Dalton’s Atomic Theory

• Matter consists of indivisible atoms.
• All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
• Compounds are formed when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.

John Dalton

Atomic Mass

• Atomic mass is the mass of the atom.
• Nowadays sophisticated techniques e.g. mass spectrometry is used to determine the atomic mass of the atom.
• But in 19^th century atomic mass was determined by calculating the mass of the atom relative to hydrogen by using stoichiometry.

History of Atomic Mass

• In 1803 John Dalton determined the atomic mass or atomic weight.
• Atomic weight was originally defined as relative to that of lightest element hydrogen taken as 1.
• Oxygen was taken as base as its atomic mass was 16(whole number).All the observations were based on stoichiometry.
• Chemists picked naturally occurring Oxygen, which is a mixture of isotopes Oxygen-16, Oxygen-17 and Oxygen-18 (Atomic mass 16.008).
• Physicists picked up carbon 12 as base  based  on mass  spectrometry & mass of one carbon -12 atom is a whole number.
• Carbon-12 was agreed to be taken as standard in 1961 because the mass of one carbon -12 atom is a whole number.
• The masses of all other atoms are taken relative to mass of 12C.

Average Atomic Mass

• Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence).

• From the above data, the average atomic mass of carbon will come out to be :
  • (0.98892) (12 u) + ( 0.01108) (13.00335 u) + (2 × 10^–12) (14.00317 u)
  • = 12.011 u

Problem:-

Calculate the atomic mass (average) of chlorine using the following data:-

Isotopes of Chlorine        % Natural Abundance                Molar Mass

³⁵Cl                         75.77                                      34.9689

³⁷Cl                       24.23                              36.9659

 Answer:-

Atomic mass of first isotope = 34.9689

Natural abundance of first isotope = 75.77% or 0.757

 Atomic mass of second isotope= 36.9659

 Natural abundance of second isotope= 24.23% or 0.242

Now average atomic mass of chlorine

= [34.9689x 0.757 + 36.9659x 0.242]/ (0.757+0.242) =35.4521

So, the average atomic mass of chlorine = 35.4527 u

Problem:-

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope        Isotopic molar mass         Abundance

³⁶Ar              35.96755 gmol^–1               0.337%

⁴⁰Ar            39.9624 gmol^–1                  99.600%

Answer:-

Molar mass of argon =

Atomic mass of ³⁶Ar = 35.96755 & abundance = 0.337

Or

Total atomic mass of ³⁶Ar = 35.96755 * 0.337= 0.121g/mol

Atomic mass of ³⁸Ar = 37.96272 & abundance = 0.063

Or

Total atomic mass of ³⁸Ar = 37.96272 * 0.063= 0.024g/mol

Atomic mass of ⁴⁰Ar =39.9624 & abundance = 99.600

Or

Total atomic mass of ⁴⁰Ar = 39.9624 * 99.600= 39.802g/mol

 Therefore molar mass of argon = total mass of ³⁶Ar+ total mass of ³⁸Ar + atomic mass of ⁴⁰Ar

= 0.121 + 0.024 + 39.802= 39.947 g/mol

Molecular Mass

• Molecular mass is the sum of atomic masses of the elements present in a molecule.
• For example:-Molecular mass of methane CH₄ = (12.011 u) + 4 (1.008 u) = 16.043 u.

Formula Mass

• Some substances such as sodium chloride do not contain discrete molecules as their constituent units. E.g. NaCl.
• In such compounds, positive (sodium) and negative (chloride) entities are arranged in a three-dimensional structure as shown in the figure.
  • For example: – Formula mass of sodium chloride (NaCl) = atomic mass of sodium (Na) + atomic mass of chlorine (Cl) = 23.0 u + 35.5 u = 58.5 u.

(Packing of Na⁺ and Cl^– ions in sodium chloride NaCl)

Problem:-

Calculate the molecular mass of the following:

(i) H₂O (ii) CO₂ (iii) CH₄

Answer:-

(i) H₂O:

The molecular mass of water, H₂O

= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)

= [2(1.0084) + 1(16.00 u)]

= 2.016 u + 16.00 u

= 18.016

= 18.02 u

(ii) CO₂:

The molecular mass of carbon dioxide, CO₂

= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)

= [1(12.011 u) + 2 (16.00 u)]

= 12.011 u + 32.00 u

= 44.01 u

(iii) CH₄:

The molecular mass of methane, CH₄

= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)

= [1(12.011 u) + 4 (1.008 u)]

= 12.011 u + 4.032 u

= 16.043 u

Question 1.
Calculate the molecular mass of the following :
(i) H₂o
(ii) CO₂
(iii) CH₄
Solution:
(i) Molecular mass of H₂O :2 × 1 + 1 × 16 = 18u
(ii) Molecular mass of CO₂ :1 × 12 + 2× 16 = 44 u
(iii) Molecular mass of CH₄ : 12 + 4 × 1 = 16 u