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% PURITY CALCULATIONS

and ASSAY CALCULATIONS

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

study examples carefully14. Other GCSE chemical calculations % PURITY OF A PRODUCT from a chemical preparation reactionstudy examples carefully

soluble salt preparation from insoluble base-acid neutralisationQuantitative 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 (91) chemistry science courses as well as Advanced A Level Chemistry courses.

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study examples carefully14. Other GCSE chemical calculationsstudy examples carefully

14.1 % purity of a product  *  14.2a % reaction yield  *  14.2b atom economy  *  14.3 dilution of solutions

14.4 water of crystallisation calculation  *  14.5 how much reactant is needed? & limiting reactant calculations

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% (to 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 110oC 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% (to 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.

  • 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

      • Mr(aspirin) = 180,  Mr(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%

    • -

  • -

 

Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Calculating relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

  8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

  9. Moles and the molar volume of a gas, Avogadro's Law

  10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

  13. Electrolysis products calculations (negative cathode and positive anode products)

  14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials

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