Molar mass molar volume formula. Molar volume

The mass of 1 mole of a substance is called molar. What is the volume of 1 mole of a substance called? Obviously, this is also called molar volume.

What is the molar volume of water? When we measured 1 mole of water, we did not weigh 18 g of water on the scales - this is inconvenient. We used measuring utensils: a cylinder or a beaker, since we knew that the density of water is 1 g/ml. Therefore, the molar volume of water is 18 ml/mol. For liquids and solids, the molar volume depends on their density (Fig. 52, a). It's a different matter for gases (Fig. 52, b).

Rice. 52.
Molar volumes (n.s.):
a - liquids and solids; b - gaseous substances

If you take 1 mole of hydrogen H2 (2 g), 1 mole of oxygen O2 (32 g), 1 mole of ozone O3 (48 g), 1 mole of carbon dioxide CO2 (44 g) and even 1 mole of water vapor H2 O (18 g) under the same conditions, for example normal (in chemistry it is customary to call normal conditions (n.s.) a temperature of 0 ° C and a pressure of 760 mm Hg, or 101.3 kPa), then it turns out that 1 mol of any of the gases will occupy the same volume, equal to 22.4 liters, and contain the same number of molecules - 6 × 10 23.

And if you take 44.8 liters of gas, then how much of its substance will be taken? Of course, 2 moles, since the given volume is twice the molar volume. Hence:

where V is the volume of gas. From here

Molar volume is a physical quantity equal to the ratio of the volume of a substance to the amount of substance.

The molar volume of gaseous substances is expressed in l/mol. Vm - 22.4 l/mol. The volume of one kilomole is called kilomolar and is measured in m 3 /kmol (Vm = 22.4 m 3 /kmol). Accordingly, the millimolar volume is 22.4 ml/mmol.

Problem 1. Find the mass of 33.6 m 3 of ammonia NH 3 (n.s.).

Problem 2. Find the mass and volume (n.v.) of 18 × 10 20 molecules of hydrogen sulfide H 2 S.

When solving the problem, let's pay attention to the number of molecules 18 × 10 20. Since 10 20 is 1000 times less than 10 23, obviously, calculations should be carried out using mmol, ml/mmol and mg/mmol.

Key words and phrases

1. Molar, millimolar and kilomolar volumes of gases.
2. The molar volume of gases (under normal conditions) is 22.4 l/mol.
3. Normal conditions.

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Questions and tasks

1. Find the mass and number of molecules at n. u. for: a) 11.2 liters of oxygen; b) 5.6 m 3 nitrogen; c) 22.4 ml of chlorine.
2. Find the volume that at n. u. will take: a) 3 g of hydrogen; b) 96 kg of ozone; c) 12 × 10 20 nitrogen molecules.
3. Find the densities (mass 1 liter) of argon, chlorine, oxygen and ozone at room temperature. u. How many molecules of each substance will be contained in 1 liter under the same conditions?
4. Calculate the mass of 5 liters (n.s.): a) oxygen; b) ozone; c) carbon dioxide CO 2.
5. Indicate which is heavier: a) 5 liters of sulfur dioxide (SO 2) or 5 liters of carbon dioxide (CO 2); b) 2 liters of carbon dioxide (CO 2) or 3 liters of carbon monoxide (CO).

The molar volume of a gas is equal to the ratio of the volume of the gas to the amount of substance of this gas, i.e.

V m = V(X) / n(X),

where V m is the molar volume of gas - a constant value for any gas under given conditions;

V(X) – volume of gas X;

n(X) – amount of gas substance X.

The molar volume of gases under normal conditions (normal pressure p n = 101,325 Pa ≈ 101.3 kPa and temperature T n = 273.15 K ≈ 273 K) is V m = 22.4 l/mol.

Ideal gas laws

In calculations involving gases, it is often necessary to switch from these conditions to normal ones or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

pV / T = p n V n / T n

Where p is pressure; V - volume; T - temperature on the Kelvin scale; the index “n” indicates normal conditions.

Volume fraction

The composition of gas mixtures is often expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

φ(X) = V(X) / V

where φ(X) is the volume fraction of component X;

V(X) - volume of component X;

V is the volume of the system.

Volume fraction is a dimensionless quantity; it is expressed in fractions of a unit or as a percentage.

Example 1. What volume will ammonia weighing 51 g occupy at a temperature of 20°C and a pressure of 250 kPa?

Example 2. Determine the volume that a gas mixture containing hydrogen, weighing 1.4 g, and nitrogen, weighing 5.6 g, will occupy under normal conditions.

Law of volumetric relations

How to solve a problem using the “Law of Volumetric Relations”?

Law of Volume Ratios: The volumes of gases involved in a reaction are related to each other as small integers equal to the coefficients in the reaction equation.

The coefficients in the reaction equations show the numbers of volumes of reacting and formed gaseous substances.

Example. Calculate the volume of air required to burn 112 liters of acetylene.

1. We compose the reaction equation:

2. Based on the law of volumetric relations, we calculate the volume of oxygen:

112 / 2 = X / 5, from where X = 112 5 / 2 = 280l

3. Determine the volume of air:

V(air) = V(O 2) / φ(O 2)

V(air) = 280 / 0.2 = 1400 l.

Names of acids are formed from the Russian name of the central atom of the acid with the addition of suffixes and endings. If the oxidation state of the central atom of the acid corresponds to the group number of the Periodic Table, then the name is formed using the simplest adjective from the name of the element: H 2 SO 4 - sulfuric acid, HMnO 4 - manganese acid. If acid-forming elements have two oxidation states, then the intermediate oxidation state is denoted by the suffix –ist-: H 2 SO 3 – sulfurous acid, HNO 2 – nitrous acid. Various suffixes are used for the names of halogen acids that have many oxidation states: typical examples are HClO 4 - chlorine n acid, HClO 3 – chlorine novat acid, HClO 2 – chlorine ist acid, HClO – chlorine novatist ic acid (oxygen-free acid HCl is called hydrochloric acid - usually hydrochloric acid). Acids can differ in the number of water molecules that hydrate the oxide. Acids containing the largest number of hydrogen atoms are called ortho acids: H 4 SiO 4 - orthosilicic acid, H 3 PO 4 - orthophosphoric acid. Acids containing 1 or 2 hydrogen atoms are called metaacids: H 2 SiO 3 - metasilicic acid, HPO 3 - metaphosphoric acid. Acids containing two central atoms are called di acids: H 2 S 2 O 7 – disulfuric acid, H 4 P 2 O 7 – diphosphoric acid.

The names of complex compounds are formed in the same way as names of salts, but the complex cation or anion is given a systematic name, that is, it is read from right to left: K 3 - potassium hexafluoroferrate(III), SO 4 - tetraammine copper(II) sulfate.

Names of oxides are formed using the word “oxide” and the genitive case of the Russian name of the central atom of the oxide, indicating, if necessary, the oxidation state of the element: Al 2 O 3 - aluminum oxide, Fe 2 O 3 - iron (III) oxide.

Names of bases are formed using the word “hydroxide” and the genitive case of the Russian name of the central hydroxide atom, indicating, if necessary, the oxidation state of the element: Al(OH) 3 - aluminum hydroxide, Fe(OH) 3 - iron(III) hydroxide.

Names of compounds with hydrogen are formed depending on the acid-base properties of these compounds. For gaseous acid-forming compounds with hydrogen, the following names are used: H 2 S – sulfane (hydrogen sulfide), H 2 Se – selan (hydrogen selenide), HI – hydrogen iodide; their solutions in water are called hydrogen sulfide, hydroselenic and hydroiodic acids, respectively. For some compounds with hydrogen, special names are used: NH 3 - ammonia, N 2 H 4 - hydrazine, PH 3 - phosphine. Compounds with hydrogen having an oxidation state of –1 are called hydrides: NaH is sodium hydride, CaH 2 is calcium hydride.

Names of salts are formed from the Latin name of the central atom of the acidic residue with the addition of prefixes and suffixes. The names of binary (two-element) salts are formed using the suffix - eid: NaCl – sodium chloride, Na 2 S – sodium sulfide. If the central atom of an oxygen-containing acidic residue has two positive oxidation states, then the highest oxidation state is denoted by the suffix – at: Na 2 SO 4 – sulf at sodium, KNO 3 – nitr at potassium, and the lowest oxidation state is the suffix - it: Na 2 SO 3 – sulf it sodium, KNO 2 – nitr it potassium To name oxygen-containing halogen salts, prefixes and suffixes are used: KClO 4 – lane chlorine at potassium, Mg(ClO 3) 2 – chlorine at magnesium, KClO 2 – chlorine it potassium, KClO – hypo chlorine it potassium

Covalent saturationsconnectionto her– manifests itself in the fact that in compounds of s- and p-elements there are no unpaired electrons, that is, all unpaired electrons of atoms form bonding electron pairs (exceptions are NO, NO 2, ClO 2 and ClO 3).

Lone electron pairs (LEP) are electrons that occupy atomic orbitals in pairs. The presence of NEP determines the ability of anions or molecules to form donor-acceptor bonds as donors of electron pairs.

Unpaired electrons are electrons of an atom, contained one in an orbital. For s- and p-elements, the number of unpaired electrons determines how many bonding electron pairs a given atom can form with other atoms through the exchange mechanism. The valence bond method assumes that the number of unpaired electrons can be increased by lone electron pairs if there are vacant orbitals within the valence electron level. In most compounds of s- and p-elements there are no unpaired electrons, since all unpaired electrons of the atoms form bonds. However, molecules with unpaired electrons exist, for example, NO, NO 2, they have increased reactivity and tend to form dimers like N 2 O 4 due to unpaired electrons.

Normal concentration – this is the number of moles equivalents in 1 liter of solution.

Normal conditions - temperature 273K (0 o C), pressure 101.3 kPa (1 atm).

Exchange and donor-acceptor mechanisms of chemical bond formation. The formation of covalent bonds between atoms can occur in two ways. If the formation of a bonding electron pair occurs due to the unpaired electrons of both bonded atoms, then this method of formation of a bonding electron pair is called an exchange mechanism - the atoms exchange electrons, and the bonding electrons belong to both bonded atoms. If the bonding electron pair is formed due to the lone electron pair of one atom and the vacant orbital of another atom, then such formation of the bonding electron pair is a donor-acceptor mechanism (see. valence bond method).

Reversible ionic reactions – these are reactions in which products are formed that are capable of forming starting substances (if we keep in mind the written equation, then about reversible reactions we can say that they can proceed in one direction or another with the formation of weak electrolytes or poorly soluble compounds). Reversible ionic reactions are often characterized by incomplete conversion; since during a reversible ionic reaction, molecules or ions are formed that cause a shift towards the initial reaction products, that is, they seem to “slow down” the reaction. Reversible ionic reactions are described using the ⇄ sign, and irreversible ones - the → sign. An example of a reversible ionic reaction is the reaction H 2 S + Fe 2+ ⇄ FeS + 2H +, and an example of an irreversible one is S 2- + Fe 2+ → FeS.

Oxidizing agents – substances in which, during redox reactions, the oxidation states of some elements decrease.

Redox duality – the ability of substances to act in redox reactions as an oxidizing or reducing agent depending on the partner (for example, H 2 O 2, NaNO 2).

Redox reactions(OVR) – These are chemical reactions during which the oxidation states of the elements of the reacting substances change.

Oxidation-reduction potential – a value characterizing the redox ability (strength) of both the oxidizing agent and the reducing agent that make up the corresponding half-reaction. Thus, the redox potential of the Cl 2 /Cl - pair, equal to 1.36 V, characterizes molecular chlorine as an oxidizing agent and chloride ion as a reducing agent.

Oxides – compounds of elements with oxygen in which oxygen has an oxidation state of –2.

Orientation interactions– intermolecular interactions of polar molecules.

Osmosis – the phenomenon of transfer of solvent molecules on a semi-permeable (permeable only to solvent) membrane towards a lower solvent concentration.

Osmotic pressure – physicochemical property of solutions due to the ability of membranes to pass only solvent molecules. Osmotic pressure from a less concentrated solution equalizes the rate of penetration of solvent molecules into both sides of the membrane. The osmotic pressure of a solution is equal to the pressure of a gas in which the concentration of molecules is the same as the concentration of particles in the solution.

Arrhenius bases – substances that split off hydroxide ions during electrolytic dissociation.

Bronsted bases - compounds (molecules or ions of the S 2-, HS - type) that can attach hydrogen ions.

Reasons according to Lewis (Lewis bases) – compounds (molecules or ions) with lone pairs of electrons capable of forming donor-acceptor bonds. The most common Lewis base is water molecules, which have strong donor properties.

Gases are the simplest object to study, therefore their properties and reactions between gaseous substances have been studied most fully. To make it easier for us to understand the decision rules calculation tasks,based on the equations of chemical reactions,it is advisable to consider these laws at the very beginning of the systematic study of general chemistry

French scientist J.L. Gay-Lussac laid down the law volumetric relations:

For example, 1 liter of chlorine connects with 1 liter of hydrogen , forming 2 liters of hydrogen chloride ; 2 l sulfur oxide (IV) connect with 1 liter of oxygen, forming 1 liter of sulfur oxide (VI).

This law allowed the Italian scientist assume that molecules of simple gases ( hydrogen, oxygen, nitrogen, chlorine, etc. ) consist of two identical atoms . When hydrogen combines with chlorine, their molecules break down into atoms, and the latter form hydrogen chloride molecules. But since two molecules of hydrogen chloride are formed from one molecule of hydrogen and one molecule of chlorine, the volume of the latter must be equal to the sum of the volumes of the original gases.
Thus, volumetric relations are easily explained if we proceed from the idea of ​​​​the diatomic nature of molecules of simple gases ( H2, Cl2, O2, N2, etc. ) - This, in turn, serves as proof of the diatomic nature of the molecules of these substances.
The study of the properties of gases allowed A. Avogadro to put forward a hypothesis, which was subsequently confirmed by experimental data, and therefore became known as Avogadro’s law:

Avogadro's law implies an important consequence: under the same conditions, 1 mole of any gas occupies the same volume.

This volume can be calculated if the mass is known 1 l gas Under normal conditions conditions, (n.s.) i.e. temperature 273К (О°С) and pressure 101,325 Pa (760 mmHg) , the mass of 1 liter of hydrogen is 0.09 g, its molar mass is 1.008 2 = 2.016 g/mol. Then the volume occupied by 1 mole of hydrogen under normal conditions is equal to 22.4 l

Under the same conditions the mass 1l oxygen 1.492g ; molar 32g/mol . Then the volume of oxygen at (n.s.) is also equal to 22.4 mol.

Hence:

The molar volume of a gas is the ratio of the volume of a substance to the amount of that substance:

Where V m - molar volume of gas (dimensionl/mol ); V is the volume of the system substance;n - the amount of substance in the system. Example entry:V m gas (Well.)=22.4 l/mol.

Based on Avogadro's law, the molar masses of gaseous substances are determined. The greater the mass of gas molecules, the greater the mass of the same volume of gas. Equal volumes of gases under the same conditions contain the same number of molecules, and therefore moles of gases. The ratio of the masses of equal volumes of gases is equal to the ratio of their molar masses:

Where m 1 - mass of a certain volume of the first gas; m 2 — mass of the same volume of the second gas; M 1 And M 2 - molar masses of the first and second gases.

Typically, gas density is determined in relation to the lightest gas - hydrogen (denoted D H2 ). The molar mass of hydrogen is 2g/mol . Therefore we get.

The molecular mass of a substance in the gaseous state is equal to twice its hydrogen density.

Often the density of a gas is determined relative to air (D B ) . Although air is a mixture of gases, they still talk about its average molar mass. It is equal to 29 g/mol. In this case, the molar mass is determined by the expression M = 29D B .

Determination of molecular masses showed that molecules of simple gases consist of two atoms (H2, F2,Cl2, O2 N2) , and molecules of inert gases are made from one atom (He, Ne, Ar, Kr, Xe, Rn). For noble gases, “molecule” and “atom” are equivalent.

Boyle-Mariotte Law: at a constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is located.From here pV = const ,
Where R - pressure, V - volume of gas.

Gay-Lussac's Law: at constant pressure and the change in gas volume is directly proportional to temperature, i.e.
V/T = const,
Where T - temperature on scale TO (kelvin)

Combined gas law of Boyle - Mariotte and Gay-Lussac:
pV/T = const.
This formula is usually used to calculate the volume of a gas under given conditions if its volume under other conditions is known. If a transition is made from normal conditions (or to normal conditions), then this formula is written as follows:
pV/T = p 0 V 0 /T 0 ,
Where R 0 ,V 0 ,T 0 -pressure, gas volume and temperature under normal conditions ( R 0 = 101 325 Pa , T 0 = 273 K V 0 =22.4 l/mol) .

If the mass and quantity of a gas are known, but it is necessary to calculate its volume, or vice versa, use Mendeleev-Clayperon equation:

Where n - amount of gas substance, mol; m — mass, g; M - molar mass of gas, g/iol ; R — universal gas constant. R = 8.31 J/(mol*K)

P1V1=P2V2, or, which is the same, PV=const (Boyle-Mariotte law). At constant pressure, the ratio of volume to temperature remains constant: V/T=const (Gay-Lussac law). If we fix the volume, then P/T=const (Charles’ law). Combining these three laws gives a universal law that states that PV/T=const. This equation was established by the French physicist B. Clapeyron in 1834.

The value of the constant is determined only by the amount of substance gas. DI. Mendeleev derived an equation for one mole in 1874. So it is the value of the universal constant: R=8.314 J/(mol∙K). So PV=RT. In the case of an arbitrary quantity gasνPV=νRT. The amount of a substance itself can be found from mass to molar mass: ν=m/M.

Molar mass is numerically equal to relative molecular mass. The latter can be found from the periodic table; it is indicated in the cell of the element, as a rule, . The molecular weight is equal to the sum of the molecular weights of its constituent elements. In the case of atoms of different valences, an index is required. On at mer, M(N2O)=14∙2+16=28+16=44 g/mol.

Normal conditions for gases at It is commonly assumed that P0 = 1 atm = 101.325 kPa, temperature T0 = 273.15 K = 0°C. Now you can find the volume of one mole gas at normal conditions: Vm=RT/P0=8.314∙273.15/101.325=22.413 l/mol. This table value is the molar volume.

Under normal conditions conditions quantity relative to volume gas to molar volume: ν=V/Vm. For arbitrary conditions you need to use the Mendeleev-Clapeyron equation directly: ν=PV/RT.

Thus, to find the volume gas at normal conditions, you need the amount of substance (number of moles) of this gas multiply by the molar volume equal to 22.4 l/mol. Using the reverse operation, you can find the amount of a substance from a given volume.

To find the volume of one mole of a substance in a solid or liquid state, find its molar mass and divide by its density. One mole of any gas under normal conditions has a volume of 22.4 liters. If conditions change, calculate the volume of one mole using the Clapeyron-Mendeleev equation.

You will need

• Periodic table of Mendeleev, table of density of substances, pressure gauge and thermometer.

Instructions

Determining the volume of one mole or solid
Determine the chemical formula of the solid or liquid you are studying. Then, using the periodic table, find the atomic masses of the elements that are included in the formula. If one is included in the formula more than once, multiply its atomic mass by that number. Add up the atomic masses and get the molecular mass of what the solid or liquid is made of. It will be numerically equal to the molar mass measured in grams per mole.

Using the table of substance densities, find this value for the material of the body or liquid being studied. After this, divide the molar mass by the density of the substance, measured in g/cm³ V=M/ρ. The result is the volume of one mole in cm³. If the substance remains unknown, it will be impossible to determine the volume of one mole of it.