JEE Chemistry Some Basic Concepts Of Chemistry Notes Set 03

Measurement, Significant Figures and Units

Every measured value is written as a number together with a unit. The digits we are sure of plus the first doubtful digit are called significant figures; a more precise instrument gives more significant figures.
Counting rules: all non-zero digits count; zeros between non-zero digits count; zeros that only locate the decimal point at the start of a number such as 0.0084 do not count; trailing zeros after a decimal point (0.00800) count; trailing zeros of a whole number like 378000 are taken as not significant unless they come from the measurement.
Changing units or moving the decimal point never changes the number of significant figures: 3050 m, 3.050 km and 3.050 × 10⁵ cm all have four.
Calculations: in addition and subtraction the answer keeps as many decimal places as the term with the fewest decimal places; in multiplication and division it keeps as many significant figures as the least precise factor. Multiplying by an exact constant does not change the count.

Rounding Off

Drop digit below 5: keep the previous digit (7.82 gives 7.8). Drop digit above 5, or 5 followed by non-zero digits: raise the previous digit by one (6.87 gives 6.9; 16.351 gives 16.4).
When the dropped digit is exactly 5 (or 5 followed only by zeros), make the previous digit even: 3.250 gives 3.2 and 3.750 gives 3.8.

SI Units, Prefixes and Handy Conversions

The seven base SI units are metre (length), kilogram (mass), second (time), kelvin (temperature), ampere (current), candela (luminous intensity) and mole (amount of substance). Derived units: newton = kg m s⁻², pascal = N m⁻², joule = N m = kg m² s⁻².
Prefixes: giga 10⁹, mega 10⁶, kilo 10³, centi 10⁻², milli 10⁻³, micro 10⁻⁶, nano 10⁻⁹, pico 10⁻¹², femto 10⁻¹⁵, atto 10⁻¹⁸, zepto 10⁻²¹, yocto 10⁻²⁴.
Useful values: 1 atm = 101325 Pa; 1 bar = 10⁵ Pa; 1 L = 1000 mL = 1 dm³; 1 Å = 10⁻¹⁰ m; 1 cal = 4.184 J; 1 eV = 1.602 × 10⁻¹⁹ J; 1 L atm = 101.3 J; °F = (9/5)°C + 32.

Laws of Chemical Combination

Law of conservation of mass (Lavoisier): the total mass of the reactants equals the total mass of the products. This is why equations must be balanced.
Law of constant (definite) proportions (Proust): a pure compound always has the same elements in the same ratio by mass, whatever its source.
Law of multiple proportions (Dalton): when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio (CO and CO₂; SO₂ and SO₃).
Law of reciprocal proportions (Richter): the masses of two elements that separately combine with a fixed mass of a third element are in the same ratio (or a simple multiple of it) in which they combine with each other.
Gay-Lussac's law of gaseous volumes: gases react, and form gaseous products, in volumes that are in simple whole-number ratios at the same temperature and pressure.

Atomic Theory and Avogadro's Hypothesis

Dalton pictured atoms as indivisible, identical for one element and unchanged in reactions. Modern knowledge changes this: atoms are divisible, atoms of one element can differ in mass (isotopes), atoms of different elements can have the same mass (isobars such as Ca-40 and Ar-40), and atoms do not always combine in small whole-number ratios (sucrose C₁₂H₂₂O₁₁).
Berzelius suggested that equal volumes of gases hold equal numbers of atoms, which fails. Avogadro corrected it: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Consequences: molecular mass = 2 × vapour density; one mole of any gas occupies 22.4 L at STP; the volume ratio of gases in a reaction equals their mole ratio.

Atomic, Molecular and Equivalent Masses

Atomic mass is the mass of an atom compared with one-twelfth of the mass of a carbon-12 atom; 1 amu = 1.66 × 10⁻²⁴ g. For an element with isotopes the average atomic mass is the abundance-weighted mean, e.g. boron (10: 19%, 11: 81%) = 10.81.
Molecular mass is the sum of the atomic masses in the formula. Gram atomic mass and gram molecular mass are these values in grams.
Equivalent mass of an element is the mass that combines with or displaces 1.008 g of hydrogen, 8 g of oxygen or 35.5 g of chlorine. Equivalent mass = atomic mass / valency. For acids, bases and salts, equivalent mass = molar mass / n-factor (basicity, acidity, or total charge); for oxidising and reducing agents n is the number of electrons gained or lost per formula unit.
Methods of atomic mass: Dulong and Petit (atomic mass × specific heat ≈ 6.4 for solid elements), the equivalent mass and vapour density of the chloride (valency = (2 × V.D.)/(E + 35.5)), and isomorphism.

The Mole Concept

One mole contains 6.022 × 10²³ particles (Avogadro's number, N_A). The mass of one mole is the molar mass in g/mol.
Number of moles = mass / molar mass = number of particles / N_A = volume at STP / 22.4 L.
Mass of one molecule = molar mass / N_A. Number of atoms in a sample = moles of molecules × atomicity × N_A.
Gas density at STP = molar mass / 22.4 g/L, and V.D. = molar mass / 2.

Percentage Composition and Formulae

Percentage of an element = (mass of the element in one mole of the compound / molar mass) × 100.
Empirical formula: divide each percentage by the atomic mass, divide all results by the smallest, and clear fractions to get whole numbers.
Molecular formula = (empirical formula)ⁿ where n = molecular mass / empirical formula mass.

Stoichiometry and Limiting Reagent

Write and balance the equation, convert given masses or volumes to moles, and use the coefficients as mole ratios.
When two reactants are given, divide each mole amount by its coefficient; the smaller value marks the limiting reagent, which decides the amount of product formed.
Percentage yield = (actual yield / theoretical yield) × 100; for impure samples, first take the pure fraction of the mass.

Concentration Terms and Titrations

Molarity M = moles of solute / litres of solution. Normality N = equivalents of solute / litres of solution = M × n-factor. Molality = moles of solute / kg of solvent.
For a solution of density d (g/mL) and mass fraction w of solute: M = (10 × w% × d) / molar mass.
Dilution: M₁V₁ = M₂V₂ (or N₁V₁ = N₂V₂). At the equivalence point of a titration, equivalents (milli-equivalents) of both reactants are equal: N₁V₁ = N₂V₂.
Common n-factors: HCl 1, H₂SO₄ 2, H₃PO₄ 3 (full neutralisation), oxalic acid 2, KMnO₄ 5 in acid, K₂Cr₂O₇ 6, Na₂S₂O₃ 1 (in iodometry), Fe²⁺ 1.

Solved Examples

Example 1. Calculate the number of molecules in 4.25 g of ammonia.
Answer:
Moles of NH₃ = 4.25/17 = 0.25 mol.
Molecules = 0.25 × 6.022 × 10²³ = 1.5 × 10²³.

Example 2. A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen and has molar mass 180. Find its molecular formula.
Answer:
Moles: C = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33.
Ratio 1 : 2 : 1, so the empirical formula is CH₂O (mass 30).
n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆.

Example 3. How many millilitres of 0.2 N HCl are needed to neutralise 20 mL of 0.1 M Ba(OH)₂?
Answer:
Normality of Ba(OH)₂ = 0.1 × 2 = 0.2 N.
N₁V₁ = N₂V₂: 0.2 × V = 0.2 × 20, so V = 20 mL.

Example 4. 2.4 g of magnesium is burnt in 1.2 g of oxygen. Which reagent is limiting and how much MgO forms?
Answer:
2Mg + O₂ → 2MgO. Moles Mg = 0.1; moles O₂ = 0.0375.
Mg needs 0.05 mol O₂ for complete reaction, but only 0.0375 mol is present, so O₂ is limiting.
MgO formed = 2 × 0.0375 = 0.075 mol = 3.0 g.