Some Basic Concepts of Chemistry Class 11: Mole Concept, Stoichiometry and Solved Numericals

What Some Basic Concepts of Chemistry is about

Weightage: This chapter is Unit I, Some Basic Concepts of Chemistry, which the syllabus gives 7 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit I Some Basic Concepts of Chemistry

Key concepts

Nature of matter and its classification

NCERT section 1.1 Importance of Chemistry; 1.2 Nature of Matter; 1.3.1 Physical and Chemical Properties

Chemistry is the study of the composition, structure and properties of matter and the changes it undergoes. It matters in daily life: fertilisers, medicines, plastics, fuels and clean water all depend on chemical knowledge.

Matter is anything that has mass and occupies space. It exists in three physical states:

• Solid: particles are packed tightly in fixed positions, so a solid has a definite shape and volume.
• Liquid: particles are close but can move past each other, so a liquid has a definite volume but takes the shape of its container.
• Gas: particles are far apart and move freely, so a gas has neither definite shape nor definite volume.

Heating or cooling (and changing pressure) changes one state into another.

At the macroscopic level, matter is classified as:

• Mixture: two or more substances in any ratio. A homogeneous mixture (sugar solution, air) has the same composition throughout; a heterogeneous mixture (sand in water, a salad) does not. Mixtures can be separated by physical methods such as filtration, crystallisation and distillation.
• Pure substance: fixed composition. An element has only one kind of atom (Na, Cu, O₂, H₂); a compound has two or more kinds of atoms joined in a fixed ratio (H₂O, CO₂, NaCl). A compound can be broken into its elements only by chemical methods, and its properties differ from those of its elements: water puts out fire although hydrogen burns and oxygen supports burning.

Physical properties (colour, odour, melting point, boiling point, density) can be measured without changing the identity of the substance. Chemical properties (acidity, combustibility, reaction with water) are seen only when a chemical change occurs.

Worked example

Classify each as an element, a compound, a homogeneous mixture or a heterogeneous mixture: (a) brass (b) dry ice (c) muddy river water (d) graphite (e) a 10% glucose solution. Give one reason for (b). (2 marks)

Try it first, then show the solutionSolution

1. (a) Brass is an alloy of copper and zinc with a uniform composition throughout: homogeneous mixture. (½)
2. (b) Dry ice is solid carbon dioxide, CO₂: a compound, because carbon and oxygen are combined in a fixed 1 : 2 ratio of atoms and can be separated only chemically. (½)
3. (c) Muddy water has visible particles that settle: heterogeneous mixture. (½)
4. (d) Graphite contains only carbon atoms: element. (e) Glucose solution is uniform throughout: homogeneous mixture. (½)

Answer: (a) homogeneous mixture (b) compound, fixed C : O ratio (c) heterogeneous mixture (d) element (e) homogeneous mixture

SI units, mass, volume, density and temperature

NCERT section 1.3.2 to 1.3.7 Measurement, SI Units, Mass and Weight, Volume, Density, Temperature

The International System of Units (SI) has seven base quantities: length (metre, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), thermodynamic temperature (kelvin, K), amount of substance (mole, mol) and luminous intensity (candela, cd). Other units are derived from these; for example, the SI unit of volume is m³ and that of density is kg m⁻³.

Prefixes scale units by powers of ten: milli (10⁻³), micro (10⁻⁶), nano (10⁻⁹), pico (10⁻¹²), femto (10⁻¹⁵); kilo (10³), mega (10⁶), giga (10⁹).

• Mass is the amount of matter in a body and stays the same everywhere. Weight is the force of gravity on it and changes with place. Mass is measured with an analytical balance.
• Volume: in the lab we use litres. 1 L = 1000 mL = 1000 cm³ = 1 dm³, and 1 m³ = 1000 L.
• Density = mass ÷ volume. It tells how closely particles are packed. In the lab it is usually given in g cm⁻³.
• Temperature is measured on three scales: Celsius (°C), Fahrenheit (°F) and Kelvin (K). The conversions are ${}^{\circ }F=\frac{9}{5}{(}^{\circ }C)+32$ and $K{=}^{\circ }C+273.15$. The kelvin scale has no negative values because 0 K is the lowest possible temperature.

The kilogram is now defined through the Planck constant, and the mole through the Avogadro constant; you need only the idea that SI units are fixed by constants of nature, not by objects kept in a vault.

Worked example

A rectangular metal block measures 2.0 cm × 3.0 cm × 5.0 cm and has a mass of 81 g. (a) Find its density in g cm⁻³ and in kg m⁻³. (b) A day in Jaisalmer reaches 45 °C. Express this in °F and in K. (3 marks)

Try it first, then show the solutionSolution

1. (a) Volume = 2.0 × 3.0 × 5.0 = 30 cm³. Density = 81 ÷ 30 = 2.7 g cm⁻³. (1)
2. 1 g cm⁻³ = $\frac{10^{-3}kg}{10^{-6}{m}^3}$ = 1000 kg m⁻³, so density = 2.7 × 1000 = 2700 kg m⁻³. (1)
3. (b) °F = $\frac{9}{5}\times 45+32=81+32=113$ °F; K = 45 + 273.15 = 318.15 K. (1)

Answer: (a) 2.7 g cm⁻³ = 2.7 × 10³ kg m⁻³ (b) 113 °F and 318.15 K

Uncertainty in measurement: scientific notation, significant figures and dimensional analysis

NCERT section 1.4 Uncertainty in Measurement (1.4.1 to 1.4.3)

Scientific notation writes any number as $N\times 10^n$ with $1\le N<10$: 0.000 064 = 6.4 × 10⁻⁵ and 232.508 = 2.32508 × 10². To multiply, multiply the N parts and add the powers; to divide, subtract the powers; to add or subtract, first make the powers equal.

Precision is how close repeated measurements are to each other; accuracy is how close they are to the true value. A student can be precise but inaccurate if the instrument is wrongly set.

Significant figures are the certain digits plus one uncertain digit. Rules:
1. All non-zero digits are significant (285 has three).
2. Zeros to the left of the first non-zero digit are not significant (0.0032 has two).
3. Zeros between non-zero digits are significant (2.005 has four).
4. Zeros at the end are significant if there is a decimal point (0.200 has three; 2.0 has two). In 100 without a decimal they are not; write 1.00 × 10² if you mean three.
5. Exact numbers from counting or definitions (2 balls, 1 kg = 1000 g) have infinite significant figures.

In calculations:

• Addition and subtraction: keep as many decimal places as the number with the fewest.
• Multiplication and division: keep as many significant figures as the least precise number.
• Rounding: if the dropped digit is more than 5, raise the previous digit; less than 5, leave it; exactly 5, round to make the previous digit even (so 4.35 → 4.4 but 4.45 → 4.4 by this rule).

Dimensional analysis (unit factor method): multiply by a fraction equal to 1, such as $\frac{1000g}{1kg}$, so the unwanted unit cancels.

Worked example

(a) How many significant figures are in 0.04060 and in 3.00 × 10⁴? (b) Report the answers to the correct significant figures: (i) 12.11 + 18.0 + 1.012 (ii) 4.52 × 2.0. (c) Convert 3 days into seconds by the unit factor method. (3 marks)

Try it first, then show the solutionSolution

1. (a) 0.04060: leading zeros do not count; 4, 0, 6 and the final 0 (after a decimal point) count, so four. 3.00 × 10⁴: three. (1)
2. (b)(i) Sum = 31.122. The fewest decimal places is one (18.0), so report 31.1. (ii) Product = 9.04; 2.0 has two significant figures, so report 9.0. (1)
3. (c) 3 d × $\frac{24h}{1d}\times \frac{60min}{1h}\times \frac{60s}{1min}$ = 259 200 s = 2.592 × 10⁵ s. (1)

Answer: (a) four and three (b) 31.1 and 9.0 (c) 2.592 × 10⁵ s

Laws of chemical combination and Dalton's atomic theory

NCERT section 1.5 Laws of Chemical Combinations; 1.6 Dalton's Atomic Theory

Five laws govern how elements combine:
1. Law of conservation of mass (Lavoisier): matter is neither created nor destroyed in a chemical or physical change. Total mass of reactants = total mass of products.
2. Law of definite proportions (Proust): a given compound always contains exactly the same proportion of elements by mass, whatever its source. Natural and lab-made copper carbonate have the same percentage of copper, carbon and oxygen.
3. Law of multiple proportions (Dalton): when the same two elements make two or more different compounds, and you keep the mass of one element fixed, the masses of the second element are in a small whole-number ratio. For example, 2 g of hydrogen combines with 16 g of oxygen in water and with 32 g in hydrogen peroxide: 16 : 32 = 1 : 2.
4. Gay Lussac's law of gaseous volumes: when gases react or are produced, their volumes (at the same temperature and pressure) are in a simple whole-number ratio. 2 volumes of hydrogen + 1 volume of oxygen give 2 volumes of water vapour.
5. Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Avogadro explained Gay Lussac's result by saying hydrogen and oxygen exist as diatomic molecules, so 2 molecules of H₂ and 1 molecule of O₂ give 2 molecules of water.

Dalton's atomic theory (1808) explained the first three laws:

• Matter consists of indivisible atoms.
• All atoms of a given element have identical properties, including mass; atoms of different elements differ in mass.
• Compounds form when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve only the reorganisation of atoms; atoms are neither created nor destroyed.

The theory could not explain Gay Lussac's law, and later discoveries (isotopes, sub-atomic particles) showed atoms are not indivisible and not all identical in mass.

Worked example

Sulphur forms two oxides. In oxide A, 32 g of sulphur is combined with 32 g of oxygen; in oxide B, 32 g of sulphur is combined with 48 g of oxygen. (a) Which law do these data illustrate? State it. (b) Suggest formulas for A and B. (S = 32, O = 16) (3 marks)

Try it first, then show the solutionSolution

1. Fix the mass of sulphur at 32 g. Masses of oxygen are 32 g and 48 g, in the ratio 32 : 48 = 2 : 3, a simple whole-number ratio. (1)
2. (a) Law of multiple proportions: when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a simple whole-number ratio. (1)
3. (b) 32 g S = 1 mol S. In A, 32 g O = 2 mol O, so A is SO₂; in B, 48 g O = 3 mol O, so B is SO₃. (1)

Answer: Law of multiple proportions (ratio 2 : 3); A = SO₂, B = SO₃.

Atomic mass, average atomic mass, molecular mass and formula mass

NCERT section 1.7 Atomic and Molecular Masses (1.7.1 to 1.7.4)

Atoms are far too light to weigh in grams, so we compare them with a standard. Since 1961 the standard is the carbon-12 atom, given a mass of exactly 12 atomic mass units. One atomic mass unit (amu, now written u, the unified mass) is one-twelfth of the mass of one carbon-12 atom: 1 u = 1.66056 × 10⁻²⁴ g. On this scale hydrogen-1 has a mass of 1.0078 u and oxygen-16 has 15.995 u.

Average atomic mass: most elements are mixtures of isotopes, so the atomic mass in the periodic table is a weighted average using natural abundances:
average atomic mass = Σ (fractional abundance × isotopic mass).
That is why chlorine is 35.5 u, not a whole number.

Molecular mass is the sum of the atomic masses of all atoms in a molecule. For CO₂: 12.011 + 2 × 16.00 = 44.011 u.

Formula mass is used for ionic compounds such as NaCl, which form a three-dimensional lattice of ions instead of separate molecules. Formula mass of NaCl = 23.0 + 35.5 = 58.5 u.

Worked example

Natural copper is 69.15% ⁶³Cu (isotopic mass 62.93 u) and 30.85% ⁶⁵Cu (isotopic mass 64.93 u). Calculate the average atomic mass of copper. (2 marks)

Try it first, then show the solutionSolution

1. Average atomic mass = (0.6915 × 62.93) + (0.3085 × 64.93). (1)
2. = 43.516 + 20.031 = 63.547 u ≈ 63.55 u. (1)
3. Check: the answer lies between 62.93 and 64.93 and closer to 62.93, the more abundant isotope, as it should.

Answer: About 63.55 u

Mole concept and molar mass

NCERT section 1.8 Mole Concept and Molar Masses

Just as a dozen means 12 and a gross means 144, chemists count particles in moles. One mole contains exactly 6.02214076 × 10²³ elementary entities (atoms, molecules, ions or electrons). This number is the Avogadro constant, $N_A$, and since 2019 it is fixed exactly by definition. Earlier the mole was defined as the number of atoms in exactly 12 g of carbon-12; the two definitions give the same number for practical work.

The molar mass of a substance is the mass of one mole of it in grams. It is numerically equal to the atomic, molecular or formula mass in u:

• H₂O: molecular mass 18.02 u, molar mass 18.02 g mol⁻¹.
• NaCl: formula mass 58.5 u, molar mass 58.5 g mol⁻¹.

The three links you will use again and again:

• $n=\frac{m}{M}$ (moles = mass ÷ molar mass)
• $N=n\times N_A$ (number of particles = moles × Avogadro constant)
• For an ideal gas at STP (273.15 K, 1 bar), one mole occupies about 22.7 L; many older books use 22.4 L at 1 atm. Use the value given in the question.

Always ask: moles of what? One mole of H₂O molecules contains two moles of H atoms and one mole of O atoms, so 3 × $N_A$ atoms in all.

Worked example

For 4.4 g of carbon dioxide, calculate (a) the number of moles, (b) the number of molecules and (c) the total number of atoms. (C = 12, O = 16, $N_A$ = 6.022 × 10²³ mol⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. (a) Molar mass of CO₂ = 12 + 2 × 16 = 44 g mol⁻¹, so $n=\frac{4.4}{44}=0.10$ mol. (1)
2. (b) Molecules = 0.10 × 6.022 × 10²³ = 6.022 × 10²² molecules. (1)
3. (c) Each CO₂ molecule has 3 atoms, so atoms = 3 × 6.022 × 10²² = 1.807 × 10²³ atoms. (1)

Answer: (a) 0.10 mol (b) 6.022 × 10²² molecules (c) about 1.81 × 10²³ atoms

Percentage composition, empirical formula and molecular formula

NCERT section 1.9 Percentage Composition (1.9.1 Empirical Formula for Molecular Formula)

Mass per cent of an element = $\frac{\text{mass of that element in 1 mol of compound}}{\text{molar mass of compound}}\times 100$. Chemists use it to check the purity of a sample: a pure sample matches the calculated percentages.

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of each kind of atom in one molecule. For glucose, the empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆.

Method for the empirical formula from percentages:
1. Take 100 g of the compound, so each percentage becomes grams.
2. Divide each mass by the atomic mass to get moles.
3. Divide every mole value by the smallest one.
4. If you get values like 1.5 or 1.33, multiply all by 2 or 3 to make whole numbers.
5. Write the empirical formula.

Then, n = molar mass ÷ empirical formula mass, and molecular formula = (empirical formula)ₙ.

Worked example

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56 g mol⁻¹. Find its empirical and molecular formulas. (C = 12, H = 1) (3 marks)

Try it first, then show the solutionSolution

1. In 100 g: C = 85.7 g, H = 14.3 g. Moles: C = 85.7 ÷ 12 = 7.14; H = 14.3 ÷ 1 = 14.3. (1)
2. Divide by the smaller value: C = 7.14 ÷ 7.14 = 1; H = 14.3 ÷ 7.14 = 2.0. Empirical formula CH₂. (1)
3. Empirical formula mass = 12 + 2 = 14. n = 56 ÷ 14 = 4, so the molecular formula is C₄H₈. (1)

Answer: Empirical formula CH₂; molecular formula C₄H₈

Stoichiometry and the limiting reagent

NCERT section 1.10 Stoichiometry and Stoichiometric Calculations; 1.10.1 Limiting Reagent

Stoichiometry is the calculation of the masses (and sometimes volumes) of reactants and products from a balanced equation. The coefficients give the mole ratio. For the combustion of methane,
$C{H}_4+2O_2\to C{O}_2+2H_{2}O$
one mole of methane reacts with two moles of oxygen to give one mole of carbon dioxide and two moles of water; equivalently, 16 g of CH₄ reacts with 64 g of O₂ to give 44 g of CO₂ and 36 g of H₂O, and the 80 g on each side shows mass is conserved.

Routine for any stoichiometry problem:
1. Write and balance the equation.
2. Convert the given quantity to moles.
3. Use the mole ratio from the coefficients.
4. Convert moles of the required substance back to grams, particles or volume.

Often the reactants are not mixed in the exact ratio of the equation. The reactant that is used up first is the limiting reagent; it decides how much product forms. The other is in excess and some of it is left over. To find the limiting reagent, divide the moles of each reactant by its coefficient: the smaller result belongs to the limiting reagent.

Worked example

3.0 g of hydrogen is ignited with 29.0 g of oxygen. $2H_2+O_2\to 2H_{2}O$. (a) Which is the limiting reagent? (b) What mass of water forms? (c) What mass of the excess reagent is left? (H = 1, O = 16) (3 marks)

Try it first, then show the solutionSolution

1. Moles: H₂ = 3.0 ÷ 2 = 1.5 mol; O₂ = 29.0 ÷ 32 = 0.906 mol.
2. (a) Divide by coefficients: H₂ 1.5 ÷ 2 = 0.75; O₂ 0.906 ÷ 1 = 0.906. Hydrogen gives the smaller value, so H₂ is limiting. (1)
3. (b) 1.5 mol H₂ gives 1.5 mol H₂O = 1.5 × 18 = 27 g. (1)
4. (c) O₂ used = 1.5 ÷ 2 = 0.75 mol = 24 g, so O₂ left = 29.0 − 24.0 = 5.0 g. Check: 3.0 + 29.0 = 27 + 5.0 = 32 g. (1)

Answer: (a) Hydrogen (b) 27 g of water (c) 5.0 g of oxygen left

Reactions in solutions: mass per cent, mole fraction, molarity and molality

NCERT section 1.10.2 Reactions in Solutions

When a reaction is done in solution we need the amount of substance in a given volume or mass. Four ways of expressing concentration:
1. Mass per cent = $\frac{\text{mass of solute}}{\text{mass of solution}}\times 100$.
2. Mole fraction of A, $x_A=\frac{n_A}{n_A+n_B}$. The mole fractions of all components add up to 1.
3. Molarity (M) = moles of solute per litre of solution. A 1 M NaOH solution has 40 g of NaOH in 1 L of solution. For dilution, $M_1{V}_1=M_2{V}_2$.
4. Molality (m) = moles of solute per kilogram of solvent.

Molarity depends on temperature because volume expands on heating. Mass per cent, mole fraction and molality depend only on masses, so they do not change with temperature.

To convert molarity into molality you need the density of the solution: mass of solution = density × volume, then subtract the mass of solute to get the mass of solvent.

Worked example

5.85 g of sodium chloride is dissolved in water and the solution is made up to 500 mL. (a) Find its molarity. (b) What volume of this solution must be diluted to make 250 mL of 0.050 M NaCl? (Na = 23, Cl = 35.5) (2 marks)

Try it first, then show the solutionSolution

1. (a) Moles = 5.85 ÷ 58.5 = 0.100 mol; M = 0.100 ÷ 0.500 L = 0.200 mol L⁻¹. (1)
2. (b) $M_1{V}_1=M_2{V}_2$: 0.200 × V₁ = 0.050 × 250, so V₁ = 62.5 mL. (1)

Answer: (a) 0.200 M (b) 62.5 mL, made up to 250 mL with water

Chemistry in real life: Some Basic Concepts of Chemistry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Mole Concept, Molar Mass and Formula Calculations
• Stoichiometry, Limiting Reagent and Solution Concentration

Some Basic Concepts of Chemistry formula sheet

Temperature scales

${}^{\circ }F=\frac{9}{5}{(}^{\circ }C)+32,K{=}^{\circ }C+273.15$

A temperature difference of 1 °C equals a difference of 1 K.

Density

$d=\frac{m}{V}$

1 g cm⁻³ = 1000 kg m⁻³.

Average atomic mass

$\bar{A}=\sum (\text{fractional abundance}\times \text{isotopic mass})$

Use fractions (69.15% = 0.6915), not percentages.

Moles from mass and particles

$n=\frac{m}{M}=\frac{N}{N_A}$

$N_A$ = 6.022 × 10²³ mol⁻¹.

Mass of one atom or molecule

$m_{particle}=\frac{M}{N_A}$

Mass of one ¹²C atom = 12 ÷ (6.022 × 10²³) = 1.993 × 10⁻²³ g.

Mass per cent of an element

$\%\text{ of element}=\frac{\text{mass of element in 1 mol}}{\text{molar mass}}\times 100$

All the percentages must add up to 100.

Molecular formula

$n=\frac{\text{molar mass}}{\text{empirical formula mass}},\text{molecular formula}=(\text{empirical formula}{)}_n$

n is always a whole number.

Mole fraction

$x_A=\frac{n_A}{n_A+n_B},x_A+x_B=1$

No unit.

Molarity and dilution

$M=\frac{n_{solute}}{V_{solution}(L)},M_1{V}_1=M_2{V}_2$

Changes with temperature.

Molality

$m=\frac{n_{solute}}{\text{mass of solvent (kg)}}$

Does not change with temperature.

Some Basic Concepts of Chemistry: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Some Basic Concepts of Chemistry

The question patterns to expect, and the method that works for each.

Tricks to remember Some Basic Concepts of Chemistry

Short memory aids. Say each one aloud once, then use it on a question the same day.

• King Henry Died By Drinking Chocolate Milk

  Kilo, Hecto, Deca, Base unit, Deci, Centi, Milli: each step to the right multiplies by 10.

  Use it when: Unit conversions and prefix MCQs.

• Zeros in the middle are 'sandwiched' and count

  Leading zeros are placeholders (don't count), sandwiched zeros always count, trailing zeros count only if a decimal point is present.

  Use it when: Significant-figure MCQs.

• Add by Decimals, Multiply by Sig figs

  A-D, M-S: addition and subtraction follow decimal places; multiplication and division follow significant figures.

  Use it when: Reporting answers correctly.

• The mole triangle

  Draw a triangle with m on top and n and M at the bottom: cover the one you want. Do the same with N on top and n, N_A at the bottom.

  Use it when: Any mass–mole–particle conversion.

• Percent to mass, mass to mole, divide by small, multiply till whole

  Chant the four steps for empirical formulas.

  Use it when: Empirical and molecular formula questions.

Some Basic Concepts of Chemistry: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Some Basic Concepts of Chemistry

The slips examiners see most often, and what to write instead.

• ✗Mistake: Finding the limiting reagent by comparing masses

  Students think the smaller mass must run out first.

  Wrong working

  3.0 g H₂ and 29 g O₂: hydrogen has the smaller mass, so it is limiting (right answer for the wrong reason, and it fails in other problems).

  Right working

  Moles ÷ coefficient: H₂ 1.5 ÷ 2 = 0.75; O₂ 0.906 ÷ 1 = 0.906; hydrogen is limiting.

  ✓Do this: Always convert to moles and divide by the coefficients.

  Usually costs: Usually the whole 2-3 marks, since every later step uses it.

• ✗Mistake: Using litres of solvent for molarity

  Students divide by the volume of water added instead of the final volume of solution.

  Wrong working

  0.1 mol NaCl added to 500 mL water: M = 0.1 ÷ 0.5 = 0.2 M exactly.

  Right working

  Molarity needs the volume of the solution after mixing; 'made up to 500 mL' gives 0.2 M, 'dissolved in 500 mL water' does not.

  ✓Do this: Underline 'made up to' or 'dissolved in' in the question.

  Usually costs: 1 mark.

• ✗Mistake: Counting trailing zeros wrongly

  Students count every zero.

  Wrong working

  2500 has four significant figures.

  Right working

  2500 has two (no decimal point); 2500. or 2.500 × 10³ has four.

  ✓Do this: Look for a decimal point before counting trailing zeros.

  Usually costs: 1 mark per item.

• ✗Mistake: Stopping at 1.5 in the empirical formula

  Students round 1.5 to 2 instead of multiplying.

  Wrong working

  Ratio Fe : O = 1 : 1.5, so FeO₂.

  Right working

  Multiply by 2: Fe : O = 2 : 3, so Fe₂O₃.

  ✓Do this: Only round values within about 0.1 of a whole number; otherwise multiply by 2, 3 or 4.

  Usually costs: 1-2 marks.

• ✗Mistake: Forgetting atoms per molecule

  Students give the number of molecules when atoms are asked.

  ✓Do this: Read the question for 'atoms', 'molecules' or 'ions' and multiply by the count per formula unit.

  Usually costs: 1 mark.

• ✗Mistake: Writing the formula without the balanced equation

  Students jump to a ratio they remember.

  ✓Do this: Write the balanced equation first; it carries a step mark and stops ratio errors.

  Usually costs: ½-1 mark.

NCERT exercise guide: Some Basic Concepts of Chemistry

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Some Basic Concepts of Chemistry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The number of significant figures in 6.0230 × 10²³ is:
   (a) 3 (b) 4 (c) 5 (d) 23

   Hint

   The power of ten does not count.

   Answer

   (c) 5: the trailing zero after the decimal point counts.

2. Q2 · 1 mark · MCQExam level

   Which of the following is the SI unit of amount of substance?
   (a) gram (b) mole (c) kilogram (d) litre

   Hint

   One of the seven base units.

   Answer

   (b) mole

3. Q3 · 2 marks · Very short answerExam level

   State the law of multiple proportions with one example. (2 marks)

   Hint

   Two compounds of the same two elements.

   Answer

   When two elements combine to form two or more compounds, then for a fixed mass of one element, the masses of the other element are in a simple whole-number ratio (1). Example: 12 g of carbon combines with 16 g of oxygen in CO and 32 g in CO₂; 16 : 32 = 1 : 2 (1).

4. Q4 · 2 marks · Very short answerExam level

   Calculate the mass of one molecule of water in grams. (H = 1, O = 16, $N_A$ = 6.022 × 10²³ mol⁻¹) (2 marks)

   Hint

   Molar mass ÷ Avogadro constant.

   Answer

   18 ÷ (6.022 × 10²³) (1) = 2.99 × 10⁻²³ g (1).

5. Q5 · 3 marks · Short answerHarder exam level

   Natural boron has two isotopes, ¹⁰B (10.01 u) and ¹¹B (11.01 u). Its average atomic mass is 10.81 u. Find the percentage abundance of each isotope. (3 marks)

   Hint

   Let the fraction of ¹⁰B be x.

   Answer

   10.01x + 11.01(1 − x) = 10.81 (1); 11.01 − 1.00x = 10.81, so x = 0.20 (1). ¹⁰B 20%, ¹¹B 80% (1).

6. Q6 · 3 marks · Short answerHarder exam level

   An oxide of iron contains 72.4% iron by mass. Find its empirical formula. (Fe = 56, O = 16) (3 marks)

   Hint

   If you get 1 : 1.33, multiply by 3.

   Answer

   In 100 g: Fe = 72.4 ÷ 56 = 1.293 mol; O = 27.6 ÷ 16 = 1.725 mol (1). Ratio = 1 : 1.334 (1). × 3 gives 3 : 4, so Fe₃O₄ (1).

7. Q7 · 2 marks · Very short answerHarder exam level

   What volume of 0.25 M H₂SO₄ is needed to neutralise exactly 50 mL of 0.40 M NaOH? $H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$ (2 marks)

   Hint

   One mole of acid uses two moles of base.

   Answer

   NaOH = 0.40 × 0.050 = 0.020 mol, so H₂SO₄ needed = 0.010 mol (1). Volume = 0.010 ÷ 0.25 = 0.040 L = 40 mL (1).

8. Q8 · 3 marks · Short answerHarder exam level

   Suppose a lab burner burns 11 g of propane, C₃H₈: $C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$. (a) What mass of oxygen is used? (b) What volume of CO₂ is produced at STP (22.7 L mol⁻¹)? (C = 12, H = 1, O = 16) (3 marks)

   Hint

   Propane is 44 g mol⁻¹.

   Answer

   Moles of propane = 11 ÷ 44 = 0.25 (1). (a) O₂ = 5 × 0.25 = 1.25 mol = 40 g (1). (b) CO₂ = 3 × 0.25 = 0.75 mol × 22.7 = 17.0 L (1).

9. Q9 · 3 marks · Short answerHarder exam level

   A town water supply contains 1.5 ppm (by mass) of fluoride ions, F⁻. A student drinks 2.0 L of this water in a day (density 1.00 g mL⁻¹). (a) What mass of fluoride does she take in? (b) How many fluoride ions is that? (F = 19, $N_A$ = 6.022 × 10²³ mol⁻¹) (3 marks)

   Hint

   1.5 ppm means 1.5 g of F⁻ in 10⁶ g of water.

   Answer

   (a) Mass of water = 2000 g; fluoride = 2000 × 1.5 ÷ 10⁶ = 3.0 × 10⁻³ g (3.0 mg) (1). (b) Moles of F⁻ = 3.0 × 10⁻³ ÷ 19 = 1.58 × 10⁻⁴ mol (1); ions = 1.58 × 10⁻⁴ × 6.022 × 10²³ = 9.5 × 10¹⁹ (1).

10. Q10 · 2 marks · Very short answerHarder exam level

    Arrange in increasing order of number of molecules: 0.5 mol of N₂, 16 g of O₂, 3.01 × 10²³ molecules of CO₂, 22 g of CO₂. Explain. (2 marks)

    Hint

    Convert all to moles.

    Answer

    N₂ 0.5 mol; O₂ 16 ÷ 32 = 0.5 mol; 3.01 × 10²³ CO₂ = 0.5 mol; 22 g CO₂ = 0.5 mol (1). All four contain the same number of molecules, 3.01 × 10²³ (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    Suppose a student burns 2.4 g of magnesium ribbon in a closed flask containing 1.12 L of oxygen (take 22.4 L mol⁻¹ for the molar volume). $2Mg+O_2\to 2MgO$. (a) Identify the limiting reagent. (b) What mass of MgO forms? (c) The student finds the flask and contents weigh the same before and after. Which law does this show, and why would an open crucible give a different reading? (Mg = 24, O = 16) (5 marks)

    Hint

    Oxygen = 1.12 ÷ 22.4 mol.

    Answer

    Mg = 2.4 ÷ 24 = 0.10 mol; O₂ = 1.12 ÷ 22.4 = 0.050 mol (1). Ratio needed 2 : 1; 0.10 ÷ 2 = 0.050 and 0.050 ÷ 1 = 0.050, so the reactants are in exact proportion and neither is in excess (1). (b) MgO = 0.10 mol × 40 = 4.0 g (1). (c) Law of conservation of mass (1). In an open crucible the oxygen comes from the air, so the solid gains mass as oxygen combines with it; the total mass of everything taking part is still conserved (1).

12. Q12 · 3 marks · Short answerHOTS / competency

    Two students make 'one molar' urea solutions. Student A dissolves 60 g of urea in 1.000 kg of water; Student B dissolves 60 g of urea in water and makes the volume up to 1.000 L. Which student made a 1 M solution and which a 1 m solution? Whose concentration would remain unchanged if both solutions were warmed from 20 °C to 40 °C, and why? (Urea = 60 g mol⁻¹) (3 marks)

    Hint

    Per kg of solvent versus per litre of solution.

    Answer

    A: 1 mol in 1 kg of solvent, so 1 m (molal) (1). B: 1 mol in 1 L of solution, so 1 M (molar) (1). On warming, B's volume expands and its molarity falls slightly; A's molality is unchanged because it depends only on masses (1).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Name the seven SI base quantities.Show answer

   Length, mass, time, electric current, thermodynamic temperature, amount of substance, luminous intensity.

2. 2.Define one atomic mass unit.Show answer

   One-twelfth of the mass of one carbon-12 atom (1.66056 × 10⁻²⁴ g).

3. 3.What is the value of the Avogadro constant?Show answer

   6.02214076 × 10²³ mol⁻¹, exactly.

4. 4.State the law of definite proportions.Show answer

   A given compound always contains exactly the same proportion of elements by mass.

5. 5.What is the difference between mass and weight?Show answer

   Mass is the amount of matter and is constant; weight is the gravitational force on it and changes with place.

6. 6.Which concentration terms are independent of temperature?Show answer

   Mass per cent, mole fraction and molality.

7. 7.What is a limiting reagent?Show answer

   The reactant that is completely used up first and so limits the amount of product.

8. 8.Give the empirical formula of glucose and of benzene.Show answer

   CH₂O and CH.

9. 9.Which part of Dalton's theory was later found wrong?Show answer

   That atoms are indivisible and that all atoms of an element have the same mass (isotopes exist).

10. 10.How many significant figures are in 0.00250?Show answer

    Three.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you convert 37 °C to °F and K?
• ☐Can you state the rules for significant figures in addition and in multiplication?
• ☐Can you show the law of multiple proportions with data for two compounds?
• ☐Can you calculate an average atomic mass from isotope data?
• ☐Can you find the number of atoms in a given mass of a compound?
• ☐Can you get an empirical and molecular formula from percentages and molar mass?
• ☐Can you find a limiting reagent and the mass of excess reagent left?
• ☐Can you convert molarity to molality using density?

Some Basic Concepts of Chemistry: frequently asked questions