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Atom Economy (Atom Utilisation)

Key Concepts

  • One of the key concepts of green chemistry is to reduce waste.

  • Waste can be reduced by designing chemical processes to incorporate the maximum amount of all the raw materials into the final product.

  • The atom economy (atom utilisation) of a chemical reaction is a measure of the percentage of the starting materials that actually end up in useful products.

  • For the general chemical reaction:

    reactants desired product + waste products

    The atom economy can be calculated in either of two ways:*


  1. % Atom economy =                   mass of desired product                  
    mass of desired product + mass of waste products
    x 100


  2. % Atom economy =   mass of desired product  
    total mass of all reactants
    x 100

    Equation (I) is identical to equation (II) because by the Law of Mass Conservation:
    total mass of all reactants = mass of desired product + mass of waste products
  • The greater the value of the %atom economy, the less the amount of waste product produced.

*% Atom economy is not the same as % yield

Example: Production of Bromoethane

  1. Bromoethane (desired product) can be produced from the reactants ethene (ethylene) and hydrogen bromide in an addition reaction.
    Calculate the % atom economy for this reaction.

    (i) Write the word equation for the reaction :
    reactants desired product + waste product
    ethene + hydrogen bromide bromoethane
    Note: no waste product is formed in this reaction.
     
    (ii) Write the balanced chemical equation for the reaction:
    C2H4 + HBr C2H5Br
     
    (iii) Assuming the reaction goes to completion, if 1 mole of each reactant is used, then 1 mole of the product will be produced
    1 mole + 1 mole 1 mole
     
    (iv) Calculate the mass of each reactant and product assuming there is a 100% yield
    (mass = moles x molar mass)
    C2H4
    moles = 1 mol
    molar mass =
        (2x12+4x1)
        = 28 g/mol
    mass =1x28 = 28 g
      HBr
    moles = 1 mol
    molar mass =
        (1+79.9)
        = 80.9 g/mol
    mass =1x80.9 = 80.9 g
      C2H5Br
    moles = 1 mol
    molar mass =
        (2x12+5x1+79.9)
        = 108.9 g/mol
    mass =1x108.9 = 108.9 g
     
     
    (v) Calculate the % atom economy:
      using equation (I)
        % Atom economy =                   mass of desired product                  
    mass of desired product + mass of waste products
    x 100
        % Atom economy =   108.9 g  
    108.9 g
    x 100 = 100%
      or using equation (II)
        % Atom economy =   mass of desired product  
    total mass of all reactants
    x 100
        % Atom economy =     108.9 g    
    28 g + 80.9 g
    x 100 = 100%
     
    (vi) Conclusion

    If no waste product is formed, the reaction goes to completion and the yield of product is 100%, then the % atom economy is 100%.

    All the atoms present in the reacting molecules are converted into the desired product.

  2. Bromoethane (desired product) and hydrogen bromide (waste product) can be prepared using the reactants ethane and bromine in a substitution reaction.
    Calculate the % atom economy for this reaction.

    (i) Write the word equation for the reaction :
    reactants desired product + waste product
    ethane + bromine bromoethane + hydrogen bromide
     
    (ii) Write the balanced chemical equation for the reaction:
    C2H6 + Br2 C2H5Br + HBr
     
    (iii) Assuming the reaction goes to completion, if 1 mole of each reactant is used, then 1 mole of the product will be produced
    1 mole + 1 mole 1 mole + 1 mole
     
    (iv) Calculate the mass of each reactant and product assuming there is a 100% yield
    (mass = moles x molar mass)
    C2H6
    moles = 1 mol
    molar mass =
        (2x12+6x1)
        = 30 g/mol
    mass =1x30
        = 30 g
      Br2
    moles = 1 mol
    molar mass =
        (2x79.9)
        = 159.8 g/mol
    mass =1x159.8
        = 159.8 g
      C2H5Br
    moles = 1 mol
    molar mass =
        (2x12+5x1+79.9)
        = 108.9 g/mol
    mass =1x108.9
        = 108.9 g
      HBr
    moles = 1
    molar mass =
        (1+79.9)
        = 80.9 g/mol
    mass =1x80.9
        = 80.9 g
     
    (v) Calculate the % atom economy:
      using equation (I)
        % Atom economy =                   mass of desired product                  
    mass of desired product + mass of waste products
    x 100
        % Atom economy =       108.9 g      
    108.9 g + 80.9 g
    x 100 = 57.4%
      or using equation (II)
        % Atom economy =   mass of desired product  
    total mass of all reactants
    x 100
        % Atom economy =       108.9 g      
    30 g + 159.8 g
    x 100 = 57.4%
     
    (vi) Conclusion

    If one or more waste products are formed, the % atom economy is less than 100%.

    Some of the atoms present in the reacting molecules are converted into the desired product, the rest go into the formation of waste products.

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