Some Basic Concepts of Chemistry ISC Class 11: Mole Concept and Stoichiometry

What Some Basic Concepts of Chemistry is about

Weightage: Some Basic Concepts of Chemistry belongs to unit 1 of the ISC Class XI Chemistry syllabus. CISCE weights units 1-6 together (Physical Chemistry) at 41 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Precision, accuracy, significant figures and units

Precision describes how close repeated measurements are to each other; accuracy describes how close they are to the true value. Significant figures are the digits known reliably plus one estimated digit. Rules: all non-zero digits count; zeros between non-zero digits count; leading zeros never count; trailing zeros count only after a decimal point (so 0.04060 has 4). In addition/subtraction, keep the fewest decimal places; in multiplication/division, keep the fewest significant figures. SI units have seven base units (m, kg, s, A, K, mol, cd). Dimensional analysis converts units by multiplying by conversion factors such as (1000 g)/(1 kg) = 1.

Worked example

(i) How many significant figures are in 0.002050? (ii) Convert a density of 3.0 g cm⁻³ into kg m⁻³.

Try it first, then show the solutionSolution

1. (i) Leading zeros do not count; 2, 0 (between), 5 and the trailing 0 after the decimal point count: 4
2. (ii) 3.0 g cm⁻³ × (1 kg/1000 g) × (100 cm/1 m)³ = 3.0 × 10⁻³ × 10⁶ kg m⁻³

Answer: 4; 3.0 × 10³ kg m⁻³

Laws of chemical combination

Conservation of mass: in a chemical change, total mass of reactants = total mass of products. Definite proportions: a pure compound always contains the same elements combined in the same fixed ratio by mass. Multiple proportions: when the same two elements make two or more compounds, fix the mass of one element; the masses of the second element combining with it then stand in a small whole-number ratio. Reciprocal proportions: if A and B each combine with C, then the ratio in which A and B combine with each other is the same as (or a simple multiple of) the ratio of the masses of A and B that combine with a fixed mass of C. Gay-Lussac's law of gaseous volumes: gases react, and gaseous products form, in volumes bearing a simple whole-number ratio, at the same temperature and pressure.

Worked example

Carbon forms CO (42.86% C) and CO₂ (27.27% C). Show that the law of multiple proportions holds.

Try it first, then show the solutionSolution

1. In CO: oxygen per gram of carbon = 57.14/42.86 = 1.333 g
2. In CO₂: oxygen per gram of carbon = 72.73/27.27 = 2.667 g
3. Ratio 1.333 : 2.667 = 1 : 2, a simple whole-number ratio

Answer: Oxygen masses per fixed mass of carbon are in the ratio 1 : 2

Dalton's atomic theory: postulates and limitations

Postulates: matter is made of tiny indivisible particles called atoms; atoms of one element are identical in mass and properties, and differ from those of other elements; compounds form when atoms of different elements combine in fixed, simple whole-number ratios; atoms are neither created nor destroyed in a chemical reaction, only rearranged. These explain conservation of mass and definite and multiple proportions. Limitations: atoms are divisible (electrons, protons, neutrons); atoms of the same element can differ in mass (isotopes), while atoms of different elements can have the same mass (isobars); it cannot explain Gay-Lussac's law (that needed Avogadro's idea of molecules) or why atoms combine at all.

Worked example

Which observation is not explained by Dalton's theory: (i) 2 volumes of hydrogen react with 1 volume of oxygen to give 2 volumes of steam (ii) water always contains hydrogen and oxygen in the ratio 1 : 8 by mass?

Try it first, then show the solutionSolution

1. (ii) follows from atoms combining in fixed ratios: explained
2. (i) would require half-atoms of oxygen if gases were single atoms: explained only after Avogadro introduced molecules such as O₂

Answer: (i), the volume relationship

Atomic mass, isotopic average and molecular mass

Atomic masses are measured relative to the carbon-12 standard: 1 u = 1/12 of the mass of one C-12 atom, so C-12 = 12.00 u. Most elements are mixtures of isotopes, so the tabulated atomic mass is the abundance-weighted average of the isotopic masses. The molecular mass is the sum of the atomic masses in one molecule; for ionic compounds the term formula mass is used.

Worked example

Chlorine is 75.77% Cl-35 (34.969 u) and 24.23% Cl-37 (36.966 u). Find its average atomic mass.

Try it first, then show the solutionSolution

1. 0.7577 × 34.969 = 26.496
2. 0.2423 × 36.966 = 8.957
3. Sum = 35.453

Answer: 35.45 u

Mole concept, Avogadro's number and gram molecular volume

One mole is the amount of substance containing 6.022 × 10²³ (Avogadro's number, N_A) elementary entities. Its mass in grams equals the molar mass: n = m/M. Number of particles N = n × N_A. One mole of any ideal gas at STP (273 K, 1 atm) occupies the gram molecular volume, 22.4 dm³. Always say which particles you are counting: 1 mol of CO₂ contains 1 mol of molecules but 3 mol of atoms.

Worked example

For 8.8 g of CO₂ (M = 44 g mol⁻¹), find the moles, number of molecules, number of oxygen atoms and volume at STP.

Try it first, then show the solutionSolution

1. n = 8.8/44 = 0.20 mol
2. Molecules = 0.20 × 6.022 × 10²³ = 1.204 × 10²³; O atoms = 2 × 1.204 × 10²³ = 2.409 × 10²³
3. Volume = 0.20 × 22.4 = 4.48 dm³

Answer: 0.20 mol; 1.204 × 10²³ molecules; 2.409 × 10²³ O atoms; 4.48 dm³

Percentage composition, empirical and molecular formula

Percentage of an element = (mass of that element in one mole/molar mass) × 100. The empirical formula gives the simplest whole-number ratio of atoms: divide each percentage by the atomic mass, divide by the smallest result, then multiply to clear any fractions such as 1.5 or 1.33. The molecular formula = (empirical formula)ₙ with n = molar mass/empirical formula mass.

Worked example

A compound contains C 40.0%, H 6.67%, O 53.33%, and its molar mass is 180 g mol⁻¹. Find its empirical and molecular formulae.

Try it first, then show the solutionSolution

1. Moles per 100 g: C 40.0/12 = 3.33; H 6.67/1 = 6.67; O 53.33/16 = 3.33
2. Divide by 3.33: C 1, H 2, O 1: empirical formula CH₂O (mass 30)
3. n = 180/30 = 6

Answer: CH₂O; C₆H₁₂O₆

Equivalent weight and n-factor

Equivalent weight (EW) = molar mass/n-factor, and one gram equivalent is EW in grams. The n-factor is: for an acid, its basicity (replaceable H⁺ per molecule: H₂SO₄ 2); for a base, its acidity (replaceable OH⁻: Ca(OH)₂ 2); for a salt, the total positive charge of the cations (Na₂CO₃ 2); for an oxidising or reducing agent, the number of electrons gained or lost per formula unit (KMnO₄ in acid: Mn +7 → +2, 5). An element can have a variable equivalent weight when it shows more than one valency: EW = atomic mass/valency (Fe: 56/2 = 28 in FeO, 56/3 = 18.67 in Fe₂O₃).

Worked example

Find the equivalent weights of H₂SO₄ (98), Na₂CO₃ (106) and KMnO₄ (158) in acidic medium.

Try it first, then show the solutionSolution

1. H₂SO₄: basicity 2 ⇒ 98/2 = 49
2. Na₂CO₃: total cation charge 2 ⇒ 106/2 = 53
3. KMnO₄ (acid): Mn +7 → +2 gains 5 electrons ⇒ 158/5 = 31.6

Answer: 49; 53; 31.6

Concentration terms: %, molarity, molality, normality, mole fraction

% (w/w) = g solute per 100 g solution; % (w/v) = g solute per 100 mL solution. Molarity M = moles of solute per litre of solution (depends on temperature, since volume changes). Molality m = moles of solute per kg of solvent (independent of temperature). Normality N = gram equivalents of solute per litre of solution, so N = M × n-factor. Mole fraction x_A = n_A/(n_A + n_B), and the mole fractions add to 1. For dilution, moles are unchanged: M₁V₁ = M₂V₂; in a titration, equivalents match: N₁V₁ = N₂V₂.

Worked example

4.9 g of H₂SO₄ is dissolved to make 500 mL of solution. Find its molarity and normality.

Try it first, then show the solutionSolution

1. n = 4.9/98 = 0.050 mol
2. M = 0.050/0.500 = 0.10 M
3. N = 0.10 × 2 = 0.20 N

Answer: 0.10 M; 0.20 N

Stoichiometry and the limiting reagent

A balanced equation gives the mole ratio of reactants and products. Steps: (1) convert given masses or volumes into moles; (2) use the mole ratio; (3) convert back into the quantity asked (mass, or volume at STP). For gases at the same T and P, volumes are in the same ratio as moles, so volume-volume problems need no molar masses. If amounts of two reactants are given, find the limiting reagent: divide each reactant's moles by its coefficient; the smallest result is limiting, and it fixes the amount of product. The other reactant is in excess.

Worked example

4 g of H₂ reacts with 16 g of O₂: $2H_2+O_2\to 2H_{2}O$. Which is limiting and what mass of water forms?

Try it first, then show the solutionSolution

1. H₂: 4/2 = 2 mol (÷2 = 1); O₂: 16/32 = 0.5 mol (÷1 = 0.5)
2. O₂ gives the smaller value, so O₂ is limiting
3. Water = 2 × 0.5 = 1 mol = 18 g; H₂ left = 2 − 1 = 1 mol (2 g)

Answer: O₂ is limiting; 18 g of 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

• Concentration Terms and Limiting Reagent
• Mole Concept and Empirical Formula

Some Basic Concepts of Chemistry formula sheet

Moles from mass

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

m in g, M in g mol⁻¹.

Particles and gas volume

$N=n{N}_A,N_A=6.022\times 10^{23};V_{\text{STP}}=n\times 22.4{\text{dm}}^3$

STP: 273 K, 1 atm.

Molecular formula

$n=\frac{\text{molar mass}}{\text{empirical formula mass}}$

Molecular formula = (empirical)ₙ.

Equivalent weight

$\text{EW}=\frac{\text{molar mass}}{n\text{-factor}}$

Basicity, acidity, cation charge or electrons transferred.

Molarity and molality

$M=\frac{n_{\text{solute}}}{V_{\text{solution}}(\text{L})};m=\frac{n_{\text{solute}}}{\text{mass of solvent (kg)}}$

Molality does not change with temperature.

Normality

$N=M\times n\text{-factor}$

Gram equivalents per litre.

Mole fraction

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

No units.

Dilution and titration

$M_1{V}_1=M_2{V}_2;N_1{V}_1=N_2{V}_2$

Equivalents of acid = equivalents of base at the end point.

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.

• Everything through the mole

  Grams, particles and gas volumes all convert to moles first.

  Use it when: Any calculation.

• Molarity: solution; molality: solvent

  molaRity → litRes of solution; molaLity → kiLograms of solvent.

  Use it when: Not confusing M and m.

• N = M × n

  Normality is molarity times the n-factor.

  Use it when: Converting concentrations.

• Divide by the coefficient

  Moles ÷ coefficient: the smallest is the limiting reagent.

  Use it when: Limiting reagent.

• Leading zeros are placeholders

  0.00xyz: the zeros only place the decimal point and never count.

  Use it when: Significant figures.

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 D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 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: Using 22.4 dm³ at conditions other than STP

  The gram molecular volume applies only at 273 K and 1 atm.

  Wrong working

  1 mol of gas at 27 °C occupies 22.4 dm³.

  Right working

  At 300 K and 1 atm it occupies 22.4 × 300/273 = 24.6 dm³.

  ✓Do this: Check the stated conditions before using 22.4.

  Usually costs: 1-2 marks.

• ✗Mistake: Choosing the limiting reagent by mass

  The smaller mass is not necessarily the smaller number of moles needed.

  Wrong working

  4 g H₂ and 16 g O₂: H₂ is limiting because 4 < 16.

  Right working

  Moles ÷ coefficient: H₂ 1, O₂ 0.5, so O₂ is limiting.

  ✓Do this: Always compare moles divided by coefficients.

  Usually costs: Full marks for the part.

• ✗Mistake: Molality calculated per kg of solution

  Students divide by the total mass.

  Wrong working

  20% NaOH: 0.5 mol/0.100 kg = 5 m.

  Right working

  Solvent is 80 g: 0.5/0.080 = 6.25 m.

  ✓Do this: Molality uses the mass of solvent only.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong n-factor for KMnO₄

  It depends on the medium.

  Wrong working

  EW of KMnO₄ = 158/7.

  Right working

  Acidic: Mn +7 → +2, n = 5, EW 31.6; neutral: +7 → +4, n = 3, EW 52.7.

  ✓Do this: Find the change in oxidation number for the given medium.

  Usually costs: 1 mark.

• ✗Mistake: Rounding too early

  Intermediate rounding piles up errors.

  Wrong working

  8.77 rounded to 9 mol before multiplying.

  Right working

  Keep extra digits in intermediate steps; round the final answer to the right significant figures.

  ✓Do this: Round only once, at the end.

  Usually costs: 1 mark.

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 mass of 1 mol of oxygen atoms is:
   (a) 8 g
   (b) 16 g
   (c) 32 g
   (d) 6.022 × 10²³ g

   Hint

   Atomic mass in grams.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   CO and CO₂ together illustrate the law of:
   (a) conservation of mass
   (b) definite proportions
   (c) multiple proportions
   (d) gaseous volumes

   Hint

   Same two elements, two compounds.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   How many moles of gas occupy 11.2 dm³ at STP?

   Hint

   ÷ 22.4.

   Answer

   0.50 mol

4. Q4 · 1 mark · Very short answerExam level

   Find the equivalent weight of Ca(OH)₂ (M = 74).

   Hint

   Acidity 2.

   Answer

   37

5. Q5 · 2 marks · Short answerExam level

   Calculate the molar mass of CuSO₄·5H₂O (Cu = 63.5).

   Hint

   Include the water.

   Answer

   63.5 + 32 + 64 + 5 × 18 = 249.5 g mol⁻¹

6. Q6 · 2 marks · Short answerHarder exam level

   Write 0.000325 in scientific notation to 2 significant figures, and convert 6.0 km into metres with the correct number of significant figures.

   Hint

   Round the 5 up.

   Answer

   3.3 × 10⁻⁴; 6.0 × 10³ m

7. Q7 · 2 marks · Short answerHarder exam level

   How many atoms are present in 0.50 mol of NH₃?

   Hint

   4 atoms per molecule.

   Answer

   0.50 × 4 × 6.022 × 10²³ = 1.204 × 10²⁴

8. Q8 · 2 marks · Short answerHarder exam level

   Find the percentage of water of crystallisation in CuSO₄·5H₂O.

   Hint

   90/249.5.

   Answer

   36.1%

9. Q9 · 3 marks · Short answerHarder exam level

   Find the equivalent weight of (i) H₃PO₄ (M = 98) in the reaction $H_{3}P{O}_4+2NaOH\to N{a}_{2}HP{O}_4+2H_{2}O$ (ii) FeSO₄ (M = 152) when it is oxidised to Fe₂(SO₄)₃ (iii) K₂Cr₂O₇ (M = 294) in acidic medium.

   Hint

   Take the n-factor from the actual change, not from the formula alone.

   Answer

   (i) Only 2 H⁺ are replaced, n = 2: 98/2 = 49 (ii) Fe +2 → +3, n = 1: 152 (iii) 2 Cr, each +6 → +3, n = 6: 294/6 = 49

10. Q10 · 3 marks · Short answerHarder exam level

    Commercial sulphuric acid is 98% H₂SO₄ by mass with density 1.84 g mL⁻¹. Find its molarity and normality.

    Hint

    Take 1 L of solution.

    Answer

    1 L has 1840 g, of which 1803.2 g is H₂SO₄ = 18.4 mol: 18.4 M; 36.8 N

11. Q11 · 3 marks · Short answerHarder exam level

    What mass of oxygen is needed to burn 4.4 g of propane completely, and what mass of CO₂ forms? $C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$

    Hint

    4.4 g = 0.10 mol.

    Answer

    O₂: 0.50 mol = 16 g; CO₂: 0.30 mol = 13.2 g

12. Q12 · 2 marks · Short answerHarder exam level

    What volume of water must be added to 200 mL of 0.50 M HCl to make it 0.20 M?

    Hint

    M₁V₁ = M₂V₂.

    Answer

    V₂ = 0.50 × 200/0.20 = 500 mL, so add 300 mL

13. Q13 · 5 marks · Long answerHarder exam level

    (a) A hydrocarbon contains C 85.71% and H 14.29%. Find its empirical formula. [2]
    (b) Its molar mass is 56 g mol⁻¹. Find its molecular formula. [1]
    (c) Find the volume of oxygen at STP needed to burn 5.6 g of it completely. [2]

    Hint

    Balance the combustion equation first.

    Answer

    (a) C 7.14, H 14.29 ⇒ 1 : 2, CH₂. (b) n = 56/14 = 4: C₄H₈. (c) $C_4{H}_8+6O_2\to 4C{O}_2+4H_{2}O$; 5.6 g = 0.10 mol needs 0.60 mol O₂ = 13.44 dm³.

14. Q14 · 3 marks · Short answerHOTS / competency

    10.0 g of a mixture of Na₂CO₃ and NaHCO₃ loses 1.55 g on strong heating: $2NaHC{O}_3\to N{a}_{2}C{O}_3+H_{2}O+C{O}_2$. Find the percentage of NaHCO₃ in the mixture.

    Hint

    168 g of NaHCO₃ loses 62 g (H₂O + CO₂).

    Answer

    Mass of NaHCO₃ = 168 × 1.55/62 = 4.2 g, i.e. 42%

15. Q15 · 3 marks · Short answerHOTS / competency

    CH₄ contains 25.0% H; CO₂ contains 72.73% O; H₂O contains 11.11% H. Show that these data illustrate the law of reciprocal proportions.

    Hint

    Fix 12 g of carbon.

    Answer

    12 g C combines with 4 g H (in CH₄) and with 32 g O (in CO₂), so H : O = 4 : 32 = 1 : 8. In water H : O = 11.11 : 88.89 = 1 : 8, the same ratio.

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.Avogadro's number?Show answer

   6.022 × 10²³ mol⁻¹.

2. 2.Gram molecular volume at STP?Show answer

   22.4 dm³.

3. 3.Standard for atomic masses?Show answer

   Carbon-12 = 12.00 u.

4. 4.Law of multiple proportions?Show answer

   Masses of one element combining with a fixed mass of another are in a simple whole-number ratio.

5. 5.Equivalent weight of a salt?Show answer

   Molar mass ÷ total positive charge of its cations.

6. 6.Molality?Show answer

   Moles of solute per kg of solvent.

7. 7.Normality and molarity?Show answer

   N = M × n-factor.

8. 8.Limiting reagent?Show answer

   The reactant with the smallest moles ÷ coefficient; it is used up first.

Quick revision checklist

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

• ☐Can you count significant figures and round a product correctly?
• ☐Can you show multiple proportions from percentage data?
• ☐Can you state two limitations of Dalton's theory?
• ☐Can you convert grams to moles to molecules to STP volume?
• ☐Can you find a molecular formula from percentages and molar mass?
• ☐Can you find the equivalent weight of an acid, base, salt and KMnO₄?
• ☐Can you convert between molarity, molality, normality and mole fraction?
• ☐Can you identify a limiting reagent and the excess left?

Some Basic Concepts of Chemistry: frequently asked questions