GCSE & A level chemical calculations notes and practise questions

% PURITY CALCULATIONS & ASSAY CALCULATIONS

Advanced UK A/AS level, IB and US grade 11-12 pre-university chemistry

14. Other GCSE chemical calculations % PURITY OF A PRODUCT from a chemical preparation reaction

Keywords: Quantitative chemistry calculations Help for problem solving in doing % percentage purity calculations. To 'assay' means to analyse a sample for its purity. This page includes of fully worked examples of percent purity calculations. How do you do assay calculations? Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses. These revision notes and practice questions on how to calculate percent purity and assay chemical calculations with worked examples are fully worked out below and should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses as well as Advanced A Level Chemistry courses.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

See also

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.2b atom economy calculations  *  14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

14.1 Percentage purity of a chemical reaction product

• % Purity is very important

  • e.g. for analytical standards in laboratories or pharmaceutical products where impurities could have dangerous side effects in a drug or medicine.

  • However in any chemical process it is almost impossible to get 100.00% purity and so samples are always analysed in industry to monitor the quality of the product.

  • The more a product is processed e.g. by distillation or crystallisation, the more costly the process, but the purer the product gets.

  • Somewhere there has to be a compromise, so it is important that before sale, the product is assayed or analysed as to its percentage purity.

  • It would not be acceptable e.g. in the pharmaceutical industry to manufacture a drug for treating us, with impurities in it, that may have harmful effects.

  • Similarly in fuels for road vehicles, which themselves have additives in to enhance engine performance, you wouldn't want other impurities that may cause engine damage.

  • You can apply the same sort of argument to thousands of domestic and industrial products from the chemical and pharmaceutical industries.

  • An assay is any procedure used to analyse and test for its purity of the % content of a specified component in a mixture of a % of an element or ion etc.

• % purity is the percentage of the material which is the actually desired chemical in a sample of it.

  	MASS of USEFUL PRODUCT
  PERCENT PURITY   =  100  x	-------------------------------------------------
  	in TOTAL MASS of SAMPLE

• Example 14.1 (Q1) Purity calculation

  • A 12.00g sample of a crystallised pharmaceutical product was found to contain 11.57g of the active drug.

  • Calculate the % purity of the sample of the drug.

  • % purity = actual amount of desired material x 100 / total amount of material

  • % purity = 11.57 x 100 / 12 = 96.4% (3 sf, 1dp)

  • -

• Example 14.1 (Q2) Purity calculation

  • Sodium chloride was prepared by neutralising sodium hydroxide solution with dilute hydrochloric acid. The solution was gently heated to evaporate most of the water and allow the salt to crystallise. The crystals were separated from any remaining solution and dried on a filter paper. However, the crystals are not necessarily completely dry.

  • The salt maybe required to be completely anhydrous, that is, not containing any water.

  • The prepared salt was analysed for water by heating a sample in an oven at 110^oC to measure the evaporation of any residual water.

  • The following results were obtained and from them calculate the % purity of the salt.

  • Mass of evaporating dish empty = 51.32g.

  • Mass of impure salt + dish = 56.47g

  • Mass of dish + salt after heating = 56.15g

  • Therefore the mass of original salt = 56.47 - 51.32 = 5.15g

  • and the mass of pure salt remaining = 56.15 - 51.32 = 4.83g

  • % salt purity = 4.83 x 100 / 5.15 = 93.8% ( (3 sf, 1dp)

  • -

• Example 14.1 (Q3) Purity calculation - an assay calculation is sketched out below for A Level students + link to others.

  • Titrations can be used to analyse the purity of a substance e.g. here an acid (aspirin) is titrated with standard sodium hydroxide solution of concentration 0.1000 mol dm^-3.

  • The aspirin is dissolved in ethanol solvent, diluted with deionised water and titrated with standardised 0.100 mol/dm³ sodium hydroxide solution using phenolphthalein indicator, the end-point is the first permanent pink colour.

  • An assay calculation is 'sketched out' below.

• See also ...

• Advanced AS Q17 & Q20 examples of purity-titration calculation based on an organic acid titrations

• Example 14.1 (4) Purity calculation - an assay calculation is sketched out below for A Level students + link to others.

  • Even at pre-A level you can do a simple titration and analyse an aspirin sample without using the mole concept in the calculation e.g. the above assay calculation could be presented via a reacting mass calculation as follows ...

  • 0.300g of aspirin was titrated with sodium hydroxide solution of concentration 4.00g/dm³.

  • If the aspirin required 16.45 cm³ of the NaOH(aq) to neutralise it, calculate the percent purity of the aspirin.

    • The simplified equation for the reaction is ...

    • C₆H₄(OCOCH₃)COOH + NaOH  ==> C₆H₄(OCOCH₃)COONa + H₂O

    • M_r(aspirin) = 180,  M_r(NaOH) = 40   (atomic masses: C = 12, H = 1, O = 16, Na = 23)

    • Therefore the reacting mass ratio is 180g aspirin reacts with 40g of sodium hydroxide.

    • The titration was 16.45 cm³, so, converting the cm³ to dm³,

    • the mass of NaOH used in the titration = 4.00 x 16.45/1000 = 0.0658g,

    • so we can scale this up to get the mass of aspirin titrated,

    • therefore the mass of aspirin titrated = 0.0658 x 180 / 40 = 0.296g

    • therefore the % purity = 100 x 0.296 / 0.300 = 98.7% (3 sf)

  • -

• -