Some Basic Concepts of Chemistry

Chemistry is the branch of science dealing with the composition, structure, properties and transformations of matter. It underlies industries such as fertilisers, pharmaceuticals, polymers and dyes, and helps address problems of health, nutrition, energy and the environment. Long before atoms could be observed directly, chemists deduced how elements combine by carefully weighing reactants and products. This gave rise to five quantitative laws of chemical combination.

• Law of Conservation of Mass (Lavoisier): in a chemical reaction, the total mass of the reactants equals the total mass of the products; matter can neither be created nor destroyed.
• Law of Definite Proportions (Proust): a given compound always contains the same elements combined in the same fixed proportion by mass, whatever its source or method of preparation.
• Law of Multiple Proportions (Dalton): when two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
• Gay Lussac's Law of Gaseous Volumes: when gases react or are produced, they do so in a simple whole-number ratio by volume, all gases being at the same temperature and pressure.
• Avogadro's Law: equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules.

To explain these laws, John Dalton proposed his atomic theory. Its main postulates state that matter consists of indivisible atoms; atoms of the same element are identical in mass and properties while atoms of different elements differ; atoms combine in small whole-number ratios to form compounds; and a chemical reaction merely rearranges atoms, it does not create, destroy, or change them. This theory correctly explained conservation of mass, definite proportions, and multiple proportions, though later discoveries of subatomic particles and isotopes showed that atoms are divisible and that atoms of the same element can have different masses.

Worked example (Law of Multiple Proportions): Carbon forms two oxides with oxygen. In the first, 3 g of carbon combines with 4 g of oxygen (carbon monoxide, CO). In the second, 3 g of carbon combines with 8 g of oxygen (carbon dioxide, CO₂). For the same fixed mass of carbon (3 g), the masses of oxygen that combine are 4 g and 8 g, giving the ratio 4 : 8 = 1 : 2, a ratio of small whole numbers, confirming the Law of Multiple Proportions.

Since atoms are extremely small, their masses are expressed relative to a standard rather than in grams directly. The modern standard is the carbon-12 isotope, assigned a mass of exactly 12 atomic mass units (u). One atomic mass unit is defined as exactly 1/12th of the mass of one atom of carbon-12, and equals 1.66056 × 10^-24 g.

Most elements occur naturally as a mixture of isotopes, atoms with the same number of protons but different numbers of neutrons. The average atomic mass of an element is the weighted average of the masses of its naturally occurring isotopes, taking their relative abundance into account. This is the value shown in the periodic table.

The molecular mass of a substance is the sum of the atomic masses of all atoms present in one molecule of that substance. For ionic (non-molecular) compounds such as NaCl, which do not exist as discrete molecules, chemists use the term formula mass, calculated the same way from the formula unit.

Worked example (average atomic mass): Naturally occurring chlorine consists of two isotopes: ³⁵Cl with mass 35 u and abundance 75%, and ³⁷Cl with mass 37 u and abundance 25%. Average atomic mass = (35 × 75/100) + (37 × 25/100) = 26.25 + 9.25 = 35.5 u, matching the value conventionally used for chlorine in NCERT problems.

Worked example (molecular mass): For sulphuric acid, H₂SO₄: atomic masses are H = 1 u, S = 32 u, O = 16 u. Molecular mass = 2(1) + 1(32) + 4(16) = 2 + 32 + 64 = 98 u.

Atoms and molecules are too small to count individually, so chemists use the mole as a counting unit, similar to using "dozen" for 12 items. One mole is defined as the amount of substance containing exactly 6.02214076 × 10²³ elementary entities (atoms, molecules, ions, or other particles). This number is the Avogadro constant (N_A), commonly rounded to 6.022 × 10²³ mol^-1.

The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g mol^-1). Numerically, the molar mass in g mol^-1 equals the atomic or molecular mass in u. For example, the molar mass of carbon is 12 g mol^-1 and of water (H₂O) is 18 g mol^-1.

For gases, one mole of any ideal gas occupies the same volume at a given temperature and pressure (Avogadro's Law). At STP (Standard Temperature and Pressure, taken as 273.15 K and 1 bar as per the current IUPAC convention), the molar volume of an ideal gas is 22.7 L mol^-1.

Worked example: Find the number of moles and the number of molecules present in 4.4 g of carbon dioxide, CO₂ (molar mass = 44 g mol^-1).

Number of moles, n = given mass / molar mass = 4.4 / 44 = 0.1 mol.

Number of molecules = n × N_A = 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules.

The percentage composition of a compound tells us what percentage of its total mass is contributed by each element:

% of element = (mass of that element in one mole of compound / molar mass of compound) × 100

The empirical formula of a compound shows the simplest whole-number ratio of atoms of each element present. The molecular formula shows the actual number of atoms of each element in one molecule, and is always a whole-number multiple (n) of the empirical formula: Molecular formula = n × (Empirical formula), where n = molar mass / empirical formula mass.

Worked example: A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass, and has a molar mass of 98.96 g mol^-1. Find its empirical and molecular formula.

Step 1 – Assume 100 g of compound, so the mass of each element equals its percentage: H = 4.07 g, C = 24.27 g, Cl = 71.65 g.

Step 2 – Convert masses to moles: moles of H = 4.07/1 = 4.07; moles of C = 24.27/12 = 2.02; moles of Cl = 71.65/35.5 = 2.02.

Step 3 – Divide by the smallest value (2.02): H ≈ 2, C = 1, Cl = 1. This gives the empirical formula CH₂Cl, with empirical formula mass = 12 + 2(1) + 35.5 = 49.5 g mol^-1.

Step 4 – Find n = molar mass / empirical formula mass = 98.96 / 49.5 ≈ 2. So the molecular formula = 2 × CH₂Cl = C₂H₄Cl₂.

Stoichiometry is the quantitative study of the relationship between the amounts of reactants and products in a balanced chemical equation. The coefficients of a balanced equation give the mole ratio in which substances react and are formed. For example, in N₂(g) + 3H₂(g) → 2NH₃(g), 1 mole of N₂ reacts with exactly 3 moles of H₂ to give 2 moles of NH₃.

In practice, reactants are often not mixed in the exact stoichiometric ratio. The limiting reagent (limiting reactant) is the reactant that is completely consumed first, and it determines the maximum amount of product that can be formed. The other reactant, present in a greater-than-required amount, is called the excess reagent, and some of it is left unreacted.

Worked example: 280 g of N₂ is reacted with 70 g of H₂ by the reaction N₂(g) + 3H₂(g) → 2NH₃(g). Identify the limiting reagent and calculate the mass of NH₃ formed. (Molar masses: N₂ = 28 g mol^-1, H₂ = 2 g mol^-1, NH₃ = 17 g mol^-1)

Step 1 – Moles available: moles of N₂ = 280/28 = 10 mol; moles of H₂ = 70/2 = 35 mol.

Step 2 – Moles of H₂ required to react completely with 10 mol N₂ = 10 × 3 = 30 mol. Since 35 mol H₂ is available, more than the 30 mol required, H₂ is in excess and N₂ is the limiting reagent.

Step 3 – Moles of NH₃ formed (based on the limiting reagent, N₂) = 10 × 2 = 20 mol.

Step 4 – Mass of NH₃ formed = 20 × 17 = 340 g.

A solution consists of a solute dissolved in a solvent. Its concentration can be expressed in several ways:

• Mass percentage (% w/w): mass of solute per 100 g of solution.
• Mole fraction (x): ratio of moles of one component to the total moles of all components in the solution; the mole fractions of all components add up to 1.
• Molarity (M): number of moles of solute dissolved per litre (dm³) of solution. It is slightly temperature-dependent because the volume of a solution changes with temperature.
• Molality (m): number of moles of solute dissolved per kilogram of solvent. Since it is based on mass, molality does not change with temperature.

Worked example (molarity): Calculate the molarity of a solution prepared by dissolving 4 g of NaOH (molar mass = 40 g mol^-1) in enough water to make 250 mL of solution.

Moles of NaOH = 4/40 = 0.1 mol. Volume of solution in litres = 250/1000 = 0.25 L.

Molarity = moles of solute / volume of solution (L) = 0.1/0.25 = 0.4 mol L^-1 (0.4 M).

Worked example (molality): 30 g of ethylene glycol (molar mass = 62 g mol^-1) is dissolved in 500 g of water. Calculate the molality of the solution.

Moles of ethylene glycol = 30/62 = 0.484 mol. Mass of solvent in kg = 500/1000 = 0.5 kg.

Molality = moles of solute / mass of solvent (kg) = 0.484/0.5 = 0.968 mol kg^-1.