CBSE Class 9 Science Atoms and Molecules Exam Notes

CBSE Class 9 Science Atoms and Molecules Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

ATOMS AND MOLECULES

INTRODUCTION

Matter is called ‘padarth’ in Hindi. Kanad was one of the first persons to propose that matter (padarth) is made up of very small particles called ‘parmanu’. John Dalton called these particles atoms. The word ‘atom’ means ‘indivisible’, something which cannot be divided further. The particles of matter (atoms or parmanu) normally exist in a combined form. This combined form of atoms is now called a molecule. All matter is made up of small particles called atoms and molecules. Different kinds of atoms and molecules have different properties, due to which different kinds of matter also show different properties.

LAWS OF CHEMICAL COMBINATION

The laws of chemical combination played a significant role in the development of Dalton’s atomic theory of matter. There are two important laws of chemical combination.

(A) Law of conservation of mass. This law was given by Lavoisier in 1774. According to it, matter is neither created nor destroyed in a chemical reaction. The substances which combine in a chemical reaction are known as reactants, whereas the new substances formed are called products. The law means that in a chemical reaction the total mass of the products is equal to the total mass of the reactants; there is no change in mass during a chemical reaction.

Example. Lavoisier showed that when mercuric oxide is heated, it produces free mercury and oxygen. The sum of the masses of mercury and oxygen was found to be equal to the mass of mercuric oxide.

Solution. Mass of the reactant = 100 g. Mass of the products = 92.6 g + 7.4 g = 100.0 g. Since the total mass of the products is equal to the total mass of the reactants, the data agree with the law of conservation of mass.

(B) Law of constant proportions. This law was discovered by the French chemists A. Lavoisier and Joseph Proust. A pure chemical compound always contains the same elements combined together in the same proportion by mass.

Example. Pure water obtained from different sources such as river, well, spring or sea always contains hydrogen and oxygen combined in the ratio 1 : 8 by mass.

Limiting reagent. In a reaction having more than one reactant, we must identify the limiting reagent. The reagent which is finished first is known as the limiting reagent.

Example. In Haber’s process to manufacture NH₃, N₂ + 3H₂ → 2NH₃. If we take 2 moles of N₂ (56 g) along with 3 moles of H₂ (6 g), only 1 mole of N₂ (28 g) is sufficient to react with 3 moles of H₂. It means 1 mole of N₂ (28 g) is in excess and H₂ will be finished when the reaction is completed, i.e. H₂ is the limiting reagent.

DALTON’S ATOMIC THEORY

On the basis of the laws of chemical combination, John Dalton, an English school teacher in Manchester, proposed in 1808 that the behaviour of matter could be explained using an atomic theory. The main points of Dalton’s atomic theory are:

• All matter is made up of very small particles called atoms.
• Atoms cannot be divided.
• Atoms can neither be created nor destroyed.
• Atoms are of various kinds; there are as many kinds of atoms as there are elements.
• All the atoms of a given element are identical in every respect, having the same mass, size and chemical properties.
• Atoms of different elements differ in mass, size and chemical properties.
• Chemical combination between two (or more) elements consists in the joining together of atoms of these elements to form molecules of compounds.
• The number and kind of atoms in a given compound is fixed.
• During chemical combination, atoms of different elements combine in small whole numbers to form compounds.
• Atoms of the same elements can combine in more than one ratio to form more than one compound.

Drawbacks of Dalton’s atomic theory

• Atoms were thought to be indivisible. We now know that under special circumstances atoms can be divided into still smaller particles called electrons, protons and neutrons.
• Dalton’s theory says that all the atoms of an element have exactly the same mass. It is now known that atoms of the same element can have slightly different masses (isotopes).
• Dalton’s theory said that atoms of different elements have different masses. It is now known that even atoms of different elements can have the same mass (isobars).
• It failed to explain how atoms of different elements differ from each other, i.e. it said nothing about the internal structure of the atom.
• It could not explain how and why atoms of different elements combine to form compound atoms or molecules.
• It failed to explain the nature of the forces that hold different atoms together in a molecule.
• It did not distinguish between the ultimate particle of an element that takes part in reactions (atom) and the ultimate particle that has independent existence (molecule).

ATOMS

All matter is made up of atoms. An atom is the smallest particle of an element that can take part in a chemical reaction. Atoms of most elements are very reactive and do not exist in the free state; they exist in combination with atoms of the same element or of another element.

Atoms are very, very small. The size of an atom is indicated by its radius, called the atomic radius, which is measured in nanometres (nm).

\[1\ \text{nm}=\dfrac1{10^9}\ \text{m}=10^{-9}\ \text{m}\]

The hydrogen atom is the smallest atom of all. Atoms cannot be viewed by simple optical microscopes. However, through modern techniques such as the scanning tunnelling microscope it is possible to produce magnified images of the surfaces of elements showing atoms.

SYMBOLS OF ELEMENTS

To represent the elements, scientists use abbreviated names instead of full lengthy names. These abbreviated names are known as symbols. A symbol may be defined as the abbreviation used for the name of an element.

Dalton’s symbols of elements. Dalton was the first scientist to use symbols to represent elements in a short way. He used drawn signs for hydrogen, carbon, oxygen, phosphorus, sulphur, platinum, iron, copper, silver, gold, lead and mercury. These symbols were difficult to draw and inconvenient to use, so they are only of historical importance and are not used at all.

Modern symbols of elements. IUPAC (International Union of Pure and Applied Chemistry) approves the names of elements. The symbols of elements are generally either the first letter, or the first two letters, or the first and the third letters of the name of the element.

Symbols derived from the first letter of the name:

Symbols derived from the first two letters of the name:

Symbols derived from the first and third letters of the name:

Symbols derived from the Latin names of the elements:

The first letter of every chemical symbol is a capital letter, but if the symbol consists of two letters, the second letter is not a capital. For example, the symbol for aluminium is Al and not AL; the symbol for lead is Pb and not PB.

Significance of the symbol of an element

• A symbol represents the name of the element.
• A symbol represents one atom of the element.
• A symbol also represents one mole of atoms of the element, i.e. \(6.022\times10^{23}\) atoms of the element.
• A symbol represents a definite mass of the element (equal to the atomic mass expressed in grams).

ATOMIC MASS

Atoms are extremely small; the heaviest atoms have masses of about \(10^{-22}\) g. Even an ultramicrobalance cannot measure the mass of a single atom. However, the relative masses of atoms of different elements can be determined. At first, the mass of the lightest atom, hydrogen, was used as the standard. In 1961 the International Union of Chemists selected the most stable isotope of carbon (carbon-12) as the standard for comparing the atomic masses of various elements. The atomic mass of an element tells us the number of times an atom of the element is heavier than \(\dfrac1{12}\) of the mass of an atom of carbon-12.

The atomic mass of an element may be defined as the average relative mass of an atom of the element as compared with the mass of an atom of carbon-12 taken as 12 u (amu).

\[\text{Atomic mass}=\dfrac{\text{mass of 1 atom of the element}}{\frac1{12}\text{ of the mass of an atom of carbon-12}}\]

Example. The atomic mass of magnesium is 24 u, which indicates that one atom of magnesium is 24 times heavier than \(\dfrac1{12}\) of a carbon-12 atom.

Atomic masses of some elements

Gram atomic mass. The gram atomic mass of an element is that quantity of the element whose mass expressed in grams is numerically equal to its atomic mass. To find it, keep the numerical value the same as the atomic mass and change the unit from u to g. For example, the atomic mass of aluminium is 27 u, so its gram atomic mass is 27 g.

Atomic mass of an element having isotopes. If \(M_1\) and \(M_2\) are the relative masses of the isotopes and \(X_1\) and \(X_2\) are their relative percentage contents, then

\[\text{Atomic mass}=\dfrac{M_1X_1+M_2X_2}{X_1+X_2}\]

Example. Chlorine contains two types of atoms having relative masses 35 and 37 and relative abundance 3 : 1. The atomic mass of the element is the average of the relative masses of the isotopes: atomic mass of chlorine \(=\dfrac{35\times3+37\times1}{4}=35.5\).

Atoms usually exist in two ways: (i) molecules, (ii) ions.

MOLECULES

A combination of atoms is called a molecule. The forces which hold the atoms together in a molecule are called covalent bonds. A molecule is the smallest particle of a substance which has the properties of that substance and can exist in the free state. There are two types of molecules.

Molecules of elements. The molecule of an element contains two (or more) similar atoms chemically combined together. Example: a molecule of hydrogen contains 2 hydrogen atoms and is written H₂.

Molecules of compounds. The molecule of a compound contains two (or more) different types of atoms chemically combined together. Example: the molecule of hydrogen chloride (HCl) contains one hydrogen atom and one chlorine atom.

Molecules of some compounds

Atomicity. The number of atoms present in one molecule of an element is called its atomicity.

• Noble gases (helium, neon, argon, krypton, etc.) have one atom in each molecule: He, Ne, Ar, Kr. So the atomicity of noble gases is 1 (monoatomic).
• Hydrogen (H₂), nitrogen (N₂), oxygen (O₂), chlorine (Cl₂), bromine (Br₂) and iodine (I₂) have 2 atoms in each molecule, so their atomicity is 2 (diatomic).

Atomicity of some common elements

IONS

An ion is a positively or negatively charged atom (or group of atoms). An ion is formed by the loss or gain of electrons by an atom, so it contains an unequal number of electrons and protons. Examples: sodium ion Na⁺, magnesium ion Mg²⁺, chloride ion Cl^– and oxide ion O^2–. There are two types of ions: cations and anions.

Cation. A positively charged ion is known as a cation. A cation is formed by the loss of one or more electrons by an atom. The ions of all metal elements are cations. A sodium atom loses 1 electron to form a sodium ion:

\[\text{Na}\;\xrightarrow{-1\ \text{electron}}\;\text{Na}^{+}\]

Anion. A negatively charged ion is known as an anion. An anion is formed by the gain of one or more electrons by an atom. It contains more electrons than protons, whereas a neutral atom contains equal numbers of protons and electrons. The ions of all non-metal elements are anions. A chlorine atom gains 1 electron to form a chloride ion:

\[\text{Cl}\;\xrightarrow{+1\ \text{electron}}\;\text{Cl}^{-}\]

Simple ions are formed from single atoms; e.g. the sodium ion Na⁺ is formed from a single sodium atom. Compound ions are formed from groups of joined atoms; e.g. the ammonium ion NH₄⁺ is made up of nitrogen and hydrogen atoms joined together.

IONIC COMPOUNDS

The compounds which are made up of ions are known as ionic compounds. In an ionic compound the positively charged ions (cations) and negatively charged ions (anions) are held together by strong electrostatic forces of attraction. These forces are known as ionic bonds or electrovalent bonds. Since an ionic compound has an equal total positive and negative charge, its overall charge is zero. Example: sodium chloride (NaCl) is made up of an equal number of Na⁺ and Cl^– ions.

Some ionic compounds

VALENCY

The combining capacity of an element is known as its valency.

Formulae and valencies of common ions – electropositive ions (cations)

Formulae and valencies of common ions – electronegative ions (anions)

Variable valency. Certain elements exhibit more than one valency in their ions or compounds. For example, iron can exist as Fe²⁺ or Fe³⁺ in its compounds. In such cases the name of the ion with the lower valency ends with the suffix ‘ous’, while that with the higher valency ends with the suffix ‘ic’.

CHEMICAL FORMULAE

A compound is represented in abbreviated form by a chemical formula. The chemical formula of a compound represents the composition of a molecule of the compound in terms of the symbols of the elements present in it. For example, water is made up of 2 atoms of hydrogen and 1 atom of oxygen, so its formula is H₂O. In H₂O the subscript 2 indicates 2 atoms of hydrogen; oxygen is written without a subscript, which indicates 1 atom of oxygen.

Formulae of elements. The chemical formula of an element states the composition of its molecule: the symbol tells us the element and the subscript tells us how many atoms are present in one molecule. One molecule of hydrogen contains two atoms, so the formula of hydrogen is H₂. 2H represents 2 separate atoms of hydrogen; H₂ represents 1 molecule of hydrogen; 2H₂ represents 2 molecules of hydrogen.

Formulae of compounds. The chemical formula of a compound states its composition: the chemical symbols tell us which elements are present and the subscripts tell us how many atoms of each element are present in one molecule of the compound. For example, the molecule of water contains 2 atoms of hydrogen and 1 atom of oxygen, so its formula is H₂O.

Rules for writing a chemical formula

1. Write the symbols of the elements which form the compound.
2. Below the symbol of each element, write down its valency.
3. Cross over the valencies of the combining atoms: with the first atom write the valency of the second atom (as a subscript), and with the second atom write the valency of the first atom (as a subscript).

Writing the formula of a compound in terms of ions

1. Write the symbols or formulae of the ions side by side, with the positive ion on the left and the negative ion on the right.
2. Enclose a polyatomic ion in a bracket.
3. Write the valency of each ion below its symbol.
4. Reduce the valency numerals to a simple ratio by dividing by a common factor, if any.
5. Cross the valencies. Do not write the positive or negative charges of the ions.

Example. Formula of barium nitrate. Step 1: Ba²⁺ and NO₃^–. Step 2: Ba²⁺ (NO₃)^–. Step 3: the valencies are 2 and 1. Step 4: not applicable, because the ratio is already simple. Step 5: cross the valencies. Thus the formula of barium nitrate is Ba(NO₃)₂.

Significance of the formula of a substance

• The formula represents the name of the substance.
• It represents one molecule of the substance.
• It also represents one mole of molecules of the substance, i.e. \(6.022\times10^{23}\) molecules.
• It gives the names of all the elements present in the molecule.
• It gives the number of atoms of each element present in one molecule.
• It represents a definite mass of the substance.

MOLECULAR MASS / FORMULA MASS

The molecular mass expresses how many times a molecule of a substance is heavier than \(\dfrac1{12}\) of the mass of an atom of carbon-12.

\[\text{Molecular mass}=\dfrac{\text{mass of a molecule}}{\frac1{12}\text{ of the mass of a carbon-12 atom}}\]

Also, molecular mass \(=2\times\) vapour density. Example: a molecule of water is 18 times heavier than \(\dfrac1{12}\) of the mass of a carbon atom, so the molecular mass of water is 18 u.

Calculation of molecular mass from atomic masses. The molecular mass is the sum of the atomic masses of all the atoms in a molecule of the substance.

Example. Ammonia, NH₃, consists of one atom of N and three atoms of H. The atomic masses of N and H are 14 and 1. Molecular mass of NH₃ = at. mass of N + 3 × at. mass of H = 14 + 3 × 1 = 17 u.

Example. Sulphuric acid, H₂SO₄, consists of 2 H, 1 S and 4 O atoms. The atomic masses of H, S and O are 1, 32 and 16. Molecular mass = (2 × 1) + (1 × 32) + (4 × 16) = 98 u.

Gram molecular mass. The molecular mass of a substance expressed in grams is called its gram molecular mass. Example: the molecular mass of oxygen, O₂, is 32 u, so its gram molecular mass is 32 g.

MOLE CONCEPT

Atoms and molecules are so small that they cannot be counted individually. Chemists use the unit mole for counting atoms, molecules or ions; it is represented by n. A mole represents \(6.022\times10^{23}\) particles. 1 mole of atoms \(=6.022\times10^{23}\) atoms; 1 mole of molecules \(=6.022\times10^{23}\) molecules. The number of particles present in 1 mole of any substance is fixed, \(6.022\times10^{23}\). This number is called the Avogadro constant or Avogadro number, represented by \(N_0\).

1 mole of atoms \(=6.022\times10^{23}\) atoms \(=\) gram atomic mass (molar mass) of the element.

\[\text{Number of moles}=\dfrac{\text{mass of element}}{\text{molar mass}}\quad\Rightarrow\quad n=\dfrac mM\]

\[\text{Number of moles}=\dfrac{\text{given number of atoms}}{\text{Avogadro number}}\quad\Rightarrow\quad n=\dfrac N{N_0}\]

Here n = number of moles, m = given mass, M = molar mass, N = given number of particles, \(N_0\) = Avogadro number of particles. These relations can be interchanged: mass of element \(m=n\times M\); number of particles \(N=n\times N_0\).

Similarly, 1 mole of molecules \(=6.022\times10^{23}\) molecules \(=\) gram molecular mass (molar mass).

\[n=\dfrac mM\qquad n=\dfrac N{N_0}\qquad m=n\times M\qquad N=n\times N_0\]

Relationship between mole, number of particles, mass and volume. 1 mole of a substance corresponds to \(N_A\) particles (atoms or molecules), a mass equal to 1 gram atom or 1 gram formula (molecular) mass, and, for a gas, a volume of 22.4 litres at STP.

Molarity (M). The number of moles of solute in one litre of solution is known as molarity.

\[\text{Molarity}=\dfrac{\text{number of moles of solute}}{\text{volume of solution in litres}}\]

Example. An ornament of silver contains 20 g of silver. Calculate the moles of silver present (atomic mass of silver = 108 u).

Solution. Mass of silver \(m=20\) g; molar mass \(M=108\) g. \(n=\dfrac mM=\dfrac{20}{108}=0.185\) mol.

Example. How many moles of CO₂ are present in 51.2 g of it?

Solution. Molecular mass of CO₂ = 12 + 2 × 16 = 44 u, so M = 44 g and m = 51.2 g. \(n=\dfrac{51.2}{44}=1.16\) mol.

Example. Calculate the mass of (i) 0.5 mole of N₂ gas, (ii) 0.5 mole of N atoms.

Solution. (i) Mass = molar mass × number of moles; M = 28 g, n = 0.5, so \(m=28\times0.5=14\) g. (ii) M = 14 g, n = 0.5 mole, so \(m=14\times0.5=7\) g.

MASS PERCENTAGE OF AN ELEMENT FROM THE MOLECULAR FORMULA

The molecular formula of a compound specifies the number of atoms of the various elements in the molecule of the compound. For example, the molecular formula of glucose is C₆H₁₂O₆: a molecule of glucose contains six atoms of carbon, twelve of hydrogen and six of oxygen. The mass percentage of each element is calculated as

\[\text{Mass percentage of element X}=\dfrac{\text{mass of X in one mole}}{\text{gram molecular mass}}\times100\]

Example. Calculate the percentage composition (by mass) of formaldehyde (CH₂O).

Solution. Molecular mass of CH₂O = 12 × 1 + 1 × 2 + 16 × 1 = 30, so the mass of one mole is 30 g. One mole of CH₂O contains 1 mole (12 g) of carbon, 2 moles (2 g) of hydrogen and 1 mole (16 g) of oxygen. Percentage of carbon \(=\dfrac{12}{30}\times100=40.0\%\); hydrogen \(=\dfrac2{30}\times100=6.7\%\); oxygen \(=\dfrac{16}{30}\times100=53.3\%\).

EMPIRICAL FORMULA

The empirical formula of a compound is the formula which gives the simplest whole-number ratio of the atoms of the various elements present in the molecule. For example, the empirical formula of glucose (C₆H₁₂O₆) is CH₂O, which shows that C, H and O are present in the simplest ratio 1 : 2 : 1.

Rules for writing the empirical formula

1. Divide the percentage of each element by its atomic mass. This gives the relative number of moles of the various elements present in the compound.
2. Divide the quotients obtained by the smallest of them, to get a simple ratio of moles of the various elements.
3. Multiply the figures so obtained by a suitable integer, if necessary, to obtain a whole-number ratio.
4. Write down the symbols of the various elements side by side and put the above numbers as subscripts at the lower right-hand corner of each symbol. This is the empirical formula of the compound.

Example. A substance on analysis gave: Na = 43.4%, C = 11.3%, O = 45.3%. Calculate its empirical formula. [Atomic masses: Na = 23, C = 12, O = 16]

Therefore, the empirical formula is Na₂CO₃.

DETERMINATION OF THE MOLECULAR FORMULA

\[\text{Molecular formula}=\text{empirical formula}\times n,\qquad n=\dfrac{\text{molecular mass}}{\text{empirical formula mass}}\]

Example. What is the simplest formula of the compound which has the percentage composition carbon 80%, hydrogen 20%? If the molecular mass is 30, calculate its molecular formula.

Solution. Empirical formula is CH₃. Empirical formula mass = 12 × 1 + 1 × 3 = 15. \(n=\dfrac{30}{15}=2\). Molecular formula = CH₃ × 2 = C₂H₆.

SOME MORE SOLVED EXAMPLES

Example. On heating a sample of CaCO₃, the volume of CO₂ evolved at NTP is 112 cc. Calculate (i) the weight of CO₂ produced, (ii) the weight of CaCO₃ taken, (iii) the weight of CaO remaining.

Solution. (i) Moles of CO₂ produced \(=\dfrac{112}{22400}=\dfrac1{200}\) mole; mass of CO₂ \(=\dfrac1{200}\times44=0.22\) g. (ii) CaCO₃ → CaO + CO₂: 1 mole of CaCO₃ gives 1 mole of CO₂, so moles of CaCO₃ \(=\dfrac1{200}\) and mass \(=\dfrac1{200}\times100=0.5\) g. (iii) Moles of CaO produced \(=\dfrac1{200}\), so mass of CaO \(=\dfrac1{200}\times56=0.28\) g. Check by conservation of mass: weight of CaO = weight of CaCO₃ taken – weight of CO₂ produced = 0.5 – 0.22 = 0.28 g.

Example. If all the iron present in 1.6 g of Fe₂O₃ is converted into FeSO₄·(NH₄)₂SO₄·6H₂O after a series of reactions, calculate the mass of the product obtained.

Solution. Moles of Fe₂O₃ \(=\dfrac{1.6}{160}=\dfrac1{100}\). Moles of Fe atoms \(=2\times\dfrac1{100}=\dfrac1{50}\). Since one atom of Fe is present in one molecule of the product, the moles of FeSO₄·(NH₄)₂SO₄·6H₂O are also \(\dfrac1{50}\). Mass \(=\dfrac1{50}\times392=7.84\) g.

Example. Calculate the mass of hydrazine N₂H₄ obtained when 1.12 litre of N₂ at NTP reacts with H₂ according to N₂ + 2H₂ → N₂H₄.

Solution. Moles of N₂ taken \(=\dfrac{1.12}{22.4}=\dfrac1{20}\). Moles of N₂H₄ formed \(=\dfrac1{20}\). Mass \(=\dfrac1{20}\times32=1.6\) g.

Example. Calculate the mass of Na₂SO₄ obtained when 100 mL of 0.2 M H₂SO₄ is completely neutralised by NaOH.

Solution. Moles of H₂SO₄ taken = molarity × volume (L) = \(0.2\times\dfrac{100}{1000}=0.02\). H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O: 0.02 mole of H₂SO₄ gives 0.02 mole of Na₂SO₄. Mass of Na₂SO₄ = 0.02 × 142 = 2.84 g.