Chapter 3 Atoms and Molecules

Law of conservation of mass: Mass can neither be created nor be destroyed in a chemical reaction.

Given reaction: sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Total mass of reactants = 5.3 + 6 = 11.3 g.

Total mass of products = 8.2 + 2.2 + 0.9 = 11.3 g 

Total mass of reactants = Total mass of products

Hence, these observation are in agreement with the law of conservation of mass.

Given,

Ratio of hydrogen and oxygen by mass = 1:8 

Mass of oxygen gas that completely reacts with 1g of hydrogen gas is equal to 8 g i.e Mass of oxygen gas that completely reacts with 3 g of hydrogen gas

= 8 x 3 = 24g. 

Hence, 24 g mass of oxygen gas is required to react completely with 3 g of hydrogen gas.

Postulate of Dalton's atomic theory: Atoms are indivisible particles which cannot be created or destroyed in a chemical reaction. It is the result of the law of conservation of mass.

The postulate of Dalton's theory, "The relative numbers and finds of atoms are content in a given compound” can explain the law of definite proportion.

Atomic mass unit: Mass unit equals to exactly 1/12 th the mass of one atom of carbon-12 is known as an atomic mass unit.

Note:

Carbon-12: Standard reference for measuring atomic Masses. 

1 amu (atomic mass unit) = 1.66x10^-24 g.

The size of an atom is very very small that we do not see with our naked eyes. Atoms of mait elemente cannot exist independently.

(i)  Na₂O (Sodium oxide)

(ii)  AlCl₂ (Aluminium chloride)

(iii)  Na₂S (Sodium sulphide)

(iv)  Mg(OH)₂ (Magnesium hyelsonide)

(i)  Aluminium sulphate
(ii)  Calcium chloride
(iii)  Potassium sulphate
(iv)  Potassium Nitrate
(v)  Calcium Carbonate

It is the symbolic representation of the compound's composition. For example, the chemical formula of Calcium Carbonate is CaCo₃.

(i)  3 atoms are present in the H₂S molecule. There are 2 atoms of hydrogen and one is sulphur.
(ii)  5 atoms are present in PO₄^3- ion. There are 4 atoms of oxygen and one in phosphorus.

Atomic mass includes the masses of three subatomic particles which make up an atom i.e. proton, neutron and electron.

Unit of atomic mass is 'u'

‘u’ = unified atomic mass unit & 1/12 th mass of single carbon-12 atom 

‘u’ = 1.66 x 10^-24 g

(Just for understanding - Atomic mass approximately equal to twice of atomic no.)

Molecular mass of a substance is the sum of atomic masses of all the atoms in a molecule of a subitance.

Unit of molecular mass = 'u’

Molecular mass of H₂ = 2 x 1 = 2u (Atomic mass of H=1)

Molecular mass of O₂ = 16 x 2 = 32 u (Atomic mass of O=16)

Molecular Mass of Cl₂ = 35.2 x 2 = 71 u (Atomic mass of Cl = 35.5)

Molecular Mass of CO₂ = 12 x 1 + 16 x 2

                = 12 + 32 = 44 u ( Atomic Mass of C = 12)

Molecular Mass of CH₄ = 12 x 1 + 4 x 1

               = 12 + 4 = 16 u

Molecular Mass of C₂H₆ = 12 x 2 + 6 x 1

               = 24 + 6 = 30 u 

Molecular Mass of C₂H₄ = 12 x 2 + 4 x 1

               = 24 + 4 = 28 u

Molecular Mass of  NH₃ = 14 X 1 + 3 x 1

               = 14 x 3 = 17 u  (Atomic Mass of N = 14)

Molecular Mass of CH₃OH = 12 x 1 + 4 x 1 + 16 x 1

               = 12 + 4 + 16 = 32 u

Formula unit mass of ZnO

= 1 x Atomic mass of Zn + 1 x Atomic mass of ‘O’ atom 

= 1 x 65 + 1 x 16

= 65 + 16

= 81 u

 

Formula unit mass of Na₂o 

 = 2 x Atomic mass of Na + 1 x Atomic mass of ‘O’ atom

= 2 x 23 + 1 x 16

= 46 + 16

= 62 u

 

Formula unit mass of K₂CO₃

= 2 x Atomic mass of K + 1 x Atomic mass of C + 3 x Atomic mass of ‘O’

= 2 x 39 + 1 x 12 + 3 x 16

= 78 + 12 + 48

= 138 u

1 mole of carbon atoms = 6.022 x 10²³

1 mole of carbon atoms weighs = 12 gram

Mass of 6.022 x 10²³ atoms of carbon = 12 gram 

1 atom of carbon = 12 / 6.022 x 10²³

    = 1.99 x 10^-23 gram

Hence, Mass of 1 atom of carbon

   = 1.99 x 10^-23 gram

Atomic mass of sodium = 23 u

Gram atomic mass of sodium = 23 g 

1 mole or 23 g of sodium has = 6.022 x 10²³ atom

100 g of sodium has = 6.022x10²³ / 23 x 100

       = 2.618 x 10²⁴ atom

Atomic mass of Ison (fe) = 56 u

Gram atomic mass of fe = 56 g

1 mole of fe has = 6.022 x 10²³ atom

56 g of fe = 6.022 x 10²³ atom

100 g of fe = 6.022 x 10²³ / 56 x 100

      = 1.075 x 10²⁴ atom

Hence, 100 gram of Na has more number of atoms than 100 gram of fe.

Mass of Boron  =  0.096 g

Маss of Oxygen = 0.144 g

Mass of sample of compound of oxygen and Boron = 0.24 g 

% of Boron by weight in the compound = 0.096 x 100 / 0.24 = 40%

% of Oxygen by weight in the compound = 0.144 x 100 / 0.24 = 60%

1 mole of C = 12 g

1 mole of O2 = 32 g

C + O₂ → CO₂

12 g  +  32 g  →  44 g
 4           4            4

[3g + 8g - 11g] → constant

8 g of O₂ will be consumed from 50g of O₂ and 42 g of O₂ will be unreactive. It shows that 12 g of carbon burns in 32 g oxygеn from 44 g of CO2 bесоuse 3 g of carbon reacts with 8 g of oxygen to form 11 g of CO₂. The same is given to us. Consequently 11.0 g of CO₂ will be formed when 3.0 g of carbon is burnt in 50 g of oxygen consuming 8 g of O₂, leaving behind 42 g of O₂. The answer governs the Law of constant Proportion.

Polyatomic Ions are that ions which have more than one atom of same kind of different kind and behave as a single unit is known as polyatomic ion.

E.g. : OH^- , SO₄^2- , PO₄^3-

(a)     MgCl₂
(b)     Cao
(c)     Cu(NO₃)₂
(d)     AlCl₃
(е)     CаCо₃

(a) Calcium oxide (Cao), There are two type of elements are present first in calcium and the other is oxygen 

(b) Chemical formula of Hydrogen Bromide is HBr. There are two types of element present, the first one in hydrogen and the other in Bromine.

(c) Chemical formula of Baking powder is NaHCO₃ (Sodium Hydrogen Carbonate). There are 4 types of element sodium, hydrogen, carbon and oxygen.

(d) Chemical formula of Potassium sulphate is K₂SO₄. There are three types of element: Potassium, sulphur and oxygen.

Molar mass: The mass of the mole of substance is known as molar mass.

Unit = gm/mol.

(a) C₂H₂ = H - C = C - H

     = 2 x Atomic mass of C + 2 x Atomic mass of H
     = 2 x 12 + 2 x 1
     = 24 + 2 = 26 g

(b) S₈ =  8 x atomic mass of sulphur.
      = 8 x 32 = 256 g
 

(c) P4 = 4 x atomic mass of phosphorus

     = 4 x 31 = 124 g

(d) HCl = 1 x atomic mass of H + 1 x atomic mass of Cl

      = 1 x 1 + 1 x 35.5
      = 1 + 35.5 = 36.5 g

(e) HNO₃ = 1 x atomic mass of H + 1 x atomic mass of N + 3 x atomic mass of O

      = 1 x 1 + 1 x 14 + 3 x 16
      = 1 + 14 + 48 = 63 g

1 mole of any particle = Relative mass of that particle in grams

1 mole of any molecule = molecular mass in grams

(a) 1 mole of nitrogen atom = 14 grams
(b) 1 mole of aluminium atom = 27 grams
     4 mole of aluminium atom = 27 x 4 = 108 grams
(c) Na2SO3 = 2 x atomic mass of Na + 1 x atomic mass of sulphur 3 x atomic mass of O
     = 2 x 23 + 1 x 32 + 3 x 16
     = 46 + 32 + 48 = 126 g

1 mole of Na₂SO₃ = 126 g 

10 moles of Na₂SO₃ = 126 x 10 = 1260 grams

Mass (in grams) = no. of moles x molecular mass (g/mol)

(a) Molar mass of O₂ = 2 x 16 = 32 g/mol
      Mass = no. of mole x molar mass
      12 = n x 32
      n = 12/3 = 3/8
      n = 0.375 mole 

(b) Given mass = 20 g
      Molar mass of H₂O = 2 x 1 + 1 x 16
                                      = 2 + 16 = 18 g/mol
      Mass = no. of mole x molar mass
      20 = n x 18
      n = 20/18 = 10/9
      n = 1.11 mole

(c) Given mass = 22g
      Molar mass of CO₂ = 1 x 12 + 2 X 16
                                      = 12 + 32 = 44 g/mol
      Mass = no. of moles x molar mass
      22 = n x 44
      n = 22/44 = ½
      n = 0.5. mole

(a) No. of moles = 0.2
     Molar mass of oxygen = 16 g/mol
     Mass = no. of moles of molar mass
     Mass = 0.2 x 16 = 3.2 g

(b) No. of moles = 0.5
     Molar mass of H₂O = 2 x 1 + 16
                                     = 2 x 16 = 18 g/mole
     Mast = no. of moles x molar mass
     Mass = 0.5 x 18 = 9.0 g

Molar mass of S₈ = 8 x 32 = 256 g

           256 g = 6.022 x 10²³ no. of molecules of S₈

No. of molecules present in 16 g of solid sulphur = 6.022x10²³ / 256 x 16 = 0.376 x 10²³

          = 3.76x10²²

Molar Mass of Al₂O₃ = 2 x Atomic mass of Al + 3 x Atomic mass of O
    = 2 x 27 + 3 x 16
    = 54 + 48 = 102 g 

102 g of Al₂O₃ contain = 2 x 6.022 x 10²³ aluminium ion
   = 0.051 g of Al₂O₃ = 2 x 6.022 x 10²³ / 102 x 0.051
   = 6.022 x 10²³ x 10^-3 = 6.022 x 10²⁰