CBSE Class 11 Chemistry Basic Concepts Of Chemistry Notes Set 03

These revision notes for CBSE Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry cover the classification of matter, physical quantities and their units, significant figures, accuracy and precision, the mole concept, empirical and molecular formulae, stoichiometry, limiting reagent and all the concentration terms (mass per cent, ppm, mole fraction, molarity and molality) with the formulae you need for numericals.

1. Matter and its Classification

All those things that have mass and occupy space (have a physical appearance) can be considered as matter.

(i) Mixture is formed by mixing two or more substances together in any proportion. It can be homogeneous (uniform composition throughout) or heterogeneous (different composition throughout).
(ii) Pure substances have fixed composition and non-variable properties. An element is a substance that contains only one type of atoms, whereas a compound is formed when atoms of different elements combine in a fixed ratio.

2. Some Physical Quantities

(i) Mass and weight: Mass (m) is the amount of matter present in a substance. Weight (w) is the force exerted by gravity on an object.
w = m × g, where m = mass and g = acceleration due to gravity.

(ii) Volume: The space occupied by matter (usually by a liquid or a gas) is called its volume.
Volume = (length)³ = m³
1 m³ = (100 cm)³ = 10⁶ cm³ = 10³ dm³ = 10³ L

(iii) Density: It is the amount of mass per unit volume and has units kg m⁻³ or g cm⁻³.
Density = Mass / Volume

(iv) Temperature: It is the degree of hotness or coldness. Its units are °C, °F and K, but its SI unit is K.
°C = \( \frac{5}{9} \) (°F − 32)  or  °F = \( \frac{9}{5} \) (°C) + 32
K = °C + 273.15

3. Significant Figures

These are the total number of digits in a number including the last digit whose value is uncertain.
Rules for counting the number of significant figures:
(i) All non-zero digits are significant.
(ii) The zeroes to the left of the first non-zero digit in a number are not significant (e.g. 0.0025 has 2 significant figures).
(iii) Zeroes between two non-zero digits are significant, and zeroes at the end (right) of a number are significant if they are on the right side of the decimal point (e.g. 0.200 has 3 significant figures).
(iv) In scientific notation every number is written as N × 10ⁿ, where N = a number with a single non-zero digit to the left of the decimal point and n = an integer.

Significant figures in calculations:
1. The result of an addition or subtraction should be reported to the same number of decimal places as that of the term with the least number of decimal places.
2. The result of a multiplication or division should be reported to the same number of significant figures as is possessed by the least precise term used in the calculation.
3. If a calculation involves a number of steps, the result should contain the same number of significant figures as that of the least precise number involved, other than exact numbers.

4. Accuracy and Precision

(i) Accuracy is the agreement of a particular value to the true value of the result.
Accuracy = Mean value − True value
(ii) Precision refers to the closeness of various measurements for the same quantity.
Precision = Individual value − Arithmetic mean value

5. Dimensional Analysis

In calculations it is often necessary to convert units from one system to another. This is done by the factor label method, unit factor method (CF) or dimensional analysis.
Information sought = Information given × Conversion factor (CF)

6. Different Masses, Mole Concept and Formula of Compounds

(i) Atomic mass = \( \frac{\text{Mass of an atom}}{\frac{1}{12} \times \text{Mass of a carbon atom } (^{12}\text{C})} \)
(ii) Average atomic mass = \( \frac{\sum p_i A_i}{100} \) = \( \frac{p_1 A_1 + p_2 A_2 + \ldots}{100} \), where p_i is the percentage abundance of the isotope having atomic mass A_i.
(iii) Gram atomic mass = Atomic mass expressed in grams = 1 gram atom
(iv) Molecular mass is the sum of the atomic masses of all the atoms present in a molecule.
(v) Gram molecular mass (gram molecule) = Molecular mass expressed in grams
(vi) Formula mass of an ionic compound = (number of cations × atomic mass of cation) + (number of anions × atomic mass of anion)
(vii) Relation between molecular mass and vapour density of a gas: Molecular mass = 2 × Vapour density

Mole Concept and Molar Masses

1. Number of moles of atoms = \( \frac{\text{Given mass of element}}{\text{Atomic mass of element}} \)
2. Number of moles of molecules = \( \frac{\text{Given mass of substance}}{\text{Molecular mass}} \)
3. Number of moles of a gas = \( \frac{\text{Volume of the gas at STP}}{22.4\ \text{L}} \)
4. 1 mole = 6.022 × 10²³ particles = Gram atomic / gram molecular mass
5. Number of molecules = Number of moles × N_A, where N_A = Avogadro's number = 6.022 × 10²³
6. Number of atoms = Number of molecules × atomicity (or number of atoms of the element in one molecule of the compound)

7. Percentage Composition

Mass % of an element = \( \frac{\text{Mass of the element in 1 mole of compound}}{\text{Mass of 1 mole of compound}} \) × 100

8. Empirical and Molecular Formula

An empirical formula represents the simplest whole number ratio of the various atoms present in a compound, whereas the molecular formula shows the exact number of the different types of atoms present in a molecule of the compound.
n = \( \frac{\text{Molecular mass}}{\text{Empirical formula mass}} \)
Molecular formula = n × Empirical formula

Short Trick to Find Empirical and Molecular Formula

Step 1: Divide the percentage of each element by its atomic mass to obtain the atomic (molar) ratio.
Step 2: Divide each value of the atomic ratio by the minimum value to obtain the simplest ratio.
Step 3: If needed, multiply the simplest ratio by a suitable integer to obtain the simplest whole number ratio.
Step 4: Write the symbols of the elements with their whole numbers as subscripts at the lower right corner of each symbol to obtain the empirical formula.
Step 5: Multiply the empirical formula by n to obtain the molecular formula, where n = \( \frac{\text{Molecular mass}}{\text{Empirical formula mass}} \).

9. Stoichiometry and Stoichiometric Calculations

The calculations involving the amounts of reactants and products in a given chemical reaction are called stoichiometry. The numbers written before the formula units or molecules to balance an equation are called stoichiometric coefficients.
In the balanced reaction
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g),
the coefficients 2 for O₂ and H₂O are the stoichiometric coefficients. Similarly, for CH₄ and CO₂ the stoichiometric coefficient is 1.

According to the above reaction:
(i) One mole of CH₄(g) reacts with two moles of O₂(g) to give one mole of CO₂(g) and two moles of H₂O(g).
(ii) One molecule of CH₄(g) reacts with 2 molecules of O₂(g) to give one molecule of CO₂(g) and 2 molecules of H₂O(g).
(iii) 22.7 L of CH₄(g) reacts with 45.4 L of O₂(g) to give 22.7 L of CO₂(g) and 45.4 L of H₂O(g) (at STP, 1 bar and 273 K).
(iv) 16 g of CH₄(g) reacts with 2 × 32 g of O₂(g) to give 44 g of CO₂(g) and 2 × 18 g of H₂O(g).

Limiting reagent: The reactant which is consumed completely in a reaction is called the limiting reactant or limiting reagent. It decides the amount of the other reactants that react and the amount of products formed.

10. Concentration Terms of Solutions

The concentration of a solution is usually expressed in the following ways:

(i) Mass per cent = \( \frac{\text{Mass of solute}}{\text{Mass of solution}} \) × 100

(ii) Volume percentage of A = \( \frac{\text{Volume of A}}{\text{Volume of A} + \text{Volume of B}} \) × 100

(iii) Parts per million (ppm) of A = \( \frac{\text{Mass of compound A}}{\text{Total mass of solution}} \) × 10⁶

(iv) Mole fraction: For a solution of two components A and B, the mole fraction of A is
χ_A = \( \frac{\text{Number of moles of A}}{\text{Total number of moles of solution}} \) = \( \frac{n_A}{n_A + n_B} \)
where n_A and n_B are the moles of A and B respectively. Also remember that χ_A + χ_B = 1.

(v) Molarity (M): It is the number of moles of solute present in 1 L of the solution.
M = \( \frac{\text{Number of moles of solute}}{\text{Volume of solution (in L)}} \)  or  M = \( \frac{\text{Mass of solute} \times 1000}{\text{Molar mass of solute} \times V \text{ (in mL)}} \)
The units of molarity are mol L⁻¹ or mol dm⁻³.
M = \( \frac{\text{Mass per cent} \times \text{specific gravity} \times 10}{\text{Molar mass}} \)
In case of dilution, M₁V₁ = M₂V₂, where M₁ and V₁ are the molarity and volume before dilution and M₂ and V₂ are the molarity and volume after dilution.

(vi) Molality (m): It is the number of moles of solute present in 1 kg of the solvent.
m = \( \frac{\text{Number of moles of solute}}{\text{Mass of solvent (in kg)}} \)  or  m = \( \frac{\text{Mass of solute} \times 1000}{\text{Molar mass of solute} \times \text{Mass of solvent (in g)}} \)
The SI unit of molality is mol kg⁻¹.