OCR AS Chemistry

Revision Notes

2.2.5 Percentage Yield & Atom Economy

Test Yourself

Percentage Yield Calculations

Percentage yield

  • In a lot of reactions, not all reactants react to form products which can be due to several factors:
    • Other reactions take place simultaneously
    • The reaction does not go to completion
    • Products are lost during separation and purification

  • The percentage yield shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get:

p e r c e n t a g e space y i e l d space equals space fraction numerator a c t u a l space y i e l d over denominator t h e o r e t i c a l space y i e l d end fraction space cross times space 100

    • The actual yield is the number of moles or mass of product obtained experimentally
    • The theoretical yield is the number of moles or mass obtained by a reacting mass calculation

Worked example

In an experiment to displace copper from copper(II) sulfate, 6.5 g of zinc was added to an excess of copper(II) sulfate solution. The resulting copper was filtered off, washed and dried. The mass of copper obtained was 4.8 g.

Calculate the percentage yield of copper.

Answer:

Step 1: The balanced symbol equation is:

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)


Step 2: Calculate the amount of zinc reacted in moles

n u m b e r space o f space m o l e s space equals space fraction numerator 6.5 space g over denominator 65.4 space g space m o l to the power of negative 1 end exponent end fraction space equals space 0.10 space m o l

Step 3: Calculate the maximum amount of copper that could be formed from the molar ratio:
   

Since the ratio of Zn(s) to Cu(s) is 1:1 a maximum of 0.10 moles can be produced


Step 4: Calculate the maximum mass of copper that could be formed (theoretical yield)

mass =   mol  x  M
mass =   0.10 mol x 63.55 g mol-1
mass =   6.4 g (2 sig figs)


Step 5: Calculate the percentage yield of copper

p e r c e n t a g e space y i e l d space equals space fraction numerator 4.8 space g over denominator 6.4 space g end fraction space cross times space 100 space equals space 75 percent sign

Atom Economy Calculations

  • The atom economy of a reaction shows how many of the atoms used in the reaction become the desired product
    • The rest of the atoms or mass is wasted

  • It is found directly from the balanced equation by calculating the Mr of the desired product

A t o m space e c o n o m y space equals space fraction numerator m o l e c u l a r space m a s s space o f space d e s i r e d space p r o d u c t over denominator s u m space o f space m o l e c u l a r space m a s s e s space o f space A L L space r e a c t a n t s end fraction space cross times space 100

  • In addition reactions, the atom economy will always be 100% because all of the atoms are used to make the desired product
    • Whenever there is only one product, the atom economy will always be 100%
    • For example, in the reaction between ethene and bromine:

CH2=CH2 + Br2 → CH2BrCH2Br

  • The atom economy could also be calculated using mass, instead or Mr
    • In this case, you would divide the mass of the desired product formed by the total mass of all reactants, and then multiply by 100
  • Questions about atom economy often asked in qualitative or quantitative terms

Worked example

Qualitative atom economy

Ethanol can be produced by various reactions, such as:

Hydration of ethene:       C2H4 + H2O → C2H5OH
Substitution of bromoethane:      C2H5Br + NaOH → C2H5OH + NaBr


Explain which reaction has a higher atom economy.

Answer

Hydration of ethene has a higher atom economy (of 100%) because all of the reactants are converted into products, whereas the substitution of bromoethane produces NaBr as a waste product

Worked example

Quantitative atom economy

The blast furnace uses carbon monoxide to reduce iron(III) oxide to iron.

Fe2O3 + 3CO → 2Fe + 3CO2


Calculate the atom economy for this reaction, assuming that iron is the desired product.

(ArMr data: Fe2O3 = 159.6, CO = 28.0, Fe = 55.8, CO2 = 44.0)

Answer

Step 1: Write the equation:
                             A t o m space e c o n o m y space equals space fraction numerator m o l e c u l a r space m a s s space o f space d e s i r e d space p r o d u c t over denominator s u m space o f space m o l e c u l a r space m a s s e s space o f space A L L space r e a c t a n t s end fraction space cross times space 100
Step 2:
Substitute values and evaluate:

A t o m space e c o n o m y space equals space fraction numerator 2 space cross times space 55.8 over denominator 159.6 space plus space left parenthesis 3 space cross times space 28.0 right parenthesis end fraction space cross times space 100 space equals space 45.8 percent sign

Exam Tip

Careful: Sometimes a question may ask you to show your working when calculating atom economy.

In this case, even if it is an addition reaction and it is obvious that the atom economy is 100%, you will still need to show your working.

Benefits of High Atom Economy

  • Chemists use percentage yield as one possible measure of how efficient a reaction is
    • A high percentage yield suggests that a process is effective at converting reactants into products
  • The estimated percentage yield for a single run of the Haber Process is around 15%
    • This is a compromise due to the cost and safety of the required conditions against the overall rate of ammonia production
    • Any unreacted materials are also recycled so it is estimated that the percentage conversion of all reactants to products is around 97%
  • Whilst a high percentage yield can be good for profits, it does not account for any waste products
    • A reaction can have a high percentage yield but low atom economy which essentially means that more waste products are produced
  • Atom economy is a measure of the percentage of reactants that become useful products and is calculated by:
P e r c e n t a g e space a t o m space e c o n o m y space equals space fraction numerator m a s s space o f space d e s i r e d space p r o d u c t over denominator t o t a l space m a s s space o f space A L L space r e a c t a n t s end fraction space cross times space 100

Atom Economy and Green Chemistry

  • Chemists will often have several choices of reaching a target molecule and those choices need to take into the principles of Green Chemistry
The Twelve Principles of green chemistry, downloadable AS & A Level Chemistry revision notes

The twelve principles of green chemistry

  • By choosing a reaction pathway that has fewer steps, you can prevent waste and reduce energy demands which is better for the environment
    • This also reduces production costs
  • One of the key ideas behind Green Chemistry is to find reaction pathways with high percentage yield and high atom economy
  • By analysing the atom economy of each step, you can select reactions that give a higher atom economy and / or select other reactions to reduce the number of steps involved in a reaction pathway
    • For example, the synthesis of ibuprofen that was patented by Boots in the 1960's was a six-step synthesis
    • Even if each step has an atom economy of 90%, a six-step synthesis would have an overall atom economy of 53%
    • The modern production of ibuprofen is a three-step synthesis, which with the same assumptions as before, gives an overall atom economy of 73%  
  • Higher atom economy means that there is less waste produced
    • This can be considered environmentally friendly even though it may not influence the reaction conditions
    • It also means that reactions are more sustainable and often use less natural / finite resource

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