Comprehensive Study Notes on Atomic Theory, Mole Concept, and Percentage Composition

Limitations of Dalton's Atomic Theory

• It could not explain why atoms of different elements differ in their mass, size, valence, and other properties.

• Discovery of isotopes: Dalton's theory assumed all atoms of an element are identical, but isotopes have the same atomic number with different mass numbers.

• Discovery of nuclear reactions: These show that mass can be converted into energy and vice-versa ($E = mc^2$), contradicting the idea of mass conservation in a strictly Daltonian sense.

• An atom is divisible: Dalton postulated atoms were indivisible, but they are composed of subatomic particles (protons, neutrons, and electrons).

• It could not explain why atoms combine to form compounds (molecules).

• Transmutation of elements: Through nuclear reactions, atoms of one element can be converted into atoms of other elements.

• Isobars: Atoms of different elements may have the same mass. For example, the atomic mass of Calcium ($Ca$) and Argon ($Ar$) is $40$. These are called isobars.

• Non-simple ratios: Atoms may not always combine in simple whole-number ratios as suggested by Dalton. For example, in sugar ($C_{12}H_{22}O_{11}$), the ratio of atoms is $12:22:11$, which is not a simple ratio.

Fundamental Concepts of Atomic and Molecular Mass

• Atomic Mass: This is the total mass of an atom, represented as the sum of the masses of its protons and neutrons. The standard unit is the atomic mass unit ($a.m.u$ or $u$).

• Molecular Mass: The sum of the atomic masses of all the atoms present in a molecule of a substance.

• Formula Mass: Defined as the sum of the atomic masses of constituent atoms in an ionic compound. This term is used for substances that do not contain discrete molecules but consist of ions in a crystal lattice (e.g., $NaCl$).

• Gram Atomic Mass (GAM): Atomic mass expressed specifically in grams. For example, for Sodium ($Na$), the atomic mass is $23 \times a.m.u$, and the $GAM$ is $23 \times g$.

• Formula Mass Calculation Example: For $NaCl$, the mass is $23 + 35.5 = 58.5$.

Heuristic (Trick) for Calculating Atomic Mass (A)

A common shortcut used for the first 20 elements based on the atomic number ($Z$):

• If the Atomic Number ($Z$) is Even: $A = Z \times 2$

• If the Atomic Number ($Z$) is Odd: $A = Z \times 2 + 1$

Exceptions and Specific Examples:

• Hydrogen ($H$): $Z = 1$. Exception: $A = 1$.

• Helium ($He$): $Z = 2$ (Even). $A = 2 \times 2 = 4 \times a.m.u$.

• Lithium ($Li$): $Z = 3$ (Odd). $A = 3 \times 2 + 1 = 7 \times a.m.u$.

• Beryllium ($Be$): $Z = 4$ (Even). Exception: $A = 9 \times a.m.u$ (Follows the odd rule despite being even).

• Oxygen ($O$): $Z = 8$ (Even). $A = 8 \times 2 = 16 \times a.m.u$.

• Sodium ($Na$): $Z = 11$ (Odd). $A = 11 \times 2 + 1 = 23 \times a.m.u$.

• Sulfur ($S$): $Z = 16$ (Even). $A = 16 \times 2 = 32 \times g$ (expressed as GAM).

• Chlorine ($Cl$): $Z = 17$. Exception: $A = 35.5 \times a.m.u$ or $35.5 \times g$.

• Potassium ($K$): $Z = 19$ (Odd). $A = 19 \times 2 + 1 = 39 \times a.m.u$ or $39 \times g$.

Molecular Mass Calculations

• Water ($H_2O$): $(1 \times 2) + 16 = 18 \times g$.

• Sulfuric Acid ($H_2SO_4$): $(1 \times 2) + 32 + (16 \times 4) = 2 + 32 + 64 = 98 \times g$.

• Methane ($CH_4$): $12 + (1 \times 4) = 16 \times g$.

• Ethane ($C_2H_6$): $(12 \times 2) + (1 \times 6) = 24 + 6 = 30 \times g$.

• Glucose ($C_6H_{12}O_6$): $(12 \times 6) + (1 \times 12) + (16 \times 6) = 72 + 12 + 96 = 180 \times g$.

Average Atomic Mass and Isotopes

• Isotopes: Atoms with the same atomic number ($Z$) but different mass numbers ($A$).

• Average Atomic Mass: The sum of the masses of an element's isotopes, each multiplied by its natural abundance (relative proportion).

• Case Study: Carbon Isotopes:

  • Carbon-12 ($^{12}C$): Mass = $12$, Relative Abundance = $98.892 \%$.

  • Carbon-13 ($^{13}C$): Mass = $13.0035$, Relative Abundance = $1.108 \%$.

  • Carbon-14 ($^{14}C$): Mass = $14.0037$, Relative Abundance = $2 \times 10^{-10} \%$.

• Calculation for Average Atomic Mass of Carbon:

  • $\frac{98.892 \times 12 + 1.108 \times 13.0035 + 2 \times 10^{-10} \times 14.0037}{100} = 12.011 \times u$.

The Mole Concept

• Definition: A mole is the amount of substance containing as many entities (atoms, molecules, ions) as there are atoms in exactly $12 \times g$ of the Carbon-12 ($^{12}C$) isotope.

• Avogadro's Number ($N_A$): $N_A = 6.022 \times 10^{23}$.

• Standard Comparisons:

  • $1 \times \text{dozen eggs} = 12 \times \text{eggs}$.

  • $1 \times \text{pair of shoes} = 2 \times \text{shoes}$.

  • $1 \times \text{mole} = 6.022 \times 10^{23} \times \text{particles}$.

• Volume at STP: A mole of any gas occupies a volume of $22.4 \times litre$ at Standard Temperature and Pressure (STP).

Importance and Application of Avogadro's Number & Mole Concept

1. Mass of a single atom or molecule can be calculated.

2. Volume of a single atom or molecule can be calculated.

3. Total atoms in a given mass of an element can be calculated.

4. Total molecules in a given mass of a compound can be calculated.

5. Total molecules in a given volume of a gas under specified conditions can be calculated.

6. Simplifies data representation in chemical calculations.

Core Formulas for Mole Calculations

1. Gram atoms (moles) of an element ($n$): $n = \frac{\text{Mass of element in grams (w)}}{\text{Atomic mass of element (M)}}$

2. Gram moles of a substance: $n = \frac{\text{Mass of substance in grams (w)}}{\text{Molecular mass of the substance (M)}}$

3. Gram moles of a gas: $n = \frac{\text{Volume in litre at STP}}{22.4}$

4. Number of atoms: $\text{Number of atoms} = \text{Moles of atoms} \times N_A$

5. Number of molecules: $\text{Number of molecules} = \text{Moles} \times N_A$

6. Number of molecules of a gas: $\text{Number of molecules} = \frac{\text{Volume in litre at STP} \times N_A}{22.4}$

7. Molar Relationships:

   • Mass of $6.022 \times 10^{23}$ molecules = Molecular Mass.

   • Mass of $22.4 \times litre$ of gas at STP = Molecular Mass.

   • Mass of $6.022 \times 10^{23}$ atoms = Atomic Mass.

Practice Questions and Discussion

• Q: Calculate the number of moles in $49 \times g$ of $H_2SO_4$.

  • Molar Mass of $H_2SO_4 = 98 \times g/mol$.

  • $n = \frac{49}{98} = 0.5 \times mol$.

• Q: Calculate the number of atoms in $0.2 \times mol$ of Sodium ($Na$).

  • $\text{Atoms} = \text{Moles} \times 6.022 \times 10^{23}$.

  • $\text{Atoms} = 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23} \times \text{atoms}$.

• Q: Calculate the number of moles in $160 \times g$ of Methane ($CH_4$).

  • Molar Mass of Methane = $16 \times g/mol$.

  • $n = \frac{160}{16} = 10 \times mol$.

• Q: Calculate the number of moles in $160 \times g$ of Ethane ($C_2H_6$).

  • Molar Mass of Ethane = $(12 \times 2) + (1 \times 6) = 30 \times g/mol$.

  • $n = \frac{160}{30} = 5.33 \times mol$.

• Q: How many grams of glucose ($C_6H_{12}O_6$) are in $5 \times moles$?

  • Molar Mass of Glucose = $180 \times g/mol$.

  • $w = n \times M = 5 \times 180 = 900 \times g$.

• Q: Calculate the number of Hydrogen atoms in $90 \times g$ of glucose.

  • Step 1: Find moles of glucose: $n = \frac{90}{180} = 0.5 \times mol$.

  • Step 2: Find number of glucose molecules: $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \times \text{molecules}$.

  • Step 3: Find H-atoms (each molecule has 12 H-atoms): $12 \times 3.011 \times 10^{23} = 36.132 \times 10^{23} = 3.6132 \times 10^{24} \times \text{atoms}$.

Percentage Composition

• Definition: It is the percentage of each element in a compound by mass.

• Universal Formula:

  • $\% \times \text{composition} = \frac{n \times M(\text{element})}{\text{Total Molecular Mass of Compound}} \times 100$

  • Or: $\% \times \text{comp} = \frac{\text{Mass of specific element}}{\text{Total mass}} \times 100$

Calculation: Percentage of Potassium ($K$) in Potassium Dichromate ($K_2Cr_2O_7$)

• Given Atomic Masses: $K = 39$, $Cr = 52$, $O = 16$.

• Total Molecular Mass of $K_2Cr_2O_7$:

  • $K_2 = 39 \times 2 = 78$

  • $Cr_2 = 52 \times 2 = 104$

  • $O_7 = 16 \times 7 = 112$

  • $\text{Total Mass} = 78 + 104 + 112 = 294 \times g/mol$.

• Percentage Composition of Potassium ($K$):

  • $\% \times K = \frac{78}{294} \times 100$

  • $\% \times K = 0.2653 \times 100 = 26.53 \%$.