Chapter - 3 Atoms and Molecules

In a reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbondioxide, and water.

Sodium + Ethanoic → Sodium + Carbon + Water

Carbonate acid ethanoate dioxide

Mass of sodium carbonate = 5.3g      (Given)

Mass of ethanoic acid = 6g                (Given)

Mass of sodium ethanoate = 8.2g     (Given)

Mass of carbon dioxide = 2.2            (Given)

Mass of water = 0.9g                         (Given)

Now, total mass before the reaction = (5.3 + 6)g = 11. 3g

and total mass after the reaction = (8.2 + 2.2 + 0.9)g =  11.3g

Therefore, Total mass before the reaction = Total mass after the reaction

Hence, the given observations are in agreement with the law of conservation of mass.

It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8. Then, the mass of oxygen gas required to react completely with 1g of hydrogen gas is 8g. Therefore, the mass of oxygen gas required to react completely with 3g of hydrogen gas is 8 × 3g = 24 g. 

The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is

“Atoms are indivisible particles, which can neither be created nor destroyed in a chemical

reaction”.

The postulate of Dalton's atomic theory which can explain the law of definite proportion is “The relative number and kind of atoms in a given compound remains constant”.

Mass unit equal to exactly one- twelfth the mass of one atom of carbon - 12 is called one atomic mass unit. It is written as 'u'.

The size of an atom is so small that it is not possible to see it with naked eyes. Also, atom of an element does not exist independently.

(i) Sodium oxide → Na₂O

(ii) Aluminium chloride → AlCl₃

(iii) Sodium suphide → Na₂S

(iv) Magnesium hydroxide → Mg(OH)₂

(i) Al(SO₄)₃→ Aluminium sulphate

(ii) CaCl₂→ Calcium chloride

(iii) K2SO₄→ Potassium sulphate

(iv) KNO₃ → Potassium Nitrate

(v) CaCO₃→ Calcium carbonate

The chemical formula of a compound means the symbolic representation of the composition of a compound. 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 come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide. 

(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 of H₂ = 2 × Atomic mass of H

= 2 × 1 = 2u

Molecular mass of O₂ = 2 × Atomic mass of O

= 2 × 16 = 32u

Molecular mass of Cl₂ = 2 × Atomic mass of Cl

= 2 × 35.5 = 71 u

Molecular mass of CO₂ = Atomic mass of C + 2 × Atomic mass of O

= 12 + 2 × 16 = 44 u

Molecular mass of CH₄ = Atomic mass of C + 4 × Atomic mass of H

= 12 + 4 × 1 = 16 u

Molecular mass of C₂H₆ = 2 × Atomic mass of C + 6 × Atomic mass of H

= 2 × 12 + 6 × 1 = 30u

Molecular mass of C₂H₄ = 2 × Atomic mass of C + 4 × Atomic mass of H

= 2 × 12 + 4 × 1 = 28u

Molecular mass of NH₃ = Atomic mass of N + 3 × Atomic mass of H

= 14 + 3 × 1 =17 u

Molecular mass of CH₃OH Atomic mass of C+4 ×Atomic mass of H+Atomic mass of O

= 12 + 4 × 1 + 16 = 32 u

Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O

= 65 + 16 = 81 u

Formula unit mass of Na₂O = 2 × Atomic mass of Na + Atomic mass of O

= 2 × 23 + 16 = 62u

Formula unit mass of K₂CO₃

= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O

= 2 × 39 + 12 + 3 × 16 = 138u

                                                    Carbon + Oxygen → Carbon dioxide

3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11g of carbon dioxide will be formed. The above answer is governed by the law of constant proportions.

A polyatomic ion is a group of atoms carrying a charge (positive or negative).

For example, ammonium ion (NH₄⁺), hydroxide ion (OH⁻), carbonate ion (CO₃²⁻), sulphateion (SO₄²⁻).

(a) Magnesium chloride →MgCl₂

(b) Calcium oxide →CaO

(c) Copper nitrate →Cu(NO₃)₂

(d) Aluminium chloride →AlCl₃

(e) Calcium carbonate →CaCO₃

(a) Molar mass of ethyne, C₂H₂ = 2 × 12 + 2 × 1 = 28g

(b) Molar mass of sulphur molecule, S₈ = 8 × 32 = 256g

(c) Molar mass of phosphorus molecule, P₄ = 4 × 31 = 124g

(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g

(e) Molar mass of nitric acid, HNO₃ = 1 + 14 + 3 × 16 = 63g

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

(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g

(c) The mass of 10 moles of sodium sulphite (Na₂SO₃) is 10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g

(a) 32 g of oxygen gas = 1 mole

Then, 12g of oxygen gas = 12/32 mole = 0.375 mole

(b) 18g of water = 1 mole

Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)

(c) 44g of carbon dioxide = 1 mole

Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole

(a) Mass of one mole of oxygen atoms = 16g

Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g

(b) Mass of one mole of water molecule = 18g

Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g

1 mole of solid sulphur (S₈) = 8 × 32g = 256g

i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules

Then, 16g of solid sulpur contains $\frac{6.022 x {10}^{23}}{256}$ x  16 molecules

= 3.76 × 10²² molecules (approx)

1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102g

i.e., 102g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃

Then, 0.051 g of Al₂O₃ contains = $\frac{6.022 \times {10}^{23}}{102}$ × 0.051 molecules

= 3.011 × 10²⁰ molecules of Al₂O₃

The number of aluminium ions (Al³⁺) present in one molecules of aluminium oxide is 2.

Therefore, The number of aluminium ions (Al³⁺) present in 

3.11 × 10²⁰ molecules (0.051g) of aluminium oxide (Al₂O₃) = 2 × 3.011 × 10²⁰

= 6.022 × 10²⁰