ما هي معادلة الأسترة وكيف يتكون الإستر؟ What is the esterification reaction equation?

تفاعل الأسترة هو تفاعل انعكاسي بين حمض كربوكسيلي وكحول في وجود حمض الكبريتيك المركز كعامل حفاز لنزع الماء: R-COOH + R2-OH ⇌ R-COO-R2 + H₂O، ونواتجه إستر وماء. [الصيغة الرمزية الموزونة: R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O]

1.1Unit 1. Structure, Bonding and Introduction to Organic Chemistry

Formulae & Equations — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 12 parts Lesson 1 of 36 1 quiz

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision

Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution

R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O

Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء

Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.

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Lesson summary

1.1 Formulae & Equations — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Use relative masses and the mole, write balanced full and ionic equations, and calculate formulae and gas quantities.Atom — the smallest part of an element that can take part in a chemical reaction.Element — a subst…

Empirical formula
Molecular formula
Structural, displayed and skeletal formulae
Relative isotopic mass
Relative atomic mass (Ar)
Relative formula mass (Mr)

Part 1 of 10Basic terms and types of formula

Basic terms and types of formula

Atom

Atom the smallest part of an element that can take part in a chemical reaction.

Element

Element a substance made of one type of atom that cannot be broken into simpler substances.

Compound

Compound a substance with two or more elements chemically joined.

Molecule

Molecule two or more atoms chemically joined together.

Ion

Ion an atom or group of atoms that has gained or lost electrons to become charged.

Empirical formula

Empirical formula the smallest whole-number ratio of atoms of each element in a compound.

Molecular formula

Molecular formula the actual number of atoms of each element in a molecule.

Structural, displayed and skeletal formulae

Structural, displayed and skeletal formulae structural shows the arrangement (CH₃CH₂OH), displayed shows every bond and skeletal shows the carbon skeleton as lines with functional groups.

Part 2 of 10Relative masses

1Standard

Standard the mass of an atom is compared with 1/12 of the mass of a carbon-12 atom, fixed at exactly 12 atomic mass units.

2Relative isotopic mass

Relative isotopic mass the mass of an isotope relative to 1/12 of a carbon-12 atom.
Accurate values are rounded for A Level, such as N = 14.0 and O = 16.0.

3Relative atomic mass (Ar)

Relative atomic mass (Ar) the weighted mean mass of an atom of an element relative to 1/12 of a carbon-12 atom.
It is calculated from the mass numbers and relative abundances of the isotopes.

4Relative formula mass (Mr)

Relative formula mass (Mr) the sum of the A_r values of all atoms in the formula.
Use relative formula mass for giant structures such as NaCl and relative molecular mass for molecules.
Relative formula mass is safe for any compound, ionic or covalent.

5Units

A_r and M_r have no units, and molar mass has the same number in g mol^-1.

6Accuracy

Accuracy use A_r values from the Periodic Table, as in K₂CO₃ = (2 x 39.1) + 12.0 + (3 x 16.0) = 138.2.

Part 3 of 10The mole and the Avogadro constant

The mole and the Avogadro constant

Avogadro constant

N_A or L = 6.02 x 10²³ mol^-1, the number of particles in one mole of atoms, molecules, ions or electrons.

Mole

Mole the amount of substance containing the same number of fundamental units as exactly 12.00 g of carbon-12.

Molar mass (M)

Molar mass (M) the mass of one mole, in g mol^-1.

Amount

Amount n = mass / molar mass.

Number of particles

Number of particles equals moles x Avogadro constant.
Take care which particle is asked for, because 1 mol of H₂O contains 3 mol of atoms.

Size of answers

Size of answers large or tiny answers are normal because Avogadro's constant is very large.

Part 4 of 10Balanced equations and formulae of ionic compounds

1Balancing

Balancing atoms are neither created nor destroyed, so each atom must balance on both sides.
Never change a formula; only add coefficients in front.
For organic combustion balance C, then H, then O.

2Steps

Steps write the formulae, count atoms, balance one at a time and add state symbols.

3State symbols

(s) solid, (l) liquid, (g) gas and (aq) aqueous.

4Ionic formulae

Ionic formulae multiply ions until the charges sum to zero.
Al³⁺ and SO₄^2- give Al₂(SO₄)₃.
The swap-the-numbers method gives the same result.

5Balanced equations show

Balanced equations show the ratio of moles, which is used to calculate masses and volumes.

Part 5 of 10Ionic equations

Dissociation

Dissociation in aqueous solution ionic compounds separate into ions.

Spectator ions

Spectator ions ions that do not take part in the reaction.

Ionic equation

Ionic equation shows only the particles that react, balanced for atoms and charge.

Write the full balanced equation with state symbols.

Split soluble ionic compounds into separate ions.

Cancel the spectator ions.

Write what remains.

Examples
- Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
- PrecipitationAg⁺(aq) + Cl^-(aq) → AgCl(s).
- Carbonate with acidCO₃^2- + 2H⁺ → CO₂ + H₂O.

Part 6 of 10Types of reaction shown by equations

Types of reaction shown by equations

Displacement

Displacement a more reactive element displaces a less reactive one, as in Br₂ + 2I^- → I₂ + 2Br^-.
Aqueous bromine is toxic, corrosive and harmful to the environment, so use a fume cupboard and nitrile gloves.
Bromine water is corrosive above 0.2 mol dm^-3, an irritant between 0.06 and 0.2 mol dm^-3 and low hazard below 0.06 mol dm^-3.

Neutralisation

Neutralisation an acid and a base give a salt and water, sometimes with CO₂.
H⁺(aq) + OH^-(aq) → H₂O(l).
Hazards depend on the concentration and strength of the acid and base, and on whether the acid is monoprotic, diprotic or triprotic.
Solid NaOH is more corrosive than a dilute solution.

Precipitation

Precipitation two aqueous solutions react to form a solid, as in Ba²⁺ + SO₄^2- → BaSO₄(s).
A solid product means purification should include filtering or decanting.

Part 7 of 10Empirical and molecular formulae

Write the mass (or percentage) of each element.

Divide each by its A_r to give moles.

Divide every mole value by the smallest.

Multiply to get whole numbers if needed, for example x2 for 1.5.

Write the empirical formula.

Divide the compound's M_r by the empirical formula mass.

Multiply every subscript by this number to give the molecular formula.

Part 8 of 10Percentage composition and water of crystallisation

Percentage by mass

(A_r x number of atoms of the element / M_r of compound) x 100.

Water of crystallisation

Water of crystallisation water that is part of a crystal structure.
A hydrated compound contains it and is written with a dot, as in CuSO₄.5H₂O.
An anhydrous compound does not contain it, as in CuSO₄.
Degrees of hydration differCoCl₂.6H₂O and CoCl₂.2H₂O.

Reversibility

Reversibility heating a hydrated salt drives off the water: CuSO₄.5H₂O ⇌ CuSO₄ + 5H₂O.

Finding x in MX.xH2O

Weigh the hydrated salt.

Heat to constant mass.

Find moles of anhydrous salt and moles of water lost from the two masses.

Divide by the smaller value to find the ratio, which gives x.

Part 9 of 10The ideal gas equation

Equation

PV = nRT.
What the symbols mean: p = pressure (Pa) · V = volume (m³) · n = amount of gas (mol) · R = 8.31 J mol⁻¹ K⁻¹ · T = temperature (K = °C + 273)
P = pressure in Pa, V = volume in m³, n = moles, R = 8.31 J mol^-1 K^-1, T = temperature in K.

Unit conversions

Unit conversions kPa to Pa multiply by 10³;
- cm³ to m³ multiply by 10^-6;
- dm³ to m³ multiply by 10^-3;
- °C to K add 273.

Uses

Uses find the moles of a gas or of a volatile liquid above its boiling point.
With the mass, the molar mass is found, and with the empirical formula the molecular formula is deduced.

Molar volume

24.0 dm³ mol^-1 at room temperature and pressure.

Working

Working show all working so marks can be given despite one error.

Part 10 of 10Reacting masses, yield and atom economy

Reacting masses

Reacting masses use the balanced equation ratio to find moles of product and then mass.

Limiting reagent

Limiting reagent the reactant that runs out first determines the theoretical yield.

Percentage yield

Percentage yield actual yield / theoretical yield x 100.

Atom economy

M_r of desired product / total M_r of reactants x 100.
High atom economy means less waste and is important for sustainable processes.

Key terms

Term	Meaning
Relative atomic mass	The weighted mean mass of an atom of an element relative to 1/12 of the mass of a carbon-12 atom.
Mole	The amount of substance containing the same number of particles as exactly 12.00 g of carbon-12.
Avogadro constant	6.02 x 10²³ mol⁻¹, the number of particles in one mole.
Empirical formula	The simplest whole-number ratio of atoms of each element in a compound.
Spectator ion	An ion present in a reaction mixture that does not take part in the reaction.
Water of crystallisation	Water molecules that are part of the structure of a crystal.
Atom economy	Mr of the desired product divided by the total Mr of the reactants, x 100.

Avoid these mistakes

Changing a formula to balance an equation — only coefficients may be changed.

Leaving spectator ions in an ionic equation — cancel them and balance for charge.

Using °C or cm³ in PV = nRT — convert to K and m³ first.

Confusing moles of molecules with moles of atoms — 1 mol of H₂O contains 3 mol of atoms.

Giving units for Ar or Mr — relative masses have no units.

Last updated: 13 October 2026