(Higher tier) Concentration in mol/dm³ = amount in mol ÷ volume in dm³. Convert cm³ to dm³ by dividing by 1000 before using c = n/V; rearrange to n = cV or V = n/c.
(Higher tier) Concentration in g/dm³ = concentration in mol/dm³ × relative formula mass. Divide a mass concentration by the relative formula mass to obtain molar concentration.
A titration finds the volume of one solution needed to react exactly with a measured volume of another. Use a balanced equation to establish the mole ratio, which is not always 1:1.
For the core practical, use a pipette and filler to transfer a fixed volume of sodium hydroxide of unknown concentration into a conical flask. Add a few drops of a suitable indicator; hydrochloric acid of known concentration goes in the burette.Read the meniscus at eye level and swirl the flask near the endpoint.
Rinse the burette with the solution it will contain, and rinse the pipette with the solution it will measure. Rinse the conical flask with distilled water; extra water there changes dilution but not the moles of alkali already pipetted.
Fill the burette tip and remove air bubbles. Remove the filling funnel before measuring. Read the bottom of a colourless meniscus at eye level and record initial and final readings; titre = final reading − initial reading.
Place the flask on a white tile, swirl as acid is added and add acid dropwise near the endpoint. The endpoint is the indicator’s lasting colour change; methyl orange or phenolphthalein can be suitable, but universal indicator gives too gradual a change.
Do a rough titration, then repeat accurately to obtain close, concordant titres using the tolerance specified by the teacher or question. Calculate a mean from the concordant accurate results, excluding the rough trial.
(Higher tier) Find moles of the known solution with cV, use the equation ratio to find moles of the unknown, then divide by the unknown solution’s volume in dm³. Carry units through each step and round only the final answer.
Wear eye protection and use a pipette filler, never mouth pipetting. Handle acid and alkali according to the risk assessment; wash spills as instructed.
Yield, atom economy and choosing pathways
Theoretical yield is the maximum product predicted from the balanced equation and limiting reactant. Actual yield is what is collected experimentally; percentage yield = actual yield ÷ theoretical yield × 100.
Yield can be reduced by incomplete reaction, unwanted side reactions and losses during separation or transfer. A measured yield above 100% suggests wet or impure product or measurement/calculation error, not creation of extra atoms.
Atom economy is the percentage of the reaction’s product mass that is the desired product. Calculate it as: total Mr of desired products ÷ total Mr of all products × 100. Multiply each Mr by its number in the balanced equation before adding.
Because mass is conserved, the denominator can also be the total relative formula mass of the reactants with their coefficients. Atom economy describes the reaction equation, while yield describes practical success; they are different quantities.Include balanced-equation coefficients in atom-economy totals.
An addition reaction with a single product can have 100% atom economy even if its experimental yield is low. A reaction giving several products can have high yield but low atom economy for the chosen product.
(Higher tier) When choosing how to make a product, compare atom economy, yield, rate, equilibrium position, energy use, raw-material costs and the usefulness or hazards of the other products (by-products). Lower atom economy may be acceptable if those other products can also be used.
(Higher tier) At room temperature and pressure, one mole of gas occupies about 24 dm³ for GCSE calculations when specified. Use V = n × molar volume with matching volume units; detailed gas-reaction calculations continue in SC15.
Solution concentrations and titration
Titration calculation — A 25.0 cm³ alkali sample reacts with 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. HCl + NaOH → NaCl + H₂O gives a 1:1 mole ratio. Acid volume is 0.0200 dm³, so acid amount = 0.100 × 0.0200 = 0.00200 mol. Alkali amount is also 0.00200 mol, so its concentration is 0.00200/0.0250 = 0.0800 mol/dm³. Using 25 instead of 0.0250 would give a thousandfold error. A different balanced equation may require a different mole ratio.
Yield, atom economy and choosing pathways
Two product percentages — For CaCO₃ → CaO + CO₂, using relative masses Ca = 40, C = 12 and O = 16, 100 g of pure calcium carbonate can theoretically form 56 g calcium oxide. If 45 g is collected, percentage yield = 45/56 × 100 ≈ 80.4%. Atom economy for calcium oxide is 56/(56 + 44) × 100 = 56%. Yield compares actual with theoretical collection; atom economy compares the desired product with all products in the equation. Neither percentage measures the other's quantity.
Mind map
Use the branches to recall the ideas and explain their connections. Check the revision notes for the full detail.
SC14 · Overview
View SC14 · Overview mind mapOpen the full-size map to zoom. Use the printable view for A4 landscape or enlarge to A3.
Mr Kidman's Classroom
