Some Basic Concepts of Chemistry Study Notes

Matter and its Physical and Chemical Classification

  • Matter: Defined as any substance that occupies space, possesses mass, and can be felt by one or more of the five senses.
  • Physical Classification of Matter: This classification is based on the physical state under ordinary conditions of temperature and pressure.
    • Solid: A substance with a definite volume and a definite shape. Examples include sugar, iron, gold, and wood.
    • Liquid: A substance with a definite volume but no definite shape. Liquids take the shape of their container. Examples include water, milk, oil, mercury, and alcohol.
    • Gas: A substance with neither a definite volume nor a definite shape, as they fill the entire container. Examples include Hydrogen (H2H_2), Oxygen (O2O_2), and Carbon dioxide (CO2CO_2).
  • Chemical Classification of Matter: Based on internal composition.
    • Pure Substance: Material containing only one type of substance. These cannot be separated into simpler substances by physical methods.
      • Element: A pure substance containing only one kind of atom. Elements are divided into:
        1. Metal: Such as Zn,Cu,Hg,Ac,Sn,PbZn, Cu, Hg, Ac, Sn, Pb.
        2. Non-metal: Such as N2,O2,Cl2,Br2,F2,P4,S8N_2, O_2, Cl_2, Br_2, F_2, P_4, S_8.
        3. Metalloids: Such as B,Si,As,TeB, Si, As, Te.
      • Compound: A pure substance containing more than one kind of element or atom combined in a fixed proportion by weight. They can be decomposed into simpler substances by chemical methods. The properties of a compound differ completely from its constituent elements. Examples include HCl,H2O,H2SO4,HClO4,HNO3HCl, H_2O, H_2SO_4, HClO_4, HNO_3.
    • Mixture: A material containing more than one type of substance mixed in any ratio by weight. The properties of a mixture reflect the properties of its components. They can be separated by simple physical methods.
      • Homogeneous Mixture: All components are present uniformly in a single phase. Examples: Water + Salt, Water + Sugar, Water + Alcohol.
        • Pure Substances: Defined by constant chemical composition (Elements and Compounds).
        • Solutions: A homogeneous mixture of two or more pure substances (e.g., air, NaClNaCl and water). Solutions do not have a definite composition.
      • Heterogeneous Mixture: Components are present non-uniformly. Examples: Water + Sand, Water + Oil, blood, petrol.

Atomic and Molecular Particles

  • Atom: Term introduced by Dalton. It is the smallest particle of matter that takes part in a chemical reaction and retains all the properties of an element.
  • Molecule: Term introduced by Avogadro. It is the smallest particle of matter that exists independently and can retain all properties of that substance. Molecules are formed by combinations of atoms.

Properties of Matter and Physical Quantities

  • Physical Property: A property measurable without changing the chemical composition, such as mass, volume, density, and refractive index.
  • Chemical Property: A property evaluated through the alteration of the matter itself. Examples include the combustible nature of hydrogen verified by burning and the sweet taste of sugar verified by consumption.
  • Physical Quantities: All quantities that can be measured (e.g., time, length, mass, force, work done).
  • Fundamental Quantities: A set of independent physical quantities through which all other quantities are expressed. These include mass, time, length, current, temperature, luminous intensity, and amount of substance.
  • International System (SI) of Units:
    1. Length: metre (mm)
    2. Mass: kilogram (kgkg)
    3. Time: second (ss)
    4. Temperature: kelvin (KK)
    5. Electric current: ampere (AA)
    6. Luminous Intensity: candela (cdcd)
    7. Amount of substance: mole (molmol)

Scientific Notation and Measurements

  • Scientific Notation: Expressed as P=A×10xP = A \times 10^x, where 1A<101 \le A < 10 and xx is the order of magnitude.
  • SI Prefixes:
    • 101210^{12}: tera (TT)
    • 10910^9: giga (GG)
    • 10610^6: mega (MM)
    • 10310^3: kilo (kk)
    • 10210^2: hecto (hh)
    • 10110^1: deca (dada)
    • 10110^{-1}: deci (dd)
    • 10210^{-2}: centi (cc)
    • 10310^{-3}: milli (mm)
    • 10610^{-6}: micro (μ\mu)
    • 10910^{-9}: nano (nn)
    • 101210^{-12}: pico (pp)
    • 101510^{-15}: femto (ff)

Significant Figures and Arithmetic Rules

  • Significant Figures (SF): Digits known with certainty plus one uncertain digit. Higher SF count indicates greater accuracy.
  • Counting Rules:
    • I: All non-zero digits are significant (19841984 has 4 SF).
    • II: Zeros between non-zeros are significant (1080610806 has 5 SF).
    • III: Zeros to the left of the first non-zero digit are not significant (0010800108 has 3 SF).
    • IV: For numbers less than 1, zeros between the decimal and first non-zero digit are not significant (0.0023080.002308 has 4 SF).
    • V: Trailing zeros with a decimal point are significant (01.08001.080 has 4 SF).
    • VI: Trailing zeros without a decimal point are not significant (010100010100 has 3 SF).
    • VII: Exponential terms do not affect SF count (e.g., 12.312.3 and 1.23×1011.23 \times 10^1 both have 3 SF).
  • Arithmetic Operations:
    • Addition/Subtraction: The result must have the same number of decimal places as the term with the fewest decimal places (e.g., 12.58712.5=0.0870.112.587 - 12.5 = 0.087 \rightarrow 0.1).
    • Multiplication/Division: The result must have the same number of SF as the factor with the fewest SF (e.g., 5.0×0.125=0.6250.625.0 \times 0.125 = 0.625 \rightarrow 0.62).
  • Unit changes: Changing units does not affect the SF count (e.g., 2.308cm=23.08mm=0.02308m2.308\,cm = 23.08\,mm = 0.02308\,m all have 4 SF).

Accuracy, Precision, and Laws of Chemical Combination

  • Accuracy: Closeness of a measured value to the true value.
  • Precision: Resolution or closeness of multiple measurements to each other.
  • Law of Conservation of Mass: Mass is neither created nor destroyed in a balanced chemical reaction. Total mass of reactants equals total mass of products plus any unreacted reactants.
  • Law of Constant Composition (Definite Proportions): A chemical compound always contains its component elements in a fixed ratio by weight, regardless of source or preparation method (e.g., H2OH_2O always has a 2:12:1 molar ratio of HH to OO).
  • Law of Multiple Proportions: If two elements form more than one compound, the masses of one element that combine with a fixed mass of the second are in a ratio of small whole numbers. (e.g., Nitrogen oxides forming ratios 1:2:3:4:51:2:3:4:5 for oxygen weights relative to 14g14\,g of nitrogen).
  • Law of Reciprocal Proportion: The ratio of masses of two elements A and B combining separately with a fixed mass of C is either the same or a simple multiple of the ratio in which A and B combine directly (CH4,CO2,H2OCH_4, CO_2, H_2O).
  • Gay-Lussac’s Law of Combining Volume: Gases combine in simple whole-number ratios by volume at constant temperature and pressure (e.g., 1vol.H2+1vol.Cl22vol.HCl1\,vol.\,H_2 + 1\,vol.\,Cl_2 \rightarrow 2\,vol.\,HCl).
  • Avogadro’s Hypothesis: Equal volumes of all gases contain an equal number of molecules at the same temperature and pressure. Volume measurement is equivalent to counting molecules.
  • STP (Standard Temperature and Pressure): Temperature = 0C0^\circ C (273K273\,K), Pressure = 1atm1\,atm (760mmHg760\,mm\,Hg). Molar volume of gas at STP = 22.4litres22.4\,litres.

Atomic and Molecular Masses

  • Relative Atomic Mass (R.A.M.): Expressed relative to 1/121/12 the mass of a carbon-12 atom. Carbon-12 is exactly 12.000u12.000\,u.
    • RAM=Mass of one atom of an element112×Mass of one C-12 atomRAM = \frac{\text{Mass of one atom of an element}}{\frac{1}{12} \times \text{Mass of one C-12 atom}}
  • Atomic Mass Unit (amu): 1/121/12 of the mass of one carbon-12 atom. 1amu=1.66×1024gm1\,amu = 1.66 \times 10^{-24}\,gm or 1.66×1027kg1.66 \times 10^{-27}\,kg. Modern symbol is 'u' (unified atomic mass). Also called One Dalton (DaDa).
  • Average Atomic Mass: Weighted average of naturally occurring isotopes.
    • Ax=a1x1+a2x2++anxn100A_x = \frac{a_1x_1 + a_2x_2 + \dots + a_nx_n}{100}, where aa is isotopic mass and xx is percentage abundance.
  • Mean Molar Mass: Is the average molar mass of substances in a container: Mavg=n1M1+n2M2+n1+n2+M_{avg} = \frac{n_1M_1 + n_2M_2 + \dots}{n_1 + n_2 + \dots}.
  • Formula Mass: Sum of atomic masses in an ionic compound (e.g., NaCl=23+35.5=58.5uNaCl = 23 + 35.5 = 58.5\,u).

Mole Concept and Gram-Mass Relationships

  • Mole: Counting unit for microscopic particles. Amount of substance containing as many entities as atoms in 0.012kg0.012\,kg (12gm12\,gm) of Carbon-12. This number is Avogadro's constant (NA=6.022×1023N_A = 6.022 \times 10^{23}).
  • Gram Atomic Mass: Atomic mass of an element expressed in grams. It is the mass of one mole of atoms.
  • Gram Molecular Mass: Molecular mass expressed in grams. It is the mass of one mole of molecules.
  • Relationship between Gram and amu: 1amu=1NAgm1\,amu = \frac{1}{N_A}\,gm.
  • Mole Calculations:
    1. n=Given no. of particlesNAn = \frac{\text{Given no. of particles}}{N_A}
    2. n=Given weightAtomic/Molecular weightn = \frac{\text{Given weight}}{\text{Atomic/Molecular weight}}
    3. n=PVRTn = \frac{PV}{RT} (Ideal gas equation, R=0.0821lit-atm/mol-KR = 0.0821\,lit\text{-}atm/mol\text{-}K)

Density and Atomicity

  • Atomicity: Number of atoms in one molecule (H2H_2: 2, CO2CO_2: 3, O3O_3: 3).
  • Density:
    • Absolute Density: Mass/VolumeMass/Volume.
    • Relative Density: Density of substanceDensity of standard substance\frac{\text{Density of substance}}{\text{Density of standard substance}}.
    • Specific Gravity: Density of substanceDensity of Water at 4\frac{\text{Density of substance}}{\text{Density of Water at 4}^℃}.
    • Density of Gas: d=PMRTd = \frac{PM}{RT}.
  • Vapour Density (V.D.): Density of gas relative to H2H_2 gas at same T and P. V.D.=Mol. wt.2\text{V.D.} = \frac{\text{Mol. wt.}}{2}.
  • Key Constants: Density of liquid water at 4C4^\circ C is 1g/mL=1g/cc=103kg/m31\,g/mL = 1\,g/cc = 10^3\,kg/m^3.

Empirical and Molecular Formula Determinations

  • Percentage Composition: % of element=Atomic weight×Number of atomsTotal molecular weight×100\text{\% of element} = \frac{\text{Atomic weight} \times \text{Number of atoms}}{\text{Total molecular weight}} \times 100.
  • Empirical Formula: Simplest whole-number ratio of atoms in a compound.
  • Molecular Formula: Actual number of atoms in a molecule. Molecularformula=Empiricalformula×nMolecular\,formula = Empirical\,formula \times n.
    • n=Molecular Formula MassEmpirical Formula Massn = \frac{\text{Molecular Formula Mass}}{\text{Empirical Formula Mass}}
  • Calculations for C and H:
    • Mass % of C=1244×mCO2mcompound×100\text{Mass \% of C} = \frac{12}{44} \times \frac{m_{CO_2}}{m_{compound}} \times 100
    • Mass % of H=218×mH2Omcompound×100\text{Mass \% of H} = \frac{2}{18} \times \frac{m_{H_2O}}{m_{compound}} \times 100

Concentration of Solutions

  • % by weight (w/w): Grams of solute in 100gm100\,gm of solution.
  • % by volume (v/v): mL of solute in 100ml100\,ml of solution.
  • Molarity (M): Moles of solute per 1000ml1000\,ml (1L1\,L) of solution.
  • Molality (m): Moles of solute per 1000gm1000\,gm (1kg1\,kg) of solvent. Independent of temperature.
  • Mole Fraction (X): Moles of a component divided by total moles. XA+XB=1X_A + X_B = 1.
  • Parts per million (ppm): Mass of soluteMass of solution×106\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6.
  • Conversion Formulas:
    • M=% (w/w) ×d×10Mol. wt. soluteM = \frac{\text{\% (w/w) } \times d \times 10}{\text{Mol. wt. solute}}
    • m=Mole fraction solute×1000Mole fraction solvent×Mol. wt. solventm = \frac{\text{Mole fraction solute} \times 1000}{\text{Mole fraction solvent} \times \text{Mol. wt. solvent}}
    • m=M×10001000dMM2m = \frac{M \times 1000}{1000d - MM_2} (where M2M_2 is molar mass of solute).

Stoichiometry and Chemical Reactions

  • Stoichiometry: Based on quantitative relationships in a balanced chemical equation.
  • Balancing: Must satisfy conservation of mass and charge. Subscripts in formulas cannot be changed.
  • Mole-Mole Analysis: For a reaction aA+bBcC+dDaA + bB \rightarrow cC + dD, the ratio is nAa=nBb=nCc=nDd\frac{n_A}{a} = \frac{n_B}{b} = \frac{n_C}{c} = \frac{n_D}{d}.
  • Types of Relationships:
    1. Mass-Mass: Relating mass of one reagent to another.
    2. Mass-Volume: Relating mass to volume of gas (utilizing 22.4L22.4\,L at STP).
    3. Volume-Volume: Relating gas volumes directly (Gay-Lussac's Law).

Limiting Reagent and Yield

  • Limiting Reagent (L.R.): The reactant completely consumed first in a reaction. It determines the maximum amount of product formed.
  • Finding L.R.: Least ratio of Given MolesStoichiometric Coefficient\frac{\text{Given Moles}}{\text{Stoichiometric Coefficient}}.
  • Percentage Yield: Actual YieldTheoretical Maximum Yield×100\frac{\text{Actual Yield}}{\text{Theoretical Maximum Yield}} \times 100.

Sequential Reactions and POAC

  • Sequential Reaction: Products of one reaction act as reactants for the next. Reactants are balanced and multiplied such that intermediate products cancel out to provide a final reaction.
  • POAC (Principle of Atom Conservation): Moles of atoms of a specific element are conserved regardless of the chemical path. For KClO3KCl+O2KClO_3 \rightarrow KCl + O_2, POAC for OO is: 3×nKClO3=2×nO23 \times n_{KClO_3} = 2 \times n_{O_2}.
  • Strength (Labelling) of Oleum: Oleum (SO3SO_3 in 100%H2SO4100\% H_2SO_4) is labelled as y%(>100%)y\% (> 100\%) where (y100)(y-100) grams is the water needed to convert free SO3SO_3 to H2SO4H_2SO_4. Weight % of free SO3=80(y100)18\text{\% of free } SO_3 = \frac{80(y-100)}{18}.

Quantitative Examples and Problem Analysis

  • Mass Conservation Example: 15.9g15.9\,g Na2CO3+20.0gNa_2CO_3 + 20.0\,g acetic acid reacts to release CO2CO_2. Residual weight 29.3g29.3\,g. Mass released =35.929.3=6.6g= 35.9 - 29.3 = 6.6\,g.
  • Molecule Counts: 5.0g5.0\,g of chlorophyll complex (3.68%Mg3.68\% Mg) contains 0.184g0.184\,g of MgMg. Calculated atoms: 0.18424×6.022×1023=4.617×1021\frac{0.184}{24} \times 6.022 \times 10^{23} = 4.617 \times 10^{21}.
  • Dilution and Mixing:
    • Dilution: M1V1=M2V2M_1V_1 = M_2V_2.
    • Mixing: MR=M1V1+M2V2V1+V2M_R = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}.
  • Vapour Density Problems: 7.5L7.5\,L of gas at STP weighs 16g16\,g. Moles =7.5/22.4= 7.5/22.4. Mass per mole =16/(7.5/22.4)=48g/mol= 16 / (7.5/22.4) = 48\,g/mol. Vapour Density =48/2=24= 48/2 = 24.
  • Isotope Problems: Chlorine (75%35Cl75\% \, ^{35}Cl and 25%37Cl25\% \, ^{37}Cl). Average mass =(75×35)+(25×37)100=35.5amu= \frac{(75 \times 35) + (25 \times 37)}{100} = 35.5\,amu.