Some Basic Concepts of Chemistry Class 11 Notes

Some Basic Concepts of Chemistry Class 11 notes — the whole of NCERT Chapter 1 on one revision page.

Matter and its classification, SI units, significant figures, the mole, stoichiometry and solution concentration are covered in the order a teacher builds them, with every definition and formula traceable to the NCERT Class 11 Chemistry Part I textbook — you can cross-check any point against the official NCERT chapter PDF.

These notes are part of our Class 11 Chemistry notes hub; read the sections in order, or jump straight to what you need.

Some Basic Concepts of Chemistry Class 11 Notes: Matter and Its Classification

Matter is anything that has mass and occupies space — the book in your hand, the air around you, the water you drink (NCERT, p. 4). Chemistry studies matter as atoms and molecules, so every later calculation in this chapter (moles, molar mass, stoichiometry) begins with knowing what kind of matter you are counting.

Matter exists in three physical states, which differ only in how their particles are arranged (NCERT, p. 5):

The three states are interconvertible by changing temperature (NCERT, p. 5):

\[ \text{Solid} \xrightarrow{\text{heat}} \text{Liquid} \xrightarrow{\text{heat}} \text{Gas} \]

Cooling reverses the process: a gas liquifies, a liquid freezes to a solid.

At the bulk level, matter is classified as a mixture or a pure substance; a pure substance is further an element or a compound (NCERT, p. 5). The flow chart below shows the full tree.

Two exam-relevant facts from this classification (NCERT, p. 6):

• A compound has a fixed and characteristic ratio of atoms; its constituents can be separated only by chemical methods, not by physical ones.
• The properties of a compound differ from those of its elements: hydrogen and oxygen are gases (hydrogen burns with a pop sound; oxygen supports combustion), yet their compound water is a liquid used as a fire extinguisher.

SI Units, Prefixes and Measuring Physical Quantities

Every quantitative measurement is a number followed by a unit. The scientific community agreed on the International System of Units (SI) for uniformity; it has seven base units (NCERT, p. 7).

Prefixes express multiples and sub-multiples of any unit (NCERT, p. 8). The ones you will actually use in numerical problems are:

Memory device: the order from smallest to largest is captured by the sentence “Please Never Make My Coffee Decent, Keep Mugs Great” — Pico, Nano, Micro, Milli, Centi, Deci, Kilo, Mega, Giga. So \( 1\ \text{nm} = 10^{-9}\ \text{m} \) and \( 1\ \text{mg} = 10^{-3}\ \text{g} \).

Two quantities students confuse: mass is the amount of matter in a substance and is constant, while weight is the gravitational force on it and varies from place to place (NCERT, p. 9). The SI unit of mass is the kilogram, but laboratories use the gram because chemical amounts are small.

Volume has the SI unit \( \text{m}^3 \), but chemists use \( \text{cm}^3 \), \( \text{dm}^3 \) and the non-SI litre (NCERT, p. 9):

\[ 1\ \text{L} = 1000\ \text{mL} = 1000\ \text{cm}^3 = 1\ \text{dm}^3 \]

Density is mass per unit volume: \( \rho = \frac{\text{mass}}{\text{volume}} \). Its SI unit is \( \text{kg m}^{-3} \), but chemists usually express it in \( \text{g cm}^{-3} \). Higher density means particles are more closely packed (NCERT, p. 10).

Temperature has three scales — Celsius, Fahrenheit and kelvin (the SI unit). The relations are (NCERT, p. 10):

\[ ^{\circ}\text{F} = \frac{9}{5}(^{\circ}\text{C}) + 32 \qquad K = ^{\circ}\text{C} + 273.15 \]

For example, \( 25\ ^{\circ}\text{C} = (25 + 273.15)\ \text{K} = 298.15\ \text{K} \). Negative kelvin is impossible because the kelvin scale ends at absolute zero. In India, the National Physical Laboratory (NPL), New Delhi maintains the national standards of measurement (NCERT, p. 7).

Scientific Notation, Significant Figures and Unit Conversion

Chemistry deals with numbers as large as \( 6.022 \times 10^{23} \) and as small as \( 1.66 \times 10^{-24} \) g. Scientific notation writes any number as \( N \times 10^n \), where \( N \) lies between 1 and 9.999… (NCERT, p. 11).

• Decimal moved left → positive exponent: \( 232.508 = 2.32508 \times 10^2 \).
• Decimal moved right → negative exponent: \( 0.00016 = 1.6 \times 10^{-4} \).
• Multiply/divide: add/subtract the exponents: \( (4.2 \times 10^6) \times (3.0 \times 10^3) = 1.26 \times 10^{10} \).
• Add/subtract: first make the exponents equal: \( 3.1 \times 10^5 + 2.0 \times 10^4 = (3.1 + 0.20) \times 10^5 = 3.3 \times 10^5 \) (NCERT, p. 12).

Significant figures are the meaningful digits in a measurement — the digits known with certainty plus one estimated digit (NCERT, p. 12).

To show how many digits 100 really means, use scientific notation: \( 1 \times 10^2 \) (one), \( 1.0 \times 10^2 \) (two), \( 1.00 \times 10^2 \) (three significant figures) (NCERT, p. 12-13).

Rounding rules (NCERT, p. 13): remove 6 → round up (\( 1.386 \rightarrow 1.39 \)); remove 4 → keep (\( 4.334 \rightarrow 4.33 \)); remove exactly 5 → round so the preceding digit becomes even: \( 6.35 \rightarrow 6.4 \) but \( 6.25 \rightarrow 6.2 \).

For addition/subtraction, the result has no more digits after the decimal than the value with the fewest; for multiplication/division, no more significant figures than the value with the fewest. Example: \( 3.2 \times 1.5 = 4.80 \), reported as \( 4.8 \) because 3.2 has only two significant figures.

Precision is the closeness of repeated measurements to each other; accuracy is the closeness of a result to the true value. In the textbook example, student A is precise but not accurate, student B is neither, and student C is both (NCERT, p. 13).

A precise but inaccurate result comes from a systematic error — the readings cluster, but around the wrong value.

Dimensional analysis (unit-factor method) converts units by multiplying by a fraction equal to 1. Choose the unit factor whose numerator carries the desired unit, then cancel units like numbers (NCERT, p. 13-14).

Worked example: convert 90 km/h into m/s

Laws of Chemical Combination and Dalton’s Atomic Theory

Five experimental laws describe how elements combine to form compounds (NCERT, pp. 14-15). They are a one-mark favourite — memorise the scientist, the year and the statement.

As Fig 1.9 shows, two volumes of H₂ and one of O₂ give two volumes of water vapour without leaving any oxygen unreacted — this is only possible if hydrogen and oxygen are diatomic molecules, which is why Avogadro’s distinction between atoms and molecules mattered (NCERT, p. 15-16).

Dalton’s atomic theory (1808) rests on four postulates (NCERT, p. 16):

• Matter consists of indivisible atoms.
• All atoms of a given element are identical in mass and properties; atoms of different elements differ in mass.
• Compounds form when atoms of different elements combine in a fixed ratio.
• Chemical reactions only reorganise atoms — atoms are neither created nor destroyed.

Dalton’s theory successfully explained the laws of chemical combination, but it could not explain the law of gaseous volumes, and it gave no reason why atoms combine (NCERT, p. 16).

Atomic Mass, Molecular Mass and Formula Mass

Atoms are far too small to weigh directly, so their masses are measured relative to a standard. Since 1961 the standard is carbon-12 (\( ^{12}\text{C} \)), assigned a mass of exactly 12 u (NCERT, p. 16-17):

\[ 1\ \text{u} = \frac{1}{12} \times \text{mass of one } ^{12}\text{C atom} = 1.66056 \times 10^{-24}\ \text{g} \]

On this scale, a hydrogen atom has mass \( 1.0080\ \text{u} \) and an oxygen-16 atom \( 15.995\ \text{u} \). The symbol u (unified mass) replaces the older amu.

Average atomic mass: most elements exist as isotopes, so the periodic-table value is a weighted average over natural abundance. For carbon, \( 98.892\% \) \( ^{12}\text{C} \) (12 u) and \( 1.108\% \) \( ^{13}\text{C} \) (13.00335 u) give the familiar \( 12.011\ \text{u} \) (NCERT, p. 17).

Molecular mass is the sum of the atomic masses of all atoms in a molecule (NCERT, p. 17). For methane:

\[ \text{CH}_4 = 12.011\ \text{u} + 4(1.008\ \text{u}) = 16.043\ \text{u} \]

Worked example: molecular mass of sucrose \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)

Formula mass is used for ionic compounds like sodium chloride, which have no discrete molecules. As Fig 1.10 shows, each \( \text{Na}^+ \) ion is surrounded by six \( \text{Cl}^- \) ions and vice versa, so NaCl is a formula unit, not a molecule (NCERT, p. 17-18):

\[ \text{Formula mass of NaCl} = 23.0\ \text{u} + 35.5\ \text{u} = 58.5\ \text{u} \]

The most common slip here: a molecule like water counts hydrogen twice — \( 2(1.008\ \text{u}) + 16.00\ \text{u} = 18.02\ \text{u} \), never \( 1.008 + 16.00 \).

Mole Concept and Molar Mass

You already count everyday objects in dozen (12), score (20) and gross (144). Chemistry’s counting unit for particles is the mole (NCERT, p. 18):

\[ 1\ \text{mol} = 6.02214076 \times 10^{23}\ \text{elementary entities} \]

This number is the Avogadro constant \( N_A \), with unit \( \text{mol}^{-1} \). The mole exists because atoms are too small to weigh individually and too numerous to count one by one — it is the bridge between the gram balance and the particle count: mass → moles → number of particles.

Always specify the entity (NCERT, p. 18):

• 1 mol of hydrogen atoms = \( 6.022 \times 10^{23} \) atoms
• 1 mol of water molecules = \( 6.022 \times 10^{23} \) molecules
• 1 mol of sodium chloride = \( 6.022 \times 10^{23} \) formula units

Molar mass is the mass of one mole of a substance in grams; it is numerically equal to the atomic/molecular/formula mass in u (NCERT, p. 18). So water is 18.02 u and its molar mass is 18.02 g mol⁻¹.

Worked example: number of molecules in 9.0 g of water

The next chapter, Structure of Atom, builds directly on this idea — it explains the sub-atomic particles you are counting with the mole.

Percentage Composition and Empirical Formula

Mass per cent tells you what a compound is made of by mass (NCERT, p. 19):

\[ \text{Mass \% of element} = \frac{\text{mass of that element in the compound}}{\text{molar mass of the compound}} \times 100 \]

For water, molar mass 18.02 g: hydrogen \( = \frac{2 \times 1.008}{18.02} \times 100 = 11.18\% \), oxygen \( = 88.79\% \).

Two formulas to keep distinct (NCERT, p. 19):

• Empirical formula — the simplest whole-number ratio of atoms in a compound.
• Molecular formula — the exact number of each type of atom in one molecule.

The five-step method for converting per cent data into a formula (NCERT, pp. 19-20):

1. Take 100 g of the compound, so each per cent becomes a mass in grams.
2. Convert each mass to moles by dividing by the atomic mass.
3. Divide every mole value by the smallest mole value.
4. If needed, multiply to reach whole numbers, then write the empirical formula.
5. Compute \( n = \frac{\text{molar mass}}{\text{empirical formula mass}} \) and multiply the empirical formula by \( n \) to get the molecular formula.

Worked example: empirical formula from 40.0% C, 6.7% H, 53.3% O

If the molar mass of this compound were 180 g mol⁻¹, then empirical formula mass \( = 12 + 2(1) + 16 = 30\ \text{u} \), so \( n = 180/30 = 6 \) and the molecular formula would be \( \text{C}_6\text{H}_{12}\text{O}_6 \) — the formula of glucose. In step 5, the ratio \( n \) is a pure number: the empirical mass is in u and the molar mass in g, but the units cancel.

Stoichiometry, Balancing Equations and the Limiting Reagent

Stoichiometry (from Greek stoicheion, element, and metron, measure) is the calculation of masses and volumes of reactants and products from a balanced chemical equation (NCERT, p. 20).

The combustion of methane shows what the coefficients mean:

\[ \text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O}\text{(g)} \]

• Moles: 1 mol CH₄ + 2 mol O₂ → 1 mol CO₂ + 2 mol H₂O
• Molecules: 1 molecule CH₄ + 2 molecules O₂ → 1 molecule CO₂ + 2 molecules H₂O
• Volumes (gases at same T and P): 22.7 L CH₄ + 45.4 L O₂ → 22.7 L CO₂ + 45.4 L H₂O
• Grams: 16 g CH₄ + 64 g O₂ → 44 g CO₂ + 36 g H₂O

To balance, adjust coefficients in steps (C, then H, then O) and verify at the end. Propane, for example, balances as \( \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \) (NCERT, p. 21-22). The iron rule: never change subscripts to balance an equation — only coefficients, because changing a subscript changes the substance itself.

Limiting reagent: when reactants are mixed in amounts different from the balanced ratio, the reactant that gets consumed first stops the reaction and fixes the amount of product — it is the limiting reagent (NCERT, p. 21).

Two-step test:

1. Convert every reactant mass to moles.
2. Compare the available moles with the mole ratio required by the balanced equation; the reactant that runs short is limiting.

Worked example: 12 g carbon burned in 24 g oxygen

Show the unit-factor lines in your answer — that visible working is what earns the mark, even if the final arithmetic slips.

Concentration of Solutions: Mass %, Mole Fraction, Molarity, Molality

Most reactions run in solution, so you need four ways to state how much solute is present (NCERT, p. 21):

The molarity–molality comparison is a favourite two-mark question:

Dilution: the number of moles of solute does not change when you add solvent, so \( M_1 V_1 = M_2 V_2 \) (NCERT, p. 23). The textbook’s stock-solution logic: taking 200 mL of 1 M NaOH and diluting to 1 L gives 0.2 M — the 0.2 mol of NaOH stays 0.2 mol, only the volume changes.

Worked example: molarity of 5.85 g NaCl made up to 500 mL

For molality, the denominator is the solvent, not the solution: if the same 0.10 mol of solute were dissolved in 500 g of water, \( m = \frac{0.10\ \text{mol}}{0.500\ \text{kg}} = 0.2\ \text{mol kg}^{-1} \) (0.2 m).

Common Mistakes Students Make

Each of these slips costs a mark you did earn the reasoning for. The third column tells you how to catch it before submitting:

Exam Notes: What Earns the Marks

The NCERT exercise set for this chapter (pp. 25-28) clusters into a few repeatable calculation families — the map below is an observed pattern from that set, not a prediction of any particular paper.

Examiner’s mindset:

• In stoichiometry, write every unit-factor conversion as a visible chain — the working earns the mark even if the final arithmetic slips.
• Attach units to every line; a number without a unit is not a chemistry answer.
• Specify the entity when counting moles — atoms, molecules or formula units.
• End with the conclusion stated separately: which reactant is limiting, how much product forms, what remains unreacted.

One-Shot Revision Summary

The whole chapter in one table, in the order it builds:

Constants to memorise: \( 1\ \text{u} = 1.66056 \times 10^{-24}\ \text{g} \); \( N_A = 6.022 \times 10^{23}\ \text{mol}^{-1} \); \( K = \,^{\circ}\text{C} + 273.15 \); \( 1\ \text{L} = 1000\ \text{cm}^3 = 1\ \text{dm}^3 \); molarity is temperature dependent, molality is not.

Need the wider picture? Browse the complete set of Class 11 notes or the full CBSE notes library.

Frequently Asked Questions

Why is molarity temperature dependent but molality is not?

Because molarity’s denominator is volume of solution, and volume changes with temperature, while molality’s denominator is mass of solvent, which does not change. Heat a solution and it expands, so the number of moles per litre drops even though the solute is unchanged; the mass of solvent, and therefore molality, stays the same (NCERT, p. 24).

What exactly is 1 atomic mass unit and why is it based on carbon-12?

One atomic mass unit is exactly one-twelfth of the mass of one carbon-12 atom: \( 1\ \text{u} = 1.66056 \times 10^{-24}\ \text{g} \) (NCERT, p. 17). Carbon-12 was agreed as the standard in 1961 because it is a stable, convenient isotope; assigning it exactly 12 u gives other elements near-whole-number relative masses — hydrogen ≈ 1.008 u, oxygen-16 ≈ 15.995 u — which makes the scale practical.

What is the difference between empirical formula and molecular formula?

The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of each atom in one molecule. In NCERT Problem 1.2, the empirical formula \( \text{CH}_2\text{Cl} \) with \( n = 2 \) gives the molecular formula \( \text{C}_2\text{H}_4\text{Cl}_2 \).

You need the molar mass to make the conversion: \( n = \frac{\text{molar mass}}{\text{empirical formula mass}} \).

How do I find the limiting reagent in a reaction?

Convert each reactant’s mass to moles, then compare those moles with the ratio in the balanced equation — the reactant that provides fewer moles than required is the limiting reagent, and it decides the amount of product.

In the worked example above, 12 g C (1.0 mol) and 24 g O₂ (0.75 mol) react 1 : 1, so oxygen runs out first and caps the CO₂ at about 33 g.

When are zeros significant in a measurement?

Zeros between non-zero digits are significant (\( 2.005 \) → 4); trailing zeros after a decimal point are significant (\( 0.200 \) → 3); leading zeros are never significant (\( 0.0052 \) → 2); and trailing zeros without a decimal point are not significant (\( 100 \) → 1). To show that trailing zeros are significant, write the number in scientific notation: \( 1.00 \times 10^2 \) has three significant figures (NCERT, p. 12-13).

Reference: NCERT Class 11 Chemistry Part I textbook, Chapter 1: Some Basic Concepts of Chemistry.