The molar mass of cao. Calcium oxide: composition and molar mass. Molar mass calculation

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Chemical formula

Molar Mass of CaO, Calcium Oxide 56.0774 g/mol

Mass fractions of elements in the compound

Using the Molar Mass Calculator

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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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H 2 S + 2NaOH \u003d Na 2 S + 2H 2 O; (1)

H 2 S + NaOH = NaHS + H 2 O. (2)

Solution acids or grounds participating in acid-base reactions, calculated by the formula

M eq (acids, bases) = ,

Where M is the molar mass of the acid or base; n- For acids is the number of hydrogen atoms substituted in this reaction for the metal; For grounds is the number of hydroxyl groups substituted in this reaction for an acid residue.

The value of the equivalent and the molar mass of the equivalents of a substance depends on the reaction in which this substance participates.

In the reaction H 2 S + 2NaOH \u003d Na 2 S + 2H 2 O (1), both hydrogen ions of the H 2 S molecule are replaced by a metal and, thus, the conditional particle ½ H 2 S is equivalent to one hydrogen ion. In this case

E(H 2 S) \u003d ½ H 2 S, and M ec (H 2 S) \u003d \u003d 17 g / mol.

In the reaction H 2 S + NaOH = NaHS + H 2 O (2) in the H 2 S molecule, only one hydrogen ion is replaced by a metal and, therefore, a real particle, the H 2 S molecule, is equivalent to one ion. In this case

E(H 2 S) = H 2 S, and M eq (H 2 S) = = 34 g / mol.

The NaOH equivalent in reactions (1) and (2) is equal to NaOH, since in both cases one hydroxyl group is replaced per acid residue. The molar mass of NaOH equivalents is

M eq (NaOH) = 40 g/mol.

Thus, the equivalent of H 2 S in reaction (1) is equal to ½ H 2 S, in reaction (2) -

1 H 2 S, the molar masses of H 2 S equivalents are 17 (1) and 34 (2) g/mol, respectively; the equivalent of NaOH in reactions (1) and (2) is equal to NaOH, the molar mass of base equivalents is 40 g/mol.

Solution. Molar mass equivalents oxide calculated by the formula

M eq (oxide) = ,

Where M is the molar mass of the oxide; n is the number of cations of the base corresponding to the oxide or the number of anions of the acid corresponding to the oxide; |c.o.| is the absolute value of the oxidation state of the cation or anion.

In the reaction P 2 O 5 + 3CaO \u003d Ca 3 (PO 4) 2, the equivalent of P 2 O 5, which forms two triply charged anions (PO 4) 3-, is 1/6 P 2 O 5, and M eq (P 2 O 5) = = 23.7 g / mol. The equivalent of CaO, giving one doubly charged cation (Ca 2+), is ½ CaO, and M eq(CaO)= = 28 g/mol.

Example 2.3. Calculate the equivalent and molar mass equivalents of phosphorus in the compounds РН 3 , Р 2 О 3 and Р 2 О 5 .

Solution. To determine the molar mass of equivalents element in conjunction, you can use the following formula:

M ek (element) = ,

Where M A is the molar mass of the element; |c.o.| is the absolute value of the oxidation state of the element.


The oxidation state of phosphorus in РН 3 , Р 2 О 3 , Р 2 О 5 is –3, +3 and +5, respectively. Substituting these values ​​into the formula, we find that the molar mass of phosphorus equivalents in the compounds PH 3 and P 2 O 3 is 31/3 = 10.3 g / mol; in P 2 O 5 - 31/5 \u003d 6.2 g / mol, and the equivalent of phosphorus in the compounds PH 3 and P 2 O 3 is 1/3 P, in the compound P 2 O 5 - 1/5 P.

Solution. The molar mass of the equivalents of a chemical compound is equal to the sum of the molar masses of the equivalents of its constituent parts:

M eq (PH 3) = M eq (P) + M eq (H) \u003d 10.3 + 1 \u003d 11 g / mol;

M eq (P 2 O 3) \u003d M eq (P) + M eq (O) \u003d 10.3 + 8 \u003d 18.3 g / mol;

M eq (P 2 O 5) \u003d M eq (P) + M eq (O) \u003d 6.2 + 8 \u003d 14.2 g / mol.

Example 2.5. The reduction of 7.09 g of metal oxide with an oxidation state of +2 requires 2.24 liters of hydrogen under normal conditions. Calculate the molar masses of oxide and metal equivalents. What is the molar mass of the metal?

Solution. The problem is solved according to the law of equivalents. Since one of the reactants is in a gaseous state, it is convenient to use the following formula:

Where V eq (gas) - the volume of one mole of gas equivalents. To calculate the volume of mole equivalents of a gas, it is necessary to know the number of moles of equivalents ( υ ) in one mole of gas: υ = . So, M(H 2) \u003d 2 g / mol; M eq (H 2) \u003d 1 g / mol. Therefore, one mole of hydrogen molecules H 2 contains υ = 2/1 = 2 mole equivalents of hydrogen. As is known, a mole of any gas under normal conditions (n.o.) ( T= 273 K, R= 101.325 kPa) occupies a volume of 22.4 liters. This means that a mole of hydrogen will occupy a volume of 22.4 liters, and since one mole of hydrogen contains 2 mole equivalents of hydrogen, the volume of one mole of hydrogen equivalents is equal to V eq (H 2) \u003d 22.4 / 2 \u003d 11.2 l. Similarly M(O 2) \u003d 32 g / mol, M eq (O 2) \u003d 8 g / mol. One mole of oxygen molecules O 2 contains υ = 32/8 = 4 mole equivalents of oxygen. One mole of oxygen equivalents under normal conditions occupies a volume V eq (O 2) \u003d 22.4 / 4 \u003d 5.6 l.

Substituting the numerical values ​​into the formula, we find that M eq(oxide) = g/mol.

The molar mass of the equivalents of a chemical compound is equal to the sum of the molar masses of the equivalents of its constituent parts. An oxide is a compound of a metal with oxygen, so the molar mass of oxide equivalents is the sum M eq(oxide) = M eq (metal) + M eq (oxygen). From here M eq(metal) = M eq (oxide) - M eq (oxygen) \u003d 35.45 - 8 \u003d 27.45 g / mol.

Molar mass equivalents of an element ( M eq) is related to the atomic mass of the element ( M A) ratio: M eq(element) = , where ½ s.o.½ is the oxidation state of the element. From here M A = M eq (metal) ∙ ½ s.o.½ = 27.45 x 2 = 54.9 g/mol.

Thus, M eq (oxide) = 35.45 g/mol; M eq (metal) = 27.45 g/mol; M A (metal) \u003d 54.9 g / mol.

Example 2.6. In the interaction of oxygen with nitrogen, 4 mol equivalents of nitric oxide (IV) were obtained. Calculate the volumes of gases that reacted under normal conditions.

Solution. According to the law of equivalents, the number of moles of equivalents of substances entering into the reaction and formed as a result of the reaction are equal to each other, i.e. υ (O 2) = υ (N 2) = υ (NO 2). Since 4 mol equivalents of nitric oxide (IV) were obtained, therefore, 4 mol equivalents of O 2 and 4 mol equivalents of N 2 entered into the reaction.

Nitrogen changes the oxidation state from 0 (in N 2) to +4 (in NO 2), and since there are 2 atoms in its molecule, together they give 8 electrons, therefore

M eq (N 2) \u003d \u003d 3.5 g / mol . We find the volume occupied by a mole of nitrogen (IV) equivalents: 28 g / mol N 2 - 22.4 l

3.5 g/mol N 2 - X

X= l.

Since 4 mol equivalents of N 2 entered into the reaction, their volume is V(N 2) \u003d 2.8 4 \u003d 11.2 liters. Knowing that a mole of oxygen equivalents under normal conditions occupies a volume of 5.6 liters, we calculate the volume of 4 mole equivalents of O 2 that have reacted: V(O 2) \u003d 5.6 ∙ 4 \u003d 22.4 l.

So, 11.2 liters of nitrogen and 22.4 liters of oxygen entered into the reaction.

Example 2.7. Determine the molar mass of metal equivalents if 88.65 g of its nitrate is obtained from 48.15 g of its oxide.

Solution. Given that M eq(oxide) = M eq (metal) + M eq (oxygen), a M eq(salts) = M eq (metal) + M eq (acid residue), we substitute the relevant data into the law of equivalents:

from here M eq (metal) = 56.2 g/mol.

Example 2.8. Calculate the oxidation state of chromium in an oxide containing 68.42% (mass.) of this metal.

Solution. Taking the weight of the oxide as 100%, we find the mass fraction of oxygen in the oxide: 100 - 68.42 = 31.58%, i.e. 68.42 parts of the mass of chromium account for 31.58 parts of the mass of oxygen, or 68.42 g of chromium account for 31.58 g of oxygen. Knowing that the molar mass of oxygen equivalents is 8 g/mol, we determine the molar mass of chromium equivalents in oxide according to the law of equivalents:

; M eq(Cr) = g/mol.

The oxidation state of chromium is found from the ratio,

from here | c. o.| = = 3.

Calcium oxide is a white crystalline compound. Other names for this substance are quicklime, calcium oxide, "kirabit", "boiling". Calcium oxide, the formula of which is CaO, and its interaction product with (H2O) water - Ca (OH) 2 ("fluff", or slaked lime) are widely used in the construction industry.

How is calcium oxide obtained?

1. The industrial method for obtaining this substance consists in thermal (under the influence of temperature) decomposition of limestone:

CaCO3 (limestone) = CaO (calcium oxide) + CO2 (carbon dioxide)

2. Calcium oxide can also be obtained through the interaction of simple substances:

2Ca (calcium) + O2 (oxygen) = 2CaO (calcium oxide)

3. The third method of calcium is the thermal decomposition of calcium hydroxide (Ca (OH) 2) and calcium salts of several oxygen-containing acids:

2Ca(NO3)2 = 2CaO (product) + 4NO2 + O2 (oxygen)

calcium oxide

1. Appearance: white crystalline compound. It crystallizes as sodium chloride (NaCl) in a cubic crystal face-centered lattice.

2. The molar mass is 55.07 grams/mol.

3. Density is 3.3 grams/centimeter³.

Thermal properties of calcium oxide

1. The melting point is 2570 degrees

2. The boiling point is 2850 degrees

3. Molar heat capacity (under standard conditions) is 42.06 J / (mol K)

4. Enthalpy of formation (under standard conditions) is -635 kJ/mol

Chemical properties of calcium oxide

Calcium oxide (formula CaO) is a basic oxide. Therefore, he can:

Dissolve in water (H2O) with the release of energy. This produces calcium hydroxide. This reaction looks like this:

CaO (calcium oxide) + H2O (water) = Ca(OH)2 (calcium hydroxide) + 63.7 kJ/mol;

React with acids and acid oxides. This forms salts. Here are examples of reactions:

CaO (calcium oxide) + SO2 (sulfur dioxide) = CaSO3 (calcium sulfite)

CaO (calcium oxide) + 2HCl (hydrochloric acid) = CaCl2 (calcium chloride) + H2O (water).

Applications of calcium oxide:

1. The main volumes of the substance we are considering are used in the production of silicate bricks in construction. In the past, quicklime was used as lime cement. It was obtained by mixing it with water (H2O). As a result, calcium oxide turned into hydroxide, which then, absorbing from the atmosphere (CO2), strongly hardened, turning into calcium carbonate (CaCO3). Despite the cheapness of this method, at present lime cement is practically not used in construction, since it has the ability to absorb and accumulate liquid well.

2. As a refractory material, calcium oxide is suitable as an inexpensive and readily available material. Fused calcium oxide is resistant to water (H2O), which allows it to be used as a refractory where the use of expensive materials is impractical.

3. In laboratories, calcium is used to dry those substances that do not react with it.

4. In the food industry, this substance is registered as a food additive under the designation E 529. It is used as an emulsifier to create a homogeneous mixture of immiscible substances - water, oil and fat.

5. In industry, calcium oxide is used to remove sulfur dioxide (SO2) from flue gases. As a rule, a 15% aqueous solution is used. As a result of the reaction, in which sulfur dioxide also interacts, gypsum CaCO4 and CaCO3 are obtained. When conducting experiments, scientists achieved an indicator of 98% of smoke removal from sulfur dioxide.

6. Used in special "self-heating" dishes. A container with a small amount of calcium oxide is located between the two walls of the vessel. When the capsule is pierced in water, a reaction begins with the release of a certain amount of heat.

Calcium oxide, the formula CaO, is often referred to as quicklime. This publication will tell you about the properties, production, and use of this substance.

Definition

Calcium oxide is a white crystalline substance. In some sources, it may be called calcium oxide, quicklime, "boiling" or kirabite. Quicklime is the most popular trivial name for this substance. It is the only and highest calcium oxide.

Properties

Oxide is a crystalline substance having a cubic face-centered crystal lattice.

It melts at a temperature of 2570 o C and boils at 2850 o C. It is a basic oxide, its dissolution in water leads to the formation of calcium hydroxide. The substance may form salts. To do this, it must be added to an acid or acid oxide.

Receipt

It can be obtained by thermal decomposition of limestone. The reaction proceeds as follows: calcium carbonate is gradually heated, and when the temperature of the medium reaches 900-1000 ° C, it decomposes into gaseous tetravalent carbon monoxide and the desired substance. Another way to obtain it is the simplest compound reaction. To do this, a small amount of pure calcium is immersed in liquid oxygen, followed by a reaction, the product of which will be the desired oxide. Also, the latter can be obtained in the process of decomposition of calcium hydroxide or calcium salts of certain oxygen-containing acids at high temperatures. For example, consider the decomposition of the latter. If you take calcium nitrate (the residue is taken from nitric acid) and heat it to 500 ° C, then the reaction products will be oxygen, nitrogen dioxide and the desired calcium oxide.

Application

Basically, this substance is used by the construction industry, where it is used to produce silicate bricks. Previously, calcium oxide was also used in the manufacture of lime cement, but soon the latter was no longer used due to the absorption and accumulation of moisture by this compound. And if it is used for laying the stove, then when heated, suffocating carbon dioxide will soar in the room. Also, the substance now discussed is known for its resistance to water. Because of this property, calcium oxide is used as a cheap and affordable refractory. This compound is indispensable in any laboratory when drying substances that do not react with it. Calcium oxide is known in one industry as food additive E529. Also, a 15% solution of this substance is needed in order to remove sulfur dioxide from some gaseous compounds. With the help of calcium oxide, "self-heating" dishes are also produced. This property is provided by the process of heat release during the reaction of calcium oxide with water.

Conclusion

That's all the basic information about this compound. As mentioned above, it is often referred to as quicklime. Did you know that the concept of lime in chemistry is very flexible? There are also slaked, bleach and soda lime.