AQA GCSE Chemistry Flashcards

Titration Calculations

  • First Step in Titration: Write a balanced chemical equation to determine the reacting ratio.
  • Next Step: Calculate moles of the known substance using the formula:
    • moles=concentration×volume (in dm3)\text{moles} = \text{concentration} \times \text{volume (in dm}^3)
  • After Finding Moles: Use the mole ratio from the balanced equation to find moles of the unknown substance.
  • Finding Concentration: Once moles and volume are known, calculate concentration with:
    • concentration=molesvolume (in dm3)\text{concentration} = \frac{\text{moles}}{\text{volume (in dm}^3)}
  • Equation Triangle: The relationship between moles, concentration, and volume can be visualized as:
    • moles−−−−−−−−−−−−−−−−−−−−−concentration×volume\text{moles} \newline --------------------- \newline \text{concentration} \times \text{volume}
  • Conversion from cm³ to dm³: Divide by 1000 (Example: 25.0 cm³ = 0.0250 dm³).

Relative Formula Mass

  • Calculating Relative Formula Mass (Mr): Sum the relative atomic masses (Ar) of all atoms in the formula.

Percentage Mass of Elements

  • % Mass Calculation: The percentage mass of an element in a compound is given by:
    • % mass=(Ar×number of atoms of elementMr of compound)×100\% \text{ mass} = \left( \frac{\text{Ar} \times \text{number of atoms of element}}{\text{Mr of compound}} \right) \times 100

Moles and Mass Calculations

  • Calculating Moles from Mass: Formula:
    • moles=massMr\text{moles} = \frac{\text{mass}}{\text{Mr}}
  • Calculating Mass from Moles: Formula:
    • mass=moles×Mr\text{mass} = \text{moles} \times \text{Mr}

Energy Changes in Reactions

  • Calculating Energy Change: The energy change can be determined using bond energies with:
    • energy change=energy needed to break bonds−energy released making bonds\text{energy change} = \text{energy needed to break bonds} - \text{energy released making bonds}

Reaction Rates

  • Rate of Reaction Equation: Two forms to calculate rate are:
    • rate=amount of reactant usedtime\text{rate} = \frac{\text{amount of reactant used}}{\text{time}} or rate=amount of product formedtime\text{rate} = \frac{\text{amount of product formed}}{\text{time}}
  • Volume of Gas at Room Temperature: Volume can be calculated using:
    • volume=moles×24 (in dm3)\text{volume} = \text{moles} \times 24 \text{ (in dm}^3)
  • Molar Volume of Gas: 1 mole of gas occupies 24 dm³ at room temperature and pressure.

Electrochemistry

  • Half-Equations:
    • Reduction of Cu²⁺: Cu2++2e−→Cu\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}
    • Oxidation of Cl⁻: 2Cl−→Cl2+2e−2\text{Cl}^{-} \rightarrow \text{Cl}_{2} + 2e^{-}
  • Voltage of Chemical Cell: Influenced by the reactivity difference between metals used in the electrodes.

Atom Economy and Yield

  • Atom Economy Calculation: Given by:
    • atom economy=(Mr of desired productsum of Mr of all reactants)×100\text{atom economy} = \left( \frac{\text{Mr of desired product}}{\text{sum of Mr of all reactants}} \right) \times 100
  • Percentage Yield Calculation: Formulated as:
    • % yield=(actual yieldtheoretical yield)×100\% \text{ yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100