Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry

Chemistry is the science of the preparation, properties, structure, and reactions of material substances.

Branches of Chemistry:

  • Inorganic Chemistry

  • Organic Chemistry

  • Physical Chemistry

  • Analytical Chemistry

  • Polymer Chemistry

  • Biochemistry

  • Medicinal Chemistry

  • Industrial Chemistry

  • Hydrochemistry

  • Electrochemistry

  • Green Chemistry

Matter

Matter is anything that occupies space and has mass.

States of Matter:

  • Solid- Particles are orderly and close together.

    • Particles cannot move freely.

    • Definite shape and volume.

  • Liquid- Particles are close but can move around.

    • Definite volume, no definite shape.

  • Gas- Particles are far apart with easy and fast movement.

    • No definite shape or volume; takes the shape and volume of the container.

  • Plasma

  • Bose-Einstein Condensate

  • Fermionic Condensate

  • Quark-Gluon Plasma

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

Classification of Matter Based on Chemical Composition

  • Pure Substances: Contain only one type of particle.- Examples: Sodium (Na), Potassium (K), Hydrogen (H), Oxygen (O), Helium (He), carbon dioxide (CO2), water (H2O), ammonia (NH3), cane sugar (C{12}H{22}O[11}).

  • Mixtures: Contain more than one type of particle. Components can be separated by physical methods like filtration, crystallization, distillation.- Examples: Solutions, gold ornaments, sea water, muddy water, air.

Pure Substances

  • Elements: Contain only one type of atom. The term was first introduced by Robert Boyle.- Monoatomic: Metals (Sodium, Potassium, Calcium), Noble Gases (Helium, Neon).

    • Diatomic: Hydrogen, Nitrogen, Oxygen.

    • Polyatomic: Phosphorus (P4), Sulphur (S8).

  • Compounds: Contain more than one type of atom, formed by the combination of two or more atoms of different elements in a definite ratio. Cannot be separated by physical methods, but can be by chemical methods.- Examples: CO2, H2O, NH3, H2SO_4.

Mixtures

  • Homogeneous: Uniform composition throughout.- Examples: Solutions, air.

  • Heterogeneous: Different compositions in different parts.- Examples: Sea water, soil, muddy water.

Physical and Chemical Properties

  • Physical Properties: Can be measured without changing the substance's composition or identity. Measurement does not require a chemical change.- Examples: Color, odor, melting point, boiling point, density, mass.

  • Chemical Properties: Can be measured only with a chemical change.- Examples: Composition, combustibility, reactivity with acids and bases.

Measurement of Physical Properties

Quantitative observations are represented by a number and units.

  • English System and Metric System were used earlier.

  • International System of Units (SI) is now used.

SI Base Units:

  • Length: metre (m)

  • Mass: kilogram (kg)

  • Time: second (s)

  • Electric Current: ampere (A)

  • Thermodynamic Temperature: kelvin (K)

  • Amount of Substance: mole (mol)

  • Luminous Intensity: candela (cd)

Mass and Weight

  • Mass: Amount of matter in a body; constant quantity. SI unit is kilogram (kg).

  • Weight: Gravitational force acting on a body; variable quantity. SI unit is newton (N).

Volume (V)

Amount of space occupied by a body. SI unit is m^3. Smaller volumes are often in cm^3, dm^3, mL, L.

  • 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/m^3, commonly expressed in g/cm^3.

Temperature (T)

Degree of hotness or coldness. Common units are degree celsius (^0C), degree fahrenheit (^0F), kelvin (K). SI unit is kelvin (K).

  • ^0F = \frac{9}{5}(^0C) + 32

  • K = ^0C + 273.15

  • K = ^0C + 273

Precision and Accuracy

  • Precision: Closeness of various measurements for the same quantity.

  • Accuracy: Agreement of a particular value to the true value of the result.

Scientific Notation

Representing a number as N × 10^n, where n is an exponent and N is a number between 1.000… and 9.999…..

  • Exponent ‘n’ is positive when the decimal is shifted to the left.

  • Exponent ‘n’ is negative when the decimal is shifted to the right.- Example: 368.9 = 3.689 × 10^2

    • Example: 0.000563 = 5.63 × 10^{-4}

Significant Figures

Meaningful digits known with certainty, plus the last uncertain digit.

Rules for Determining Significant Figures:

  1. All non-zero digits are significant.- Example: 285 cm (3 significant figures), 0.25 mL (2 significant figures).

  2. Zeros preceding the first non-zero digit are not significant. They indicate the position of the decimal point.- Example: 0.03 (1 significant figure), 0.0052 (2 significant figures).

  3. Zeros between two non-zero digits are significant.- Example: 2.005 (4 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; otherwise, they are not significant.- Example: 0.200 g (3 significant figures).

  5. Exact numbers have an infinite number of significant figures.- Example: 2 balls = 2.000000…

  6. In scientific notation, all digits are significant.- Example: 4.01 × 10^2 (3 significant figures), 8.256 × 10^{-3} (4 significant figures).

Rounding off a Number

  1. If the rightmost digit to be removed is more than 5, the preceding number is increased by one.- Example: 1.386 rounded to 3 significant figures becomes 1.39.

  2. If the rightmost digit to be removed is less than 5, the preceding number is not changed.- Example: 4.334 rounded to 3 significant figures becomes 4.33.

  3. If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.- Example: 6.35 rounded becomes 6.4, while 6.25 rounded becomes 6.2.

Laws of Chemical Combinations

  1. Law of Conservation of Mass (Antoine Lavoisier): Matter can neither be created nor destroyed. In a chemical reaction, the total mass of reactants equals the total mass of products.- Example: 2H2 + O2 → 2H2O (4 g of H2 combines with 32 g of O2 to form 36 g of H2O).

  2. Law of Definite Proportions (Joseph Proust): A given compound always contains the same proportion of elements by weight. The same compound always contains the same elements combined in a fixed ratio by mass.- Example: Carbon dioxide (CO_2) always contains Carbon and Oxygen in a mass ratio of 3:8.

  3. Law of Multiple Proportions (John Dalton): If two elements can combine to form more than one compound, the different masses of one of the elements that combine with a fixed mass of the other element are in small whole number ratios.- Example: Hydrogen and oxygen form water (H2O) and hydrogen peroxide (H2O_2). The masses of oxygen (16 g and 32 g) which combine with a fixed mass of hydrogen (2g) bear a simple ratio of 1:2.

  4. Gay Lussac’s Law of Gaseous Volumes: When gases combine to form gaseous products, their volumes are in simple whole number ratios at constant temperature and pressure.- Example: 2H2(g) + O2(g) → 2H_2O(g). 100 mL of hydrogen combines with 50 mL of oxygen to form 100 mL of water vapor (2:1 ratio).

  5. Avogadro’s Law: Equal volumes of all gases at the same temperature and pressure should contain equal numbers of moles or molecules.- Example: 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

The term atom was coined by John Dalton. Key postulates include:

  1. Matter is made up of minute and indivisible particles called atoms.

  2. Atoms can neither be created nor be destroyed.

  3. Atoms of the same element are identical in their properties and mass. Atoms of different elements have different properties and mass.

  4. Atoms combine to form compound atoms called molecules.

  5. When atoms combine, they do so in a fixed ratio by mass.

Atoms and Molecules

  • Atom: Smallest particle of an element.

  • Molecule: Smallest particle of a substance with all the properties of that substance.

Types of Molecules:

  • Homonuclear: Contains only one type of atom. E.g., H2, O2, N2, O3 (ozone).

  • Heteronuclear: Contains different types of atoms. E.g., CO2, H2O, C6H{12}O6, NH3.

Based on the number of atoms:

  • Monoatomic: Contains one atom. E.g., Metals, Noble gases (He, Ne, Ar).

  • Diatomic: Contains two atoms. E.g., H2, O2, N2, halogens (F2, Cl2, Br2, I_2).

  • Polyatomic: Contains more than two atoms. E.g., Ozone (O3), Phosphorus (P4), Sulphur (S_8).

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 C^{12} atom.

  • Example: Atomic mass of Nitrogen is 14, meaning one N atom is 14 times greater than 1/12th the mass of a C^{12} atom.

Atomic Mass Unit (amu):

1/12th the mass of a C^{12} atom.

  • 1 amu = \frac{1}{12} × mass \ of \ a \ C^{12} \ atom

  • = 1.66 × 10^{-24} g

  • = 1.66 × 10^{-27} kg

‘amu’ is replaced by ‘u’ (unified mass).

Average Atomic Mass:

Calculated considering the atomic mass of isotopes and their relative abundance.

  • Example: Chlorine has two isotopes ^{35}Cl and ^{37}Cl in the ratio 3:1. The average atomic mass of Cl is \frac{3 × 35 + 1 × 37}{4} = 35.5.

Molecular Mass

Sum of atomic masses of elements in a molecule.

  • Example: Molecular mass of H2SO4 is (2 × 1 + 32 + 4 × 16 = 98 u).

Formula Mass

Used for ionic compounds (like NaCl) where there are no discrete molecules; it's calculated using the molecular formula.

Mole Concept

Unit of amount of substance. It is defined as the amount of substance that contains as many particles as there are atoms in exactly 12 g of the C^{12} isotope.

  • 1 mole of any substance contains 6.022 × 10^{23} atoms (Avogadro number or Avogadro constant NA or N0).

  • 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 \ gram \ (w)}{Molar \ mass \ (M)}

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

Molar Mass:

Mass of one mole of a substance in grams (gram molecular mass). Numerically equal to molecular mass in u.

  • Molar mass of oxygen = 32g

  • Molar mass of hydrogen = 2g

Molar Volume:

Volume of 1 mole of any substance. At standard temperature and pressure (STP), molar volume of any gas = 22.4 L (or, 22400 mL).

  • 22.4 L of any gas at STP contains 1 mole of the gas or 6.022 × 10^{23} molecules of the gas and its mass = molar mass.

  • 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 \ gram}{Molar \ mass} = \frac{180}{18} = 10 \ mol

Percentage Composition

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

Applications: Check purity and determine empirical and molecular formulas.

Empirical and Molecular Formulae

  • Empirical formula: Simplest formula giving the ratio of elements in a compound.

  • Molecular formula: Actual formula giving the exact number of different elements in the sample.

  • Example: Glucose - Empirical formula is CH2O but molecular formula is C6H{12}O6.

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 Carbon = 40%, Hydrogen = 6.66% and oxygen = 53.34%. Calculate its molecular formula if its molecular mass is 180.

Answer:

Element

Percentage

Atomic Mass

Percentage / Atomic Mass

Simple Ratio

Simplest Whole No. Ratio

C

40

12

40/12 = 3.33

3.33/3.33 = 1

1

H

6.66

1

6.66/1 = 6.66

6.66/3.33 = 2

2

O

53.34

16

53.34/16 = 3.33

3.33/3.33 = 1

1

Empirical Formula = CH_2O

Empirical Formula Mass (EFM) = 12 + 2 + 16 = 30

Molar mass (MM) = 180

n = MM/EFM = 180/30 = 6

Molecular formula = Empirical formula x n = (CH2O) × 6 = C6H{12}O6

Stoichiometry and Stoichiometric Calculations

Deals with calculations involving masses or volumes of reactants and products.

Chemical Equation

Representation of a chemical reaction by symbols and formulae with reactants on the left and products on the right.

A balanced chemical equation provides:

  1. Reactants, products, and their physical states.

  2. Masses of reactants and products.

  3. Number of moles and molecules of reactants and products.

  4. Volumes of reactants and products at STP.

Limiting Reagent (Limiting Reactant)

The reagent that is completely consumed in a chemical reaction.

  • Example: 2SO2(g) + O2(g) → 2SO_3(g)

2 moles of SO2 reacts completely with 1 mole of O2 to form 2 moles of SO3. If we take 10 moles each of SO2 and O2, we get only 10 moles of SO3 because 10 moles of SO2 require only 5 moles of O2. SO SO2 is the limiting reagent and 5 moles of O2 remains unreacted.

Example Question:

A reaction mixture contains 250 g of N2 gas and 50 g of H2 gas. Identify the limiting reactant and calculate the mass of NH_3 gas produced.

Answer:

N2(g) + 3H2(g) → 2NH_3(g)

28g 6g 34g

28g N2 requires 6g H2 for the complete reaction.

250g N2 requires, \frac{6 × 250}{28} = 53.57g \ H2. But there is only 50g H2. So, consider the reverse: 6g H2 requires 28g N2. 50g H2 requires \frac{28 × 50}{6} = 233.33g \ N2. Here H2 is the limiting reagent.

Amount of ammonia formed = 50 + 233.33 = 283.33 g

Reactions in Solutions

Solutions are homogeneous mixtures containing 2 or more components with the component in larger quantity called solvent and the other components called solutes.

Binary solution: Solution containing only 2 components. If the solvent is water, it is called an aqueous solution.

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

Ways to Express Concentration:

  1. Mass percent (w/w or m/m): Number of parts solute present in 100 parts by mass of solution.

Mass \ % \ of \ a \ component = \frac{Mass \ of \ solute}{Mass \ of \ solution} × 100

  1. Mole fraction: Ratio of the number of moles of a particular component to the total number of moles of 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, if the number of moles of A and B are nA and nB respectively, then

Mole fraction of A (χA) = \frac{nA}{nA + nB}

Mole fraction of B (χB) = \frac{nB}{nA + nB}

χA + χB = \frac{nA}{nA + nB} + \frac{nB}{nA + nB} = 1

The sum of mole fractions of all components in a solution is always equal to 1.

If there are 1,2,3, …….. i components, then χ1 + χ2 + χ3 + ………………………. + χi = 1

  1. Molarity (M): Number of moles of solute dissolved per liter of solution.

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

1 M NaOH solution means 1 mole (40 g) of NaOH is present in 1 L of solution.

When a solution is diluted, its new concentration is M1V1 = M2V2, where M1 is the initial molarity, M2 is the final molarity, V1 is the initial volume and V2 is the final volume.

  1. Molality (m): Number of moles of solute present per kilogram (kg) of the solvent.

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

Among the concentration terms, molarity depends on temperature because of its relation to volume, which changes with temperature. All the others are temperature independent.

Example Questions:

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

    Mass of solute (NaOH) = 8 g, Volume of solution = 500 mL = 0.5 L

    Molar mass of NaOH = 40 g mol-1

    No. of moles of NaOH = \frac{mass \ of \ NaOH \ in \ gram}{Molar \ mass \ of \ NaOH} = \frac{8}{40} = 0.2 \ mol

    Molarity = \frac{Number \ of \ moles \ of \ solute \ (n)}{Volume \ of \ solution \ in \ litre \ (V)} = \frac{0.2}{0.5} = 0.4 \ M

  2. Calculate the mass of oxalic acid dihydrate (H2C2O4.2H2O) required to prepare 0.1M, 250 ml of its aqueous solution.

    Molarity of solution = 0.1M, Volume of solution = 250 mL = 0.25 L

    No. of moles of oxalic acid = Molarity x volume of solution in litre = 0.1 x 0.25 = 0.025 mol

    Molar mass of Oxalic acid dihydrate = 126 g/mol

    Mass of oxalic acid = No. of moles x Molar mass of oxalic acid = 0.025 x 126 = 3.15 g

  3. Calculate the amount of CO2(g) produced by the reaction of 32g of CH4 (g) and 32g of O_2 (g).

    CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)

    16g 64g 44g 36g

    64g O2 requires 16g CH4 for the complete reaction. So, 32g O2 requires 8g CH4.

    16g CH4 combines with 64g O2 to form 44g CO2. Therefore, 8g CH4 combines with 32g Oxygen to form 22g CO_2

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

    Density of methanol = 0.793 kg L-1, Mass of 1 L of methanol = 0.793 kg = 793 g

    No. of moles of methanol (CH_3OH) = \frac{Mass \ in \ gram}{Molar \ mass} = \frac{793}{32} = 24.78 \ mol

    Molarity = no. of moles of solute per L of solution = 24.78 M

    To prepare 2.5L (V2) of 0.25M (M2) methanol solution from 24.78M (M1) methanol, use M1V1 = M2V_2