Measurement Of Matter

A. Positive radicals

Positively charged ions are called cations.

Examples:- NH₄⁺ ,K ⁺ ,Fe ^{+ 2},Al ^{+ 3}

B. Basic radicals

The cationic radicals are called basic radicals. Basic radicals are formed by removal of electrons from the atoms of metals.

Example: Na ⁺ , Cu^{2 +} ,Ag ⁺ ,Mg ^{+ 2},Au ^{+ 3}

C. Composite radicals

When a radical is a group of atoms carrying a charge, it is called composite radical.

Examples:-

NO₃^-, ClO₃^- , SO₄^2-, NH₄⁺ , MnO₄^-

D. Metals with variable valency

Under different conditions, the atoms of some elements give away or take up different numbers of electrons. In such cases, those elements exhibit more than one valency. This property of elements is called variable valency.

Examples:-

Iron (Fe) exhibits the variable valencies 2 and 3: Fe^{2 +} and Fe^{3 +}

Gold(Au) exhibits the variable valencies 1 and 3: Au ⁺ and Au^{3 +}

Copper(Cu) exhibits the variable valencies 1 and 2: Cu ⁺ and Cu^{2 +}

Mercury(Hg) exhibits the variable valencies 1 and 2: Hg ⁺ and Hg^{2 +}

E. Bivalent acidic radicals

Bivalent refers to an element whose atom can replace 2 atoms of hydrogen or another univalent element. It also refers to a radical that has the same valence as a bivalent atom.

Examples: SO₄^2-, CO₃^2-, C₂O₄^2-

F. Trivalent basic radicals

A trivalent basic radical is a cation that has a valency of three. Examples of trivalence basic radicals are Al^{3 +} ,Cr^{3 +} , Bi^{3 +} and Fe ^{+ 2}

Sodium Sulphate-

Formula: Na₂SO₄

Steps to obtain the chemical formula:

Step 1: To write the symbols of the radicals. (Basic radical on the left.)

Na SO₄

Step 2: To write the valency below the respective radical. Na SO₄

1 2

Step 3: To cross-multiply as shown by the arrows the number of the radicals.

Step 4: To write down the chemical formula of the compound.

Na₂SO₄

Potassium Nitrate-

Formula: KNO₃

Step 1: To write the symbols of the radicals. (Basic radical on the left.)

K NO₃

Step 2: To write the valency below the respective radical. K NO₃

1 1

Step 3: To cross-multiply as shown by the arrows the number of the radicals.

Step 4: To write down the chemical formula of the compound.

KNO₃

Ferric Phosphate-

Formula:- FePO₄

Step 1: To write the symbols of the radicals.(Basic radical on the left.)

Fe PO₄

Step 2: To write the valency below the respective radical. Fe PO₄

3 3

Step 3: To cross-multiply as shown by the arrows the number of the radicals.

Fe PO₄

3 3

Step4: To write down the chemical formula of the compound.

FePO₄

Calcium Oxide-

Formula:- CaO

Step 1 :

To write the symbols of the radicals. (Basic radical on the left.)

Ca O

Step 2: To write the valency below the respective radical.

Ca O

+ 2 -2

Step 3: To cross-multiply as shown by the arrows the number of the radicals.

Ca O

+ 2 -2

Step4: To write down the chemical formula of the compound in the simplest form.

CaO

Aluminum Hydroxide-

Formula: -Al(OH)₃

Step 1:

To write the symbols of the radicals. (Basic radical on the left.)

Al OH

Step 2: To write the valency below the respective radical.

Al OH

3 1

Step 3: To cross-multiply as shown by the arrows the number of the radicals.

Step 4 To write down the chemical formula of the compound. Al(OH)₃

A. The atomic number of Sodium is 11. Sodium has 1 electron in its outermost shell and to acquire noble gas configuration it will lose one electron or it needs to gain 7 electrons. However, the former is possible. Hence it will loose 1 electron and so the element sodium is monovalent.

M is bivalent which means it has a valency of 2.

(i)For sulphate

Step 1: To write the symbols of the radicals. (Basic radical on the left.)

M SO₄

Step 2: To write the valency below the respective radical.

M SO₄

+ 2 -2

Step 3: To cross-multiply as shown by the arrows the number of the radicals.

Step4: To write down the chemical formula of the compound in the simplest form.

MSO₄

(ii) for phosphate

Step 1:To write the symbols of the radicals. (Basic radical on the left.)

M PO₄

Step 2: To write the valency below the respective radical.

M PO₄

+ 2 -3

Step 3: To cross-multiply as shown by the arrows the number of the radicals.

Step4: To write down the chemical formula of the compound in the simplest form.

M₃(PO₄)₂

It was not possible for scientists of the 19th century to measure atomic mass accurately. Therefore, the concept of ‘relative mass of an atom’ was put forth. A reference atom was required for expressing the relative mass of an atom. The hydrogen atom is the lightest was initially chosen as the reference atom. The relative mass of a hydrogen atom which has only one proton in its nucleus was accepted as one (1).

Therefore, the magnitude of the relative atomic masses of various atoms became equal to their atomic mass number (p + n).

In the course of time, different atoms were chosen as reference

atoms. Finally in 1961, the carbon atom was selected as the reference atom. In this scale, the relative mass a carbon atom was accepted as 12. The relative atomic mass of one hydrogen

atom compared to the carbon atom becomes 12 x 1/12, that is 1. The mass of one proton and of one neutron on the scale of relative atomic masses is approximately one.

The unified atomic mass unit or Dalton is a standard unit of mass that quantifies mass on an atomic or molecular scale. One unified atomic mass unit is approximately the mass of one nucleon and is numerically equivalent to 1 g/mol.

1 mole of a compound is the mass of that substance in grams equal in magnitude to its molecular mass. The SI unit is mol.

Number of moles of a substance (n) =

(Mass of substance in grams)/(Molecular mass of substance)

the molecular mass of oxygen is 32u, and therefore 32g oxygen is 1mole of oxygen. The molecular mass of water is 18u. Therefore, 18g of water make 1 mole of water.

i. Na₂SO₄ : Sodium Sulphate

Steps for finding the molecular mass:

Molecular mass of Na₂SO₄ = 2(Atomic mass of Na) + 1(Atomic mass of S) + 4(Atomic mass of O)

⇒ Molecular Mass = 2(23) + 1(32) + 4(16)

= 46 + 32 + 64

= 142 g/mol

ii.K₂CO₃: Potassium Carbonate

Steps for finding the molecular mass:

Molecular mass of K₂CO₃ = 2× (Atomic mass of K) + 1× (Atomic mass of C) + 3× (Atomic mass of O)

Molecular Mass = 2× (39) + 1× (12) + 3× (16)

= 78 + 12 + 48

= 138 g/mol

iii.CO₂:- Carbondioxide

Molecular mass of CO₂ = 1× (Atomic mass of C) + 2× (Atomic mass of O)

Molecular Mass = 1× (12) + 2× (16)

= 12 + 32

= 44 g/mol

iv.NaOH:Sodium Hydroxide

Molecular mass of NaOH = 1× (Atomic mass of Na) + 1× (Atomic mass of O) + 1× (Atomic mass of H)

Molecular Mass = 1× (23) + 1× (16) + 1× (1)

= 23 + 16 + 1

= 40 g/mol

v.MgCl₂:Magnesium Chloride

Molecular mass of MgCl₂ = 1× (Atomic mass of Mg) + 2× (Atomic mass of Cl)

Molecular Mass = 1× (24) + 2× (35.5)

= 24 + 70

= 94 g/mol

vi.AlPO₄:-Aluminium Phosphate

Molecular mass of AlPO4 = 1× (Atomic mass of Al) + 1× (Atomic mass of P) + 4× (Atomic mass of O)

Molecular Mass = 1× (27) + 1× (31) + 4× (16)

= 27 + 31 + 64

= 122 g/mol

vii.NaHCO₃: Sodium Bicarbonate

Molecular mass of NaHCO₃ = 1× (Atomic mass of Na) + 1× (Atomic mass of H) + 1× (Atomic mass of C) + 3× (Atomic mass of O)

Molecular Mass = 1× (23) + 1× (1) + 1× (12) + 3× (16)

= 23 + 1 + 12 + 48

= 84 g/mol

Slaked lime = calcium hydroxide {Ca(OH)₂}

The law of constant proportions states that “The proportion by weight of the constituent elements in the various samples of a compound is fixed,”

The proportion by weight of oxygen and calcium in slaked lime(in sample m) is 2:5 i.e. 0.4:1

The proportion by weight of oxygen and calcium in slaked lime(in sample n) is 0.4:1.0

Since the proportions of constituent elements remain fixed it follows the law of constant proportion.

(a) 32g of oxygen

Moles = Mass of subtance in grams/Molecular mass

Molecular mass of oxygen = 16

Mass of a substance = 32g

Moles(n) = 32/16 = 2

n = 2 mol

2 mol of oxygen contains 6.022 x 10²³ molecules.

2 mol oxygen contains 2x6.022x10²³ molecules = 12.046 x 10²³ molecules.

(b) 90g water

Moles = Mass of subtance in grams/Molecular mass

Molecular mass of water = 18

Mass of a substance = 90g

Moles(n) = 90/18 = 5

n = 5 mol

5 mol of oxygen contains 6.022 x 10²³ molecules.

5 mol oxygen contains 5x6.022x10²³ molecules = 30.115× 10²³ molecules.

(c) 8.8g carbon dioxide

Moles = Mass of subtance in grams/Molecular mass

Molecular mass of CO₂ = 44

Mass of a substance = 8.8g

Moles(n) = 8.8/44 = 0.2

n = 0.2 mol

0.2 mol of oxygen contains 6.022 x 10²³ molecules.

0.2 mol oxygen contains 0.2x6.022x10²³ molecules = 1.2046 x 10²³ molecules.

(d) 7.1g chlorine.

Moles = Mass of subtance in grams/Molecular mass

Molecular mass of chlorine = 35.5

Mass of a substance = 7.1g

Moles(n) = 7.1/35.5 = 0.2

n = 0.2 mol

0.2 mol of oxygen contains 6.022 x 10²³ molecules.

0.2 mol oxygen contains 0.2x6.022x10²³ molecules = 1.2046 x 10²³ molecules.

(a) Sodium Chloride

Moles = Mass of subtance in grams /Molecular Mass(in g/mol)

Molecular mass of NaCl = 1× (Atomic mass of Na) + 1× (Atomics mass of Cl)

= 1× (23) + 1× (35.5)

= 55.5

Mole = 0.2 mol(given)

0.2 = given mass/55.5

Given mass = 0.2× 55.5

= 11.1g

(b) Magnesium Oxide

Moles = Mass of subtance in grams /Molecular Mass(in g/mol)

Molecular mass of MgO = 1× (Atomic mass of Mg) + 1× (Atomics mass of O)

= 1× (24) + 1× (16)

= 40

Mole = 0.2 mol(given)

0.2 = given mass/40

Given mass = 0.2× 40

= 8g

(c) Calcium Carbonate

Moles = Mass of subtance in grams /Molecular Mass(in g/mol)

Molecular mass of CaCO₃ = 1× (Atomic mass of Ca) + 1× (Atomics mass of C) + 3× (atomic mass of O)

= 1× (40) + 1× (12) + 3× (16)

= 100

Mole = 0.2 mol(given)

0.2 = given mass/100

Given mass = 0.2× 100

= 20g