Short Notes: Basic Concepts of Chemistry

1.1 Fundamental Laws of Chemical Combination

Chemical reactions follow basic, experimentally established laws that describe how matter combines. These laws provide the foundation for atomic theory and stoichiometry. Key laws are given below with simple explanations and illustrative examples.

Law of Conservation of Mass

The total mass of reactants in a chemical reaction equals the total mass of products. Mass is neither created nor destroyed in an ordinary chemical change.

Example: In the reaction between hydrogen and oxygen to form water, the mass of hydrogen plus the mass of oxygen before reaction equals the mass of water formed.

Chemical equation:

2H2 + O2 → 2H2O

Law of Definite Proportions (Proust)

A chemical compound always contains the same elements in a fixed proportion by mass, regardless of its source or method of preparation.

Example: Pure water always contains hydrogen and oxygen in a mass ratio of approximately 1:8 (or 11.11% H and 88.89% O by mass).

Law of Multiple Proportions (Dalton)

When two elements form more than one compound, the different masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers.

Example: Carbon and oxygen form CO and CO₂. For a fixed mass of carbon, the masses of oxygen that combine are in a 1:2 ratio.

Gay-Lussac's Law of Gaseous Volumes

When gases react together at constant temperature and pressure, the volumes of reacting gases and the volumes of gaseous products (if measured in the gaseous state) are in simple whole-number ratios.

Example: Hydrogen and oxygen react according to the volume ratio 2 : 1 to give water (as per the balanced molecular equation for gaseous reactants):

2 volumes of H2 : 1 volume of O2 → 2 volumes of H2O (gases if water remains gaseous)

Avogadro's Law (Hypothesis)

Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. This leads to the important concept of the mole and molecular mass.

Mathematical form:

\( \dfrac{V}{n} = \text{constant} \)

where V is volume and n is the amount of substance in moles.

Dalton's Atomic Theory (Basic Points)

• Atoms are indivisible particles that make up elements.
• All atoms of a given element are identical in mass and properties (later modified by isotopes).
• Compounds are formed by combination of atoms in simple whole-number ratios.
• Chemical reactions rearrange atoms but do not create or destroy them.

Limitations: The theory does not explain isotopes or subatomic structure; it was refined after discovery of electrons, protons and neutrons.

Applications of Fundamental Laws

• Establish stoichiometry for quantitative calculations in chemical reactions.
• Determine empirical and molecular formulas from mass or percentage data.
• Predict volumes in gaseous reactions using Avogadro's principle.

1.2 Mole Concept

The mole is the central unit for quantitative chemistry. It links microscopic particles (atoms, molecules, ions) to macroscopic amounts that can be weighed.

Definition of Mole

Mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, electrons) as there are atoms in exactly 0.012 kilogram of carbon-12.

Avogadro constant: N_A = 6.022 × 10²³ entities per mole.

Molar Mass and Relative Masses

Molar mass (M) is the mass of one mole of a substance expressed in grams per mole (g mol^-1). Numerically, molar mass in g mol^-1 equals the relative molecular or atomic mass in unified atomic mass units (u).

• Relative atomic mass: average mass of atoms of an element compared to 1/12 of carbon-12.
• Relative molecular mass (M_r): sum of relative atomic masses of atoms in a molecule.

Relation between Mass, Moles and Molar Mass

Number of moles, mass and molar mass are related as follows.

\( n = \dfrac{m}{M} \)

where n is amount in moles, m is mass in grams, and M is molar mass in g mol^-1.

Example - Simple Mole Calculation

Problem: Find the number of moles in 18 g of water (H2O).

Mass of one mole of water (molar mass) is 2 × 1.008 + 16.00 = 18.016 g mol^-1.

\( n = \dfrac{m}{M} \)

\( n = \dfrac{18\,\text{g}}{18.016\,\text{g mol}^{-1}} \)

Therefore, n ≈ 0.999 mol (approximately 1.00 mol).

Empirical and Molecular Formula

Empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. Molecular formula gives the actual number of atoms of each element in a molecule.

To determine empirical formula from percentage composition:

• Assume 100 g of compound so that percentages become masses in grams.
• Convert masses to moles using \( n = \dfrac{m}{M} \).
• Divide by smallest number of moles to obtain simplest whole-number ratio.
• If necessary, multiply ratios by an integer to get whole numbers.

Worked Example - Empirical Formula

Problem: A compound contains 40.00% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Determine its empirical formula.

Assume 100 g sample:

Mass of C = 40.00 g

Mass of H = 6.67 g

Mass of O = 53.33 g

Convert to moles using atomic masses (C = 12.01, H = 1.008, O = 16.00):

\( n_{\text{C}} = \dfrac{40.00}{12.01} \)

\( n_{\text{H}} = \dfrac{6.67}{1.008} \)

\( n_{\text{O}} = \dfrac{53.33}{16.00} \)

Divide each by the smallest of these mole values and obtain the simplest whole-number ratio.

The ratio is approximately C : H : O = 1 : 2 : 1, so empirical formula = CH2O.

Molecular Formula from Empirical Formula

Find the empirical formula mass (EFM). Then

\( \text{Molecular formula} = (\text{empirical formula}) \times \dfrac{\text{molar mass}}{\text{EFM}} \)

The ratio must be an integer.

Stoichiometry and Limiting Reagent

Stoichiometry uses mole ratios from balanced equations to calculate amounts of reactants and products.

Limiting reagent is the reactant that gets completely consumed first and thus limits the amount of product formed. To find it, calculate the amount of product each reactant can produce and identify the smaller value.

Percentage Yield

Percentage yield is the efficiency of a reaction and is defined as:

\( \%\text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \)

Important Numerical Facts and Conversions

• 1 mole of any substance contains 6.022 × 10²³ entities (Avogadro constant).
• Molar volume of an ideal gas at STP (0 °C, 1 atm) is approximately 22.4 L mol^-1. At other conditions use the ideal gas law to convert.
• Use the ideal gas equation when required: \( PV = nRT \). Place the equation on its own line when used.

Summary

The fundamental laws establish the quantitative relationships in chemical changes and led to the mole concept, which connects measurable masses to numbers of particles. Mastery of the mole concept, molar masses, empirical/molecular formulas, stoichiometry, limiting reagents and percentage yield is essential for quantitative problems in chemistry.