixture of gases is conveniently found in bulk shares and molar masses of individual gases:
M (mixtures) \u003d m 1 φ 1 + m 2 φ 2 + m 3 φ 3 + ... + m n φ n. (1.28)
Obviously:
φ 1 + φ 2 + φ 3 + ... + φ n \u003d 1.
For a mixture of two gases (φ 1 + φ 2 \u003d 1) φ 2 \u003d 1 - φ 1. Then
M (mixtures) \u003d m 1 φ 1 + m 2 φ 2 \u003d m 1 φ 1 + m 2 (1 - φ 1). (1.29)
Example 1.7. Find the molar mass of the gas mixture (N.U.), consisting of nitrogen volume (N.U.) 1.12 DM 3 and oxygen weighing 5.76 g.
Decision . According to formulas (1.12) and (1.17) we find the chemical amount of gases and mixtures:
n (O 2) \u003d M (O 2) M (O 2) \u003d 5.76 32 \u003d 0.18 (mol),
n (n 2) \u003d V (n 2) v m \u003d 1.12 22.4 \u003d 0.05 (mol).
In this way,
n (mixtures) \u003d n (o 2) + n (n 2) \u003d 0.05 + 0.18 \u003d 0.23 (mol).
By formula (1.25) we find bulk shares of gases in the mixture:
φ (n 2) \u003d 0.05 0.23 \u003d 0.217,
φ (O 2) \u003d 0.18 0.23 \u003d 0,783
or (since the mixture consists of two gases):
φ (O 2) \u003d 1 - 0.217 \u003d 0.783.
By formula (1.29) we find the molar mass of the mixture:
M (mixtures) \u003d m (o 2) φ (o 2) + m (n 2) φ (n 2);
M (mixtures) \u003d 32 ⋅ 0,783 + 28 ⋅ 0,217 \u003d 31.2 (g / mol).
Answer: 31.2 g / mol.
1. The molar mass of the gas mixture is between the molar weight values \u200b\u200bof the easiest and most heavy gas mixture. For example, the molar mass of the mixture of NH 3 (M \u003d 17 g / mol) and CO 2 (m \u003d 44 g / mol), depending on the volume fractions of gases, can take values \u200b\u200b17< M (смеси) < 44 (г/моль).
2. If the molar masses of gases in the mixture are the same, the molar mass of the mixture does not depend on the volume fractions of individual gases. For example, the molar mass of the mixture of CO, C 2 H 2 and N 2 is always equal to 28 g / mol regardless of the volume fractions of the components.
3. If gas is added to the gas mixture, M of which is more than M of the heavy gas itself, then M (mixtures) increases. For example, if the mixtures of N 2 and O 2 are different to add CO 2, then M (mixtures) will increase.
4. If gas is added to the mixture of gases, the m of which is less than M of the easiest gas mixture, then M (mixtures) of the mixture decreases. For example, if a different mixture of NE and AR mixtures add HE, then M (mixtures) will decrease.
5. In the equality of volumetric fractions of gases in the mixture, the molar mass of the mixture is equal to the medium-graded molar masses of individual gases. For example, for a mixture of equal volumes of CO 2 and O 2:
M (mixtures) \u003d m (o 2) + m (CO 2) 2 \u003d 32 + 44 2 \u003d 38 g / mol.
The most typical processes carried out in chemistry are chemical reactions, i.e. The interactions between some kind of source substances leading to the formation of new substances. Substances react in certain quantitative relations that need to be taken into account in order to obtain the desired products to spend the minimum amount of starting materials and not to create useless production waste. To calculate the masses of the reactant substances, another physical value is necessary, which characterizes the portion of the substance from the point of view of the number of structural units contained in it. In itself, the ego number is extremely great. This is obvious, in particular, from Example 2.2. Therefore, in practical calculations, the number of structural units is replaced by a special value called number Substances.
The amount of substance is a measure of the number of structural units determined by the expression
where N (x) - number of structural units of substance X. In real or mentally taken portion of the substance, N a \u003d. 6.02 10 23 - constant (number) of the Avogadro, widely used in science, one of the fundamental physical constants. If necessary, it is possible to use a more accurate value of the Avogadro constant 6.02214 10 23. Portion of a substance containing N A structural units is a single amount of substance - 1 mol. Thus, the amount of substance is measured in moles, and the Avogadro constant has a unit of measurement 1 / mol, or in another mole -1 record.
For all sorts of reasoning and calculations associated with the properties of the substance and chemical reactions, the concept number of substances Fully replaces the concept the number of structural units. Thanks to this, it disappears the need to use large numbers. For example, instead of saying "6.02 10 23 structural units (molecules) of water", we will say, "take 1 mol of water".
Any portion of the substance is characterized by both the mass and the amount of substance.
The ratio of mass of matterX. The amount of substance is called molar massM (x):
The molar mass is numerically equal to mass 1 mol of substance. This is an important quantitative characteristic of each substance depending only on the mass of structural units. The Avogadro number is established such that the molar mass of the substance, expressed in g / mol, is numerically coincided with the relative molecular weight Mr. For water molecule Mg \u003d. 18. This means that the molar mass of water M (H 2 0) \u003d 18 g / mol. Using the data of the Mendeleev table, you can calculate more accurate values. Mr. and M (x), But in class tasks in chemistry it is usually not required. Of all this, it is clear how easy it is to calculate the molar mass of the substance - it is enough to add atomic masses in accordance with the formula of the substance and put a unit of measurement of g / mol. Therefore, formula (2.4) is practically used to calculate the amount of substance:
Example 2.9.Calculate the molar mass of drinking soda NaHC0 3.
Decision. According to the formula of substances Mg \u003d. 23 + 1 + 12 + 3 16 \u003d 84. Hence, by definition, M (Naiic0 3) \u003d 84 g / mol.
Example 2.10.What amount of substance make up 16.8 g of drinking soda? Decision. M (NaHC0 3) \u003d 84 g / mol (see above). By Formula (2.5)
Example 2.11.How many Tolik (structural units) drinking soda is located in 16.8 g of substance?
Decision. Converting formula (2.3), we find:
AT (NaHC0 3) \u003d N A N (NaHC0 3);
tT (NaHC0 3) \u003d 0.20 mol (see example 2.10);
N (NaHC0 3) \u003d 6.02 10 23 mol "1 0.20 mol \u003d 1,204 10 23.
Example 2.12.How many atoms are located in 16.8 g of drinking soda?
Decision. Drinking soda, NaHC0 3, consists of sodium, hydrogen atoms, carbon and oxygen. In total, in the structural unit of substance 1 + 1 + 1+ 3 \u003d 6 atoms. How was found in Example 2.11, this mass Drinking soda consists of 1.204 10 23 structural units. Therefore, the total number of atoms in the substance is
