Dalton's Atomic Theory (1808)

The idea that matter is made of tiny indivisible particles isn't new — it goes all the way back to Democritus, a Greek philosopher (460–370 BC), who called these particles "a-tomio" (meaning indivisible). But it was John Dalton who, in 1808, put this idea on a proper scientific footing in his book A New System of Chemical Philosophy.

Dalton's Postulates

  1. Matter consists of indivisible atoms. Atoms are the smallest particles and cannot be further divided.
  2. All atoms of a given element are identical — they have the same mass and same chemical properties. Atoms of different elements have different masses.
  3. Compounds are formed when atoms of different elements combine in a fixed ratio. This explains why compounds have fixed composition.
  4. Chemical reactions involve reorganisation of atoms. Atoms are neither created nor destroyed — they simply rearrange.

What Dalton's Theory Explained

Dalton's atomic theory beautifully explained three laws of chemical combination:

  • Law of Conservation of Mass — atoms are conserved (postulate 4)
  • Law of Definite Proportions — atoms combine in fixed ratios (postulate 3)
  • Law of Multiple Proportions — different compounds have different atom ratios (postulates 2 & 3)

Limitation

Dalton's theory could NOT explain Gay Lussac's Law of Gaseous Volumes. This is because Dalton believed that atoms of the same element cannot combine with each other — so he rejected the idea of diatomic molecules like H2\text{H}_2 and O2\text{O}_2. This limitation was resolved by Avogadro.

[Board Important] Dalton's postulates and their connection to the laws of chemical combination is a very frequently tested topic.

Key Point: Dalton's theory: atoms are indivisible, identical within an element, combine in fixed ratios, and are conserved in reactions. It explains conservation, definite proportions, and multiple proportions — but NOT gaseous volumes.

Atomic Mass

Atoms are incredibly tiny — a single hydrogen atom has a mass of just 1.6736×10−241.6736 \times 10^{-24} g. Working with such small numbers directly would be impractical, so chemists use a relative scale instead.

The Atomic Mass Unit (amu / u)

The current system (adopted in 1961) uses Carbon-12 (12C^{12}\text{C}) as the standard:

1 u=112×mass of one 12C atom1 \text{ u} = \frac{1}{12} \times \text{mass of one } ^{12}\text{C atom}

1 u=1.66056×10−24 g1 \text{ u} = 1.66056 \times 10^{-24} \text{ g}

The term "amu" has been replaced by "u" (unified mass), though both are used interchangeably.

Some Important Atomic Masses

Atom Absolute Mass (g) Mass in u
Hydrogen (1H^1\text{H}) 1.6736×10−241.6736 \times 10^{-24} 1.008 u
Carbon (12C^{12}\text{C}) 1.9927×10−231.9927 \times 10^{-23} 12.000 u (exact, by definition)
Oxygen (16O^{16}\text{O}) 2.6567×10−232.6567 \times 10^{-23} 15.995 u

Historical Note

Before the C-12 standard, hydrogen (mass = 1) was used as the reference. The switch to C-12 was made because it gave more consistent and accurate relative masses for all elements.

[JEE Tip] Know the exact definition: 1 u = 112\frac{1}{12} the mass of one 12C^{12}\text{C} atom = 1.66056×10−241.66056 \times 10^{-24} g. This conversion is used frequently in numerical problems.

Key Point: 1 u (atomic mass unit) = 112\frac{1}{12} mass of 12C^{12}\text{C} = 1.66056×10−241.66056 \times 10^{-24} g. This relative scale makes atomic-level calculations practical.

Average Atomic Mass

Many elements exist as a mixture of isotopes — atoms of the same element with different masses (different number of neutrons). The atomic mass listed in the periodic table is actually the weighted average of the masses of all naturally occurring isotopes.

Formula

Average atomic mass=∑(fractional abundancei×massi)\text{Average atomic mass} = \sum (\text{fractional abundance}_i \times \text{mass}_i)

Example: Carbon

Carbon has three isotopes:

Isotope Relative Abundance (%) Atomic Mass (u)
12C^{12}\text{C} 98.892 12.000
13C^{13}\text{C} 1.108 13.00335
14C^{14}\text{C} 2×10−102 \times 10^{-10} 14.00317

Average=(0.98892)(12)+(0.01108)(13.00335)+(2×10−12)(14.00317)\text{Average} = (0.98892)(12) + (0.01108)(13.00335) + (2 \times 10^{-12})(14.00317) =11.867+0.144+negligible=12.011 u= 11.867 + 0.144 + \text{negligible} = 12.011 \text{ u}

This is why the periodic table shows carbon's atomic mass as 12.011 u, not exactly 12.

Example: Chlorine

Chlorine has two isotopes: 35Cl^{35}\text{Cl} (75.77%) and 37Cl^{37}\text{Cl} (24.23%) Average=(0.7577)(35)+(0.2423)(37)=26.52+8.97=35.49≈35.5 u\text{Average} = (0.7577)(35) + (0.2423)(37) = 26.52 + 8.97 = 35.49 \approx 35.5 \text{ u}

This is why chlorine's atomic mass is 35.5 — it's not a whole number because it's an average of two isotopes.

[NEET Important] Understanding why atomic masses in the periodic table are not whole numbers (because they're averages of isotope masses) is frequently tested.

Key Point: The atomic mass in the periodic table is a weighted average of all isotopic masses. That's why most elements have non-integer atomic masses.

Molecular Mass

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

Molecular mass=∑(number of atoms of element×atomic mass of element)\text{Molecular mass} = \sum (\text{number of atoms of element} \times \text{atomic mass of element})

Examples

Methane (CH4\text{CH}_4): =1(12.011)+4(1.008)=12.011+4.032=16.043 u= 1(12.011) + 4(1.008) = 12.011 + 4.032 = 16.043 \text{ u}

Water (H2O\text{H}_2\text{O}): =2(1.008)+1(16.00)=2.016+16.00=18.02 u= 2(1.008) + 1(16.00) = 2.016 + 16.00 = 18.02 \text{ u}

Glucose (C6H12O6\text{C}_6\text{H}_{12}\text{O}_6): =6(12.011)+12(1.008)+6(16.00)=72.066+12.096+96.00=180.162 u= 6(12.011) + 12(1.008) + 6(16.00) = 72.066 + 12.096 + 96.00 = 180.162 \text{ u}

Formula Mass

Some substances — especially ionic compounds — don't exist as discrete molecules. For example, sodium chloride (NaCl) exists as a 3D crystal lattice of Na+\text{Na}^+ and Cl−\text{Cl}^- ions, where each Na+\text{Na}^+ is surrounded by 6 Cl−\text{Cl}^- and vice versa.

For such substances, we use formula mass instead of molecular mass:

Formula mass of NaCl=23.0+35.5=58.5 u\text{Formula mass of NaCl} = 23.0 + 35.5 = 58.5 \text{ u}

When to Use Which?

Term Used For Example
Molecular mass Covalent/molecular compounds H2O\text{H}_2\text{O} (18.02 u), CO2\text{CO}_2 (44.01 u)
Formula mass Ionic compounds NaCl (58.5 u), CaCO3\text{CaCO}_3 (100 u)

[JEE Tip] In practice, both terms are often used interchangeably. But in theory questions, remember: NaCl has a "formula mass" (not "molecular mass") because it doesn't exist as discrete molecules.

Key Point: Molecular mass = sum of atomic masses of all atoms in a molecule. Formula mass is used for ionic compounds that exist as crystal lattices, not discrete molecules.

Solved Examples

Example 1: Absolute Mass of an Atom

The atomic mass of oxygen is 16 u. Calculate the absolute mass of one oxygen atom in grams.

Solution:

  1. Given: Atomic mass of O = 16 u, and 1 u = 1.66056×10−241.66056 \times 10^{-24} g
  2. Calculate: Mass of 1 O atom=16×1.66056×10−24 g\text{Mass of 1 O atom} = 16 \times 1.66056 \times 10^{-24} \text{ g} =26.569×10−24 g=2.6569×10−23 g= 26.569 \times 10^{-24} \text{ g} = 2.6569 \times 10^{-23} \text{ g}

Final Answer: Mass of one oxygen atom = 2.6569×10−232.6569 \times 10^{-23} g

Takeaway: To convert from u to grams, multiply by 1.66056×10−241.66056 \times 10^{-24}.

Example 2: Molecular Mass of Glucose

Calculate the molecular mass of glucose (C6H12O6\text{C}_6\text{H}_{12}\text{O}_6).

Solution: Atomic masses: C = 12.011 u, H = 1.008 u, O = 16.00 u

Molecular mass=6(12.011)+12(1.008)+6(16.00)\text{Molecular mass} = 6(12.011) + 12(1.008) + 6(16.00) =72.066+12.096+96.00=180.162 u= 72.066 + 12.096 + 96.00 = 180.162 \text{ u}

Final Answer: Molecular mass of glucose = 180.162 u ≈ 180 u

Takeaway: For large organic molecules, calculate each element's contribution separately, then add.

Example 3: Average Atomic Mass of Boron

Boron has two naturally occurring isotopes: 10B^{10}\text{B} (mass = 10.013 u, abundance = 19.9%) and 11B^{11}\text{B} (mass = 11.009 u, abundance = 80.1%). Calculate the average atomic mass.

Solution: Average=(0.199)(10.013)+(0.801)(11.009)\text{Average} = (0.199)(10.013) + (0.801)(11.009) =1.993+8.818=10.811 u= 1.993 + 8.818 = 10.811 \text{ u}

Final Answer: Average atomic mass of boron = 10.811 u (periodic table shows 10.81) ✓

Takeaway: The average is always closer to the more abundant isotope. Since 11B^{11}\text{B} is 80.1% abundant, the average (10.81) is much closer to 11 than to 10.

Example 4: Finding Isotope Abundance

Copper has two isotopes: 63Cu^{63}\text{Cu} (mass = 62.93 u) and 65Cu^{65}\text{Cu} (mass = 64.93 u). The average atomic mass is 63.55 u. Find the percentage abundance of each.

Solution:

  1. Let fraction of 63Cu^{63}\text{Cu} = xx, fraction of 65Cu^{65}\text{Cu} = (1−x)(1-x)
  2. Equation: 63.55=x(62.93)+(1−x)(64.93)63.55 = x(62.93) + (1-x)(64.93)
  3. Solve: 63.55=62.93x+64.93−64.93x63.55 = 62.93x + 64.93 - 64.93x 63.55=64.93−2.00x63.55 = 64.93 - 2.00x 2.00x=1.38  ⟹  x=0.692.00x = 1.38 \implies x = 0.69

Final Answer: 63Cu^{63}\text{Cu}: 69%, 65Cu^{65}\text{Cu}: 31%

Takeaway: Given the average mass and individual isotope masses, set up a linear equation with abundance as the unknown.

Example 5: Formula Mass of Ionic Compounds

Calculate the formula mass of: (a) CaCO3\text{CaCO}_3, (b) Na2SO4\text{Na}_2\text{SO}_4.

Solution:

(a) CaCO3\text{CaCO}_3: =40+12+3(16)=40+12+48=100 u= 40 + 12 + 3(16) = 40 + 12 + 48 = 100 \text{ u}

(b) Na2SO4\text{Na}_2\text{SO}_4: =2(23)+32+4(16)=46+32+64=142 u= 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 \text{ u}

Takeaway: Formula mass calculation is identical to molecular mass calculation — just add up the atomic masses.

Example 6: Molecular Mass of Sulphuric Acid

Calculate the molecular mass of H2SO4\text{H}_2\text{SO}_4.

Solution: =2(1.008)+32.07+4(16.00)=2.016+32.07+64.00=98.086 u≈98 u= 2(1.008) + 32.07 + 4(16.00) = 2.016 + 32.07 + 64.00 = 98.086 \text{ u} \approx 98 \text{ u}

Takeaway: The molecular mass of H2SO4\text{H}_2\text{SO}_4 (≈ 98 u) is one of the most commonly used values. Memorise it!

Example 7: Dalton's Theory and Conservation of Mass

Using Dalton's atomic theory, explain why 4 g of hydrogen always combines with 32 g of oxygen to form 36 g of water.

Solution: 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}

  1. Fixed ratios (Postulate 3): Water always has formula H2O\text{H}_2\text{O} — 2 H atoms per 1 O atom.
  2. Mass calculation: 4 u of H (from 4 H atoms) + 32 u of O (from 2 O atoms) = 36 u of water (2 molecules)
  3. Conservation (Postulate 4): Reactant mass (4 + 32 = 36) = Product mass (36). Atoms rearrange, not created or destroyed.

Takeaway: Dalton's theory connects microscopic atom behaviour with macroscopic mass observations.

Example 8: Average Atomic Mass of Chlorine

Chlorine has two isotopes: 35Cl^{35}\text{Cl} (75.77%, mass = 34.97 u) and 37Cl^{37}\text{Cl} (24.23%, mass = 36.97 u). Calculate the average atomic mass and explain why it's 35.5 u.

Solution: Average=(0.7577)(34.97)+(0.2423)(36.97)=26.50+8.96=35.46 u≈35.5 u\text{Average} = (0.7577)(34.97) + (0.2423)(36.97) = 26.50 + 8.96 = 35.46 \text{ u} \approx 35.5 \text{ u}

Why 35.5? No individual Cl atom weighs 35.5 u — this is a statistical average reflecting 75.77% of atoms being mass-35 and 24.23% being mass-37. The average is closer to 35 because that isotope is more abundant.

Takeaway: Non-integer atomic masses always indicate a mixture of isotopes.

Example 9: Molecular Mass of Ethanol

Calculate the molecular mass of ethanol (C2H5OH\text{C}_2\text{H}_5\text{OH}).

Solution: First, note the molecular formula: C2H5OH=C2H6O\text{C}_2\text{H}_5\text{OH} = \text{C}_2\text{H}_6\text{O}

Molecular mass=2(12)+6(1)+16=24+6+16=46 u\text{Molecular mass} = 2(12) + 6(1) + 16 = 24 + 6 + 16 = 46 \text{ u}

Takeaway: When given a structural formula like C2H5OH\text{C}_2\text{H}_5\text{OH}, first count the total atoms: 2C + 6H (5 from C2H5\text{C}_2\text{H}_5 + 1 from OH) + 1O.

Example 10: Ordering by Molecular Mass

Arrange in order of increasing molecular/formula mass: H2O\text{H}_2\text{O}, CO2\text{CO}_2, NaCl, CaCO3\text{CaCO}_3, H2SO4\text{H}_2\text{SO}_4

Solution:

  1. H2O\text{H}_2\text{O}: 2(1)+16=182(1) + 16 = 18 u
  2. CO2\text{CO}_2: 12+2(16)=4412 + 2(16) = 44 u
  3. NaCl: 23+35.5=58.523 + 35.5 = 58.5 u
  4. H2SO4\text{H}_2\text{SO}_4: 2(1)+32+4(16)=982(1) + 32 + 4(16) = 98 u
  5. CaCO3\text{CaCO}_3: 40+12+3(16)=10040 + 12 + 3(16) = 100 u

Order: H2O\text{H}_2\text{O} (18) < CO2\text{CO}_2 (44) < NaCl (58.5) < H2SO4\text{H}_2\text{SO}_4 (98) < CaCO3\text{CaCO}_3 (100)

Takeaway: Quick molecular mass calculations are an essential chemistry skill — practise until they become second nature.