CBSE Class 11 Chemistry Basic Concepts Of Chemistry Notes Set 04

These Class 11 CBSE Chemistry notes for Chapter 1 Some Basic Concepts of Chemistry cover the classification of matter, SI units, significant figures, the laws of chemical combination, Dalton's atomic theory, the mole concept, empirical and molecular formulae, stoichiometry with limiting reagent and the common ways of expressing concentration. Nine solved examples show the calculations step by step.

Important Notes and Formulas

Nature and Classification of Matter

1. Matter is anything that has mass and occupies space. It exists in three physical states: solid (definite shape and volume), liquid (definite volume, no definite shape) and gas (neither definite shape nor definite volume).
2. At the macroscopic level matter is classified into mixtures and pure substances.
3. A mixture contains two or more substances in any ratio, and the components keep their identity. A homogeneous mixture has uniform composition throughout, e.g. air, sea water, brass and sugar solution. A heterogeneous mixture has visibly different parts (phases), e.g. sand and salt, chalk powder in water.
4. A pure substance has fixed composition and characteristic properties. It is either an element or a compound.
5. An element contains only one kind of atom and cannot be split into simpler substances (metals, non-metals and metalloids). A compound consists of two or more elements combined in a fixed ratio by mass, and its properties differ from those of its elements; it can be broken down only by chemical methods.

Properties of Matter and Their Measurement

1. Physical properties can be measured or observed without changing the identity or composition of the substance, e.g. colour, odour, melting point, boiling point and density.
2. Chemical properties need a chemical change, e.g. acidity or basicity and combustibility.
3. Every measurement of a physical quantity has two parts: a number and a unit.

SI System of Units

The International System of Units (SI) has seven base units.

Decimal multiples and sub-multiples are written with prefixes:

Mass, Weight, Volume, Density and Temperature

1. Mass is the amount of matter in a substance and is constant. Weight is the force exerted by gravity on the object and varies from place to place. SI unit of mass: kg; mass is measured accurately with an analytical balance.
2. Volume has the unit (length)³. SI unit: m³. The common unit litre (L) is not an SI unit. 1 L = 1000 mL, 1000 cm³ = 1 dm³ = 1 L, and 1 m³ = 10³ dm³ = 10⁶ cm³.
3. Density = mass ÷ volume. SI unit: kg m⁻³; chemists commonly use g cm⁻³ (1 g cm⁻³ = 10³ kg m⁻³).
4. Temperature scales: °C, °F and K (SI unit). K = °C + 273.15, and °F = (9/5)(°C) + 32. A negative temperature is not possible on the Kelvin scale.

Scientific Notation, Precision and Accuracy

1. In scientific notation a number is written as N × 10ⁿ, where N lies between 1 and 10 and n is a positive or negative integer. For example 232.508 = 2.32508 × 10² and 0.00016 = 1.6 × 10⁻⁴.
2. In addition and subtraction the numbers are first written with the same power of 10. In multiplication the N values are multiplied and the exponents added; in division the exponents are subtracted.
3. Precision is the closeness of various measurements of the same quantity. Accuracy is the agreement of a measured value with the true value.

Significant Figures

Significant figures are the meaningful digits in a measurement: all the certain digits plus one doubtful digit.

1. All non-zero digits are significant: 285 cm has three and 0.25 mL has two.
2. Zeros between two non-zero digits are significant: 2.005 has four.
3. Zeros at the beginning of a number are not significant: 0.002 has one and 0.0045 has two.
4. Zeros at the end of a number are significant if the number has a decimal point: 16.0 has three and 16.00 has four. Zeros at the end of a number without a decimal point are ambiguous.
5. Counting numbers of objects and exact numbers (such as 20 pens, or 1 m = 100 cm) have an infinite number of significant figures.
6. In scientific notation, the digits in N are the significant figures: 4.5 × 10⁻⁴ has two.

Calculations. In addition or subtraction the result is reported to the same number of decimal places as the term with the fewest decimal places. In multiplication or division the result has the same number of significant figures as the quantity with the fewest significant figures.

Rounding off. If the digit to be dropped is more than 5, increase the preceding digit by one; if it is less than 5, leave the preceding digit unchanged. If it is exactly 5, increase the preceding digit by one only when it is odd, and leave it unchanged when it is even: 6.35 rounds to 6.4 and 6.25 rounds to 6.2.

Dimensional Analysis (Unit Factor Method)

To convert a quantity from one unit to another, multiply it by a unit factor, a ratio equal to 1 in which the units cancel. For example, with 1 inch = 2.54 × 10⁻² m, the height 5 feet 2 inches = 62 inches = 62 × (2.54 × 10⁻² m ÷ 1 inch) = 1.58 m.

Laws of Chemical Combination

Dalton's Atomic Theory

1. Matter consists of indivisible atoms.
2. All the atoms of a given element have identical properties, including identical mass; atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve reorganisation of atoms. Atoms are neither created nor destroyed in a chemical reaction.

The theory explains the laws of chemical combination. Later work showed that atoms are divisible and that atoms of the same element can have different masses (isotopes), so the first two postulates were modified.

Atomic Mass, Molecular Mass and Formula Mass

1. One atomic mass unit (u, formerly amu) is exactly one-twelfth the mass of one carbon-12 atom: 1 u = 1.66056 × 10⁻²⁴ g.
2. Atomic mass of an element is its average mass relative to the mass of carbon-12 taken as 12 u. Because many elements have isotopes, it is the average over the naturally occurring isotopes: average atomic mass = Σ (fractional abundance × isotopic mass). Chlorine: 0.755 × 34.97 + 0.245 × 36.97 = 35.46 u.
3. Molecular mass is the sum of the atomic masses of all the atoms in a molecule. H₂O: 2(1.008) + 16.00 = 18.016 u.
4. Formula mass is the sum of the atomic masses of all the atoms in a formula unit of an ionic compound. NaCl: 23.0 + 35.5 = 58.5 u.

Mole Concept and Molar Mass

1. One mole is the amount of a substance that contains as many particles (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12. This number is Avogadro's constant, N_A = 6.022 × 10²³ mol⁻¹.
2. The molar mass is the mass of one mole of a substance in grams. It is numerically equal to the atomic, molecular or formula mass in u (units: g mol⁻¹). H₂O: 18.02 g mol⁻¹; CO₂: 44 g mol⁻¹.
3. The molar volume of any gas at 273.15 K and 1 atm (STP) is 22.4 L mol⁻¹. At 273.15 K and 1 bar the value is 22.7 L mol⁻¹. Problems in this chapter use 22.4 L.

Percentage Composition

Mass % of an element = (mass of that element in one mole of the compound ÷ molar mass of the compound) × 100. The percentages of all elements add up to 100.

Empirical Formula and Molecular Formula

The empirical formula shows the simplest whole-number ratio of the atoms of the elements in a compound; the molecular formula shows the exact number of atoms of each element in a molecule. Molecular formula = (empirical formula)_n, where n = molar mass ÷ empirical formula mass. Example: CH is the empirical formula of benzene and C₆H₆ is its molecular formula.

Steps to find the empirical formula:

1. Convert the percentage of each element to grams by taking a 100 g sample.
2. Divide each mass by the atomic mass to get the number of moles.
3. Divide all the mole values by the smallest one.
4. If a ratio is not a whole number, multiply all by the smallest integer that makes them whole (e.g. 1.5 by 2, 1.33 by 3).
5. Write the formula with these whole numbers as subscripts.

Stoichiometry and Limiting Reagent

1. Stoichiometry deals with the quantitative relations between reactants and products. The coefficients in a balanced equation give the mole ratio (and the volume ratio for gases), not the mass ratio.
2. A chemical equation must be balanced: the number of atoms of each element is the same on both sides. For N₂ + 3H₂ → 2NH₃, 1 mol (28 g) of N₂ reacts with 3 mol (6 g) of H₂ to give 2 mol (34 g) of NH₃.
3. The limiting reagent is the reactant that is consumed first and so limits the amount of product. The other reactants are in excess. To find it, divide the moles of each reactant by its coefficient: the smallest value belongs to the limiting reagent.
4. The amount of product is calculated from the limiting reagent only.

Reactions in Solutions: Concentration Terms

1. Molarity depends on temperature because the volume of a solution changes with temperature; mass per cent, mole fraction and molality do not.
2. Moles of solute = M × V (in L). On dilution the moles of solute stay the same: M₁V₁ = M₂V₂.
3. Mass of solution = volume × density. This is used to convert between molarity and molality.

Solved Examples

Example 1. Classify the following substances into elements, compounds and mixtures: air, diamond, LPG, dry ice, graphite, steel, marble, smoke, glucose, laughing gas.
Answer:
Elements: diamond and graphite (both carbon).
Compounds: dry ice (CO₂), marble (CaCO₃), glucose and laughing gas (N₂O).
Mixtures: air, LPG, steel and smoke.

Example 2. In three moles of ethane (C₂H₆), calculate (i) the number of moles of carbon atoms, (ii) the number of moles of hydrogen atoms and (iii) the number of molecules of ethane.
Answer:
(i) 1 mol of C₂H₆ contains 2 mol of C atoms, so 3 mol contain 6 mol of C atoms.
(ii) 1 mol of C₂H₆ contains 6 mol of H atoms, so 3 mol contain 18 mol of H atoms.
(iii) Molecules = 3 × 6.022 × 10²³ = 1.8066 × 10²⁴ molecules (18.07 × 10²³).

Example 3. Calculate the mass per cent of the different elements present in sodium sulphate (Na₂SO₄).
Answer:
Molar mass of Na₂SO₄ = 2(23.0) + 32.0 + 4(16.0) = 142 g mol⁻¹.
Mass % of Na = (46 ÷ 142) × 100 = 32.39%
Mass % of S = (32 ÷ 142) × 100 = 22.54%
Mass % of O = (64 ÷ 142) × 100 = 45.07%

Example 4. Boron has two isotopes, boron-10 and boron-11, with percentage abundances 19.6% and 80.4% respectively. What is the average atomic mass of boron?
Answer:
Average atomic mass = (10 × 19.6 + 11 × 80.4) ÷ 100 = (196 + 884.4) ÷ 100 = 10.804 u.

Example 5. Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air, (ii) 1 mole of carbon is burnt in 16 g of dioxygen, (iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
The balanced equation is C(s) + O₂(g) → CO₂(g): 1 mol (12 g) of C and 1 mol (32 g) of O₂ give 1 mol (44 g) of CO₂.
(i) Air has plenty of oxygen, so carbon is limiting and 1 mol of CO₂ = 44 g is formed.
(ii) 16 g of O₂ is 0.5 mol, which can burn only 0.5 mol of carbon. O₂ is the limiting reagent, so 0.5 mol of CO₂ = 22 g is formed.
(iii) Again O₂ (16 g) is limiting, so 22 g of CO₂ is formed.

Example 6. Calculate the empirical formula of a compound that contains 26.6% potassium, 35.4% chromium and 38.1% oxygen. (K = 39.1, Cr = 52.0, O = 16.0)
Answer:
Moles per 100 g: K = 26.6 ÷ 39.1 = 0.68; Cr = 35.4 ÷ 52.0 = 0.68; O = 38.1 ÷ 16.0 = 2.38.
Dividing by 0.68: K : Cr : O = 1 : 1 : 3.5.
Multiplying by 2 gives 2 : 2 : 7, so the empirical formula is K₂Cr₂O₇.

Example 7. 250 cm³ of a sulphuric acid solution contains 24.5 g of H₂SO₄. If the density of the solution is 1.98 g cm⁻³, determine (i) its molarity and (ii) its molality.
Answer:
Molar mass of H₂SO₄ = 98 g mol⁻¹, so moles = 24.5 ÷ 98 = 0.25 mol.
(i) Volume = 250 cm³ = 0.250 L, so molarity = 0.25 ÷ 0.250 = 1 M.
(ii) Mass of solution = 250 × 1.98 = 495 g. Mass of solvent = 495 − 24.5 = 470.5 g = 0.4705 kg.
Molality = 0.25 ÷ 0.4705 = 0.53 m.

Example 8. Calculate the mass of potassium chlorate (KClO₃) needed to produce 1.23 g of oxygen, and the mass of KCl formed with it. (K = 39, Cl = 35.5, O = 16)
Answer:
2KClO₃ → 2KCl + 3O₂. 2 mol (245 g) of KClO₃ gives 2 mol (149 g) of KCl and 3 mol (96 g) of O₂.
Mass of KClO₃ = (245 ÷ 96) × 1.23 = 3.139 g.
Mass of KCl = (149 ÷ 245) × 3.139 = 1.909 g. Check: 1.909 + 1.23 = 3.139 g, so mass is conserved.

Example 9. Express the result of 12.11 + 18.0 + 1.012 to the correct number of significant figures.
Answer:
Sum = 12.11 + 18.0 + 1.012 = 31.122.
The term with the fewest decimal places is 18.0 (one decimal place), so the answer is rounded to one decimal place: 31.1.