Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

4.5Chemical Reactions

Stoichiometry — AP CHEMISTRY Revision Notes & Topical Questions

5 min read 8 parts Lesson 8 of 12 1 quiz

📚 AP CHEMISTRY Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read
Lesson summary

4.5 Stoichiometry — Notes — AP CHEMISTRY notes: Use balanced equations, moles, gas laws and molarity to calculate amounts in reactions.Atoms are conserved, so product amounts can be calculated from reactant amounts and vice versa.Coefficients in a balanced equation give the mole …

  • What the symbols mean
  • Limiting reactant
  • Theoretical yield
  • Percent yield
  • Mole ratio
  • Ideal gas law

Part 1 of 6Principles

Principles
Atoms are conserved, so product amounts can be calculated from reactant amounts and vice versa.
Coefficients in a balanced equation give the mole ratio in which substances react and form, not a mass ratio.
Mass is conserved because atoms are rearranged, so an equation must be balanced before use.

Part 2 of 6Amount equations

Amount equations
Solids
  • Solids n = m/M, with n in mol, m in g and M in g/mol.
    What the symbols mean: n = amount of substance (mol) · m = mass (g) · M or Mr = molar mass (g mol−1, numerically equal to Mr)
Particles
  • N = n × NA, with NA = 6.022 × 1023 mol−1.
    What the symbols mean: N = number of particles · n = amount (mol) · NA = Avogadro constant (6.02 × 1023 mol−1)
Solutions
  • Solutions n = M × V, with molarity in mol/L and volume in L.
Gases
  • Gases n = PV/RT, with R = 0.08206 L atm mol−1 K−1 and T in kelvin.
    • (K = °C + 273.15)

Part 3 of 6Calculation route

1
Convert the given quantity to moles of the given substance.
2
Use the mole ratio from the balanced equation.
3
Obtain moles of the wanted substance.
4
Convert to the wanted quantity.
StartConvert withThenEnd
Mass of A÷ MA× ratio× MB gives mass of B
Volume of solution A× M× ratio÷ M gives volume of B
Gas volume at STP÷ 22.4 L× ratioMoles of B
Particles÷ NA× ratioMoles of B

Part 4 of 6Limiting reactant and yield

Limiting reactant
Limiting reactant the reactant completely consumed first, determining the maximum product.
  • Compare moles available ÷ coefficient for each reactant; the smaller value is limiting.
Theoretical yield
Theoretical yield the maximum mass of product calculated from the limiting reactant.
Percent yield
(actual yield ÷ theoretical yield) × 100.
A percent yield above 100% suggests an impure or wet product.

Part 5 of 6Particle diagrams

  • Counting particles before and after a reaction shows which reactant runs out and how many of each remain.

Part 6 of 6More from the specification

More from the specification
Significant figures
  • The digits in a value that are reliable and carry meaning.
Carbon dioxide
  • A gas that turns limewater cloudy.
Analyte
  • Solution whose amount is being determined.
Concentration
  • Moles ÷ volume in dm³, or g/dm³.
Solute
  • A substance that dissolves in a solvent.
Hydrogen
  • A gas that gives a squeaky pop with a lighted splint.
Titration
  • Method for finding the volume of one solution that reacts exactly with another.
Titrant
  • Solution of known concentration added from the burette.
Oxygen
  • A gas that relights a glowing splint.

Key terms

TermMeaning
Mole ratioRatio of coefficients in a balanced equation.
Limiting reactantReactant that runs out first.
Theoretical yieldMaximum product predicted from the limiting reactant.
Percent yieldActual yield divided by theoretical yield, times 100.
MolarityMoles of solute per litre of solution.
Ideal gas lawPV = nRT.

Avoid these mistakes

1
Using masses rather than moles in the mole ratio
2
Reading coefficients as masses
3
Forgetting to convert mL to L or °C to K
4
Taking the reactant with the smaller mass as limiting without comparing moles

Last updated: