The relative masses of atoms and molecules are determined using the D.I. Mendeleev values of atomic masses. At the same time, when carrying out calculations for educational purposes, the values of the atomic masses of the elements are usually rounded to integers (with the exception of chlorine, whose atomic mass is assumed to be 35.5).
Example 1 Relative atomic mass of calcium And r (Ca)=40; relative atomic mass of platinum And r (Pt)=195.
The relative mass of a molecule is calculated as the sum of the relative atomic masses of the atoms that make up this molecule, taking into account the amount of their substance.
Example 2. Relative molar mass of sulfuric acid:
M r (H 2 SO 4) \u003d 2A r (H) + A r (S) + 4A r (O) \u003d 2 · 1 + 32 + 4· 16 = 98.
The absolute masses of atoms and molecules are found by dividing the mass of 1 mole of a substance by the Avogadro number.
Example 3. Determine the mass of one atom of calcium.
Decision. The atomic mass of calcium is And r (Ca)=40 g/mol. The mass of one calcium atom will be equal to:
m (Ca) \u003d A r (Ca) : N A \u003d 40: 6.02 · 10 23 = 6,64· 10 -23 years
Example 4 Determine the mass of one molecule of sulfuric acid.
Decision. The molar mass of sulfuric acid is M r (H 2 SO 4) = 98. The mass of one molecule m (H 2 SO 4) is:
m (H 2 SO 4) \u003d M r (H 2 SO 4) : N A \u003d 98: 6.02 · 10 23 = 16,28· 10 -23 years
2.10.2. Calculation of the amount of matter and calculation of the number of atomic and molecular particles from known values of mass and volume
The amount of a substance is determined by dividing its mass, expressed in grams, by its atomic (molar) mass. The amount of a substance in the gaseous state at n.o. is found by dividing its volume by the volume of 1 mol of gas (22.4 l).
Example 5 Determine the amount of sodium substance n(Na) in 57.5 g of metallic sodium.
Decision. The relative atomic mass of sodium is And r (Na)=23. The amount of a substance is found by dividing the mass of metallic sodium by its atomic mass:
n(Na)=57.5:23=2.5 mol.
Example 6 . Determine the amount of nitrogen substance, if its volume at n.o. is 5.6 liters.
Decision. The amount of nitrogen substance n(N 2) we find by dividing its volume by the volume of 1 mol of gas (22.4 l):
n(N 2) \u003d 5.6: 22.4 \u003d 0.25 mol.
The number of atoms and molecules in a substance is determined by multiplying the number of atoms and molecules in the substance by Avogadro's number.
Example 7. Determine the number of molecules contained in 1 kg of water.
Decision. The amount of water substance is found by dividing its mass (1000 g) by the molar mass (18 g / mol):
n (H 2 O) \u003d 1000: 18 \u003d 55.5 mol.
The number of molecules in 1000 g of water will be:
N (H 2 O) \u003d 55.5 · 6,02· 10 23 = 3,34· 10 24 .
Example 8. Determine the number of atoms contained in 1 liter (n.o.) of oxygen.
Decision. The amount of oxygen substance, the volume of which under normal conditions is 1 liter is equal to:
n(O 2) \u003d 1: 22.4 \u003d 4.46 · 10 -2 mol.
The number of oxygen molecules in 1 liter (N.O.) will be:
N (O 2) \u003d 4.46 · 10 -2 · 6,02· 10 23 = 2,69· 10 22 .
It should be noted that 26.9 · 10 22 molecules will be contained in 1 liter of any gas at n.o. Since the oxygen molecule is diatomic, the number of oxygen atoms in 1 liter will be 2 times greater, i.e. 5.38 · 10 22 .
2.10.3. Calculation of the average molar mass of the gas mixture and volume fraction
the gases it contains
The average molar mass of a gas mixture is calculated from the molar masses of the constituent gases of this mixture and their volume fractions.
Example 9 Assuming that the content (in volume percent) of nitrogen, oxygen and argon in the air is 78, 21 and 1, respectively, calculate the average molar mass of air.
Decision.
M air = 0.78 · M r (N 2)+0.21 · M r (O 2)+0.01 · M r (Ar)= 0.78 · 28+0,21· 32+0,01· 40 = 21,84+6,72+0,40=28,96
Or approximately 29 g/mol.
Example 10. The gas mixture contains 12 l of NH 3 , 5 l of N 2 and 3 l of H 2 measured at n.o. Calculate the volume fractions of gases in this mixture and its average molar mass.
Decision. The total volume of the mixture of gases is V=12+5+3=20 l. Volume fractions j of gases will be equal:
φ(NH 3)= 12:20=0.6; φ(N 2)=5:20=0.25; φ(H 2)=3:20=0.15.
The average molar mass is calculated on the basis of the volume fractions of the constituent gases of this mixture and their molecular masses:
M=0.6 · M (NH 3) + 0.25 · M(N2)+0.15 · M (H 2) \u003d 0.6 · 17+0,25· 28+0,15· 2 = 17,5.
2.10.4. Calculation of the mass fraction of a chemical element in a chemical compound
The mass fraction ω of a chemical element is defined as the ratio of the mass of an atom of a given element X contained in a given mass of a substance to the mass of this substance m. Mass fraction is a dimensionless quantity. It is expressed in fractions of a unit:
ω(X) = m(X)/m (0<ω< 1);
or in percentage
ω(X),%= 100 m(X)/m (0%<ω<100%),
where ω(X) is the mass fraction of the chemical element X; m(X) is the mass of the chemical element X; m is the mass of the substance.
Example 11 Calculate the mass fraction of manganese in manganese (VII) oxide.
Decision. The molar masses of substances are equal: M (Mn) \u003d 55 g / mol, M (O) \u003d 16 g / mol, M (Mn 2 O 7) \u003d 2M (Mn) + 7M (O) \u003d 222 g / mol. Therefore, the mass of Mn 2 O 7 with the amount of substance 1 mol is:
m(Mn 2 O 7) = M(Mn 2 O 7) · n(Mn 2 O 7) = 222 · 1= 222
From the formula Mn 2 O 7 it follows that the amount of substance of manganese atoms is twice the amount of substance of manganese oxide (VII). Means,
n(Mn) \u003d 2n (Mn 2 O 7) \u003d 2 mol,
m(Mn)= n(Mn) · M(Mn) = 2 · 55 = 110 g.
Thus, the mass fraction of manganese in manganese(VII) oxide is:
ω(X)=m(Mn) : m(Mn 2 O 7) = 110:222 = 0.495 or 49.5%.
2.10.5. Establishing the formula of a chemical compound by its elemental composition
The simplest chemical formula of a substance is determined on the basis of the known values of the mass fractions of the elements that make up this substance.
Suppose there is a sample of a substance Na x P y O z with a mass m o g. Consider how its chemical formula is determined if the quantities of the substance of the atoms of the elements, their masses or mass fractions in the known mass of the substance are known. The formula of a substance is determined by the ratio:
x: y: z = N(Na) : N(P) : N(O).
This ratio does not change if each of its terms is divided by Avogadro's number:
x: y: z = N(Na)/N A: N(P)/N A: N(O)/N A = ν(Na) : ν(P) : ν(O).
Thus, to find the formula of a substance, it is necessary to know the ratio between the amounts of substances of atoms in the same mass of a substance:
x: y: z = m(Na)/M r (Na) : m(P)/M r (P) : m(O)/M r (O).
If we divide each term of the last equation by the mass of the sample m o , then we get an expression that allows us to determine the composition of the substance:
x: y: z = ω(Na)/M r (Na) : ω(P)/M r (P) : ω(O)/M r (O).
Example 12. The substance contains 85.71 wt. % carbon and 14.29 wt. % hydrogen. Its molar mass is 28 g/mol. Determine the simplest and true chemical formulas of this substance.
Decision. The ratio between the number of atoms in a C x H y molecule is determined by dividing the mass fractions of each element by its atomic mass:
x: y \u003d 85.71 / 12: 14.29 / 1 \u003d 7.14: 14.29 \u003d 1: 2.
Thus, the simplest formula of a substance is CH 2. The simplest formula of a substance does not always coincide with its true formula. In this case, the formula CH 2 does not correspond to the valency of the hydrogen atom. To find the true chemical formula, you need to know the molar mass of a given substance. In this example, the molar mass of the substance is 28 g/mol. Dividing 28 by 14 (the sum of atomic masses corresponding to the formula unit CH 2), we obtain the true ratio between the number of atoms in a molecule:
We get the true formula of the substance: C 2 H 4 - ethylene.
Instead of the molar mass for gaseous substances and vapors, the density for any gas or air can be indicated in the condition of the problem.
In the case under consideration, the gas density in air is 0.9655. Based on this value, the molar mass of the gas can be found:
M = M air · D air = 29 · 0,9655 = 28.
In this expression, M is the molar mass of gas C x H y, M air is the average molar mass of air, D air is the density of gas C x H y in air. The resulting value of the molar mass is used to determine the true formula of the substance.
The condition of the problem may not indicate the mass fraction of one of the elements. It is found by subtracting from unity (100%) the mass fractions of all other elements.
Example 13 An organic compound contains 38.71 wt. % carbon, 51.61 wt. % oxygen and 9.68 wt. % hydrogen. Determine the true formula of this substance if its oxygen vapor density is 1.9375.
Decision. We calculate the ratio between the number of atoms in the molecule C x H y O z:
x: y: z = 38.71/12: 9.68/1: 51.61/16 = 3.226: 9.68: 3.226= 1:3:1.
The molar mass M of a substance is:
M \u003d M (O 2) · D(O2) = 32 · 1,9375 = 62.
The simplest formula of a substance is CH 3 O. The sum of atomic masses for this formula unit will be 12+3+16=31. Divide 62 by 31 and get the true ratio between the number of atoms in the molecule:
x:y:z = 2:6:2.
Thus, the true formula of the substance is C 2 H 6 O 2. This formula corresponds to the composition of dihydric alcohol - ethylene glycol: CH 2 (OH) -CH 2 (OH).
2.10.6. Determination of the molar mass of a substance
The molar mass of a substance can be determined on the basis of its gas vapor density with a known molar mass.
Example 14 . The vapor density of some organic compound in terms of oxygen is 1.8125. Determine the molar mass of this compound.
Decision. The molar mass of an unknown substance M x is equal to the product of the relative density of this substance D by the molar mass of the substance M, according to which the value of the relative density is determined:
M x = D · M = 1.8125 · 32 = 58,0.
Substances with the found value of the molar mass can be acetone, propionaldehyde and allyl alcohol.
The molar mass of a gas can be calculated using the value of its molar volume at n.c.
Example 15. Mass of 5.6 liters of gas at n.o. is 5.046 g. Calculate the molar mass of this gas.
Decision. The molar volume of gas at n.s. is 22.4 liters. Therefore, the molar mass of the desired gas is
M = 5.046 · 22,4/5,6 = 20,18.
The desired gas is neon Ne.
The Clapeyron–Mendeleev equation is used to calculate the molar mass of a gas whose volume is given under non-normal conditions.
Example 16 At a temperature of 40 ° C and a pressure of 200 kPa, the mass of 3.0 liters of gas is 6.0 g. Determine the molar mass of this gas.
Decision. Substituting the known quantities into the Clapeyron–Mendeleev equation, we obtain:
M = mRT/PV = 6.0 · 8,31· 313/(200· 3,0)= 26,0.
The gas under consideration is acetylene C 2 H 2.
Example 17 Combustion of 5.6 l (N.O.) of hydrocarbon produced 44.0 g of carbon dioxide and 22.5 g of water. The relative density of the hydrocarbon with respect to oxygen is 1.8125. Determine the true chemical formula of the hydrocarbon.
Decision. The reaction equation for the combustion of hydrocarbons can be represented as follows:
C x H y + 0.5 (2x + 0.5y) O 2 \u003d x CO 2 + 0.5 y H 2 O.
The amount of hydrocarbon is 5.6:22.4=0.25 mol. As a result of the reaction, 1 mol of carbon dioxide and 1.25 mol of water are formed, which contains 2.5 mol of hydrogen atoms. When a hydrocarbon is burned with a quantity of a substance of 1 mole, 4 moles of carbon dioxide and 5 moles of water are obtained. Thus, 1 mol of hydrocarbon contains 4 mol of carbon atoms and 10 mol of hydrogen atoms, i.e. chemical formula of hydrocarbon C 4 H 10 . The molar mass of this hydrocarbon is M=4 · 12+10=58. Its relative oxygen density D=58:32=1.8125 corresponds to the value given in the condition of the problem, which confirms the correctness of the found chemical formula.
Relative atomic and relative molecular weight. Moth. Avogadro's number
Modern research methods make it possible to determine extremely small masses of atoms with great accuracy. So, for example, the mass of a hydrogen atom is 1.674 10 27 kg, oxygen - 2.667 x 10 -26 kg, carbon - 1.993 x 10 26 kg. In chemistry, not absolute values of atomic masses are traditionally used, but relative ones. In 1961, the atomic mass unit (abbreviated a.m.u.) was adopted as a unit of atomic mass, which is 1/12 of the mass of an atom of the carbon isotope "C". Most chemical elements have atoms with different masses. Therefore, the relative atomic mass A, a chemical element, is a value equal to the ratio of the average mass of an atom of the natural isotopic composition of the element to 1/12 of the mass of the carbon atom 12C. The relative atomic masses of the elements are denoted A, where the index r is the initial letter of the English word relative - relative. The entries Ar(H), Ar(0), Ar(C) mean: the atomic mass of hydrogen, the atomic mass of oxygen, and the atomic mass of carbon. For example, Ar(H) = 1.6747x 10-27 = 1.0079; 1/12 x 1.993 x 10 -26Relative atomic mass is one of the main characteristics of a chemical element. The relative molecular weight M of a substance is a value equal to the ratio of the average mass of a molecule of the natural isotopic composition of a substance to 1/12 of the mass of a 12C carbon atom. Instead of the term "attributes atomic mass", the term "atomic mass" can be used. The relative molecular weight is numerically equal to the sum of the relative atomic masses of all the atoms that make up the molecule of the substance. It is easily calculated by the formula of the substance. For example, Mg(H2O) is composed of 2Ar(H)=2 1.00797=2.01594 Ar(0)=1x15, 9994=15.9994
Mr (H2O) \u003d 18.01534 This means that the ratio of the molecular weight of water is 18.01534, rounded, 18. The ratio of the molecular weight shows how much the mass of a molecule of a given substance is more than 1/12 of the mass of an atom C +12. So, the molecular weight of water is 18. This means that the mass of a water molecule is 18 times greater than 1/12 of the mass of a C +12 atom. Molecular weight refers to one of the main characteristics of a substance. Moth. Molar mass. The International System of Units (SI) uses the mole as the unit of quantity of a substance. A mole is the amount of a substance containing as many structural units (molecules, atoms, ions, electrons, and others) as there are atoms in 0.012 kg of the carbon isotope C +12. Knowing the mass of one carbon atom (1.993 10-26 kg), you can calculate the number of NA atoms in 0.012 kg of carbon: NA \u003d 0.012 kg / mol \u003d 1.993 x 10-26 kg 6.02 x 1023 units / mol.
This number is called the Avogadro constant (designation HA, dimension 1/mol), shows the number of structural units in a mole of any substance. Molar mass is a value equal to the ratio of the mass of a substance to the amount of a substance. It has the units of kg/mol or g/mol; usually it is denoted by the letter M. The molar mass of a substance is easy to calculate, knowing the mass of the molecule. So, if the mass of a water molecule is 2.99x10-26, kg, then the molar mass Mr (H2O) \u003d 2.99 10-26 kg 6.02 1023 1 / mol \u003d 0.018 kg / mol, or 18 g / mol. In general, the molar mass of a substance, expressed in g/mol, is numerically equal to the relative atomic or relative molecular mass of that substance. -For example, the relative atomic and molecular masses of C, Fe, O, H 2 O are respectively 12, 56, 32.18, and their molar masses are respectively 12 g / mol, 56 g / mol, 32 g / mol, 18 g / mol. Molar mass can be calculated for substances in both molecular and atomic states. For example, the relative molecular mass of hydrogen Mr (H 2) \u003d 2, and the atomic mass of hydrogen A (H) \u003d 1 refers. The amount of substance determined by the number of structural units (H A) is the same in both cases - 1 mol. However, the molar mass of molecular hydrogen is 2 g/mol, and the molar mass of atomic hydrogen is 1 g/mol. One mole of atoms, molecules or ions contains the number of these particles equal to the Avogadro constant, for example
1 mole of C atoms +12 = 6.02 1023 C atoms +12
1 mol of H 2 O molecules \u003d 6.02 1023 H 2 O molecules
1 mol of S0 4 2- ions = 6.02 1023 S0 4 2- ions
Mass and quantity of a substance are different concepts. Mass is expressed in kilograms (grams), and the amount of a substance is expressed in moles. There are simple relationships between the mass of a substance (t, g), the amount of a substance (n, mol) and the molar mass (M, g / mol): m=nM, n=m/M M=m/n Using these formulas, it is easy to calculate the mass of a certain the amount of a substance, or to determine the amount of a substance in a known assay of it, or to find the molar mass of a substance.
To measure the mass of an atom, the relative atomic mass is used, which is expressed in atomic mass units (a.m.u.). The relative molecular mass is the sum of the relative atomic masses of substances.
Concepts
To understand what relative atomic mass is in chemistry, it should be understood that the absolute mass of an atom is too small to be expressed in grams, and even more so in kilograms. Therefore, in modern chemistry, 1/12 of the mass of carbon is taken as an atomic mass unit (amu). The relative atomic mass is equal to the ratio of the absolute mass to 1/12 of the absolute mass of carbon. In other words, the relative mass reflects how many times the mass of an atom of a particular substance exceeds 1/12 of the mass of a carbon atom. For example, the relative mass of nitrogen is 14, i.e. the nitrogen atom contains 14 a. e. m. or 14 times more than 1/12 of a carbon atom.
Rice. 1. Atoms and molecules.
Among all the elements, hydrogen is the lightest, its mass is 1 unit. The heaviest atoms have a mass of 300 amu. eat.
Molecular weight - a value showing how many times the mass of a molecule exceeds 1/12 of the mass of carbon. Also expressed in a. e. m. The mass of a molecule is made up of the mass of atoms, therefore, to calculate the relative molecular mass, it is necessary to add the masses of the atoms of a substance. For example, the relative molecular weight of water is 18. This value is the sum of the relative atomic masses of two hydrogen atoms (2) and one oxygen atom (16).
Rice. 2. Carbon in the periodic table.
As you can see, these two concepts have several common characteristics:
- the relative atomic and molecular masses of a substance are dimensionless quantities;
- relative atomic mass is denoted A r , molecular mass - M r ;
- the unit of measurement is the same in both cases - a. eat.
The molar and molecular masses coincide numerically, but differ in dimension. Molar mass is the ratio of the mass of a substance to the number of moles. It reflects the mass of one mole, which is equal to Avogadro's number, i.e. 6.02 ⋅ 10 23 . For example, 1 mol of water weighs 18 g / mol, and M r (H 2 O) \u003d 18 a. e.m. (18 times heavier than one atomic mass unit).
How to calculate
To express the relative atomic mass mathematically, one should determine that 1/2 part of carbon or one atomic mass unit is equal to 1.66⋅10 −24 g. Therefore, the formula for the relative atomic mass is as follows:
A r (X) = m a (X) / 1.66⋅10 −24 ,
where m a is the absolute atomic mass of the substance.
The relative atomic mass of chemical elements is indicated in the periodic table of Mendeleev, so it does not need to be calculated independently when solving problems. Relative atomic masses are usually rounded to integers. The exception is chlorine. The mass of its atoms is 35.5.
It should be noted that when calculating the relative atomic mass of elements that have isotopes, their average value is taken into account. The atomic mass in this case is calculated as follows:
A r = ΣA r,i n i ,
where A r,i is the relative atomic mass of isotopes, n i is the content of isotopes in natural mixtures.
For example, oxygen has three isotopes - 16 O, 17 O, 18 O. Their relative mass is 15.995, 16.999, 17.999, and their content in natural mixtures is 99.759%, 0.037%, 0.204%, respectively. Dividing the percentages by 100 and substituting the values, we get:
A r = 15.995 ∙ 0.99759 + 16.999 ∙ 0.00037 + 17.999 ∙ 0.00204 = 15.999 amu
Referring to the periodic table, it is easy to find this value in an oxygen cell.
Rice. 3. Periodic table.
Relative molecular weight - the sum of the masses of the atoms of a substance:
Symbol indices are taken into account when determining the relative molecular weight value. For example, the calculation of the mass of H 2 CO 3 is as follows:
M r \u003d 1 ∙ 2 + 12 + 16 ∙ 3 \u003d 62 a. eat.
Knowing the relative molecular weight, one can calculate the relative density of one gas from the second, i.e. determine how many times one gaseous substance is heavier than the second. For this, the equation D (y) x \u003d M r (x) / M r (y) is used.
What have we learned?
From the 8th grade lesson, we learned about the relative atomic and molecular mass. The unit of relative atomic mass is 1/12 of the mass of carbon, equal to 1.66⋅10 −24 g. To calculate the mass, it is necessary to divide the absolute atomic mass of a substance by the atomic mass unit (a.m.u.). The value of the relative atomic mass is indicated in the periodic system of Mendeleev in each cell of the element. The molecular weight of a substance is the sum of the relative atomic masses of the elements.
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Atomic-molecular doctrine
The concept of atoms as the smallest indivisible particles originated in ancient Greece. The foundations of modern atomic and molecular science were first formulated by M.V. Lomonosov (1748), but his ideas, set out in a private letter, were unknown to most scientists. Therefore, the English scientist J. Dalton, who formulated (1803–1807) his main postulates, is considered the founder of modern atomic and molecular theory.
1. Each element consists of very small particles - atoms.
2. All atoms of one element are the same.
3. Atoms of different elements have different masses and have different properties.
4. Atoms of one element do not turn into atoms of other elements as a result of chemical reactions.
5. Chemical compounds are formed as a result of a combination of atoms of two or more elements.
6. In a given compound, the relative numbers of atoms of various elements are always constant.
These postulates were first indirectly proven by a set of stoichiometric laws. Stoichiometry - part of chemistry that studies the composition of substances and its change in the course of chemical transformations. This word is derived from the Greek words "stechion" - element and "metron" - measure. The laws of stoichiometry include the laws of conservation of mass, constancy of composition, multiple ratios, volumetric ratios, Avogadro's law and the law of equivalents.
1.3. Stoichiometric laws
The laws of stoichiometry are considered integral parts of AMU. Based on these laws, the concept of chemical formulas, chemical equations and valency was introduced.
The establishment of stoichiometric laws made it possible to assign a strictly defined mass to the atoms of chemical elements. The masses of atoms are extremely small. So, the mass of a hydrogen atom is 1.67∙10 -27 kg, oxygen - 26.60∙10 -27 kg, carbon - 19.93∙10 -27 kg. It is very inconvenient to use such numbers for various calculations. Therefore, since 1961, 1/12 of the mass of the carbon isotope 12 C - atomic mass unit (a.m.u.). Previously, it was called the carbon unit (c.u.), but now this name is not recommended.
A.m.u. mass is 1.66. 10 -27 kg or 1.66. 10–24
Relative atomic mass of an element (Ar) is the ratio of the absolute mass of an atom to 1/12 of the absolute mass of an atom of the carbon isotope 12 C. In other words, A r shows how many times the mass of an atom of a given element is heavier than 1/12 of the mass of an atom 12 C. For example, the value of A r oxygen rounded to a whole number is 16; this means that the mass of one oxygen atom is 16 times greater than 1/12 of the mass of a 12 C atom.
The relative atomic masses of the elements (Ar) are given in the Periodic Table of Chemical Elements by D.I. Mendeleev.
Relative molecular weight (M r) a substance is called the mass of its molecule, expressed in amu. It is equal to the sum of the atomic masses of all the atoms that make up the molecule of the substance and is calculated by the formula of the substance. For example, the relative molecular mass of sulfuric acid H 2 SO 4 is composed of the atomic masses of two hydrogen atoms (1∙2 = 2), the atomic mass of one sulfur atom (32) and the atomic mass of four oxygen atoms (4∙16 = 64). It is equal to 98.
This means that the mass of a sulfuric acid molecule is 98 times greater than 1/12 of the mass of a 12 C atom.
Relative atomic and molecular masses are relative quantities, and therefore dimensionless.
