Comprehensive Notes on Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry

Chemistry is the branch of science that studies the preparation, properties, structure, and reactions of material substances. Key branches include inorganic, organic, physical, analytical, polymer, biochemistry, medicinal, industrial, hydrochemistry, electrochemistry, and green chemistry.

Matter

Matter is anything that occupies space and has mass. It exists in several states:

1. Solid
2. Liquid
3. Gas
4. Plasma
5. Bose-Einstein condensate
6. Fermionic condensate
7. Quark-Gluon Plasma

On Earth, matter predominantly exists in solid, liquid, and gaseous states.

States of Matter

• Solids: Particles are orderly arranged and closely packed, restricting movement. They have definite shape and volume.
• Liquids: Particles are close but can move, allowing liquids to have definite volume but not definite shape.
• Gases: Particles are far apart with easy and fast movement, resulting in no definite shape or volume. Gases take the shape and volume of their container.

The states of matter are interconvertible by changing temperature and pressure conditions.

Classification of Matter

Based on chemical composition, matter is divided into pure substances and mixtures.

Pure Substances

Contain only one type of particle. Examples include sodium (Na), potassium (K), hydrogen (H), oxygen (O), helium (He), carbon dioxide (CO2), water (H2O), ammonia (NH3), and cane sugar (C12H22O11).

Pure substances are classified into elements and compounds.

• Elements: Pure substances containing only one type of atom. Robert Boyle introduced the term 'element'. There are 118 known elements, from hydrogen ($1H$) to oganesson ($118Og$). Elements can be monoatomic (e.g., metals like sodium, noble gases like helium), diatomic (e.g., hydrogen, nitrogen, oxygen), or polyatomic (e.g., phosphorus ($P4$), sulfur ($S8$)).
• Compounds: Pure substances containing more than one type of atom, formed by combining two or more elements in a definite ratio. They can only be separated by chemical methods. Examples: CO2, H2O, NH3, H2SO4.

Mixtures

Contain more than one type of particle, separable by physical methods like filtration, crystallization, and distillation. Examples include solutions, gold ornaments, sea water, muddy water, and air.

Mixtures are either homogeneous or heterogeneous.

• Homogeneous Mixtures: Have uniform composition throughout (e.g., solutions, air).
• Heterogeneous Mixtures: Have different compositions in different parts (e.g., sea water, soil, muddy water).

Physical and Chemical Properties

Matter has physical and chemical properties.

• Physical Properties: Measurable without changing the substance's composition. Measurement does not require a chemical change (e.g., color, odor, melting point, boiling point, density, mass).
• Chemical Properties: Measurable only with a chemical change (e.g., composition, combustibility, reactivity with acids and bases).

Measurement of Physical Properties

Quantitative observations are represented by a number and a unit. The International System of Units (SI) is the standard, featuring seven base units:

Mass and Weight

• Mass: Amount of matter in a body; a constant quantity (SI unit: kilogram (kg)).
• Weight: Gravitational force acting on a body; a variable quantity (SI unit: newton (N)).

Volume (V)

Amount of space occupied by a body (SI unit: m3). Smaller units like cm3, dm3, mL, and L are commonly used in Chemistry.

$1 m^3 = 10^6 cm^3$
$1 L = 10^3 cm^3 (mL)$
$1 cm^3 = 1 mL$
$1 dm^3 = 10^3 cm^3$
$1 dm^3 = 1 L$

Density (d)

Mass per unit volume. $density = \frac{mass}{volume}$. SI unit is kg/m3, commonly expressed as g/cm3.

Temperature (T)

Degree of hotness or coldness (commonly in °C). Other units include °F and K. SI unit is kelvin (K).

$°F = \frac{9}{5}(°C) + 32$
$K = °C + 273.15$
$K = °C + 273$

Precision and Accuracy

• Precision: Closeness of multiple measurements of the same quantity.
• Accuracy: Agreement of a measurement to the true value.

Scientific Notation

Expressing numbers as $N × 10^n$, where $n$ is an exponent (positive or negative) and $N$ is a number between 1.000… and 9.999….. If the decimal is shifted to the left $n$ is positive and if the decimal is shifted to the right, $n$ is negative.

• Ex: 368.9 = $3.689 × 10^2$
• Ex: 0.000563 = $5.63 × 10^{-4}$

Significant Figures

Meaningful digits known with certainty plus one uncertain digit.

Rules for determining significant figures:

1. All non-zero digits are significant (e.g., 285 cm has three significant figures).
2. Zeros preceding the first non-zero digit are not significant (e.g., 0.03 has one significant figure).
3. Zeros between two non-zero digits are significant (e.g., 2.005 has four significant figures).
4. Zeros at the end or right of a number are significant if they are on the right side of the decimal point (e.g., 0.200 g has three significant figures).
5. Exact numbers have an infinite number of significant figures (e.g., 2 balls = 2.000000).
6. In scientific notation, all digits are significant (e.g., $4.01 × 10^2$ has three significant figures).

Rules for rounding off numbers:

1. If the rightmost digit to be removed is more than 5, the preceding number is increased by one (e.g., 1.386 rounds to 1.39).
2. If the rightmost digit to be removed is less than 5, the preceding number is not changed (e.g., 4.334 rounds to 4.33).
3. If the rightmost digit to be removed is 5:
   • If the preceding number is even, it remains unchanged (e.g., 6.25 rounds to 6.2).
   • If the preceding number is odd, it is increased by one (e.g., 6.35 rounds to 6.4).

Laws of Chemical Combinations

1. Law of Conservation of Mass

Proposed by Antoine Lavoisier, stating that matter is neither created nor destroyed. In a chemical reaction, the total mass of the reactants equals the total mass of the products. Chemical equations are balanced according to this law.

Illustration: $2H<em>2 + O</em>2 → 2H_2O$ (4 g of H2 + 32 g of O2 = 36 g of H2O)

2. Law of Definite Proportions

Proposed by Joseph Proust, stating that a given compound always contains the same proportion of elements by weight, i.e., the same elements combined in a fixed ratio by mass.

Illustration: Carbon dioxide (CO2) always contains carbon and oxygen in a mass ratio of 3:8.

3. Law of Multiple Proportions

Proposed by John Dalton, stating that if two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other element are in a small whole number ratio.

Illustration: Hydrogen and oxygen form water (H2O) and hydrogen peroxide (H2O2).

$H<em>2 + \frac{1}{2}O</em>2 → H_2O$ (2g H2 + 16g O2 = 18g H2O)

$H<em>2 + O</em>2 → H<em>2O</em>2$ (2g H2 + 32g O2 = 34g H2O2)

The masses of oxygen (16g and 32g) that combine with 2g of hydrogen are in a simple ratio of 1:2.

4. Gay-Lussac’s Law of Gaseous Volumes

Proposed by Gay-Lussac, stating that when gases combine to form gaseous products, their volumes are in simple whole number ratios at constant temperature and pressure.

Illustration: $2H<em>2(g) + O</em>2(g) → 2H_2O(g)$. 100 mL of hydrogen combines with 50 mL of oxygen to produce 100 mL of water vapor, demonstrating a 2:1 ratio between hydrogen and oxygen volumes.

5. Avogadro’s Law

Proposed by Amedeo Avogadro, stating that equal volumes of all gases at the same temperature and pressure contain equal numbers of moles or molecules.

Illustration: 10L each of NH3, N2, O2, and CO2 at the same temperature and pressure contain the same number of moles and molecules.

Dalton’s Atomic Theory

John Dalton used the term 'atom' (from the Greek word 'a-tomio,' meaning indivisible) and proposed the first atomic theory:

1. Matter is made up of minute, indivisible particles called atoms.
2. Atoms cannot be created or destroyed.
3. Atoms of the same element are identical in properties and mass, while atoms of different elements differ.
4. Atoms combine to form compound atoms called molecules.
5. Atoms combine in fixed ratios by mass.

Dalton’s theory explained the laws of chemical combination.

Atoms and Molecules

• Atom: Smallest particle of an element.
• Molecule: Smallest particle of a substance that retains its properties.

Types of Molecules

Based on the type of atoms, molecules are divided into homonuclear and heteronuclear molecules.

• Homonuclear Molecules: Contain only one type of atom (e.g., H2, O2, N2, O3).
• Heteronuclear Molecules: Contain different types of atoms (e.g., CO2, H2O, C6H12O6, NH3).

Based on the number of atoms, molecules are monoatomic, diatomic, or polyatomic.

• Monoatomic Molecules: Contain one atom (e.g., metals, noble gases like He, Ne, Ar).
• Diatomic Molecules: Contain two atoms (e.g., H2, O2, N2, halogens like F2, Cl2, Br2, I2).
• Polyatomic Molecules: Contain more than two atoms (e.g., ozone (O3), phosphorus (P4), sulfur (S8)).

Atomic Mass

Atomic mass (relative atomic mass) of an element expresses how many times the mass of an atom of the element is greater than 1/12th the mass of a C12 atom. The atomic mass unit (amu) is 1/12th the mass of a C12 atom.

$1 amu = \frac{1}{12} × mass \, of \, C^{12} atom$
$= 1.66 × 10^{-24} g$
$= 1.66 × 10^{-27} kg$

'amu' is now replaced by 'u' (unified mass).

Average Atomic Mass

Calculated by considering the atomic masses and relative abundances of an element's isotopes.

Example: Chlorine has isotopes 35Cl and 37Cl in a 3:1 ratio. The average atomic mass of Cl is:$(\frac{3 × 35 + 1 × 37}{4} = 35.5)$

Molecular Mass

Molecular mass is the sum of the atomic masses of the elements in a molecule and can be calculated by multiplying the atomic mass of each element by the number of its atoms and adding them together.

Example: Molecular mass of H2SO4:

$(2 × 1 + 32 + 4 × 16 = 98 u)$

Formula Mass

For ionic compounds (like NaCl) without discrete molecules, formula mass is calculated using the compound's molecular formula, considering the three-dimensional arrangement of ions.

Mole Concept

A mole is the unit of amount of substance, defined as the amount of substance containing as many particles as there are atoms in exactly 12 g of the C12 isotope.

1 mole contains $6.022 × 10^{23}$ particles (Avogadro's number or Avogadro's constant, $N<em>A$ or $N</em>0$).

Illustration:

• 1 mol of hydrogen atoms = $6.022 × 10^{23}$ atoms
• 1 mol of water molecules = $6.022 × 10^{23}$ water molecules
• 1 mol of sodium chloride = $6.022 × 10^{23}$ formula units of sodium chloride

$No. \, of \, moles \, (n) = \frac{Given \, mass \, in \, grams \, (w)}{Molar \, mass \, (M)}$

$No. \, of \, molecules = no. \, of \, moles × 6.022 × 10^{23}$

Molar Mass

The mass of one mole of a substance in grams; numerically equal to the molecular mass in atomic mass units. Molar mass of oxygen ($O<em>2$) = 32g. Molar mass of hydrogen ($H</em>2$) = 2g.

Molar Volume

The volume of 1 mole of any substance. At standard temperature and pressure (STP), the molar volume of any gas = 22.4 L (or 22400 mL), and contains $6.022 × 10^{23}$ molecules.

Example: 22.4 L of hydrogen gas = 1 mole of H2 = $6.022 × 10^{23}$ molecules of hydrogen = 2 g of H2.

Example Question:

How many moles of water molecules are present in 180 g of water?

Answer:

$No. \, of \, moles = \frac{Given \, mass \, in \, grams}{Molar \, Mass} = \frac{180}{18} = 10 \, mol$

Percentage Composition

The percentage of each element present in 100g of a substance.

$Percentage \, composition \, of \, an \, element = \frac{Mass \, of \, that \, element \, in \, the \, compound}{Molar \, mass \, of \, the \, compound} × 100$

It helps in checking purity and determining empirical and molecular formulas.

Empirical and Molecular Formulae

• Empirical Formula: Simplest formula showing the ratio of different elements in a compound.
• Molecular Formula: Actual formula showing the exact number of different elements in a compound.

Example: Glucose has an empirical formula of $CH<em>2O$ and a molecular formula of $C</em>6H<em>{12}O</em>6$.

$Molecular \, formula \, (M.F) = Empirical \, formula \, (E.F) × n$

Where $\, n = \frac{Molecular \, mass \, (MM)}{Empirical \, formula \, mass \, (EFM)}$

Example Question:

An organic compound contains 40% carbon, 6.66% hydrogen, and 53.34% oxygen. Its molecular mass is 180. Calculate its molecular formula.

Answer:

Empirical formula = $CH_2O$

Empirical formula mass (EFM) = 12 + 2 + 16 = 30

Molar mass (MM) = 180

$n = \frac{MM}{EFM} = \frac{180}{30} = 6$

Molecular formula = $(CH<em>2O) × 6 = C</em>6H<em>{12}O</em>6$

Stoichiometry and Stoichiometric Calculations

Stoichiometry deals with calculations involving masses or volumes of reactants and products in chemical reactions.

Chemical Equation

A representation of a chemical reaction using symbols and formulae. Reactants are on the left, products on the right. Equations must be balanced, and physical states are indicated in brackets.

A chemical equation provides:

1. Reactants, products, and their physical states.
2. Masses of reactants and products.
3. Moles and molecules of reactants and products.
4. Volumes of reactants and products at STP.

Limiting Reagent

The reagent that is completely consumed in a chemical reaction and limits further reaction.

Example:

In the reaction: $2SO<em>2(g) + O</em>2(g) → 2SO_3(g)$

If 10 moles each of $SO<em>2$ and $O</em>2$ are taken, only 10 moles of $SO<em>3$ form because 10 moles of $SO</em>2$ require only 5 moles of $O<em>2$. Here, $SO</em>2$ is the limiting reagent, and 5 moles of $O_2$ remain unreacted.

Example Question:

A reaction mixture for the production of $NH<em>3$ gas contains 250 g of $N</em>2$ gas and 50 g of $H<em>2$ gas under suitable conditions. Identify the limiting reactant, if any, and calculate the mass of $NH</em>3$ gas produced.

Answer:

$N<em>2(g) + 3H</em>2(g) → 2NH_3(g)$

28g 6g 34g

28g $N<em>2$ requires 6g $H</em>2$ for complete reaction.

250g $N<em>2$ requires $(\frac{6 × 250}{28} = 53.57g)$ $H</em>2$.

Since there is only 50g $H<em>2$, consider the reverse: 6g $H</em>2$ requires 28g $N_2$.

50g $H<em>2$ requires $(\frac{28 × 50}{6} = 233.33g)$ $N</em>2$.

Hydrogen is completely consumed and is the limiting reagent.

Amount of ammonia formed

$= 50 + 233.33 = 283.33 g$

Reactions in Solutions

Solutions are homogeneous mixtures containing two or more components. The component in larger quantity is the solvent; the dissolved substance is the solute. A binary solution has only two components, and an aqueous solution uses water as the solvent.

The composition of a solution is expressed in terms of concentration, defined as the amount of solute in a given volume of solution.

1. Mass Percent (w/w or m/m)

The number of parts of solute per 100 parts by mass of solution.

$Mass \,$

2. Mole Fraction

The ratio of the number of moles of a component to the total number of moles in the solution.

$Mole \, fraction \, of \, a \, component = \frac{Number \, of \, moles \, of \, the \, component}{Total \, number \, of \, moles \, of \, all \, the \, components}$

In a binary solution with nA moles of A and nB moles of B:

Mole fraction of A: $\chi<em>A = \frac{n</em>A}{n<em>A + n</em>B}$

Mole fraction of B: $\chi<em>B = \frac{n</em>B}{n<em>A + n</em>B}$

$\chi<em>A + \chi</em>B = 1$ (Sum of mole fractions in a solution is always 1).

If there are $i$ components: $\chi<em>1 + \chi</em>2 + \chi<em>3 + … + \chi</em>i = 1$

3. Molarity (M)

The number of moles of solute per liter of solution.

$Molarity \, (M) = \frac{Number \, of \, moles \, of \, solute \, (n)}{Volume \, of \, solution \, in \, litre \, (V)}$

1 M NaOH solution contains 1 mole (40 g) of NaOH in 1 L of solution.

When diluting a solution: $M<em>1V</em>1 = M<em>2V</em>2$ ($M<em>1$: initial molarity, $M</em>2$: final molarity, $V<em>1$: initial volume, $V</em>2$: final volume).

4. Molality (m)

The number of moles of solute per kilogram (kg) of solvent.

$Molality \, (m) = \frac{Number \, of \, moles \, of \, solute}{Mass \, of \, solvent \, in \, kg}$

Molarity depends on temperature (due to volume changes), while molality and mole fraction are temperature independent.

Example Questions:

1. Calculate the molarity of a solution containing 8 g of NaOH in 500 mL of water.

Answer:

• Mass of NaOH = 8 g
• Volume of solution = 500 mL = 0.5 L
• Molar mass of NaOH = 40 g/mol
• $No. \, of \, moles \, of \, NaOH = \frac{8}{40} = 0.2 \, mol$
• $Molarity = \frac{0.2}{0.5} = 0.4 \, M$

1. Calculate the mass of oxalic acid dihydrate ($H<em>2C</em>2O<em>4 \cdot 2H</em>2O$) required to prepare 0.1M, 250 ml of its aqueous solution.

Answer:

• Molarity = 0.1M
• Volume = 250 mL = 0.25 L
• $No. \, of \, moles = 0.1 × 0.25 = 0.025 \, mol$
• Molar mass of $H<em>2C</em>2O<em>4 \cdot 2H</em>2O$ = 126 g/mol
• $Mass \, of \, oxalic \, acid = 0.025 × 126 = 3.15 \, g$

1. Calculate the amount of $CO<em>2(g)$ produced by the reaction of 32g of $CH</em>4(g)$ and 32g of $O_2(g)$.

Answer:

$CH<em>4(g) + 2O</em>2(g) → CO<em>2(g) + 2H</em>2O(g)$

16g 64g 44g 36g

64g $O<em>2$ requires 16g $CH</em>4$, so 32g $O<em>2$ requires 8g $CH</em>4$.

16g $CH<em>4$ combines with 64g $O</em>2$ to form 44g $CO<em>2$, so 8g $CH</em>4$ combines with 32g $O<em>2$ to form 22g $CO</em>2$.

4. If the density of methanol is 0.793 kg $L^{-1}$, what volume is needed for making 2.5 L of its 0.25 M solution?

Answer:

• Density = 0.793 kg $L^{-1}$, so 1 L = 793 g
• $No. \, of \, moles \, of \, CH_3OH = \frac{793}{32} = 24.78 \, mol$
• Molarity = 24.78 M
• Using $M<em>1V</em>1 = M<em>2V</em>2$:
• $24.78 × V_1 = 0.25 × 2.5$
• $V_1 = 0.025 L = 25 mL$
• Volume of methanol required = 25 mL