CBSE Class 9 Science Atoms and Molecules Notes

ATOMS AND MOLECULES

ATOMS AND MOLECULES

LAWS OF CHEMICAL COMBINATION

Before Dalton, the concept of the atom was merely philosophical. Dalton explained the atom on the basis of the laws of chemical combination. There are three laws of chemical combination: (1) law of conservation of mass, (2) law of constant proportions, (3) law of multiple proportions.

Law of conservation of mass. Antoine L. Lavoisier, a French scientist, established the law of conservation of mass. It states: mass can neither be created nor destroyed in a chemical reaction. According to this law the mass of an isolated system remains constant over time: when matter is enclosed in a system and none is allowed in or out, its quantity never changes. The total mass of the products is always equal to the total mass of the reactants. As there is no loss of mass of substances, i.e. mass is conserved, Lavoisier called this the law of conservation of mass.

All matter in the universe exists in three states. There are two ways of classifying matter: (1) according to physical state, as solid, liquid or gas; (2) according to composition, as element, compound or mixture.

Reactants and products. In a chemical reaction the substances that combine or react are known as reactants, and the new substance or substances formed are called the product or products. A chemical reaction can be written in general as: reactants → products.

Example. When calcium oxide is dissolved in water, calcium hydroxide is formed: CaO + H₂O → Ca(OH)₂. Here calcium oxide and water are the reactants and calcium hydroxide is the product. 74 g of calcium hydroxide is obtained when 56 g of calcium oxide reacts with 18 g of water, as proved by experiment. The total mass of the reactants (56 g + 18 g = 74 g) is equal to the mass of the product (74 g). This proves the law of conservation of mass.

Calcium oxide + Water → Calcium hydroxide (reactants: calcium oxide and water; product: calcium hydroxide).

Calcium oxide + Water → Calcium hydroxide: 56 g + 18 g = 74 g → 74 g.

Law of constant proportions. This law states that a chemical compound always contains exactly the same proportion of elements by mass. It is also known as the law of definite proportions or Proust’s law, after Joseph Louis Proust, who gave it.

Explanation. Compounds are formed by the combination of two or more elements. In a compound the ratio by mass of the elements always remains the same, irrespective of the source of the compound. If the ratio of the masses of the constituent elements is altered, a new compound is formed.

Examples. Water is formed by the combination of hydrogen and oxygen. The ratio of the masses of hydrogen and oxygen in water is always 1 : 8, whether the water is collected from a well, river, pond or anywhere else. Nitrogen and oxygen form three different compounds, each with its own fixed ratio by mass:

2H₂ + O₂ → 2H₂O; masses 4 and 32; ratio of hydrogen to oxygen = 1 : 8 (by mass).

These three compounds are formed by the combination of the same elements, but because the elements combine in different ratios by mass, different compounds are formed.

DALTON’S ATOMIC THEORY

John Dalton, a British chemist, gave the atomic theory in 1808. It is popularly known as Dalton’s atomic theory in his honour. He gave the theory on the basis of the laws of chemical combination, and it explains them properly.

Main postulates of Dalton’s atomic theory

1. Elements are made of extremely small particles called atoms.
2. Atoms of a given element are identical in size, mass and other properties.
3. Atoms of different elements differ in size, mass and other properties.
4. Atoms cannot be subdivided, created or destroyed.
5. Atoms of different elements combine in simple whole-number ratios to form chemical compounds.
6. In chemical reactions, atoms are combined, separated or rearranged.

Dalton’s Atomic Theory (figure, continued)

3. Atoms of one element cannot be changed into atoms of a different element by chemical reactions; atoms are neither created nor destroyed in chemical reactions. (Oxygen atom ⊘ Nitrogen atom: an oxygen atom cannot change into a nitrogen atom.)

4. Compounds are formed when atoms of more than one element combine; a given compound always has the same relative number and kind of atoms. (Elements: N + O → Compound: NO.)

INTEXT QUESTIONS (PAGE 32)

In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. (sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water)

Answer: Sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide and water.

Total mass before the reaction = 5.3 g + 6 g = 11.3 g.

Total mass after the reaction = 8.2 g + 2.2 g + 0.9 g = 11.3 g.

Total mass before the reaction = total mass after the reaction. Hence the given observations are in agreement with the law of conservation of mass.

Hydrogen and oxygen combine in the ratio 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Answer: 24 g.

The ratio of hydrogen to oxygen by mass in water is 1 : 8, so 1 g of hydrogen reacts completely with 8 g of oxygen. Therefore 3 g of hydrogen reacts with 8 × 3 g = 24 g of oxygen.

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer: Atoms can neither be created nor destroyed in a chemical reaction. Since atoms are only rearranged, the total mass stays the same.

Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer: The postulate that the relative number and kinds of atoms are constant in a given compound. Since every atom of an element has a fixed mass and the atoms combine in a fixed whole-number ratio, the ratio of masses of the elements in a compound is always the same.

ATOMS

On the basis of Dalton’s atomic theory, the atom may be defined as the smallest particle of matter. Characteristics of atoms:

• The atom is the smallest particle of matter.
• All elements are made of tiny particles called atoms.
• Atoms are very small in size and cannot be seen with the naked eye.
• An atom does not usually exist in the free state in nature, but it takes part in a chemical reaction.
• The properties of matter depend upon the characteristics of its atoms.
• Atoms are the building blocks of an element, like bricks which combine to make a building.
• The size of an atom is indicated by its radius.
• In ancient times atoms were considered indivisible.

SYMBOLS OF ATOMS OF ELEMENTS

Dalton was the first scientist to use symbols for elements in a very specific sense. When he used a symbol for an element he also meant a definite quantity of that element, that is, one atom of that element. Berzelius suggested that the symbols of elements be made from one or two letters of the name of the element. Many symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (uppercase) and the second letter as a small letter (lowercase).

For convenience, elements are represented by unique symbols. For example, hydrogen is ‘H’, oxygen is ‘O’, nitrogen is ‘N’ and iron is ‘Fe’. His work proved a boon to science. For his marvellous work Berzelius, together with John Dalton, Antoine Lavoisier and Robert Boyle, is considered the Father of Modern Chemistry. The symbols of many elements are taken from their English names, while those of many others are taken from their Greek or Latin names. Several elements are named after the place where they were discovered, such as ‘copper’, which was taken from Cyprus. Some elements are named after their colour, such as ‘gold’, which means yellow.

Symbols of some elements derived from their Latin names

ATOMIC MASS

The mass of an atom is called its atomic mass. Since atoms are very small, the actual mass of an atom is very small. For example, the actual mass of one atom of hydrogen is about \(1.673\times10^{-24}\) g, i.e. 0.000000000000000000000001673 g. It is difficult to deal with such small numbers, so for convenience relative atomic mass is used.

Carbon-12 (an isotope of carbon) is taken as the standard for calculating atomic masses. The relative masses of all atoms are found with respect to C-12. One atomic mass unit \(=\dfrac1{12}\) of the mass of one atom of C-12. Thus atomic mass is the relative mass of an atom with respect to \(\dfrac1{12}\) of the mass of a carbon-12 atom. ‘amu’ is the abbreviation of atomic mass unit, but now it is denoted simply by ‘u’.

The atomic mass of hydrogen is 1 u. This means one hydrogen atom is 1 time as heavy as \(\dfrac1{12}\) of a carbon atom. The atomic mass of oxygen is 16 u, which means one atom of oxygen is 16 times as heavy as \(\dfrac1{12}\) of a carbon atom.

Absolute (actual) atomic mass. The actual mass of a carbon-12 atom is \(1.9926\times10^{-23}\) g. Thus, by multiplying the relative atomic mass by \(1.6605\times10^{-24}\) g (the mass of 1 u), we get the absolute or actual mass of an atom.

Example 1. Find the absolute mass of oxygen. Atomic mass of oxygen = 16 u; 1 u = \(1.6605\times10^{-24}\) g. Absolute mass \(=1.6605\times10^{-24}\times16=26.568\times10^{-24}\) g \(=2.6568\times10^{-25}\) g.

Example 2. Find the absolute mass of sodium. Atomic mass = 23 u. Absolute mass \(=1.6605\times10^{-24}\times23=38.191\times10^{-24}\) g \(=3.8191\times10^{-25}\) g.

Example 3. Calculate the absolute mass of a hydrogen atom. Atomic mass = 1 u. Absolute mass \(=1.6605\times10^{-24}\times1=1.6605\times10^{-24}\) g.

Example 4. Calculate the actual mass of a nitrogen atom. Atomic mass = 14 u. Absolute mass \(=1.6605\times10^{-24}\times14=23.247\times10^{-24}\) g \(=2.3247\times10^{-25}\) g.

EXISTENCE OF ATOMS

Atoms of most elements exist in the form of molecules or ions, since they are very reactive; for example hydrogen, oxygen and chlorine. However, atoms of some elements which are non-reactive exist in the free state in nature; for example helium, neon and argon. Usually atoms exist in two forms: (i) in the form of molecules, (ii) in the form of ions.

INTEXT QUESTIONS (PAGE 35)

Define the atomic mass unit.

Answer: The mass unit equal to exactly one-twelfth (\(\dfrac1{12}\)) of the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’.

Why is it not possible to see an atom with the naked eye?

Answer: The size of an atom is so small that it is not possible to see it with the naked eye. Also, the atom of an element usually does not exist independently.

MOLECULE

A molecule is the smallest particle of an element or a compound that can exist in the free state and shows all the properties of that substance. Atoms exist in the free state in the form of molecules.

• A molecule may be formed by the combination of two or more similar atoms of an element: the oxygen molecule is formed by two oxygen atoms and the hydrogen molecule by two hydrogen atoms.
• A molecule may be formed by the combination of atoms of two or more different elements. For example, the molecule of water is formed by two atoms of hydrogen and one atom of oxygen; the molecule of nitric oxide (nitrogen monoxide) is formed by one nitrogen atom and one oxygen atom.
• A molecule takes part in a chemical reaction.

Most atoms exist in the form of molecules. Examples: hydrogen (H₂), oxygen (O₂), nitrogen (N₂). Molecules are of two kinds: molecules of elements and molecules of compounds.

Molecules of elements. When two or more atoms of the same element combine to form a molecule, it is called a molecule of an element. Examples: hydrogen molecule (H₂), formed from two hydrogen atoms; oxygen molecule (O₂), from two oxygen atoms; sulphur molecule (S₈), from eight sulphur atoms; phosphorus molecule (P₄), from four phosphorus atoms.

Molecules of some non-reactive elements are formed by a single atom, for example helium, neon and argon. Metals form big clusters of atoms and are represented simply by their symbols, e.g. iron (Fe), copper (Cu), zinc (Zn). These are known as giant molecules. Carbon is a non-metal, but it also exists as a giant molecule and is represented by its symbol ‘C’.

ATOMICITY

• Monoatomic: the molecule is formed by a single atom. Generally noble gases form monoatomic molecules: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn).
• Diatomic: the molecule is formed by two atoms: hydrogen (H₂), oxygen (O₂), nitrogen (N₂), chlorine (Cl₂).
• Triatomic: the molecule is formed by three atoms: ozone (O₃).
• Tetra-atomic: the molecule is formed by four atoms: phosphorus (P₄).
• Polyatomic: the molecule is formed by more than two atoms: sulphur (S₈).

• ATOMICITY (Molecule of an Element)
  • Monoatomic — Mono-atomic [1 atom element]: He, Ne, Ar, Kr
  • Diatomic — Di-atomic [2 atom molecule]: O₂, N₂, Cl₂, H₂
  • Triatomic — Tri-atomic [3 atom molecule]: O₃
  • Tetraatomic / Poly-atomic [more than 3 atom molecule]: P₄, S₈

MOLECULES OF COMPOUNDS

When a molecule is formed by the combination of two or more atoms of different elements, it is called a molecule of a compound. Example: the molecule of water (H₂O) is formed by two hydrogen atoms and one oxygen atom.

IONS

Atoms of several elements exist in the form of ions. Atoms or groups of atoms with a negative or positive charge are called ions.

• Cations: ions having a positive charge, e.g. sodium ion (Na⁺), potassium ion (K⁺).
• Anions: ions having a negative charge, e.g. chloride ion (Cl^–), fluoride ion (F^–).
• Monoatomic ions: ions formed by one atom, e.g. Na⁺, K⁺, Cl^–, F^–.
• Polyatomic ions: ions formed by two or more atoms. These are groups of atoms of different elements which behave as single units, e.g. ammonium ion (NH₄⁺), hydroxide ion (OH^–).

WRITING CHEMICAL FORMULAE

The chemical formula of a compound is the symbolic representation of its composition. To write the chemical formula of a compound, the symbols and valencies of the constituent elements must be known. The valency of an atom of an element can be thought of as the hands or arms of that atom.

Points to remember

• The symbols or formulae of the component radicals of the compound are written side by side.
• Positive radicals are written on the left and negative radicals on the right.
• The valencies of the radicals are written below the respective symbols.
• The criss-cross method is applied: the numerical value of the valency of each radical is exchanged and written as a subscript of the other radical.
• A polyatomic radical is enclosed in a bracket and the subscript is placed outside, at the lower right corner.
• The common factor is removed.
• If the subscript of a radical is one, it is omitted.

Rules to follow while writing a chemical formula

• The valencies or charges on the ions must balance.
• When a compound consists of a metal and a non-metal, the name or symbol of the metal is written first. Examples: calcium oxide (CaO), sodium chloride (NaCl), iron sulphide (FeS), copper oxide (CuO), where oxygen, chlorine and sulphur are non-metals and are written on the right, whereas calcium, sodium, iron and copper are metals and are written on the left.
• In compounds formed with polyatomic ions, the ion is enclosed in a bracket before writing the number that indicates the ratio.

The simplest compounds, made up of two different elements, are called binary compounds. While writing the chemical formula, we write the constituent elements with their valencies and then cross over the valencies of the combining atoms. The formulae of ionic compounds are simply the whole-number ratio of the positive to the negative ions in the structure. For magnesium chloride, we write the symbol of the cation (Mg²⁺) first, followed by the symbol of the anion (Cl^–), and then criss-cross their charges to get MgCl₂.

Examples of formulae obtained by the criss-cross method

INTEXT QUESTIONS (PAGE 39)

Write down the formulae of (i) sodium oxide, (ii) aluminium chloride, (iii) sodium sulphide, (iv) magnesium hydroxide.

Answer: (i) Na₂O; (ii) AlCl₃; (iii) Na₂S; (iv) Mg(OH)₂.

Write down the names of the compounds represented by the following formulae: (i) Al₂(SO₄)₃, (ii) CaCl₂, (iii) K₂SO₄, (iv) KNO₃, (v) CaCO₃.

Answer: (i) Aluminium sulphate; (ii) calcium chloride; (iii) potassium sulphate; (iv) potassium nitrate; (v) calcium carbonate.

What is meant by the term chemical formula?

Answer: The chemical formula of a compound is the symbolic representation of its composition. From the chemical formula of a compound we can know the number and kinds of atoms of different elements that constitute the compound.

For example, from the chemical formula CO₂ of carbon dioxide we know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound.

How many atoms are present in (i) an H₂S molecule and (ii) a PO₄^3– ion?

Answer: (i) In an H₂S molecule three atoms are present: two of hydrogen and one of sulphur. (ii) In a PO₄^3– ion five atoms are present: one of phosphorus and four of oxygen.

MOLECULAR MASS

Atomic mass: the atomic mass of an element is the mass of one atom of that element in atomic mass units (u).

Atomic mass unit (amu): \(\dfrac1{12}\) of the mass of an atom of carbon-12 is called the atomic mass unit. It is a unit of mass used to express atomic masses and molecular masses.

Molar mass: the molar mass of an element is equal to the numerical value of its atomic mass, but the unit changes from ‘u’ to ‘g’. The molar mass of an atom is also known as the gram atomic mass. For example, the atomic mass of carbon is 12 u, so the gram atomic mass of carbon is 12 g.

Molecular mass of a molecule: the sum of the atomic masses of all the atoms in a molecule of a substance is called the molecular mass of the molecule. Generally the relative atomic masses of atoms are used for calculating the molecular mass of 1 mole of any molecular or ionic substance.

Example. Molecular mass of H₂SO₄: atomic mass of hydrogen = 1, sulphur = 32, oxygen = 16. Molecular mass = 2 × 1 + 32 + 4 × 16 = 98 u.

Molecular mass of hydrogen chloride = atomic mass of hydrogen + atomic mass of chlorine = 1 + 35.5 = 36.5 u.

FORMULA UNIT MASS

The formula unit mass of a substance is the sum of the atomic masses of all the atoms in a formula unit of a compound. The term ‘formula unit’ is used for substances which are made up of ions. Formula unit mass of NaCl = 1 × 23 + 1 × 35.5 = 58.5 u. Formula unit mass of ZnO = 1 × 65 + 1 × 16 = 81 u.

INTEXT QUESTIONS (PAGE 40)

Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Answer: H₂ = 2 u; O₂ = 32 u; Cl₂ = 71 u; CO₂ = 44 u; CH₄ = 16 u; C₂H₆ = 30 u; C₂H₄ = 28 u; NH₃ = 17 u; CH₃OH = 32 u.

H₂ = 2 × 1 = 2 u.

O₂ = 2 × 16 = 32 u.

Cl₂ = 2 × 35.5 = 71 u.

CO₂ = 12 + 2 × 16 = 44 u.

CH₄ = 12 + 4 × 1 = 16 u.

C₂H₆ = 2 × 12 + 6 × 1 = 30 u.

C₂H₄ = 2 × 12 + 4 × 1 = 28 u.

NH₃ = 14 + 3 × 1 = 17 u.

CH₃OH = 12 + 4 × 1 + 16 = 32 u.

Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u.

Answer: ZnO = 81 u; Na₂O = 62 u; K₂CO₃ = 138 u.

ZnO = 65 + 16 = 81 u.

Na₂O = 2 × 23 + 16 = 62 u.

K₂CO₃ = 2 × 39 + 12 + 3 × 16 = 138 u.

MOLE CONCEPT

The mole is the unit used in chemistry to express the amount of a chemical substance. One mole is defined as the amount of substance of a system which contains as many entities (atoms, molecules or ions) as there are atoms in 12 g of carbon-12.

Avogadro number. The number of particles present in one mole of any substance is \(6.022\times10^{23}\). This is called Avogadro’s number or Avogadro’s constant. 1 mole of hydrogen atoms represents \(6.022\times10^{23}\) hydrogen atoms; 1 mole of hydrogen molecules represents \(6.022\times10^{23}\) hydrogen molecules; 1 mole of water molecules represents \(6.022\times10^{23}\) water molecules.

Conversion of moles to mass and vice versa. The key concept is that a mole of any substance has a mass equal to its gram formula mass (molecular mass) in grams. For example, the molecular mass of hydrogen is 2 u, so the mass of 1 mole of hydrogen, its molar mass, is 2 g. Similarly, 2 moles of hydrogen weigh 2 × 2 = 4 g.

Mole concept calculations. Most of the time the moles or the number of atoms or molecules are given and the mass is to be calculated. In the other cases the mass is given and the moles or number are needed. In both, proceed in two steps: Step 1: establish the relationship between the molar mass and the number (\(N_A\)) or moles of that particular entity (atom, molecule or ion). Step 2: use the unitary method to calculate what is asked.

Note: when we say oxygen gas weighs 32 g we mean that 1 mole of oxygen molecules (O₂) weighs 32 g and not 1 mole of oxygen atoms (O), because oxygen exists naturally as O₂ molecules.

1 Mole

• In terms of particles: 6.023 × 10²³ particles. 1 mole of oxygen atoms = 6.023 × 10²³ atoms; 1 mole of oxygen molecules = 6.023 × 10²³ molecules.
• In terms of volume: 22.4 L of a gas at N.T.P.
• In terms of mass:
  • 1 gram atom of elements: 12 g C, 23 g Na
  • 1 gram molecule of substance: 18 g H₂O, 44 g CO₂
  • 1 gram formula weight of substance: 58.5 g NaCl, 100 g CaCO₃

PROBLEMS BASED ON THE MOLE CONCEPT

1. When the mass of the substance is given:

\[\text{Number of moles}=\dfrac{\text{given mass}}{\text{atomic mass}}\]

Example 1. Calculate the number of moles in 81 g of aluminium. Number of moles \(=\dfrac{81}{27}=3\) moles of aluminium.

Example 2. Calculate the mass of 0.5 mole of iron. Mass = atomic mass × number of moles = 55.9 × 0.5 = 27.95 g.

2. Calculation of the number of particles when the mass of the substance is given:

\[\text{Number of particles}=\dfrac{\text{Avogadro number}\times\text{given mass}}{\text{gram molecular mass}}\]

Example. Calculate the number of molecules in 11 g of CO₂. Gram molecular mass of CO₂ = 44 g. Number of molecules \(=\dfrac{6.023\times10^{23}\times11}{44}=1.51\times10^{23}\) molecules.

3. Calculation of mass when the number of particles of a substance is given:

\[\text{Mass of a substance}=\dfrac{\text{gram molecular mass}\times\text{number of particles}}{6.023\times10^{23}}\]

Example 1. Calculate the mass of \(18.069\times10^{23}\) molecules of SO₂. Gram molecular mass of SO₂ = 64 g. Mass \(=\dfrac{64\times18.069\times10^{23}}{6.023\times10^{23}}=192\) g.

Example 2. Calculate the mass of glucose in \(2\times10^{24}\) molecules. Gram molecular mass of glucose = 180 g. Mass \(=\dfrac{180\times2\times10^{24}}{6.023\times10^{23}}=597.7\) g. (The source wrote \(2\times10^{23}\) in the numerator, which does not give 597.7 g.)

4. Calculation of the number of moles when the number of atoms or molecules is given:

\[\text{Number of moles}=\dfrac{\text{number of particles}}{\text{Avogadro number}}\]

Example. Calculate the number of moles in \(12.046\times10^{22}\) atoms of copper. Number of moles \(=\dfrac{12.046\times10^{22}}{6.023\times10^{23}}=0.2\) mole.

INTEXT QUESTIONS (PAGE 42)

If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?

Answer: \(1.9926\times10^{-23}\) g.

One mole of carbon atoms weighs 12 g, i.e. \(6.022\times10^{23}\) carbon atoms have a mass of 12 g. Therefore the mass of 1 atom of carbon \(=\dfrac{12}{6.022\times10^{23}}\) g \(=1.9926\times10^{-23}\) g.

Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Answer: 100 g of sodium contains more atoms than 100 g of iron.

Gram atomic mass of Na = 23 g, and 23 g of Na contains \(6.022\times10^{23}\) atoms. So 100 g of Na contains \(\dfrac{6.022\times10^{23}}{23}\times100=2.6182\times10^{24}\) atoms.

Gram atomic mass of Fe = 56 g, and 56 g of Fe contains \(6.022\times10^{23}\) atoms. So 100 g of Fe contains \(\dfrac{6.022\times10^{23}}{56}\times100=1.0753\times10^{24}\) atoms.

Therefore 100 g of sodium contains more atoms than 100 g of iron.

EXERCISE QUESTIONS (PAGES 43, 44)

A 0.24 g sample of a 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.

Answer: Boron 40%, oxygen 60%.

Mass of boron = 0.096 g, mass of oxygen = 0.144 g, mass of sample = 0.24 g.

Percentage of boron by weight \(=\dfrac{0.096}{0.24}\times100=40\%\).

Percentage of oxygen by weight \(=\dfrac{0.144}{0.24}\times100=60\%\).

When 3.0 g 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 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Answer: 11 g of carbon dioxide; the law of constant proportions.

Carbon + oxygen → carbon dioxide. 3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide. If 3 g of carbon is burnt in 50 g of oxygen, the 3 g of carbon will still react with only 8 g of oxygen. The remaining 42 g of oxygen is left unreacted. So only 11 g of carbon dioxide is formed. This is governed by the law of constant proportions.

What are polyatomic ions? Give examples.

Answer: A polyatomic ion is a group of atoms carrying a charge (positive or negative). Examples: ammonium ion (NH₄⁺), hydroxide ion (OH^–), carbonate ion (CO₃^2–), sulphate ion (SO₄^2–).

Write the chemical formulae of the following: (a) magnesium chloride, (b) calcium oxide, (c) copper nitrate, (d) aluminium chloride, (e) calcium carbonate.

Answer: (a) MgCl₂; (b) CaO; (c) Cu(NO₃)₂; (d) AlCl₃; (e) CaCO₃.

Give the names of the elements present in the following compounds: (a) quick lime, (b) hydrogen bromide, (c) baking powder, (d) potassium sulphate.

Answer: (a) Quick lime, CaO: calcium and oxygen. (b) Hydrogen bromide, HBr: hydrogen and bromine. (c) Baking powder (sodium hydrogencarbonate), NaHCO₃: sodium, hydrogen, carbon and oxygen. (d) Potassium sulphate, K₂SO₄: potassium, sulphur and oxygen.

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.

Answer: (a) 26 g; (b) 256 g; (c) 124 g; (d) 36.5 g; (e) 63 g.

(a) C₂H₂ = 2 × 12 + 2 × 1 = 26 g.

(b) S₈ = 8 × 32 = 256 g.

(c) P₄ = 4 × 31 = 124 g.

(d) HCl = 1 + 35.5 = 36.5 g.

(e) HNO₃ = 1 + 14 + 3 × 16 = 63 g.

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)?

Answer: (a) 14 g; (b) 108 g; (c) 1260 g.

(a) The mass of 1 mole of nitrogen atoms is 14 g.

(b) 4 × 27 g = 108 g.

(c) 10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g.

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

Answer: (a) 0.375 mole; (b) 1.11 moles (approx.); (c) 0.5 mole.

(a) 32 g of oxygen gas = 1 mole, so 12 g \(=\dfrac{12}{32}=0.375\) mole.

(b) 18 g of water = 1 mole, so 20 g \(=\dfrac{20}{18}=1.11\) moles (approx.).

(c) 44 g of carbon dioxide = 1 mole, so 22 g \(=\dfrac{22}{44}=0.5\) mole.

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

Answer: (a) 3.2 g; (b) 9 g.

(a) Mass of one mole of oxygen atoms = 16 g, so 0.2 mole = 0.2 × 16 = 3.2 g.

(b) Mass of one mole of water molecules = 18 g, so 0.5 mole = 0.5 × 18 = 9 g.

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

Answer: \(3.76\times10^{22}\) molecules (approx.).

1 mole of solid sulphur (S₈) = 8 × 32 g = 256 g, i.e. 256 g of sulphur contains \(6.022\times10^{23}\) molecules. Then 16 g contains \(\dfrac{6.022\times10^{23}}{256}\times16=3.76\times10^{22}\) molecules (approx.).

Calculate the number of aluminium ions present in 0.051 g 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 = 27 u.)

Answer: \(6.022\times10^{20}\) aluminium ions.

1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102 g, i.e. 102 g of Al₂O₃ contains \(6.022\times10^{23}\) molecules. Then 0.051 g contains \(\dfrac{6.022\times10^{23}}{102}\times0.051=3.011\times10^{20}\) molecules of Al₂O₃.

One molecule of aluminium oxide contains 2 aluminium ions (Al³⁺). Therefore the number of Al³⁺ ions in 0.051 g of aluminium oxide \(=2\times3.011\times10^{20}=6.022\times10^{20}\).

2. Law of constant or definite proportion----In a chemical substance the elements are always present in definite proportion by mass.

7. A negatively charged ion is called anion and positively charged ion is called cation.

8. Combining capacity of an element is called valency.

• a) Symbol of the metal is written first followed by nonmetal.
• b) Write the valencies of these under it.
• c) Charges are criss crossed to get the formula.

\( 1\text{ mole} = 6.022 \times 10^{23}\text{ particles} \)

1 Mole Relations:

• \( 6.022 \times 10^{23}\text{ atoms} \)
• \( 6.022 \times 10^{23}\text{ molecules} \)
• \( 22.4\text{ Litre at S.T.P} \)
• It's gram atomic mass / It's molar mass

11. \( \text{Number of moles (n)} = \frac{\text{given mass (m)}}{\text{Molar mass (M)}} \)

\( \text{Number of moles(n)} = \frac{\text{Given number of particles (N)}}{\text{Avogadro number } (N_0)} \)

LAWS OF CHEMICAL COMBINATIONS

Before Dalton concept of atom was mere philosophical. Dalton explained about atom on the basis of Laws of Chemical Combinations.

Nitrogen dioxide is a compound, which is formed by the combination of nitrogen and oxygen. The ratio of nitrogen and oxygen by mass in nitrogen dioxide is in 7:16.
Nitrous oxide is a compound which is also formed by the combination of nitrogen and oxygen. The ratio of nitrogen and oxygen in nitrous oxide is in 28:16.
Nitric oxide is a compound, which is also formed by the combination of nitrogen and oxygen. The ratio of nitrogen and oxygen in nitric oxide is in 7:8.
From the above three examples it is clear that if the ratio of the atoms by mass is altered then the new compound is formed, such as in the case of nitrogen dioxide, nitrous oxide, nitric oxide. These three compounds are formed by the combination of same atoms but because of combination of the constituent atoms in different ratios by mass new compound is formed.

• Atoms are the building block of an element similar to a brick which combine together to make a building.

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