A complete, exam-focused lecture built from Karachi coaching-centre notes (Adamjee numericals + concept notes, BIEK / Sindh Board pattern). Covers every definition, law, formula, rule and worked numerical you need for the written board exam.
1 — What is stoichiometry?
The word comes from the Greek stoicheion (element) and metron (measurement).
- Stoichiometry — the study of the quantitative relationship between reactants and products in a chemical reaction, using a balanced chemical equation. (Equivalently: calculating the mass or volume of reactants/products from a balanced equation.)
Assumptions of stoichiometry
- Reactant molecules are completely converted into products.
- The reaction is irreversible.
- There is no side reaction.
Exam point: because of these assumptions, stoichiometric calculations cannot be applied to reversible reactions.
Qualitative vs quantitative analysis
- Qualitative analysis — tells us which substances (elements or compounds) are present in a sample.
- Quantitative analysis — tells us how much of each is present. Stoichiometry is the tool of quantitative chemistry.
Applications
Used to solve problems in the laboratory, chemical industries, engineering, food manufacturing, pharmaceuticals and many other fields of science.
The three types of stoichiometric relationship
| Relationship | What you find |
|---|
| Mass–Mass | Unknown mass of a reactant/product from a given mass of a substance. |
| Mass–Volume | Unknown mass or volume from a given mass or volume of a substance. |
| Volume–Volume | Unknown volume of a reactant/product from a given volume of a substance. |
2 — Laws of chemical combination
Every stoichiometric calculation rests on two classical laws. They are the reason a balanced equation can be trusted to predict masses.
- Law of Conservation of Mass (Antoine Lavoisier, 1789) — in a chemical reaction matter is neither created nor destroyed; the total mass of the products equals the total mass of the reactants.
- Law of Definite (Constant) Proportions (Joseph Proust, 1799) — a pure compound always contains the same elements combined in the same fixed ratio by mass, regardless of its source or how it was made.
- Law of Multiple Proportions (John Dalton) — when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.
Conservation of mass — check
2H₂ + O₂ → 2H₂O
reactants: 2(2) + 32 = 36 g
products: 2(18) = 36 g ✓ mass is conserved
Definite proportions in action: pure water is always 11.1 % hydrogen and 88.9 % oxygen by mass (a 1 : 8 mass ratio) — whether it comes from a river, the sea, or a chemistry lab.
3 — Atomic, molecular & formula mass
- Atomic mass — the mass of one atom of an element compared with the mass of one atom of carbon-12 (C¹²). e.g. atomic mass of H = 1 a.m.u, O = 16 a.m.u.
- Molecular mass — the sum of the atomic masses of all atoms present in a molecule, as shown by its molecular formula. e.g. H₂O = 2 + 16 = 18 a.m.u.
- Formula mass — the sum of the atomic masses given in the simplest (empirical) formula of an ionic substance. e.g. NaCl = 23 + 35.5 = 58.5 a.m.u.
a.m.u = the dalton (Da): one atomic mass unit is 1/12 the mass of a carbon-12 atom = 1.66 × 10⁻²⁴ g. "a.m.u" and "dalton" mean exactly the same thing — a common MCQ trap.
Tip: elements whose molecules are diatomic (H₂, O₂, N₂, Cl₂) have a molecular mass double their atomic mass. A monatomic gas like neon has atomic mass = molecular mass.
4 — The mole & Avogadro's number
- Mole — the gram atomic mass, gram molecular mass, or gram formula mass of any substance (atoms, molecules or ions) that contains 6.02 × 10²³ particles.
- Avogadro's number (Nₐ) — the number of particles (atoms, ions, molecules) present in one mole of any substance = 6.02 × 10²³. One mole of any substance always contains this many particles, regardless of its chemical nature.
Key formulas
no. of moles = given mass (g) / molar mass
no. of particles = no. of moles × 6.02 × 10²³
| Substance | 1 mole equals |
|---|
| C (atomic mass 12) | 12 g C |
| CO₂ (molecular mass 44) | 44 g CO₂ |
| NaCl (formula mass 58.5) | 58.5 g NaCl |
| Na (atomic mass 23) | 23 g Na |
| H₂O (molecular mass 18) | 18 g H₂O |
5 — Molar mass & molar volume
- Molar mass — the mass in grams of 1 mole of any pure substance (g/mol).
- Molar volume — the volume of 1 mole of a gas at standard temperature (273 K) and standard pressure (1 atm). For all ideal gases at STP it is 22.4 dm³.
Molar volume
molar volume = molar mass / density
e.g. 2 g/mol H₂ (1 mole) occupies 22.4 dm³ at STP
28 g/mol N₂ (1 mole) occupies 22.4 dm³ at STP
Volume units (MCQ trap): 1 L = 1 dm³ = 1000 cm³ = 1000 mL. So the molar volume 22.4 dm³ is the same as 22 400 cm³.
Avogadro's law — equal volumes, not equal mass: equal volumes of all gases at the same temperature and pressure contain an equal number of molecules (not equal mass). This is why 1 mole of any gas occupies the same 22.4 dm³ at STP.
6 — The master relationship table
This single table ties together moles, mass, number of particles and volume. Memorise it — most numericals are an application of one row.
| 1 mole of… | Mass | No. of particles | Volume (gas, STP) |
|---|
| 1 mol H (atoms) | 1 g | 6.02 × 10²³ atoms | — |
| 1 mol H₂ (molecules) | 2 g | 6.02 × 10²³ molecules | 22.4 dm³ |
| 1 mol CH₄ | 16 g | 6.02 × 10²³ molecules | 22.4 dm³ |
| 1 mol NaCl | 58.5 g | 6.02 × 10²³ formula units | — |
| 1 mol Na⁺ | 23 g | 6.02 × 10²³ cations | — |
| 1 mol Cl⁻ | 35.5 g | 6.02 × 10²³ anions | — |
The three conversion formulas
moles & mass: moles = mass / molar mass
moles & particles: moles = no. of particles / (6.02 × 10²³)
moles & volume: moles = volume at STP / 22.4 dm³
7 — Mole / mass / particle problems
Example 1 — moles from mass
Calculate the number of moles in 2400 g of CO₂.
moles = mass / molar mass = 2400 / 44 = 54.5 mol
Example 2 — mass from moles
Find the mass of 0.2 mole of CH₄.
mass = moles × molar mass = 0.2 × 16 = 3.2 g
Example 3 — atoms from mass
The tip of a carbon pencil has mass 0.3 g. How many carbon atoms does it contain?
no. of atoms = (mass × 6.02 × 10²³) / atomic mass
= (0.3 × 6.02 × 10²³) / 12
= 1.505 × 10²² atoms
Example 4 — mass from molecules
Find the mass of 3.01 × 10²³ molecules of water.
moles = particles / (6.02 × 10²³) = 3.01 × 10²³ / 6.02 × 10²³ = 0.5 mol
mass = moles × molar mass = 0.5 × 18 = 9 g
Example 5 — formula units
Calculate the number of formula units in 5.85 g NaCl.
formula units = (mass / formula mass) × Nₐ
= (5.85 / 58.5) × 6.02 × 10²³
= 6.02 × 10²² formula units
8 — Percentage composition
- Percentage composition — the mass percentage contributed by each element in a compound.
Percentage of an element
% of element = (mass of that element in 1 mole / molar mass of compound) × 100
Worked example 1 — water (H₂O)
Find the percentage composition of H₂O. (Molar mass = 18)
% H = (2 / 18) × 100 = 11.1 %
% O = (16 / 18) × 100 = 88.9 %
Worked example 2 — carbon dioxide (CO₂)
Find the percentage composition of CO₂. (Molar mass = 44)
% C = (12 / 44) × 100 = 27.3 %
% O = (32 / 44) × 100 = 72.7 %
Check: the percentages of all elements in a compound must add up to 100 %.
10 — Exponential (scientific) notation
A number is written as N × 10ⁿ, where N (the coefficient) is between 1 and 9.99, and n (the exponent) is an integer power of 10.
- If the number is greater than one → the exponent is positive.
- If the number is less than one → the exponent is negative.
Standard ⇄ exponential
| Standard | | Exponential |
|---|
| 400000000 | = | 4.0 × 10⁸ |
| 0.000045 | = | 4.5 × 10⁻⁵ |
Operations
| Operation | Rule | Example → answer |
|---|
| Add / Subtract | Convert all to the same exponent, then add/subtract coefficients | 2.03 × 10³ + 4.05 × 10² → 2.030 × 10³ + 0.405 × 10³ = 2.435 × 10³ |
| Multiply | Multiply coefficients, add exponents | (2.03 × 10³)(4.05 × 10²) = 8.22 × 10⁵ |
| Divide | Divide coefficients, subtract exponents | (2.03 × 10³)/(4.05 × 10²) = 0.50 × 10¹ = 5.0 |
| Power | Apply power to coefficient, multiply exponent by the power | (2.03 × 10³)² = 4.12 × 10⁶ |
| Root | Adjust exponent to be divisible by the root, take root of coefficient, divide exponent by root | √(2.03 × 10³) = √(20.3 × 10²) = 4.5 × 10¹ (≈ 45) |
11 — Significant figures
- Significant figures — the reliable digits of a number, i.e. the digits whose accuracy is needed to express a quantity (certain digits plus one estimated digit).
Rules
| Rule | Examples |
|---|
| Non-zero digits are significant | 183 → 3 s.f. · 0.12 → 2 s.f. |
| Zeros between non-zero digits are significant | 1008 → 4 s.f. · 10204 → 5 s.f. |
| Zeros just after the decimal point in a number < 1 are not significant | 0.00122 → 3 s.f. · 0.00002 → 1 s.f. |
| Zeros after a decimal point in a number > 1 are significant | 3.0000 → 5 s.f. · 9.000 → 4 s.f. |
| A final zero in a large number with no decimal point is not significant | 4000 → 1 s.f. · 564200 → 4 s.f. |
12 — Rounding off data
- Rounding off — reducing a number to the desired number of significant figures and adjusting the last reported digit.
| If the digit to drop is… | Do this | Example |
|---|
| greater than 5 | drop it and add 1 to the previous digit | 3.768 → 3.77 |
| less than 5 | simply drop it | 6.823 → 6.82 |
| exactly 5, previous digit even | drop the 5 (leave even digit) | 4.865 → 4.86 |
| exactly 5, previous digit odd | add 1 to make it even, drop the 5 | 4.835 → 4.84 |
Worked rule-of-s.f. example: 1.32 × 4.421 = 5.83572, but the answer is limited to 3 s.f. (the least precise value, 1.32) → 5.84.
13 — Stoichiometric calculations
Every calculation rests on the mole ratio read from a balanced equation. There are two standard methods — the factor method and the mole method — both give the same answer.
- Mole ratio — the ratio of the number of moles of any two substances in a reaction, given by the coefficients of the balanced equation. e.g. in 2H₂ + O₂ → 2H₂O the mole ratio H₂ : O₂ : H₂O = 2 : 1 : 2.
Mass–Mass — factor method
Calculate the mass of KCl and O₂ obtained on heating 49 g KClO₃.
2KClO₃ → 2KCl + 3O₂
2(122.5) = 245 g KClO₃ → 2(74.5) = 149 g KCl & 3(32) = 96 g O₂
O₂: 245 g KClO₃ → 96 g O₂, so 49 g → (96 × 49)/245 = 19.2 g O₂
KCl: 245 g KClO₃ → 149 g KCl, so 49 g → (149 × 49)/245 = 29.8 g KCl
Mass–Mass — mole method (same problem)
moles KClO₃ = 49 / 122.5 = 0.4 mol
2 mol KClO₃ → 3 mol O₂, so 0.4 mol → (3/2)(0.4) = 0.6 mol O₂
mass O₂ = 0.6 × 32 = 19.2 g
2 mol KClO₃ → 2 mol KCl, so 0.4 mol → 0.4 mol KCl
mass KCl = 0.4 × 149 = 29.8 g
Mass–Mass — finding a reactant
Calculate the mass of NH₃ required to produce 100 g of NO.
4NH₃ + 5O₂ → 4NO + 6H₂O → 68 g NH₃ gives 120 g NO
100 g NO needs (68/120) × 100 = 56.7 g NH₃
Mass–Volume
Calculate the volume of NH₃ (at STP) required to produce 100 g of NO.
4 mol NH₃ → 4 mol NO → 4(22.4) = 89.6 dm³ NH₃ gives 120 g NO
100 g NO needs (89.6/120) × 100 = 74.66 dm³ NH₃
Decomposition with all four quantities
50 g of CaCO₃ is heated: CaCO₃ → CaO + CO₂. Find the mass of CaO produced.
1 mol CaCO₃ (100 g) → 1 mol CaO (56 g)
50 g CaCO₃ → (56/100) × 50 = 28 g CaO
14 — Theoretical, practical & percentage yield
- Theoretical yield — the maximum amount of product expected from a balanced equation, calculated using the limiting reactant.
- Practical (actual) yield — the amount of product actually obtained after the experiment.
- Percentage yield — the ratio of practical to theoretical yield; it measures the efficiency of the process.
Percentage yield
% yield = (practical yield / theoretical yield) × 100
Why practical yield is always less than theoretical yield
- Some reactant does not react.
- The reactants form side products.
- The reaction is reversible.
- Product is lost in physical processes (distillation, crystallisation, filtration).
Worked example — % yield
40 g KClO₃ is fully decomposed (2KClO₃ → 2KCl + 3O₂). The O₂ collected weighs 14.9 g. Find the theoretical yield and the percentage yield.
moles KClO₃ = 40 / 122.5 = 0.326 mol
0.326 mol KClO₃ → (3/2)(0.326) = 0.489 mol O₂
theoretical yield = 0.489 × 32 = 15.648 g O₂
% yield = (14.9 / 15.648) × 100 = 95.2 %
15 — Limiting reactant
- Limiting reactant — the reactant that is entirely consumed first; it gives the least moles of product and so limits how much product forms.
- Excess reactant — the reactant that is not completely consumed.
Steps to determine the limiting reactant
- Write a balanced chemical equation.
- Find the moles of each reactant from the given amounts.
- Find the moles of product each reactant would make.
- The reactant giving the least moles of product is the limiting reactant.
Simple illustration: in 2H₂ + O₂ → 2H₂O, if you take 2 mol of each, hydrogen is used up first and the reaction stops while oxygen remains. So H₂ is the limiting reactant and O₂ is in excess. (This is exactly what the Reaction Lab lets you see.)
Worked example — DDT synthesis
2C₆H₅Cl + C₂HOCl₃ → C₁₄H₉Cl₅ + H₂O. 1.142 kg C₆H₅Cl is mixed with 485 g C₂HOCl₃. Which is limiting, and how much DDT forms?
moles C₆H₅Cl = 1142 / 112.5 = 10.15 mol → makes 10.15/2 = 5.075 mol DDT
moles C₂HOCl₃ = 485 / 147.5 = 3.28 mol → makes 3.28 mol DDT
C₂HOCl₃ gives fewer moles of product → C₂HOCl₃ is the limiting reactant
mass DDT = 3.28 × 354.5 = 1.162 kg
Now try it yourself
You've seen the mole ratio and the limiting reactant on paper. Open the Reaction Lab and watch 2H₂ + O₂ → 2H₂O play out — change the amounts and see which reactant runs out first.
16 — Exam short questions
- Define stoichiometry and state its three assumptions.
- State the Law of Conservation of Mass and the Law of Definite Proportions.
- Differentiate between qualitative and quantitative analysis.
- Name the three types of stoichiometric relationships.
- Define mole, Avogadro's number and molar volume (with values).
- Differentiate atomic mass, molecular mass and formula mass with an example of each.
- Define percentage composition and write its formula.
- Differentiate between empirical and molecular formula with an example. How are the two related?
- State the rules of significant figures and rounding off.
- Define limiting and excess reactant. Why does the limiting reactant control the product?
- Define theoretical, practical and percentage yield. Why is practical yield always less?
- Why can stoichiometry not be applied to a reversible reaction?
Practice: the practice page has 75+ board MCQs and solved numericals from past papers — use it after this lecture.