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GCSE & A level Chemistry: Calculating percentage yield, why never 100% in practice

Check out what is available? Study the different examples then try the Quizzes!

Theoretical % yield calculations and

reasons why you never get 100% yield

See also ATOM ECONOMY notes too

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

study examples carefully14. Other GCSE chemical calculations


and why you can't get a 100% yield in practicestudy examples carefully

soluble salt preparation from insoluble base-acid neutralisationQuantitative chemistry calculations Help for problem solving in doing % yield calculations. How do you calculate % yield? How to calculate the percentage yield of a chemical reaction is explained with worked out examples. Explaining what we mean by % yield calculations based on theoretical yield versus actual yield. 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 do percentage yield chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (91) chemistry science courses.

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

See also

14.1 % purity of a product and assay calculations

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

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

14.2a Percentage yield of the product of a reaction

Even though no atoms are gained or lost in a chemical reaction (law of conservation of mass), unfortunately it is not always possible to obtain the calculated amount of a product (i.e. 100% yield) because the reaction may not go to completion because it may be reversible or some of the product may be lost when it is separated from the reaction mixture (eg via crystallisation or distillation) or some of the reactants may react in ways different to the expected reaction (side reactions).

The amount of a product you actually obtain is known as the yield.

When compared with the maximum theoretical amount calculated as a percentage, it is called the percentage yield.

You should be able to calculate the percentage yield of a product from the actual yield of a reaction or calculate the theoretical mass of a product from a given mass of reactant given the balanced equation for the reaction.

In industrial processes of the chemical industry, the reactions with highest yield are going to be the most economic and produce the least waste that has to be dealt with.

  • YIELD The actual yield is the mass of useful product you get from a chemical reaction and this actual yield can be compared with the maximum theoretical yield if everything could be done perfectly, which you can't!

  • The % yield of a reaction is the percentage of the product obtained compared to the theoretical maximum (predicted) yield calculated from the balanced equation.

    • You get the predicted maximum theoretical yield from a reacting mass calculation (see examples further down).

    • The comparison of the actual yield and the theoretical maximum yield can be expressed as the percentage yield.

                        ACTUAL YIELD (e.g. in grams, kg, tonnes)


      100 x ---------------------------------------------------------------------------------------------
                      PREDICTED theoretical YIELD (same mass units as above)
    • In carrying out a chemical preparation, the aim is to work carefully and recover as much of the desired reaction product as you can, and as pure as is possible and practicable.

      • Despite the law of conservation of mass, i.e. no atoms lost or gained, in real chemical preparations things cannot work out completely according to chemical theory, often for quite simple, physical or sometimes chemical reasons,

      • and it doesn't matter if its a small scale school laboratory preparation or a large scale industrial manufacturing process, the percent yield is never 100%.

      • You may need to use the rearranged equation i.e. change of equation subject ...

      • e.g. 100 x Actual Yield = % Yield X Theoretical Yield

      • so, Actual Yield (mass) = % Yield X Theoretical Yield (mass) / 100

    • In industry, you want the highest possible yield to reduce costs and have less uneconomic waste.

  • REASONS why you never get 100 percent yield of the desired product in chemical reaction preparations

  • LOSSES So, in any chemical process, it is almost impossible to get 100% of the product because of many reasons: Three reasons why you do not get a 100% yield in a chemical reaction are explained and discussed.

    1. The reaction might not be 100% completed because it is reversible reaction and an equilibrium is established (note  thesign in the equation below.

      • Both reactants and products co-exist in the same reaction mixtures (solutions or gases) i.e. the reaction can never go to completion.

        • A good example of this is the preparation of an ester, you only get 2/3rds conversion for this incomplete reaction, and then there will be other losses in isolating and purifying the product.

        • ethanoic acid + ethanol ethyl ethanoate + water
        • + + H2O
        • For more details see Ester Preparation
      • In the Haber Synthesis of ammonia the conversion of hydrogen and nitrogen to ammonia is only about 6-8%.
        • N2(g) + 3H2(g) (c) doc b 2NH3(g)
        • BUT, it isn't all bad news, the ammonia is quite easily condensed out from the reaction mixture and the nitrogen and hydrogen recycled through the reactor, so there is very little waste.
      • -
    2. You always get losses of the desired product as it is separated from the reaction mixture by filtration, distillation, crystallisation or whatever method is required e.g.

      • Bits of solid or droplets are left behind on the sides of the apparatus or reactor vessel e.g. in the reaction flask, filter funnel and paper when recovering the product from the reaction mixture or transferring a liquid from one container to another.

      • Small amounts of liquid will be left in distillation units or solid particles on the surface of filtration units.

      • In fact, whenever you have to manipulate or transfer the product in some way, there are bound to be residual losses somewhere in the laboratory apparatus or a full-scale chemical plant.

      • If the product is a volatile liquid, there will be losses due to evaporation.

      • You cannot avoid losing traces of product in all stages of the manufacturing process.

      • See Four techniques used in a particular and separation and purification procedure,

      • DISTILLATION  and  Separating funnel, solvent extraction, centrifuging

      • where these sort of losses may be encountered.

      • -

    3. Some of the reactants may react in another way to give a different product to the one you want (so-called by-products).

      • By-products are very common in organic chemistry reactions

      • (i) A + B ==> C + D

        • (i) The main reaction to give the main desired products C or D, or both.

      • (ii) A + B ==> E + F

        • A con-current reaction, maybe just involving a few % of the reactants to give the minor, and often undesirable, by-products of E + F.

        • Sometimes the by-products can be separated as a useful product and sold to help the economics of the chemical process overall.

      • -

  • The aim is to work as carefully as possible and recover as much of the desired reaction product, and as pure as is possible and practicable

    • If a chemical reaction gives a low percentage yield of useful product, research would be undertaken to find ways the reaction can been improved to increase the yield of useful product.
    • It might be possible to find another synthetic route to produce a particular chemical that gives a higher percentage yield and less waste (see also Atom Economy).
  • % yield = actual amount of desired chemical obtained x 100 / maximum theoretical amount that could be formed

    • If the reaction doesn't work the yield is zero or 0%.

    • If the reaction works perfectly and you obtain all the product, the yield is 100%, BUT this never happens in reality (as already discussed above).

    • The theoretical yield can be calculated from the balanced equation by doing a reacting mass ratio calculation.

  • Yield and industrial production

    • The higher the yield of a reaction, the more economic is the process.

    • There will be less waste to deal with and dispose of, which involves extra cost.

    • Waste is of no commercial value and may be harmful to people and the environment.

    • High yields means less energy used, saving money.

    • Research chemists in the chemical industry are always looking for the most efficient (cost effective) of making a particular product and the main criteria being ...

      • A high yield reaction, particularly if the raw materials are expensive, and the resource may not be infinite!

      • that goes as fast as possible - a good economic rate,

      • all of the products are commercially viable, so can even by-products be converted to some saleable chemical as well as the main desired product.

  • % yield calculation Example 14.2a (1)

    • CuO  + H2  ==>  Cu  +  H2O
    • Copper(II) oxide can be reduced to copper by heating the oxide carefully in a stream of oxygen.
    • 5.0 g of copper(II) oxide where heated in oxygen until there appeared to be no more change in colour.
    • Any unreacted copper(II) oxide was dissolved in dilute sulfuric acid and the copper filtered off, dried and weighed.
    • Using the atomic masses: Cu = 63.5,  H = 1,  O =16
    • Formula masses: CuO = 79.5,  H2 = 2,  Cu = 63.5,  H2O = 18
    • (a) Calculate the maximum quantity of copper you could obtain.
      • The relevant ratio is CuO ==> CuO
      • which is 79.5 ==> 63.5
      • Therefore scaling down to 5 g of the oxide ...
      • the maximum yield of Cu = 5 x 63.5/79.5 = 3.99 g (2 dp, 3 sf)
      • -
    • (b) If 3.91 g of copper was actually obtained, calculate the percentage yield of the reaction.
      • % yield = 100 x actual yield / theoretical maximum yield
      • % yield = 100 x 3.91 / 3.99
      • % yield of Cu = 98.0% (3 sf, 1 dp)
      • -
  • % yield calculation Example 14.2a (2)

    • Magnesium metal dissolves in hydrochloric acid to form magnesium chloride.

    • Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)

    • Atomic masses : Mg = 24 and Cl = 35.5, and formula mass MgCl2 = 24 + (2 x 35.5) = 95

    • (a) What is the maximum theoretical mass of anhydrous magnesium chloride which can be made from 12g of magnesium?

      • Reacting mass ratio calculation from the balanced equation:

      • 1 Mg ==> 1 MgCl2, so 24g ==> 95g or 12g ==> 47.5g MgCl2

    • (b) If only 47.0g of purified anhydrous magnesium chloride was obtained after crystallising the salt from the solution and heating it to drive off the water of crystallisation, what is the % yield from the salt preparation?

      • % yield = actual amount obtained x 100 / maximum theoretical amount possible

      • % yield = 47.0 x 100 / 47.5 = 98.9% (to 1dp)

    • More examples of % yield and atom economy calculations in section 6.

    • -

  • % yield calculation Example 14.2a (3)

    • 2.8g of iron was heated with excess sulphur to form iron sulphide.

    • Fe + S ==> FeS

    • The excess sulphur was dissolved in a solvent and the iron sulphide filtered off, washed with clean solvent and dried.

    • If 4.1g of purified iron sulphide was finally obtained, what was the % yield of the reaction?

    • 1st a reacting mass calculation of the maximum amount of FeS that can be formed:

      • Relative atomic/formula masses: Fe =56, FeS = 56 + 32 = 88

      • This means 56g Fe ==> 88g FeS, or by ratio, 2.8g Fe ==> 4.4g FeS

      • because 2.8 is 1/20th of 56, so theoretically you can get 1/20th of 88g of FeS or 4.4g.

    • 2nd the % yield calculation itself.

      • % yield = actual amount obtained x 100 / maximum theoretical amount possible

      • % yield = 4.1 x 100 / 4.4 = 93.2% (to 1dp)

    • More examples of % yield and atom economy calculations in section 6.

    • -

  • % yield calculation Example 14.2a (4)

    • (a) Theoretically how much iron can be obtained from 1000 tonne of pure haematite ore, formula Fe2O3 in a blast furnace?

    • If we assume the iron(III) oxide ore (haematite) is reduced by carbon monoxide, the equation is:

    • Fe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g)

    • (atomic masses: Fe = 56, O = 16)

    • For every Fe2O3 ==> 2Fe can be extracted, formula mass of ore = (2 x 56) + (3 x 56) = 160

    • Therefore reacting mass ratio is: 160 ==> 112 (from 2 x 56)

    • so, solving the ratio, 1000 ==> 112 / 160 = 700 tonne copper = max. can be extracted

  • % yield calculation Example 14.2a (5)

    • Given the atomic masses: Mg = 24 and O = 16,

    • and the reaction between magnesium to form magnesium oxide is given by the symbol equation

    • 2Mg(s) + O2(g) ==> 2MgO(s)

    • (a) What mass of magnesium oxide can be made from 1g of magnesium?

      • 2Mg ==> 2MgO

      • in terms of reacting masses (2 x 24) ==> {2 x (24 +16)}

      • so 48g Mg ==> 80g MgO (or 24g ==> 40g, its all the same)

      • therefore solving the ratio

      • 1g Mg ==> w g MgO, using the ratio 48 : 80

      • w = 1 x 80 / 48 = 1.67g MgO

    • (b) Suppose the % yield in the reaction is 80%. That is only 80% of the magnesium oxide formed is actually recovered as useful product. How much magnesium needs to be burned to make 30g of magnesium oxide?

      • This is a bit tricky and needs to done in two stages and can be set out in several ways.

      • Now 48g Mg ==> 80g MgO (or any ratio mentioned above)

      • so y g Mg ==> 30g MgO

      • y = 30 x 48 / 80 = 18g Mg

      • BUT you only get back 80% of the MgO formed,

      • so therefore you need to take more of the magnesium than theoretically calculated above.

      • Therefore for practical purposes you need to take, NOT 18g Mg, BUT ...

      • ... since you only get 80/100 ths of the product ...

      • ... you need to use 100/80 ths of the reactants in the first place

      • therefore Mg needed = 18g x 100 / 80 = 22.5g Mg

      • CHECK: reacting mass calculation + % yield calculation CHECK:

        • 22.5 Mg ==> z MgO, z = 22.5 x 80 / 48 = 37.5g MgO,

        • but you only get 80% of this,

        • so you actually get 37.5 x 80 / 100 = 30g

        • This means in principle that if you only get x% yield ...

        • ... you need to take 100/x quantities of reactants to compensate for the losses.

    • -

  • Below is an example of a more advanced level percentage yield calculation

  • Rather than here, I've added more % yield calculations to the Reacting mass ratio calculations of reactants and products from equations page.


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



  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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