Ammonium Hydroxide Composition and Molar Mass. Ammonia water: obtaining, formula, application Molar mass of elements and compounds

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Chemical formula

Molar mass of NH 4 OH, ammonium hydroxide 35.0458 g/mol

14.0067+1.00794 4+15.9994+1.00794

Mass fractions of elements in the compound

Using the Molar Mass Calculator

Chemical formulas must be entered case sensitive
Indexes are entered as regular numbers
The dot on the midline (multiplication sign), used, for example, in the formulas of crystalline hydrates, is replaced by a regular dot.
Example: instead of CuSO₄ 5H₂O, the converter uses the spelling CuSO4.5H2O for ease of entry.

Molar mass calculator

mole

All substances are made up of atoms and molecules. In chemistry, it is important to accurately measure the mass of substances entering into a reaction and resulting from it. By definition, the mole is the SI unit for the amount of a substance. One mole contains exactly 6.02214076×10²³ elementary particles. This value is numerically equal to the Avogadro constant N A when expressed in units of moles⁻¹ and is called Avogadro's number. Amount of substance (symbol n) of a system is a measure of the number of structural elements. A structural element can be an atom, a molecule, an ion, an electron, or any particle or group of particles.

Avogadro's constant N A = 6.02214076×10²³ mol⁻¹. Avogadro's number is 6.02214076×10²³.

In other words, a mole is the amount of a substance equal in mass to the sum of the atomic masses of the atoms and molecules of the substance, multiplied by the Avogadro number. The mole is one of the seven basic units of the SI system and is denoted by the mole. Since the name of the unit and its symbol are the same, it should be noted that the symbol is not declined, unlike the name of the unit, which can be declined according to the usual rules of the Russian language. One mole of pure carbon-12 equals exactly 12 grams.

Molar mass

Molar mass is a physical property of a substance, defined as the ratio of the mass of that substance to the amount of the substance in moles. In other words, it is the mass of one mole of a substance. In the SI system, the unit of molar mass is kilogram/mol (kg/mol). However, chemists are accustomed to using the more convenient unit g/mol.

molar mass = g/mol

Molar mass of elements and compounds

Compounds are substances made up of different atoms that are chemically bonded to each other. For example, the following substances, which can be found in the kitchen of any housewife, are chemical compounds:

salt (sodium chloride) NaCl
sugar (sucrose) C₁₂H₂₂O₁₁
vinegar (acetic acid solution) CH₃COOH

The molar mass of chemical elements in grams per mole is numerically the same as the mass of the element's atoms expressed in atomic mass units (or daltons). The molar mass of compounds is equal to the sum of the molar masses of the elements that make up the compound, taking into account the number of atoms in the compound. For example, the molar mass of water (H₂O) is approximately 1 × 2 + 16 = 18 g/mol.

Molecular mass

Molecular weight (the old name is molecular weight) is the mass of a molecule, calculated as the sum of the masses of each atom that makes up the molecule, multiplied by the number of atoms in this molecule. The molecular weight is dimensionless a physical quantity numerically equal to the molar mass. That is, the molecular weight differs from the molar mass in dimension. Although the molecular mass is a dimensionless quantity, it still has a value called the atomic mass unit (amu) or dalton (Da), and is approximately equal to the mass of one proton or neutron. The atomic mass unit is also numerically equal to 1 g/mol.

Molar mass calculation

The molar mass is calculated as follows:

determine the atomic masses of the elements according to the periodic table;
determine the number of atoms of each element in the compound formula;
determine the molar mass by adding the atomic masses of the elements included in the compound, multiplied by their number.

For example, let's calculate the molar mass of acetic acid

It consists of:

two carbon atoms
four hydrogen atoms
two oxygen atoms

carbon C = 2 × 12.0107 g/mol = 24.0214 g/mol
hydrogen H = 4 × 1.00794 g/mol = 4.03176 g/mol
oxygen O = 2 × 15.9994 g/mol = 31.9988 g/mol
molar mass = 24.0214 + 4.03176 + 31.9988 = 60.05196 g/mol

Our calculator does just that. You can enter the formula of acetic acid into it and check what happens.

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A colorless gas with a pungent odor, ammonia NH 3 not only dissolves well in water with the release of heat. The substance actively interacts with H 2 O molecules to form a weak alkali. The solution has received several names, one of them is ammonia water. The compound has amazing properties, which are the method of formation, composition and

Formation of the ammonium ion

The formula of ammonia water is NH 4 OH. The substance contains the NH 4 + cation, which is formed by non-metals - nitrogen and hydrogen. The N atoms in the ammonia molecule are used to form only 3 of the 5 outer electrons, and one pair remains unclaimed. In a strongly polarized water molecule, hydrogen protons H + are weakly bound to oxygen, one of them becomes a donor of a free nitrogen electron pair (acceptor).

An ammonium ion is formed with one positive charge and a special type of weak covalent bond - donor-acceptor. By its size, charge, and some other features, it resembles a potassium cation and behaves like a chemically unusual compound reacts with acids, forms salts of great practical importance. Names that reflect the features of the preparation and properties of the substance:

ammonium hydroxide;
ammonia hydrate;
caustic ammonium.

Precautionary measures

Care must be taken when working with ammonia and its derivatives. Important to remember:

Ammonia water has an unpleasant odor. The released gas irritates the mucous surface of the nasal cavity, eyes, and causes coughing.
When stored in loosely closed vials, ampoules, ammonia is released.
It can be detected without instruments, only by smell, even a small amount of gas in solution and air.
The ratio between molecules and cations in solution changes at different pH.
At a value of about 7, the concentration of toxic gas NH 3 decreases, the amount of NH 4 + cations less harmful to living organisms increases

Obtaining ammonium hydroxide. Physical Properties

When ammonia is dissolved in water, ammonia water is formed. The formula of this substance is NH 4 OH, but in fact ions are present at the same time

NH 4 +, OH -, NH 3 and H 2 O molecules. In the chemical reaction of ion exchange between ammonia and water, an equilibrium state is established. The process can be reflected using a diagram in which oppositely directed arrows indicate the reversibility of phenomena.

In the laboratory, obtaining ammonia water is carried out in experiments with nitrogen-containing substances. When ammonia is mixed with water, a clear, colorless liquid is obtained. At high pressures, the solubility of the gas increases. Water releases more ammonia dissolved in it as the temperature rises. For industrial needs and agriculture on an industrial scale, a 25% substance is obtained by dissolving ammonia. The second method involves the use of reaction with water.

Chemical properties of ammonium hydroxide

Upon contact, two liquids - ammonia water and hydrochloric acid - are covered with clouds of white smoke. It consists of particles of the reaction product - ammonium chloride. With such a volatile substance as hydrochloric acid, the reaction takes place directly in the air.

Weak-alkaline chemical properties of ammonia hydrate:

The substance reversibly dissociates in water to form an ammonium cation and a hydroxide ion.
In the presence of an NH 4 + ion, a colorless solution of phenolphthalein turns crimson, as in alkalis.
Chemical with acids leads to the formation of ammonium and water salts: NH 4 OH + HCl \u003d NH 4 Cl + H 2 O.
Ammonia water enters into ion exchange reactions with metal salts, which correspond to the formation of a water-insoluble hydroxide: 2NH 4 OH + CuCl 2 \u003d 2NH 4 Cl + Cu (OH) 2 (blue precipitate).

Ammonia water: application in various sectors of the economy

An unusual substance is widely used in everyday life, agriculture, medicine, and industry. Technical ammonia hydrate is used in agriculture, production of soda ash, dyes and other products. Liquid fertilizer contains nitrogen in a form easily digestible by plants. The substance is considered the cheapest and most effective for application in the pre-sowing period for all crops.

Three times less money is spent on the production of ammonia water than on the production of solid granular nitrogen fertilizers. Hermetically sealed steel tanks are used for storage and transportation of liquids. Some types of hair dyes and bleaches are made using caustic ammonium. In every medical institution there are preparations with ammonia - a 10% ammonia solution.

Ammonium salts: properties and practical significance

Substances that are obtained by the interaction of ammonium hydroxide with acids are used in economic activities. Salts decompose when heated, dissolve in water, undergo hydrolysis. They enter into chemical reactions with alkalis and other substances. Chlorides, nitrates, sulfates, phosphates and

It is very important to follow the rules and safety measures when working with substances that contain the ammonium ion. When stored in warehouses of industrial and agricultural enterprises, in subsidiary farms, there should be no contact of such compounds with lime and alkalis. If the seals of the packages are broken, a chemical reaction will begin with the release of toxic gas. Everyone who has to work with ammonia water and its salts must know the basics of chemistry. If the safety requirements are observed, the substances used will not harm people and the environment.

Equivalent can be called a real or conditional particle of a substance that can replace, add, or be in any other way equivalent to one hydrogen ion in acid-base or ion-exchange reactions or one electron in redox reactions.

The molar mass of the equivalent in most exchange reactions (going without changing the oxidation states of the elements involved in them) can be calculated as the ratio of the molar mass of a substance to the number of bonds that are broken or formed per atom or one molecule during a chemical reaction.

The molar mass of the equivalent of the same substance can be different in different reactions.

The molar mass of the equivalent in redox reactions (going with a change in the oxidation states of the elements involved in them) can be calculated as the ratio of the molar mass of a substance to the number of donated or accepted electrons per atom or one molecule during a chemical reaction.

To find the equivalent mass of a substance in a solution, simple relationships are used:

For acid H n A m:

E to \u003d M / n, where n is the number of H + ions in acid. For example, the equivalent mass of hydrochloric acid HCl is: e k=M/1, i.e. numerically equal to the molar mass; the equivalent mass of phosphoric acid H 3 RO 4 is: e k=M/3, i.e. 3 times less than its molar mass.

For base K n (OH) m:

E main \u003d M / m, where m is the number of hydroxide-ones OH - in the base formula. For example, the equivalent mass of ammonium hydroxide NH 4 OH is equal to its molar mass: E main=M/1; the equivalent mass of copper hydroxide (II) Cu (OH) 2 is 2 times less than its molar mass: E main=M/2.

For K n A m salt:

E s \u003d M / (n × m), where n and m, respectively, the amount of cations and anions of the salt. For example, the equivalent mass of aluminum sulfate Al 2 (SO 4) 3 is: E s=M/(2×3)=M/6.

The law of equivalents - for 1 equivalent of one substance in a reaction, there is 1 equivalent of another substance.

It follows from the law of equivalents that the masses (or volumes) of the reacting and formed substances are proportional to the molar masses (molar volumes) of their equivalents. For any two substances related by the law of equivalents, we can write:

where m 1 and m 2 – masses of reactants and (or) reaction products, g;

E 1, E 2 are the molar masses of the equivalents of the reactants and (or) reaction products, g/mol;

V 1 , V 2 – volumes of reagents and (or) reaction products, l;

EV 1 , EV 2 are the molar volumes of the equivalents of the reactants and (or) reaction products, l/mol.

Gaseous substances, in addition to the molar mass equivalent, have molar volume equivalent (EV -volume occupied by molar mass equivalent or volume of one mole equivalent). At n.o. EV (O 2) \u003d 5.6 l/mol , EV (H 2) \u003d 11.2 l/mol ,

Task 1. The combustion of a mass of 12.4 g of an unknown element consumed a volume of 6.72 liters of oxygen. Calculate the equivalent of the element and determine which element was taken in this reaction.

According to the law of equivalents

EV (O 2) - equivalent volume of oxygen equal to 5.6 l

E (element) \u003d \u003d 10.3 g / mol-eq

To determine an element, you need to find its molar mass. The valency of the element (B), the molar mass (M) and the equivalent (E) are related by the relation E \u003d, hence M \u003d E ∙ V, (where B is the valency of the element).

In this problem, the valency of the element is not indicated, therefore, when solving, it is necessary to use the selection method, taking into account the rules for determining the valence - an element located in the odd (I, III, V, VII) group of the periodic table can have a valence equal to any odd number, but no more than group number; an element located in an even (II, IV, VI, VIII) group of the periodic table can have a valency equal to any even number, but not more than the group number.

M \u003d E ∙ B \u003d 10.3 ∙ I \u003d 10.3 g / mol

M \u003d E ∙ B \u003d 10.3 ∙ II \u003d 20.6 g / mol

There is no element with an atomic mass of 10.3 in the periodic table, so we continue the selection.

M \u003d E ∙ B \u003d 10.3 ∙ III \u003d 30.9 g / mol

This is the atomic mass of element number 15, this element is phosphorus (P).

(Phosphorus is located in group V of the periodic table, the valence of this element can be equal to III).

Answer: the element is phosphorus (P).

Task 2. 5.6 g of potassium hydroxide was used to dissolve 3.269 g of the unknown metal. Calculate the metal equivalent and determine which metal was taken for this reaction.

According to the law of equivalents:

The base equivalent is defined as the ratio of its molar mass to the number of OH groups - in the base: M (KOH) \u003d Ar (K) + Ar (O) + Ar (H) \u003d 39 + 16 + 1 \u003d 56 g / mol

E(KOH) = = =56 g/mol

Metal equivalent E(Me) = = = 32.69 g/mol-eq

There is no element with an atomic mass of 10.3 in the periodic table, so we continue the selection.

M \u003d E ∙ B \u003d 32.69 ∙ II \u003d 65.38 g / mol.

This is the molar mass of the element zinc (Zn).

Answer: metal - zinc, Zn

Task 3. The metal forms an oxide, in which the mass fraction of the metal is 70%. Determine what metal is included in the composition of the oxide.

Let's take the mass of the oxide equal to 100 g, then the mass of the metal will be equal to 70 g (i.e. 70% of 100 g), and the mass of oxygen will be equal to:

m (O) \u003d m (oxide) -m (Me) \u003d 100 - 70 \u003d 30 g

Let's use the law of equivalents:

, where E(O) = 8 g.

E(Me) = = = 18.67 g/mol-eq

M (Me) \u003d E ∙ B \u003d 18.69 ∙ I \u003d 18.69 g / mol

M \u003d E ∙ B \u003d 18.69 ∙ II \u003d 37.34 g / mol.There is no element with such a molar mass in the periodic table, so we continue the selection.

M \u003d E ∙ B \u003d 18.69 ∙ III \u003d 56 g / mol.

This is the molar mass of the element Iron (Fe).

Answer: metal - Iron (Fe).

Task 4. The dibasic acid contains 2.04% hydrogen, 32.65% sulfur and 65.31% oxygen. Determine the valency of sulfur in this acid.

Let's take the mass of acid equal to 100 g, then the mass of hydrogen will be equal to 2.04 g (i.e. 2.04% of 100 g), the mass of sulfur will be 32.65 g, the mass of oxygen will be 65.31 g.

We find the oxygen equivalent of sulfur using the law of equivalents:

, where E(O) = 8 g.

E (S) = = = 4 g/mol-eq

The valency of sulfur in the event that all oxygen atoms are attached to sulfur will be equal to:

B \u003d \u003d \u003d 8, therefore, oxygen atoms form eight chemical bonds in this acid. The acid is dibasic, which means that two bonds formed by oxygen atoms fall on a compound with two hydrogen atoms. Thus, out of eight oxygen bonds per compound with sulfur, six bonds are used, i.e. the valency of sulfur in this acid is VI. One oxygen atom forms two bonds (valencies), so the number of oxygen atoms in an acid can be calculated as follows:

n(O) = = 4.

Accordingly, the acid formula will be H 2 SO 4.

The valency of sulfur in acid is VI, the formula of the acid is H 2 SO 4 (sulfuric acid).

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