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4.2.2Chemical Reactions

Mole Calculations — AP CHEMISTRY Revision Notes & Topical Questions

5 min read 5 parts Lesson 15 of 18 Unit quiz

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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.
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Part 1 of 5The Mole Bridge

Because equations are written in moles, masses must be converted to moles and back again.

Molar mass
  • Molar mass The mass of one mole of a substance in g/mol is found by adding the atomic masses of all the atoms in its formula.
Mole ratio
  • Mole ratio The coefficients of a balanced equation give the ratio in which moles of substances react and form.
Mass to moles
  • Mass to moles Moles equal mass divided by molar mass, which puts quantities on a common basis.
Moles to mass
  • Moles to mass Multiplying moles by molar mass converts the answer back to grams.

Part 2 of 5Standard Procedure

1
Write a balanced equation for the reaction.
2
Convert the given mass of the known substance to moles with its molar mass.
3
Multiply by the mole ratio, wanted over given, taken from the coefficients.
4
Convert the moles of the wanted substance to mass, volume or particles as the question requires.
5
Round the answer to the correct number of significant figures and include units.

Part 3 of 5Extending the Idea

Extending the Idea
Limiting reactant
  • Limiting reactant The reactant that gives the smaller amount of product when each is converted separately, and it controls how much forms.
Percent yield
  • Percent yield Percent yield is (actual yield ÷ theoretical yield) × 100.
Gases and solutions
  • Gases and solutions Moles can also be found from PV = nRT or from M × V when substances are gases or solutions.
Particles
  • Particles One mole contains 6.022 × 10²³ particles, so moles convert to numbers of atoms or molecules.
ConversionTool
Mass ↔ molesMolar mass
Moles A ↔ moles BCoefficient ratio
Moles ↔ particles6.022 × 10²³ per mole
Moles ↔ solution volumeMolarity

Part 4 of 5Worked Example

Question In N₂(g) + 3H₂(g) → 2NH₃(g), what mass of ammonia forms from 10.0 g of hydrogen, assuming nitrogen is in excess?

  1. Molar mass of H₂ = 2.016 g/mol, so n(H₂) = 10.0 ÷ 2.016 = 4.960 mol.
  2. Mole ratio NH₃H₂ = 2 : 3, so n(NH₃) = 4.960 × 2/3 = 3.307 mol.
  3. Molar mass of NH₃ = 14.01 + 3(1.008) = 17.03 g/mol.
  4. Mass of NH₃ = 3.307 mol × 17.03 g/mol = 56.3 g.

Answer About 56.3 g of NH₃ forms.

Part 5 of 5Exam Checklist & Tips

4.2.2 – Key Points for the Exam

  • Mole The amount of substance containing 6.022 × 10²³ particles.
  • Molar mass The mass in grams of one mole of a substance.
  • Mole ratio The ratio of coefficients of two substances in a balanced equation.
  • Theoretical yield The maximum amount of product predicted by stoichiometry.

Common Mistakes

  • Using grams directly in a ratio from the coefficients without first converting to moles.
  • Using an unbalanced equation to find the mole ratio.
  • Placing the mole ratio upside down, so the wanted substance is not on top.
  • Using the molar mass of an atom instead of the whole molecule, for example 14.01 for N₂ instead of 28.02.

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