Iron(III) Oxide Composition and Molar Mass

DEFINITION

Iron is the twenty-sixth element of the Periodic Table. Designation - Fe from the Latin "ferrum". Located in the fourth period, VIIIB group. Refers to metals. The nuclear charge is 26.

Iron is the most common metal after aluminum the globe: it is 4% (wt.) earth's crust. Iron occurs in the form of various compounds: oxides, sulfides, silicates. AT free state iron is found only in meteorites.

The most important ores of iron include magnetic iron ore Fe 3 O 4 , red iron ore Fe 2 O 3 , brown iron ore 2Fe 2 O 3 ×3H 2 O and spar iron ore FeCO 3 .

Iron is a silvery (Fig. 1) ductile metal. It lends itself well to forging, rolling and other types of machining. The mechanical properties of iron strongly depend on its purity - on the content of even very small amounts of other elements in it.

Rice. 1. Iron. Appearance.

Atomic and molecular weight of iron

Relative molecular weight of a substance(M r) is a number showing how many times the mass of a given molecule is greater than 1/12 of the mass of a carbon atom, and relative atomic mass of an element(A r) - how many times the average mass of atoms of a chemical element is greater than 1/12 of the mass of a carbon atom.

Since in the free state iron exists in the form of monatomic Fe molecules, the values of its atomic and molecular masses are the same. They are equal to 55.847.

Allotropy and allotropic modifications of iron

Iron forms two crystalline modifications: α-iron and γ-iron. The first of them has a cubic body-centered lattice, the second - a cubic face-centered one. α-Iron is thermodynamically stable in two temperature ranges: below 912 o C and from 1394 o C to the melting point. The melting point of iron is 1539 ± 5 o C. Between 912 o C and 1394 o C, γ-iron is stable.

The temperature ranges of stability of α- and γ-iron are due to the nature of the change in the Gibbs energy of both modifications with a change in temperature. At temperatures below 912 o C and above 1394 o C, the Gibbs energy of α-iron is less than the Gibbs energy of γ-iron, and in the range 912 - 1394 o C - more.

Isotopes of iron

It is known that iron can occur in nature in the form of four stable isotopes 54Fe, 56Fe, 57Fe, and 57Fe. Their mass numbers are 54, 56, 57 and 58, respectively. The nucleus of an atom of the iron isotope 54 Fe contains twenty-six protons and twenty-eight neutrons, and the remaining isotopes differ from it only in the number of neutrons.

There are artificial isotopes of iron with mass numbers from 45 to 72, as well as 6 isomeric states of nuclei. The most long-lived among the above isotopes is 60 Fe with a half-life of 2.6 million years.

iron ions

The electronic formula showing the distribution of iron electrons over the orbits is as follows:

1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 .

As a result of chemical interaction, iron gives up its valence electrons, i.e. is their donor, and turns into a positively charged ion:

Fe 0 -2e → Fe 2+;

Fe 0 -3e → Fe 3+.

Molecule and atom of iron

In the free state, iron exists in the form of monatomic Fe molecules. Here are some properties that characterize the atom and molecule of iron:

iron alloys

Until the 19th century, iron alloys were mainly known for their alloys with carbon, which received the names of steel and cast iron. However, in the future, new iron-based alloys containing chromium, nickel and other elements were created. At present, iron alloys are divided into carbon steels, cast irons, alloy steels and steels with special properties.

In technology, iron alloys are usually called ferrous metals, and their production is called ferrous metallurgy.

Examples of problem solving

EXAMPLE 1

Exercise

The elemental composition of the substance is as follows: the mass fraction of the iron element is 0.7241 (or 72.41%), the mass fraction of oxygen is 0.2759 (or 27.59%). Derive the chemical formula.

Solution

The mass fraction of the element X in the molecule of the HX composition is calculated by the following formula:

ω (X) = n × Ar (X) / M (HX) × 100%.

Let us denote the number of iron atoms in the molecule as "x", the number of oxygen atoms as "y".

Find the corresponding relative atomic masses elements of iron and oxygen (the values of relative atomic masses taken from the Periodic Table of D.I. Mendeleev are rounded to integers).

Ar(Fe) = 56; Ar(O) = 16.

We divide the percentage of elements by the corresponding relative atomic masses. Thus, we will find the relationship between the number of atoms in the molecule of the compound:

x:y= ω(Fe)/Ar(Fe) : ω(O)/Ar(O);

x:y = 72.41/56: 27.59/16;

x:y = 1.29: 1.84.

We take the smallest number as one (i.e., we divide all numbers by smallest number 1,29):

1,29/1,29: 1,84/1,29;

Therefore, the simplest formula for the combination of iron with oxygen is Fe 2 O 3.

Answer

Fe2O3

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Chemical formula

Molar mass of Mn 2 O 7 , manganese(VII) oxide 221.871898 g/mol

54.938049 2+15.9994 7

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
Point on middle line(multiplication sign), used, for example, in the formulas of crystalline hydrates, is replaced by a regular dot.
Example: instead of CuSO₄ 5H₂O in the converter, the spelling is used for ease of entry CuSO4.5H2O.

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, a mole is the amount of a substance that contains as many structural elements (atoms, molecules, ions, electrons and other particles or their groups) as there are atoms in 12 grams of a carbon isotope with a relative atomic mass of 12. This number is called a constant or number Avogadro is equal to 6.02214129(27)×10²³ mol⁻¹.

Avogadro's number N A = 6.02214129(27)×10²³ mol⁻¹

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 inflected, unlike the name of the unit, which can be declined according to the usual rules of the Russian language. By definition, one mole of pure carbon-12 is exactly 12 grams.

Molar mass

Molar mass - physical property 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 (solution acetic acid) 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 2 × 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;

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Calculations by chemical formulas

Required Skills

1. Calculation of the relative molecular weight of a substance (Mr)

Exercise: Calculate the relative molecular weight of a molecule of sulfuric acid (H2SO4)

Sequencing	Taking Actions
1. Write down the molecular formula of sulfuric acid.
2. Write down the formula for finding the relative molecular weight of a substance	Mr (in-va) \u003d Ar (el.1) n1 + Ar (el.2) n2 + Ar (el.3) n3
	Mr(H2SO4) = Ar(H) 2 + Ar(S) 1 + Ar(O) 4 = 1 2 + 32 + 16∙4=98
4. Write down the answer.	Answer: Mr(H2SO4) = 98.

2. Calculation of mass ratios of elements in a complex substance

Exercise: Find mass ratios of elements in sulfur oxide (IV ) SO2.

3. Calculation of mass fractions of elements in a complex substance

Exercise: Determine the mass fractions of elements in iron(III) oxide Fe2O3.

Sequencing	Taking Actions
1. Write down a formula for calculating the mass fractions of elements in a complex substance	w(element) = Ar(element)∙n:Mr(in-va)
2. Determine the relative molecular weight of iron(III) oxide	Mr (Fe2O3) = 56∙2+16∙3 = 160
3. Calculate the mass fractions of iron and oxygen by substituting the values of the atomic masses of the elements, their indices and the relative molecular weight of the substance	w(Fe) = 56∙2:160 = 0.7(70%) w(O) \u003d 16 ∙ 3: 160 \u003d 0.3 (30%)

Additional skills

4. Determination of the simplest formula of a substance by mass fractions of elements and relative molecular weight of a substance

Exercise: Determine the formula of a substance that contains 40% sulfur and 60% oxygen. The relative molecular weight of the substance is 80.

5. Finding the simplest formula of a compound by mass fractions of elements

Exercise : What is the simplest formula of a substance in which the mass fractions of sulfur, iron and oxygen are 24, 28 and 48%, respectively.

Sequencing	Taking Actions
1. We write down the formula for determining the indices of elements by mass fractions	n1:n2:n3= w(el.1)/Ar(el.1): w(el.2)/Ar(el.2): w(el.3)/Ar(el.3)
2. Substitute in the formula the value of mass fractions and relative atomic masses of sulfur, iron and oxygen	n(S): n(Fe): n(O) = 24/32: 28/56: 48/16 =
3. We bring the obtained indices of elements to integers by multiplying by "4"	n(S): n(Fe): n(O) = 3:2:12
	S3Fe2O12 or Fe2(SO4)3

6. Derivation of the simplest formula of a compound by the ratio of the masses of elements in a complex substance

Exercise : Magnesium combines with nitrogen to form magnesium nitride, in a mass ratio of 18:7. Derive the compound formula .

Sequencing	Taking Actions
1. We write down the formula for determining the indices of elements by mass ratios	n1: n2 \u003d m (el. 1) / Ar (el. 1): m (el. 2) / Ar (el. 2)
2. Substitute in the formula the value of mass ratios and relative atomic masses of magnesium and nitrogen	n(Mg): n(N) = 18/24: 7/14 = 0.75:0.5
3. We bring the obtained indices of elements to integers by multiplying by "4"	n(Mg): n(N) = (0.75:0.5) 4 = 3:2
4. We write down the simplest formula of a substance

7. Derivation of the compound formula based on its combustion products

Exercise: When burning a hydrocarbon weighing 8.316 g, 26.4 g of CO2 was formed. The density of the substance under normal conditions is 1.875 g / ml. Find its molecular formula.

Sequencing	Taking Actions
1. Find the molar mass of a hydrocarbon based on its density	M \u003d 1.875 g / ml 22. 4 l / mol \u003d 42 g / mol
2. Determine the mass fraction of carbon in carbon monoxide and its mass	w(C) = 12g/mol/44g/mol = 0.27 m(C) = m(CO2) w(C) \u003d 26.4 g 0.27 \u003d 7.128 g
	m (H) \u003d 8.316 g-7.128 g \u003d 1.188 g
4. Define the simplest formula of a substance	n (C): n (H) \u003d 7.128 g / 12 g / mol: 1.188 g / 1 g / mol \u003d 0.594: 1.188 \u003d 1: 2, i.e., the simplest formula of the substance CH2
5. We determine the molar mass of the simplest substance and compare it with the molar mass of hydrocarbon, calculated on the basis of its density	M(CH2) = 14 g/mol x = 42g/mol: 14g/mol =3
6. We triple the indices of elements in the simplest formula of a substance, since its molar mass is 3 times less than the calculated molar mass of a hydrocarbon	Molecular formula of hydrocarbon:

Tasks for self-control.

1. Calculate mass ratios and mass fractions of elements using chemical formulas :

a) sulfur oxide SO2

b) ethane С2Н6

c) copper sulfate CuSO4

Sample execution see paragraphs 2 and 3

2. Determine the empirical formula for the compound of aluminum with carbon, in which the mass fraction of aluminum is 75%.

Sample execution see item 5

3. Determine the formula of a substance consisting of 70.9% potassium and 29.1% oxygen. The relative molecular weight of the substance is 110.

Sample execution see item 4

4. Determine the simplest formula of oxide, knowing that 3.2 g of oxide contains 2.24 g of iron.

Sample execution see item 6

Homework:

Calculate the relative molecular masses of substances;

Mass ratios of elements in a complex substance;

Mass fractions of elements in a complex substance.

2. Tasks 2, 4 p.31, 5.9 p. 32

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