Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

1.6.5Physical Chemistry

Amount of Substance Calculations — Edexcel AS Chemistry Revision Notes & Topical Questions

5 min read 6 parts Lesson 29 of 42

📚 Edexcel AS Chemistry Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
0%read

Section 1 of 6The Three Core Equations

Solid or liquid
  • Solid or liquid n = m ÷ M, where m is mass in grams and M is molar mass in g mol⁻¹.
Solution
  • Solution n = c × V, where c is concentration in mol dm⁻³ and V is volume in dm³.
Gas at RTP
  • Gas at RTP n = V ÷ 24.0, where V is the gas volume in dm³ at room temperature and pressure.
Choosing the equation
  • Choosing the equation Pick the one that matches the data supplied, then use the balanced equation to link different substances.

Section 2 of 6Concentration

Concentration
Moles per volume
  • Moles per volume Concentration is the amount of solute dissolved in one dm³ of solution.
Mass concentration
  • Mass concentration Sometimes given in g dm⁻³; divide by the molar mass to convert to mol dm⁻³.
Dilution
  • Dilution The amount of solute does not change on dilution, so c₁V₁ = c₂V₂.
Unit conversion
  • Unit conversion 1 dm³ equals 1000 cm³, so divide cm³ by 1000 to get dm³.
Volume in cm³Volume in dm³
25.00.0250
2500.250
10001.00

Section 3 of 6Multi-Step Strategy

1
Write the balanced equation and underline the substances that appear in the question.
2
Convert the given data for the known substance to moles.
3
Apply the mole ratio from the equation to find moles of the target substance.
4
Convert moles to the quantity requested, with correct units.
5
Round to the same number of significant figures as the least precise data.

Section 4 of 6Percentage Composition and Purity

Percentage by mass
  • Percentage by mass Mass of element ÷ total mass of compound × 100.
Purity
  • Purity A sample is rarely 100% pure, so only the pure fraction takes part in the reaction.
Water of crystallisation
  • Water of crystallisation Hydrated salts contain water molecules in the lattice, which must be included in the molar mass.
Checking
  • Checking Percentages of all elements in a compound should add up to about 100%.

Section 5 of 6Worked Example

Question

  • 4.00 g of sodium hydroxide, NaOH, is dissolved in water and made up to 250 cm³.
  • Calculate the concentration in mol dm⁻³.
  • (Ar Na = 23.0, O = 16.0, H = 1.0)

  1. Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹.
  2. Moles of NaOH = 4.00 ÷ 40.0 = 0.100 mol.
  3. Volume = 250 cm³ = 0.250 dm³.
  4. Concentration = 0.100 ÷ 0.250 = 0.400 mol dm⁻³.

Answer 0.400 mol dm⁻³

Section 6 of 6Exam Checklist & Tips

1.6.5 – Key Points for the Exam

  • Concentration Amount of solute per dm³ of solution.
  • Molar gas volume Volume of one mole of any gas, 24.0 dm³ at RTP.
  • Dilution Adding solvent so that concentration decreases while moles of solute stay constant.
  • Aqueous solution A solution in which water is the solvent.

Common Mistakes

  • Using cm³ directly in n = c × V without converting.
  • Forgetting that molar gas volume applies only at the stated conditions.
  • Not using the mole ratio when two substances have different coefficients.
  • Quoting concentration in g dm⁻³ when the question asks for mol dm⁻³.

Last updated:

\n