Class 11 · Chemistry · Chapter 1

Some Basic Concepts of Chemistry

This chapter lays the foundation of quantitative chemistry: units and significant figures, the laws of chemical combination, Dalton's atomic theory, and the mole concept that lets you convert between mass, number of particles and gas volume. Every numerical you will ever do in chemistry — in Class 11, Class 12, JEE or NEET — rests on the mole concept and stoichiometry built here.

10 sections12 NCERT questions23 board questions15-question quiz

Before you startWhat the board asks from this chapter

Class 11 has no board exam, but this chapter carries real weight in the school final (roughly 3–5 marks, mostly numerical) and it is the base skill for every stoichiometry, solution and equilibrium numerical in Class 12 boards, JEE and NEET.

Question typeMarksUsually asked
MCQ1Definitions, SI units, significant figures, and quick mole/mass conversions.
Assertion–Reason1Tests whether you can distinguish empirical vs molecular formula, or explain why a law or definition holds.
Very short answer2Define a term, state a law, or do a one-step mole calculation.
Short answer3Mass percentage, molarity/molality of a solution, or a short numerical using the mole concept.
Long answer5Full limiting-reagent numerical, or empirical/molecular formula determination from percentage composition.
Case-based4A passage on the mole concept or an industrial reaction with sub-questions on moles, mass, and limiting reagent.

Hot topics, in order

  1. Mole concept: mass–mole–number–volume conversions
  2. Limiting reagent numericals
  3. Empirical and molecular formula from percentage composition
  4. Molarity and molality of solutions
  5. Laws of chemical combination with examples
  6. Significant figures and their rules
  7. Percentage composition of a compound
  8. Dalton's atomic theory postulates

The whole chapter, in orderNotes

01 1.1 Importance of Chemistry

Chemistry is the branch of science that studies the composition, properties and transformation of matter. NCERT opens the chapter by pointing out how wide its reach is — chemistry gives us fertilisers, medicines, polymers, dyes, alloys and the fuels that run daily life, and it sits at the centre of biology, medicine, agriculture, geology and engineering.

You don't need to memorise a long list of applications for the exam, but you should be able to name two or three areas (health, energy, materials, agriculture) and give one concrete example in each, because 1-mark and 2-mark 'importance of chemistry' questions are common as openers.

Exam tip If asked to justify why chemistry is called a 'central science', mention that it links physics (structure, energy) to biology (life processes) — that one line earns the mark.

02 1.2 Nature of Matter

Matter is anything that has mass and occupies space. It exists in three physical states — solid, liquid and gas — which differ in how closely their particles are packed and how strongly they attract one another. Solids have definite shape and volume, liquids have definite volume but take the shape of the container, and gases have neither definite shape nor definite volume; they expand to fill any container.

Matter is also classified chemically into pure substances and mixtures. A pure substance has a fixed composition throughout and cannot be separated into simpler substances by physical methods; it is further divided into elements (made of only one type of atom, e.g. Na, Cu, Ag) and compounds (two or more elements combined in a fixed ratio, e.g. water, carbon dioxide). A compound's properties are entirely different from its constituent elements. Mixtures contain two or more substances in variable proportion and can be separated by physical means; they are homogeneous (uniform composition throughout, e.g. sugar solution, air) or heterogeneous (non-uniform, e.g. sand in water, oil and water).

Exam tip A one-line distinguishing test the examiner likes: compounds have a fixed composition by mass and fixed properties; mixtures do not.

03 1.3 Properties of Matter and Their Measurement

Every property of matter is either physical (measurable without changing chemical composition — colour, melting point, density) or chemical (describes how a substance transforms into a different substance — flammability, reactivity with acid). Physical properties are further extensive (depend on the amount of matter, e.g. mass, volume, length) or intensive (independent of amount, e.g. density, temperature, molar mass, refractive index).

Since chemistry is a quantitative science, measurement needs a common system of units. India, like most of the world, follows the International System of Units (SI), built on seven base units: metre (length), kilogram (mass), second (time), ampere (electric current), kelvin (temperature), mole (amount of substance) and candela (luminous intensity). Mass and weight are often used loosely in everyday speech but are different: mass is the amount of matter (constant everywhere, SI unit kg), while weight is the force of gravity on that mass (varies with location, unit newton).

Density = mass/volume
unitkg m−3 in SI; g cm−3 or g mL−1 commonly used in the lab
Exam tip If a question asks 'is X extensive or intensive', check: does it change if you take double the amount of substance? If yes, extensive; if no, intensive.

04 1.4 Uncertainty in Measurement

No measurement is perfectly exact — every reading carries some uncertainty from the precision of the instrument. Scientists express this using significant figures: all the digits that are known with certainty plus one estimated (uncertain) digit. The rules examiners test most:

  • All non-zero digits are significant (285 has 3 significant figures).
  • Zeros between two non-zero digits are significant (2005 has 4).
  • Leading zeros (before the first non-zero digit) are never significant — they only locate the decimal point (0.0025 has 2 significant figures).
  • Trailing zeros in a number with a decimal point are significant (0.0500 has 3; 2.500 has 4).
  • Trailing zeros in a number without a decimal point are ambiguous (100 could have 1, 2 or 3); this ambiguity is removed by writing the number in scientific notation, e.g. 1.00 × 102 clearly has 3.
  • Exact numbers (from counting, or defined conversion factors like 1 dozen = 12) have infinite significant figures and never limit a calculation.

When adding or subtracting, the result is rounded to the least number of decimal places among the terms. When multiplying or dividing, the result is rounded to the least number of significant figures among the terms. Rounding follows the standard rule: if the digit to be dropped is more than 5, the preceding digit is raised by 1; if less than 5, it is left unchanged; if exactly 5, the preceding digit is raised by 1 if it is odd and left unchanged if it is even.

Exam tip The single most-tested trap: leading zeros are NOT significant, but a zero sandwiched between non-zero digits or a trailing zero after a decimal point IS significant.

05 1.5 Laws of Chemical Combination

Quantitative experiments on chemical reactions led to five laws that Dalton later explained with his atomic theory.

  • Law of conservation of mass (Lavoisier): matter can neither be created nor destroyed in a chemical reaction — the total mass of reactants equals the total mass of products in a closed system. Lavoisier verified this by carrying out reactions in sealed vessels and weighing before and after heating.
  • Law of definite (constant) proportions (Proust): a given compound always contains the same elements combined in a fixed ratio by mass, regardless of its source or method of preparation. Pure water, whether from a river or made in the lab, is always 11.11% hydrogen and 88.89% oxygen by mass.
  • Law of multiple proportions (Dalton): when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. Example: carbon and oxygen form CO (12 g C : 16 g O) and CO2 (12 g C : 32 g O); for a fixed 12 g of carbon, the oxygen masses 16 g and 32 g are in the ratio 1:2.
  • Gay Lussac's law of gaseous volumes: when gases react, the volumes of reactants and gaseous products, measured at the same temperature and pressure, are in a ratio of small whole numbers. 2 volumes of hydrogen react with 1 volume of oxygen to give 2 volumes of water vapour.
  • Avogadro's law: equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules. This law reconciled Gay Lussac's observation with Dalton's atomic theory by introducing the idea of a molecule as a group of atoms.
Exam tip Learn one worked numerical example for each law — 'state the law and illustrate with an example' is one of the most repeated 2–3 mark questions from this chapter.

06 1.6 Dalton's Atomic Theory

In 1808 John Dalton proposed an atomic theory to explain the laws of chemical combination. Its main postulates:

  • Matter consists of indivisible particles called atoms.
  • Atoms of a given element are identical in mass and properties; atoms of different elements differ in mass and properties.
  • Atoms cannot be created or destroyed in a chemical reaction (this explains the law of conservation of mass).
  • Compounds are formed when atoms of different elements combine in fixed, simple whole-number ratios (this explains the law of definite proportions and the law of multiple proportions).
  • The relative number and kind of atoms in a given compound are constant.

Dalton's theory explained the first three laws of chemical combination successfully but could not explain Gay Lussac's law of gaseous volumes — it had no concept of a molecule as a group of atoms that could itself be divided during a reaction. Avogadro's hypothesis (introducing the molecule) later filled this gap.

Exam tip A frequent short-answer question: 'Which law of chemical combination could Dalton's theory not explain, and why?' — answer is Gay Lussac's law of gaseous volumes, because the theory treated the atom as the smallest indivisible unit and did not allow for polyatomic molecules.

07 1.7 Atomic and Molecular Masses

Atoms are far too light to weigh in grams conveniently, so atomic masses are expressed relative to a standard. The modern standard is the carbon-12 isotope: one atomic mass unit (amu, symbol u) is defined as exactly 1/12th the mass of one atom of carbon-12 in its ground state. One amu equals 1.66 × 10−24 g (1.66 × 10−27 kg).

The atomic mass of an element is its mass relative to this standard. Most elements exist as a mixture of isotopes, so the atomic mass quoted on the periodic table is the average atomic mass — the weighted average of the masses of all naturally occurring isotopes, weighted by their percentage abundance. For example, carbon occurs as 98.892% 12C (mass 12.000 u) and 1.108% 13C (mass 13.003 u); the weighted average works out to 12.011 u, the value on the periodic table.

Molecular mass is the sum of the atomic masses of all atoms in a molecule, e.g. the molecular mass of water (H2O) is 2(1.008) + 16.00 = 18.02 u. For ionic compounds, which do not exist as discrete molecules, the term formula mass is used instead — the sum of atomic masses of all atoms in one formula unit, e.g. NaCl has formula mass 23.0 + 35.5 = 58.5 u.

Average atomic mass = Σ(isotopic mass × fractional abundance)
e.g. for carbon0.98892(12.000) + 0.01108(13.003) = 12.011 u
Exam tip Don't confuse 'molecular mass' (covalent molecules like H2O, CO2) with 'formula mass' (ionic compounds like NaCl, CaCO3) — examiners sometimes ask you to name which term applies.

08 1.8 Mole Concept and Molar Masses

Because atoms and molecules are far too small to count or handle individually, chemists use a counting unit called the mole, the SI unit for amount of substance. One mole is defined as the amount of a substance that contains as many elementary entities (atoms, molecules, ions or other particles) as there are atoms in exactly 12 g of carbon-12. This number is the Avogadro constant, NA = 6.022 × 1023 mol−1.

The molar mass of a substance is the mass of one mole of it, expressed in g mol−1; numerically it equals the atomic mass, molecular mass or formula mass expressed in grams (e.g. molar mass of water = 18.02 g mol−1). This gives three interconvertible routes to the same mole value: mass, number of particles, and (for a gas) volume. For a gas, one mole occupies a fixed volume at a given temperature and pressure regardless of its identity — this is the molar volume. At STP (standard temperature and pressure, now defined as 273.15 K and 1 bar), one mole of any ideal gas occupies 22.7 L; the older convention (1 atm, 273.15 K, sometimes called NTP) gives 22.4 L, which many competitive-exam books still use — always check which convention a question uses.

n = m/M
nnumber of molesmgiven mass (g)Mmolar mass (g mol−1)
n = N/NA
Nnumber of particles (atoms/molecules/ions)NA6.022 × 1023 mol−1
n = V/22.7 (at STP, gas only)
Vvolume of gas in litres at 273.15 K and 1 bar
Exam tip Draw the 'mole triangle' — mass at one corner, number of particles at another, gas volume at the third, mole (n) in the centre — and practise converting between any two corners; this single skill answers most numericals in the chapter.

09 1.9 Percentage Composition

The percentage composition of a compound tells you what fraction of its total mass is contributed by each element. It is found by dividing the total mass of that element in one mole of the compound by the molar mass of the compound, and multiplying by 100.

Percentage composition data (usually from combustion or elemental analysis) is used to work backward to the empirical formula — the simplest whole-number ratio of atoms of each element in a compound. The standard method: assume 100 g of the compound, convert each element's percentage directly to grams, divide each mass by that element's atomic mass to get moles, then divide every mole value by the smallest one to get the simplest ratio (multiplying through by a small integer if the ratios aren't whole numbers). Once you have the empirical formula, the molecular formula is obtained using the compound's experimentally known molar mass: molecular formula = n × (empirical formula), where n = molar mass ÷ empirical formula mass. For ionic compounds, which exist as extended lattices rather than discrete molecules, the empirical formula is itself the standard way of writing the compound (NaCl, MgO).

% of element = (mass of element in 1 mole of compound / molar mass of compound) × 100
n = molar mass / empirical formula mass
molecular formula= n × empirical formula
Exam tip This is the single most common 3–5 mark numerical from the chapter. Practise the 4-step method (100 g → grams → moles → simplest ratio) until it's automatic; the marking scheme awards a mark for each correct step even if the final answer is wrong.

10 1.10 Stoichiometry and Stoichiometric Calculations

Stoichiometry is the calculation of the quantitative relationships between reactants and products in a balanced chemical equation. A balanced equation tells you the mole ratio in which substances react and are formed; from that ratio, and the moles of any one substance, you can calculate the moles (and hence mass or volume) of every other substance in the reaction.

In practice, reactants are rarely mixed in the exact stoichiometric ratio. The reactant that gets used up first, stopping the reaction and limiting the amount of product formed, is the limiting reagent; the other reactant, left over at the end, is said to be in excess. To find the limiting reagent, convert every given reactant mass to moles, divide by its coefficient in the balanced equation, and the reactant with the smallest resulting value is limiting — all product calculations must then be based on it, never on the reactant in excess.

Chemists also express the concentration of a solution in several standard ways: mass percentage (mass of solute per 100 g of solution), mole fraction (moles of one component divided by total moles of all components), molarity (moles of solute per litre of solution — the most commonly used unit in the lab, though it is temperature-dependent since volume expands with heat), molality (moles of solute per kilogram of solvent — temperature-independent, since mass doesn't change with temperature), and parts per million (ppm) for very dilute solutions.

Molarity, M = nsolute/Vsolution(L)
unitmol L−1
Molality, m = nsolute/masssolvent(kg)
unitmol kg−1
Mole fraction, xA = nA/(nA + nB + …)
propertysum of all mole fractions in a mixture = 1
Exam tip In a limiting-reagent numerical, always divide moles by the stoichiometric coefficient before comparing reactants — comparing raw mole values without dividing by coefficients is the single most common mistake.

VocabularyKey terms

MoleThe SI unit of amount of substance; the amount that contains 6.022 × 1023 elementary entities, the same number of atoms as in exactly 12 g of carbon-12.
Avogadro constant (N<sub>A</sub>)6.022 × 1023 mol−1; the number of elementary entities in one mole of any substance.
Molar massThe mass of one mole of a substance, in g mol−1; numerically equal to the atomic, molecular or formula mass.
Atomic mass unit (u)1/12th the mass of one atom of carbon-12; equals 1.66 × 10−24 g.
Empirical formulaThe simplest whole-number ratio of atoms of each element present in a compound.
Molecular formulaThe actual number of atoms of each element in one molecule of a compound; equals a whole-number multiple of the empirical formula.
Limiting reagentThe reactant that is completely consumed first in a reaction, thereby limiting the amount of product that can be formed.
MolarityMoles of solute dissolved per litre of solution; unit mol L−1, changes with temperature because volume changes.
MolalityMoles of solute per kilogram of solvent; unit mol kg−1, independent of temperature because mass does not change with heat.
Mole fractionThe ratio of moles of one component to the total moles of all components in a mixture; always dimensionless and always sums to 1.
Percentage compositionThe percentage by mass that each element contributes to the total mass of a compound.
Significant figuresAll the digits in a measured quantity that are known with certainty plus one estimated digit; a measure of the precision of a measurement.
Law of conservation of massIn a chemical reaction, matter is neither created nor destroyed; total mass of reactants equals total mass of products.
Law of definite proportionsA given compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source.
Law of multiple proportionsWhen two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a ratio of small whole numbers.
StoichiometryThe calculation of quantitative relationships (mole, mass, volume) between reactants and products from a balanced chemical equation.
Molar volumeThe volume occupied by one mole of any gas at a given temperature and pressure; 22.7 L at STP (273.15 K, 1 bar).
IsotopesAtoms of the same element (same number of protons) that have different mass numbers due to a different number of neutrons.

Keep this on one pageFormula sheet

NameExpressionNotes
Moles from massn = m/Mm = given mass in g, M = molar mass in g mol⁻¹
Moles from particle countn = N/NANA = 6.022 × 10²³ mol⁻¹
Moles of a gas at STPn = V/22.7V in litres, STP = 273.15 K and 1 bar (use 22.4 L only if the question specifies 1 atm/NTP)
MolarityM = nsolute/Vsolution(L)mol L⁻¹; temperature-dependent
Molalitym = nsolute/masssolvent(kg)mol kg⁻¹; temperature-independent
Mole fractionxA = nA/(nA + nB)dimensionless; all mole fractions in a mixture add up to 1
Mass percentage of an element% element = (mass of element in 1 mol compound/molar mass of compound) × 100used to build up empirical formula
Molecular formula from empirical formulaMolecular formula = n × empirical formula, n = molar mass/empirical formula massn is always a positive integer
Average atomic massAverage atomic mass = Σ(isotopic mass × fractional abundance)weighted average over all naturally occurring isotopes
Parts per millionppm = (mass of solute/mass of solution) × 106used for trace/dilute concentrations, e.g. pollutants in water

Practise with a pencilDiagrams that earn marks

The mole triangle (mass–number–volume)

Draw a triangle with 'Mass (g)' at the top-left vertex, 'Number of particles' at the top-right vertex, and 'Gas volume at STP (L)' at the bottom vertex; write 'Mole (n)' in the centre with double-headed arrows connecting the centre to each vertex, labelling each arrow with the conversion formula (n = m/M, n = N/NA, n = V/22.7). Note under the diagram that the volume vertex only applies to gases.

Flowchart for empirical and molecular formula

Draw five boxes connected by downward arrows: (1) '% composition given', (2) 'Assume 100 g → convert % to grams', (3) 'Divide by atomic mass → moles of each element', (4) 'Divide by smallest mole value → simplest ratio → empirical formula', (5) 'n = molar mass ÷ empirical formula mass → molecular formula = n × empirical formula'. Label the final box in a different colour to show it needs the experimentally known molar mass.

Lavoisier's sealed-tube experiment for conservation of mass

Draw a sealed glass tube/flask on a balance pan, containing a reactant (e.g. red mercury(II) oxide) before heating; draw a second identical setup after heating showing the same total mass but the substance decomposed inside the still-sealed vessel. Label both pans 'mass before = mass after' to show that mass is conserved as long as the system stays closed.

Limiting reagent bar comparison

Draw two vertical bars side by side, one for each reactant, with height representing moles available divided by the stoichiometric coefficient in the balanced equation. Draw a dashed horizontal line at the height of the shorter bar and label it 'limiting reagent — determines the product formed'; label the taller bar 'excess reagent — some left over unreacted'.

TextbookNCERT questions the board keeps asking

1.1 Calculate the molar mass of glucose, C6H12O6.2 m

Molar mass = 6 × (atomic mass of C) + 12 × (atomic mass of H) + 6 × (atomic mass of O)

= 6(12.0) + 12(1.0) + 6(16.0) = 72.0 + 12.0 + 96.0 = 180.0 g mol−1

1.2 Calculate the mass percent of different elements present in sodium sulphate, Na2SO4.3 m

Molar mass of Na2SO4 = 2(23.0) + 32.0 + 4(16.0) = 46.0 + 32.0 + 64.0 = 142.0 g mol−1

% Na = (46.0/142.0) × 100 = 32.39%

% S = (32.0/142.0) × 100 = 22.54%

% O = (64.0/142.0) × 100 = 45.07%

Check: 32.39 + 22.54 + 45.07 ≈ 100%.

1.3 An oxide of iron contains 69.9% iron and 30.1% oxygen by mass. Determine the empirical formula of the oxide (atomic mass Fe = 55.85, O = 16.0).3 m

Assume 100 g of compound: 69.9 g Fe and 30.1 g O.

Moles of Fe = 69.9/55.85 = 1.2516; Moles of O = 30.1/16.0 = 1.8813

Divide by smaller value (1.2516): Fe = 1.00, O = 1.503 ≈ 1.5

Multiply both by 2 to clear the fraction: Fe = 2, O = 3

Empirical formula = Fe2O3

1.4 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.5 m

C + O2 → CO2 (1 mol C reacts with 1 mol O2 to give 1 mol CO2, molar mass CO2 = 44 g mol−1)

(i) Air supplies more than enough O2, so all 1 mol C reacts: CO2 formed = 1 mol = 44 g.

(ii) 16 g O2 = 16/32 = 0.5 mol O2. This needs only 0.5 mol C, but 1 mol C is present, so O2 is limiting: CO2 formed = 0.5 mol = 22 g (0.5 mol C remains unreacted).

(iii) Again only 0.5 mol O2 is available, so O2 is still limiting even though 2 mol C is present: CO2 formed = 0.5 mol = 22 g (1.5 mol C remains unreacted).

1.5 50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g). Calculate the mass of NH3 formed. Identify the limiting reagent and calculate the mass of the reactant left unreacted.5 m

N2 + 3H2 → 2NH3 (molar masses: N2 = 28, H2 = 2, NH3 = 17 g mol−1)

Moles of N2 = 50000/28 = 1785.7 mol; Moles of H2 = 10000/2 = 5000 mol

N2 needed for all the H2 = 5000/3 = 1666.7 mol, which is less than the 1785.7 mol N2 available, so H2 is the limiting reagent.

Moles of NH3 formed = (2/3) × 5000 = 3333.3 mol → mass = 3333.3 × 17 = 56666.7 g ≈ 56.67 kg NH3

N2 left unreacted = 1785.7 − 1666.7 = 119.0 mol → mass = 119.0 × 28 = 3333 g ≈ 3.33 kg N2 remains unreacted.

1.6 How many moles and how many grams of NaOH are needed to prepare 250 mL of 0.4 M NaOH solution?3 m

Moles required, n = M × V(L) = 0.4 × 0.250 = 0.1 mol

Molar mass NaOH = 23 + 16 + 1 = 40 g mol−1

Mass = n × M = 0.1 × 40 = 4 g NaOH (0.1 mol)

1.7 Calculate the mole fraction of ethanol and water in a sample that contains 46 g of ethanol (C2H5OH) and 54 g of water.2 m

Molar mass ethanol = 46 g mol−1; molar mass water = 18 g mol−1

Moles ethanol = 46/46 = 1 mol; Moles water = 54/18 = 3 mol

Total moles = 1 + 3 = 4

xethanol = 1/4 = 0.25; xwater = 3/4 = 0.75 (they add up to 1, as expected).

1.8 Density of methanol is 0.793 kg L−1. What volume of methanol is needed to prepare 2.5 L of its 0.25 M solution? (molar mass CH3OH = 32 g mol−1)3 m

Moles needed = M × V = 0.25 × 2.5 = 0.625 mol

Mass needed = 0.625 × 32 = 20 g

Volume = mass/density = 20 g / 0.793 g mL−1 = 25.2 mL (approx.)

1.9 Calculate the number of moles and the number of atoms present in 52 g of helium (atomic mass He = 4 u).2 m

Moles of He = 52/4 = 13 mol

Number of atoms = 13 × 6.022 × 1023 = 7.83 × 1024 atoms

1.10 Calculate the empirical formula and molecular formula of a compound that contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass; its molar mass is 98.96 g mol−1 (atomic masses: H = 1, C = 12, Cl = 35.5).5 m

Assume 100 g of compound: 4.07 g H, 24.27 g C, 71.65 g Cl

Moles: H = 4.07/1 = 4.07; C = 24.27/12 = 2.0225; Cl = 71.65/35.5 = 2.018

Divide by smallest (2.018): H = 2.02 ≈ 2, C = 1.002 ≈ 1, Cl = 1

Empirical formula = CH2Cl, empirical formula mass = 12 + 2 + 35.5 = 49.5

n = molar mass/empirical formula mass = 98.96/49.5 ≈ 2

Molecular formula = 2 × CH2Cl = C2H4Cl2 (ethylene dichloride)

E1 Round off 6.658 to three significant figures, and state the number of significant figures in 0.0450.2 m

6.658 rounded to 3 significant figures: the digit to drop is 8 (> 5), so the preceding digit rounds up: 6.66.

0.0450: the leading zeros are not significant; 4, 5, and the trailing zero after the decimal are significant → 3 significant figures.

E2 State the law of multiple proportions and illustrate it with an example involving carbon and oxygen.3 m

Law of multiple proportions: when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

Example: carbon forms two oxides — CO (12 g C combines with 16 g O) and CO2 (12 g C combines with 32 g O). For the same 12 g of carbon, the oxygen masses 16 g and 32 g are in the ratio 1:2, a ratio of small whole numbers, confirming the law.

Previous years and frequently askedBoard questions with model answers

MCQ The SI unit of amount of substance is:
  1. kilogram
  2. mole
  3. candela
  4. kelvin
Frequently asked1 m
Answer: (b)
The mole is the SI base unit for amount of substance, defined via the Avogadro constant.
MCQ The empirical formula of glucose, C6H12O6, is:
  1. CH2O
  2. C2H4O2
  3. C3H6O3
  4. CHO
Frequently asked1 m
Answer: (a)
Divide all subscripts (6,12,6) by their HCF, 6, to get the simplest ratio 1:2:1, i.e. CH2O.
MCQ The law which states that the masses of one element that combine with a fixed mass of another element, in different compounds, are in a ratio of small whole numbers is the:
  1. law of conservation of mass
  2. law of definite proportions
  3. law of multiple proportions
  4. Gay Lussac's law of gaseous volumes
Frequently asked1 m
Answer: (c)
This is the exact statement of the law of multiple proportions, proposed by Dalton.
MCQ At STP (273.15 K, 1 bar), one mole of any ideal gas occupies a volume of:
  1. 22.4 L
  2. 22.7 L
  3. 24.0 L
  4. 11.35 L
Frequently asked1 m
Answer: (b)
Under the current IUPAC definition of STP (1 bar, 273.15 K), the molar volume of an ideal gas is 22.7 L mol⁻¹ (22.4 L applies at the older standard of 1 atm).
MCQ How many significant figures does the number 0.006025 have?
  1. 4
  2. 7
  3. 2
  4. 3
Frequently asked1 m
Answer: (a)
Leading zeros are not significant; the digits 6, 0, 2, 5 are all significant (the zero is sandwiched between non-zero digits), giving 4 significant figures.
MCQ The approximate mass of one mole of electrons is:
  1. 9.1 × 10−28 g
  2. 5.48 × 10−4 g
  3. 6.022 × 10−23 g
  4. 1.008 g
Frequently asked1 m
Answer: (b)
Mass of one electron ≈ 9.1 × 10⁻²⁸ g; multiplying by NA (6.022 × 10²³) gives 5.48 × 10⁻⁴ g for one mole of electrons.
MCQ Which of the following 1 g samples contains the largest number of atoms? (atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5)
  1. 1 g of gold
  2. 1 g of sodium
  3. 1 g of lithium
  4. 1 g of chlorine gas (Cl2)
Frequently asked1 m
Answer: (c)
Moles = mass/atomic mass; lithium has by far the smallest atomic mass (7), so 1 g of lithium gives the most moles and therefore the most atoms.
Assertion–Reason Assertion (A): The empirical formula and molecular formula of a compound can be the same. Reason (R): Molecular formula = n × empirical formula, where n is a positive integer that can equal 1.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
When n = 1, the molecular formula equals the empirical formula (e.g. H2O); R correctly explains why A is possible.
Assertion–Reason Assertion (A): The number of significant figures in 100 is ambiguous. Reason (R): Trailing zeros in a number written without a decimal point are not counted as significant unless the number is expressed in scientific notation.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
100 could represent 1, 2 or 3 significant figures depending on precision; writing it as 1 × 10², 1.0 × 10² or 1.00 × 10² removes the ambiguity, which is exactly what R states.
Assertion–Reason Assertion (A): The atomic mass unit is defined using hydrogen as the reference atom. Reason (R): One atomic mass unit (u) is exactly 1/12th the mass of one atom of carbon-12.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (d)
A is false — the modern reference is carbon-12, not hydrogen; R correctly states the actual definition of 1 u.
Very short answer Define 'limiting reagent'. Why is it important in stoichiometric calculations?Frequently asked2 m

The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thereby limiting the amount of product that can be formed; any other reactant remains partly unreacted (in excess). It is important because the amount of product must always be calculated from the limiting reagent, not from a reactant present in excess — using the wrong reactant overestimates the product.

Very short answer 1 g of hydrogen gas (H2) and 1 g of oxygen gas (O2) are taken separately. Which sample contains the greater number of atoms? Show your working.Frequently asked2 m

Moles of H2 = 1/2 = 0.5 mol → atoms of H = 2 × 0.5 = 1.0 mol = 6.022 × 1023 atoms.

Moles of O2 = 1/32 = 0.03125 mol → atoms of O = 2 × 0.03125 = 0.0625 mol = 3.76 × 1022 atoms.

1 g of hydrogen contains far more atoms than 1 g of oxygen, because hydrogen has a much smaller atomic mass.

Very short answer State the law of multiple proportions with one example.Frequently asked2 m

When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers. Example: nitrogen forms N2O (28 g N to 16 g O) and NO2 (28 g N to 64 g O); for the same 28 g N, the oxygen masses 16 g and 64 g are in the ratio 1:4.

Very short answer Differentiate between molarity and molality of a solution.Frequently asked2 m

Molarity (M) is the number of moles of solute dissolved per litre of solution; its unit is mol L−1, and it changes with temperature because the volume of a solution expands or contracts with temperature. Molality (m) is the number of moles of solute per kilogram of solvent; its unit is mol kg−1, and it does not change with temperature because mass is unaffected by temperature.

Short answer State any three postulates of Dalton's atomic theory.Frequently asked3 m

Any three of: (i) Matter is made of indivisible particles called atoms. (ii) All atoms of a given element are identical in mass and properties, while atoms of different elements differ. (iii) Atoms can neither be created nor destroyed in a chemical reaction. (iv) Compounds are formed when atoms of different elements combine in fixed, simple whole-number ratios. (v) The relative number and kind of atoms in a given compound is constant.

Short answer A vessel contains 1.6 g of dioxygen gas at STP (273.15 K, 1 bar). Calculate (i) the number of moles of O2, (ii) the number of molecules of O2, (iii) the volume occupied by the gas.Frequently asked3 m

Molar mass O2 = 32 g mol−1

(i) Moles = 1.6/32 = 0.05 mol

(ii) Molecules = 0.05 × 6.022 × 1023 = 3.011 × 1022 molecules

(iii) Volume = 0.05 × 22.7 = 1.135 L

Short answer Calculate the molality of a solution containing 18 g of glucose (C6H12O6) dissolved in 500 g of water.Frequently asked3 m

Molar mass of glucose = 180 g mol−1

Moles of glucose = 18/180 = 0.1 mol

Mass of solvent = 500 g = 0.5 kg

Molality = 0.1/0.5 = 0.2 m

Short answer State the law of conservation of mass. How did Lavoisier verify it experimentally?Frequently asked3 m

The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction — the total mass of the reactants equals the total mass of the products, provided the system is closed. Lavoisier verified this by carrying out reactions (such as heating red mercury(II) oxide) inside sealed vessels and carefully weighing the vessel and its contents before and after the reaction; the mass stayed the same as long as no matter entered or left the sealed system, confirming the law.

Long answer 50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce ammonia. Calculate the mass of NH3 formed, identify the limiting reagent, and calculate the mass of the excess reagent left unreacted.Frequently asked5 m

N2 + 3H2 → 2NH3

Moles N2 = 50000/28 = 1785.7 mol; Moles H2 = 10000/2 = 5000 mol

N2 required for all 5000 mol H2 = 5000/3 = 1666.7 mol, which is less than the 1785.7 mol N2 available, so H2 is the limiting reagent.

NH3 formed = (2/3) × 5000 = 3333.3 mol × 17 g mol−1 = 56.67 kg NH3

N2 used = 1666.7 mol; N2 left = 1785.7 − 1666.7 = 119.0 mol × 28 = 3.33 kg N2 unreacted.

Long answer A compound of nitrogen and oxygen contains 30.4% nitrogen and 69.6% oxygen by mass. Its molar mass is 92 g mol−1. Determine its empirical formula and molecular formula (atomic masses N = 14, O = 16).Frequently asked5 m

Assume 100 g compound: 30.4 g N, 69.6 g O

Moles N = 30.4/14 = 2.171; Moles O = 69.6/16 = 4.35

Divide by smaller (2.171): N = 1.00, O = 2.004 ≈ 2

Empirical formula = NO2, empirical formula mass = 14 + 32 = 46

n = 92/46 = 2

Molecular formula = 2 × NO2 = N2O4 (dinitrogen tetroxide)

Long answer Drinking water is considered unsafe if the chloroform (CHCl3) content exceeds 15 ppm by mass. Calculate the molality of chloroform in water at this limit (atomic masses: C = 12, H = 1, Cl = 35.5; assume the mass of solvent ≈ mass of solution for a dilute solution).Frequently asked5 m

15 ppm by mass means 15 mg of CHCl3 per kg of solution ≈ per kg of water (since the solution is extremely dilute).

Molar mass CHCl3 = 12 + 1 + 3(35.5) = 119.5 g mol−1

Moles CHCl3 = 0.015 g/119.5 g mol−1 = 1.255 × 10−4 mol

Mass of solvent (water) ≈ 1 kg

Molality = 1.255 × 10−4 mol/1 kg = 1.255 × 10−4 m

Case-based Read the passage and answer the questions that follow.

A student is given a sample of pure calcium carbonate (CaCO3, molar mass 100 g mol−1) and asked to study it using the mole concept. She weighs out 5 g of the sample.

(i) How many moles of CaCO3 are present in the 5 g sample?
(ii) How many formula units of CaCO3 are present?
(iii) What is the mass percentage of calcium in CaCO3 (atomic mass Ca = 40)?
(iv) If the sample is heated and fully decomposes as CaCO3 → CaO + CO2, what mass of CO2 (molar mass 44 g mol−1) is released?
Frequently asked4 m

(i) Moles = 5/100 = 0.05 mol

(ii) Formula units = 0.05 × 6.022 × 1023 = 3.011 × 1022

(iii) % Ca = (40/100) × 100 = 40%

(iv) Moles CaCO3 = moles CO2 released = 0.05 mol → mass = 0.05 × 44 = 2.2 g CO2

Case-based Read the passage and answer the questions that follow.

In the industrial synthesis of ammonia, nitrogen gas is reacted with hydrogen gas: N2 + 3H2 → 2NH3. In one trial run, 28 g of N2 is reacted with 9 g of H2 (molar masses: N2 = 28, H2 = 2, NH3 = 17 g mol−1).

(i) Calculate the moles of N2 and H2 taken.
(ii) Identify the limiting reagent.
(iii) Calculate the mass of NH3 formed.
(iv) Calculate the mass of the excess reagent left unreacted.
Frequently asked4 m

(i) Moles N2 = 28/28 = 1 mol; Moles H2 = 9/2 = 4.5 mol

(ii) N2 needs 3 × 1 = 3 mol H2 to react fully, and 4.5 mol H2 is available (more than enough), so N2 is the limiting reagent.

(iii) NH3 formed = 2 × 1 = 2 mol × 17 = 34 g NH3

(iv) H2 used = 3 mol; H2 left = 4.5 − 3 = 1.5 mol × 2 = 3 g H2 unreacted

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Confusing molarity with molality — molarity depends on the volume of solution and changes with temperature; molality depends on the mass of solvent and does not.
  • Comparing raw mole values of reactants to find the limiting reagent without first dividing each by its stoichiometric coefficient in the balanced equation.
  • Treating leading zeros as significant figures (e.g. counting 0.0025 as having 5 significant figures instead of 2).
  • Forgetting that molecular formula = n × empirical formula, and reporting the empirical formula as the final answer when the question asks for the molecular formula.
  • Using an unbalanced chemical equation for stoichiometric calculations — the mole ratio is only valid once the equation is balanced.
  • Rounding intermediate values too early in a multi-step numerical, which causes the final answer to drift outside the accepted range.
  • Mixing up molar volume conventions — using 22.4 L (1 atm, older NTP) when the question specifies STP at 1 bar (22.7 L), or vice versa.
  • Not converting all quantities to consistent units (kg vs g, mL vs L) before applying a mole-concept or molarity formula.

How to prepareStrategy for this chapter

This chapter is entirely about calculation fluency, not memorisation, so build your preparation around numericals rather than definitions. Start by getting completely comfortable with the mole triangle — converting between mass, number of particles and gas volume — because every later numerical (percentage composition, limiting reagent, molarity, molality) is just this same conversion applied inside a bigger problem.

Memorise only a short list: the five laws of chemical combination with one example each, Dalton's postulates, the definitions of molarity/molality/mole fraction, and the significant-figure rules. Everything else should come from practice, not recall.

For the exam, practise the empirical-and-molecular-formula method and the limiting-reagent method as fixed four-to-five step procedures, and write out every step even when you can do the arithmetic in your head — CBSE marking schemes give a mark per correct step, so an unlabelled final answer alone loses marks. Keep a formula card with molar masses of common compounds (water, glucose, NaOH, NaCl, CaCO3, NH3) so you don't lose time recomputing them mid-exam. Finally, double-check every numerical answer for reasonable magnitude and correct significant figures before moving on.

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