Atoms and Molecules

Table of Contents

In the fascinating world of chemistry, the study of atoms and molecules forms the foundation of our understanding of matter and its transformations. Several fundamental laws and theories govern the behavior of these tiny particles, unraveling the mysteries of the composition and behavior of elements and compounds.

Law of Conservation of Mass:

The law of conservation of mass states that in any chemical reaction, the total mass of the reactants remains the same as the total mass of the products. This principle implies that mass cannot be created or destroyed during a chemical reaction, but it can change its form.

Law of Constant Proportion:

The law of constant proportion, proposed by Joseph Proust, states that a chemical compound always contains the same elements combined in the same proportion by mass. This law highlights the fixed composition of chemical compounds, regardless of their source or origin.

Law of Multiple Proportions:

John Dalton’s law of multiple proportions describes how elements can combine in multiple ratios to form different compounds. The masses of one element that combine with a fixed mass of another element bear simple whole-number ratios. This observation provides valuable insights into the atomic nature of matter.

Dalton’s Atomic Theory:

Dalton’s atomic theory revolutionized our understanding of matter by proposing that all matter, whether an element, compound, or mixture, is composed of indivisible particles called atoms. According to this theory, atoms are indestructible, retain their identity during chemical reactions, and combine in specific ratios to form compounds.

The Atom:

An atom is the smallest unit of an element that retains its chemical properties. It comprises three subatomic particles: protons and neutrons in the nucleus and electrons in orbits surrounding the nucleus. The atomic symbol of an element contains information about its protons and neutrons (mass number) and its electrons (atomic number).

Valency:

Valency refers to the ability of an atom to form chemical bonds by gaining, losing, or sharing valence electrons to achieve a stable electron configuration. Valence electrons are the electrons in the outermost energy level of an atom, and their interactions determine the chemical properties of elements.

Symbols for some elements

Element	Symbol
Aluminium	Al
Copper	Cu
Nitrogen	N
Argon	Ar
Fluorine	F
Oxygen	O
Barium	Ba
Gold	Au
Potassium	K
Boron	B
Hydrogen	H
Silicon	Si
Bromine	Br
Iodine	I
Silver	Ag
Calcium	Ca
Iron	Fe
Sodium	Na
Carbon	C
Lead	Pb
Sulphur	S
Chlorine	Cl
Magnesium	Mg
Uranium	U
Neon	Ne
Zinc	Zn

Molecule and Compound:

A molecule is a group of atoms held together by chemical bonds. When atoms of different elements combine in fixed ratios, they form compounds. Compounds have unique properties distinct from their constituent elements, and their formation involves the rearrangement of atoms.

Ions:

An ion is an atom or molecule that has an unequal number of protons and electrons, resulting in a net positive or negative charge. Ions play a crucial role in chemical reactions, as they are involved in the transfer of electrons between atoms.

Element	Atomic Mass (u)
Hydrogen	1
Carbon	12
Nitrogen	14
Oxygen	16
Sodium	23
Magnesium	24
Sulphur	32
Chlorine	35.5
Calcium	40

FAQ

A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

To calculate the percentage composition of the compound by weight, we need to determine the percentage of boron and oxygen in the compound based on their masses.

Given: Mass of the compound = 0.24 g Mass of boron = 0.096 g Mass of oxygen = 0.144 g

Percentage of boron = (Mass of boron / Mass of the compound) × 100 Percentage of boron = (0.096 g / 0.24 g) × 100 Percentage of boron = 40%

Percentage of oxygen = (Mass of oxygen / Mass of the compound) × 100 Percentage of oxygen = (0.144 g / 0.24 g) × 100 Percentage of oxygen = 60%

So, the percentage composition of the compound by weight is 40% boron and 60% oxygen.

When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.

Given

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

Find

We need to find out the mass of carbon dioxide that will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen.

Solution

First, let us write the reaction taking place here.

C + O2 → CO2

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

The total mass of reactants = mass of carbon + mass of oxygen

=3g+8g

=11g

The total mass of reactants = Total mass of products

Therefore, the law of conservation of mass is proved.

Then, it also depicts that carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

Thus, it further proves the law of constant proportions.

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

Remaining 42 g of oxygen will be left un-reactive. In this case, too, only 11 g of carbon dioxide will be formed

Answer is governed by the law of constant proportions.

What are polyatomic ions? Give examples.

Polyatomic ions are groups of two or more atoms that are covalently bonded together and carry a net electrical charge. These ions behave as a single unit in chemical reactions and are considered as one entity with a specific charge.

Here are some examples of polyatomic ions:

Ammonium ion (NH4+): This ion consists of one nitrogen atom and four hydrogen atoms, bonded together with a positive charge.
Hydroxide ion (OH-): This ion contains one oxygen atom and one hydrogen atom, bonded together with a negative charge.
Carbonate ion (CO3^2-): It is composed of one carbon atom and three oxygen atoms, bonded together with a double negative charge.

Write the chemical formula of the following. (a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate

(a) Magnesium chloride: MgCl2

(b) Calcium oxide: CaO

(d) Aluminium chloride: AlCl3

(e) Calcium carbonate: CaCO3

Give the names of the elements present in the following compounds. (a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate

The following are the names of the elements present in the following compounds:

(a) Quick lime – Calcium and oxygen (CaO)

(b) Hydrogen bromide – Hydrogen and bromine (HBr)

(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K₂SO₄)

Calculate the molar mass of the following substances. (a) Ethyne, C2H2 (b) Sulphur molecule, S8 (c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31) (d) Hydrochloric acid, HCl (e) Nitric acid, HNO3

To calculate the molar mass of the substances, we need to sum up the atomic masses of all the atoms present in their chemical formulas.

(a) Ethyne, C2H2: Molar mass of ethyne = (2 * Atomic mass of carbon) + (2 * Atomic mass of hydrogen) Molar mass of ethyne = (2 * 12.01 g/mol) + (2 * 1.01 g/mol) = 24.02 g/mol + 2.02 g/mol = 26.04 g/mol

(b) Sulphur molecule, S8: Molar mass of sulfur molecule = (8 * Atomic mass of sulfur) Molar mass of sulfur molecule = 8 * 32.06 g/mol = 256.48 g/mol

(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31): Molar mass of phosphorus molecule = (4 * Atomic mass of phosphorus) Molar mass of phosphorus molecule = 4 * 31.00 g/mol = 124.00 g/mol

(d) Hydrochloric acid, HCl: Molar mass of hydrochloric acid = Atomic mass of hydrogen + Atomic mass of chlorine Molar mass of hydrochloric acid = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol

(e) Nitric acid, HNO3: Molar mass of nitric acid = Atomic mass of hydrogen + Atomic mass of nitrogen + (3 * Atomic mass of oxygen) Molar mass of nitric acid = 1.01 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 1.01 g/mol + 14.01 g/mol + 48.00 g/mol = 63.02 g/mol

So, the molar masses are: (a) Ethyne, C2H2: 26.04 g/mol (b) Sulphur molecule, S8: 256.48 g/mol (c) Phosphorus molecule, P4: 124.00 g/mol (d) Hydrochloric acid, HCl: 36.46 g/mol (e) Nitric acid, HNO3: 63.02 g/mol

What is the mass of (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (Atomic mass of aluminium =27)? (c) 10 moles of sodium sulphite (Na2SO3)?

(a) Mass of 1 mole of nitrogen atoms = 14 g
(b) 4 moles of aluminium atoms
Mass of 1 mole of aluminium atoms = 27 g
∴ Mass of 4 moles of aluminium atoms = 27 x 4 = 108 g
(c) 10 moles of sodium sulphite (Na₂SO₃)
Mass of 1 mole of Na₂SO₃ = 2 x 23 + 32 + 3 x 16 = 46 + 32 + 48 = 126 g
∴ Mass of 10 moles of Na₂SO₃ = 126 x 10 = 1260 g

Convert into a mole. (a) 12g of oxygen gas (b) 20g of water (c) 22g of carbon dioxide

(a) Given mass of oxygen gas = 12 g
Molar mass of oxygen gas (O₂) = 32 g
Mole of oxygen gas 12/32 = 0.375 mole
(b) Given mass of water = 20 g
Molar mass of water (H₂O) = (2 x 1) + 16 = 18 g
Mole of water = 20/18 = 1.12 mole
(c) Given mass of Carbon dioxide = 22 g
Molar mass of carbon dioxide (CO₂) = (1 x 12) + (2 x 16)
= 12 + 32 = 44 g
∴ Mole of carbon dioxide = 22/44 = 0.5 mole

What is the mass of: (a) 0.2 mole of oxygen atoms? (b) 0.5 mole of water molecules?

(a) Mole of Oxygen atoms = 0.2 mole
Molar mass of oxygen atoms = 16 g
Mass of oxygen atoms = 16 x 0.2 = 3.2 g
(b) Mole of water molecule = 0.5 mole
Molar mass of water molecules = 2 x 1 + 16= 18 g .
Mass of H₂O = 18 x 0.5 = 9 g

Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.

To calculate the molecular mass of sulphur,

Molecular mass of Sulphur (S₈) = 8xMass of Sulphur = 8×32 = 256g

Mass given = 16g

Number of moles = mass given/ molar mass of sulphur

= 16/256 = 0.0625 moles

To calculate the number of molecules of sulphur in 16g of solid sulphur,

Number of molecules = Number of moles x Avogadro number

= 0.0625 x 6.022 x 10²³ molecules

= 3.763 x 10²² molecules

Calculate the number of aluminium ions present in 0.051g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)

To calculate the number of aluminium ions in 0.051g of aluminium oxide,

1 mole of aluminium oxide = 6.022 x 10²³ molecules of aluminium oxide

1 mole of aluminium oxide (Al₂O₃) = 2 x Mass of aluminium + 3 x Mass of oxygen

= (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of aluminium oxide = 102g = 6.022 x 10²³ molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has = 6.022 x 10²³ / 102 x 0.051

= 3.011 x 10²⁰ molecules of aluminium oxide

One molecule of aluminium oxide has 2 aluminium ions; hence, the number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 10²⁰ molecules of aluminium oxide.

= 6.022 x 10²⁰

Atoms and Molecules | Class 9 Science | Chapter 3