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2.2.9 Atom Economy page 164-165

2.2.9 Atom Economy page 164-165 • Define the atom economy of a reaction and describe the benefits of developing chemical processes with a high atom economy. • Explain that addition reactions have an atom economy of 100%, whereas substitution reactions are less efficient.

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2.2.9 Atom Economy page 164-165

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  1. 2.2.9 Atom Economy page 164-165 • Define the atom economy of a reaction and describe the benefits of developing chemical processes with a high atom economy. • Explain that addition reactions have an atom economy of 100%, whereas substitution reactions are less efficient. • Carry out calculations to determine the atom economy of a reaction. • Explain that a reaction may have a high percentage yield but a low atom economy. • Perform calculations to determine the percentage yield of a reaction.

  2. Sustainable Development and Atom Economy Today, to reduce the global impact of all industrial processes, scientists aim to design reactions which have the highest possible atom economy, thereby reducing the raw materials used and the amount of energy needed to produce any product.

  3. Calculating Atom Economy The term comes from the introduction of so called ‘green chemistry’, in which the amount of reactants in a chemical reaction that ends up in the final useful product is taken into consideration. In an ideal reaction, all the atoms of the reactants would end up as useful product. Such a reaction would produce no waste at all, but this is rarely possible. The atom economy (also called atom utilisation) of a reaction, is a measure of the percentage of the starting materials that actually end up as useful products. The atom economy can be calculated in the following way: % atom economy = mass desired product(s) x 100% total mass of reactants

  4. The atom economy of a reaction is . . . . . . . . . . . .a theoretical measure of the amount of starting materials that ends up as 'desired' reaction product. The greater the atom economy of a reaction the . . . . . . . . . . more 'efficient' or 'economic' it is likely to be, though this is a gross simplification when complex and costly chemical synthesis are looked at. It can be defined numerically in words in several ways, all of which amount to the same theoretical % number

  5. Calculating Atom Economy In the production of ammonium nitrate... ammonia + nitric acid  ammonium nitrate NH3(g) + HNO3(aq) NH4NO3(aq) NH3= 17g HNO3 = 63g NH4NO3 = 80g Calculate the atom economy for this reaction: …17g of NH3 and 63g of HNO3 produce 80g of NH4NO3 Atom economy = 80g x 100 = 100% 80g As there are no waste products in this reaction, it has an atom economy of 100%.

  6. Calculating Atom Economy In the smelting of iron: iron oxide + carbon  iron + carbon dioxide 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g) 2Fe2O3= 320g 3C= 36g 4Fe= 224g 3CO2 = 132g …for every 320g of iron oxide 224g of iron is produced. Calculate the atom economy for this reaction: Atom economy = 224g x 100 = 63% 320 + 36g As the reaction produces carbon dioxide as a waste product, the reaction can not have an atom economy of 100%. The atom economy of this reaction could be improved, if a use could found for the waste carbon dioxide.

  7. What is the atom economy for making hydrogen by reacting coal with steam ? The other product is carbon dioxide. Write the balanced equation: C(s) + 2H2O(g)    →    CO2(g) + 2H2(g) Write out the Ar and Mr values underneath: C(s) + 2H2O(g)    →    CO2(g) + 2H2(g) 12   2 × 18            44     2 × 2 Remember that the Ar or Mr in grams is one mole, so: total mass of products = 44 + 4 = 48g (note that this is the same as the reactants: 12 + 36 = 48g) mass of desired product (H2) = 4g % atom economy = 4 ⁄ 48 × 100 = 8.3% This process has a low atom economy and is therefore an inefficient way to make hydrogen. It also uses a non-renewable resource: coal.

  8. Fe2O3(s) + 3CO(g) ===> 2Fe(l) + 3CO2(g) Using the atomic masses of Fe = 56, C = 12, O = 16, we can calculate the atom economy for extracting iron. the reaction equation can be expressed in terms of theoretical reacting mass units [(2 x 56) + (3 x 16)] + [3 x (12 + 16)] ===> [2 x 56] + [3 x (12 + 16 + 16)] [160 of Fe2O3] + [84 of CO] ===> [112 of Fe] + [132 of CO2] there are a total of 112 mass units of the useful/desired product iron, Fe out of a total mass of reactants or products of 112 + 132 = 244. Therefore the atom economy = 112 x 100 / 244 = 45.9%

  9. Equation 1 C = 12 , H = 1 , Na = 23 , O = 16, S = 32, Br = 80 This can be done by taking the ratio of the mass of the utilized atoms (137) to the total mass of the atoms of all the reactants (74 + 98 + 103 = 275) and multiplying by 100.  As shown below this reaction has only about 50% atom economy.           % Atom Economy = (Mr of atoms utilized/ Mr of all reactants) X 100                                        = (137 / 275) X 100 = 49.8 % Thus at best (if the reaction produced 100% yield) then only half of the mass of the reactants would be incorporated into the desired product while the rest would be wasted in unwanted side products.

  10. Equation 2 Economy = (Mr of atoms utilized/Mr of all reactants) X 100                                        = ( 56/ 205 ) X 100 = 27.3 % Equation 3 Economy = (Mr of atoms utilized/Mr of all reactants) X 100                                        = ( 137 / 137 ) X 100 = 100 %

  11. Equation 4 The reaction between propene and bromine

  12. Equation 5 Preparation of butan-1-ol from 1-bromobutane

  13. Atom Economy A sample of magnetite iron ore contains 76% of the iron oxide compound Fe3O4 and 24% of waste silicate minerals. (a) What is the maximum theoretical mass of iron that can be extracted from each tonne (1000 kg) of magnetite ore by carbon reduction? [ Atomic masses: Fe = 56, C = 12 and O = 16 ] Fe3O4 + 2C ==> 3Fe + 2CO2 (b) What is the atom economy of the carbon reduction reaction? (c) Will the atom economy be smaller, the same, or greater, if the reduction involves carbon monoxide (CO) rather than carbon (C)? explain? Fe3O4 + 4CO ==> 3Fe + 4CO2

  14. A sample of magnetite iron ore contains 76% of the iron oxide compound Fe3O4 and 24% of waste silicate minerals. (a) What is the maximum theoretical mass of iron that can be extracted from each tonne (1000 kg) of magnetite ore by carbon reduction? [ Atomic masses: Fe = 56, C = 12 and O = 16 ] The reduction equation is: Fe3O4 + 2C ==> 3Fe + 2CO2 Before doing the reacting mass calculation, you need to do simple calculation to take into account the lack of purity of the ore. 76% of 1 tonne is 0.76 tonne (760 kg). For the reacting mass ratio:  1 Fe3O4 ==> 3 Fe (you can ignore rest of equation) Therefore in reacting mass units: (3 x 56) + (4 x 16) ==> 3 x 56 so, from the reacting mass equation:  232 Fe3O4 ==> 168 Fe 0.76 Fe3O4 tonne  ==> x tonne Fe x = 0.76 /232 x 3 x 56 = 0.55 = 0.55 tonne Fe (550 kg)/tonne (1000 kg) of magnetite ore

  15. (b) What is the atom economy of the carbon reduction reaction? You can use some of the data from part (a). % atom economy = total mass of useful product x 100 total mass of reactants 168 x 100 = 168 x 100 (232 + 2x12) 256 = 65.6%

  16. (c) Will the atom economy be smaller, the same, or greater, if the reduction involves carbon monoxide (CO) rather than carbon (C)? explain? Fe3O4 + 4CO ==> 3Fe + 4CO2 The atom economy will be smaller because CO is a bigger molecular/reactant mass than C and 4 molecules would be needed per 'molecule' of Fe3O4, so the mass of reactants is greater for the same product mass of iron (i.e. bottom line numerically bigger, so % smaller). This is bound to be so because the carbon in CO is already chemically bound to some oxygen and can't remove as much oxygen as carbon itself. so the atom economy = 168 x 100 / (232 + 4x28) = 48.8 %

  17. 2.2 9 Atom Economy Page 164-165 Questions 1 and 2 Key Definition : atom economy The atom economy of a reaction is . . . . . . . . . . . . a theoretical measure of the amount of starting materials that ends up as 'desired' reaction product

  18. Exam Question – Jan 2011 ( extract ) Butyl ethanoate is an ester used as a flavouring. This ester can be synthesised from butan-1-ol by two different processes. Process 1 is a one-step process that involves a reversible reaction. CH3CH2CH2CH2OH + CH3COOH  CH3COOCH2CH2CH2CH3 + H2O 6.25 g of butan-1-ol forms 6.57 g of butyl ethanoate. Calculate • The percentage yield for process 1 • The atom economy for process 1 Process 2 is a two-step process. CH3COOH + SOCl2 → CH3COCl + SO2 + HCl CH3CH2CH2CH2OH + CH3COCl → CH3COOCH2CH2CH2CH3 + HCl • 5.450 grms of ethanoic acid produces 9.806 g of Butyl ethanoate • Calculate The overall percentage yield for process 2 The overall atom economy for process 2

  19. Exam Question – Jan 2011 ( extract ) Butyl ethanoate is an ester used as a flavouring. This ester can be synthesised from butan-1-ol by two different processes. Process 1 is a one-step process that involves a reversible reaction. CH3CH2CH2CH2OH + CH3COOH  CH3COOCH2CH2CH2CH3 + H2O 6.25 g of butan-1-ol forms 6.57 g of butyl ethanoate. Calculate • The percentage yield for process 1 • The atom economy for process 1 Process 2 is a two-step process. CH3COOH + SOCl2 → CH3COCl + SO2 + HCl CH3CH2CH2CH2OH + CH3COCl → CH3COOCH2CH2CH2CH3 + HCl • 5.450 grms of ethanoic acid produces 9.806 g of Butyl ethanoate • Calculate The overall percentage yield for process 2 is The overall atom economy for process 2 is 67.1%. 86.6%. 93.3%. 45.8%.

  20. Explain why process 2 has a high percentage yield but a low atom economy. ........................................................................................................................ [2] Suggest two reasons why butyl ethanoate is manufactured by process 1 rather than by process 2. ..............................................................................................................................................[2]

  21. Exam Question – Jan 2011 ( extract ) Butyl ethanoate is an ester used as a flavouring. This ester can be synthesised from butan-1-ol by two different processes. Process 1 is a one-step process that involves a reversible reaction. CH3CH2CH2CH2OH + CH3COOH  CH3COOCH2CH2CH2CH3 + H2O 6.25 g of butan-1-ol forms 6.57 g of butyl ethanoate. Calculate • The percentage yield for process 1 • The atom economy for process 1 67.1%. 86.6%. CH3CH2CH2CH2OH + CH3COOH → CH3COOCH2CH2CH2CH3 + H2O 6.57 g 6.25 g 74 116 moles 6.25 / 74 6.57/116 = 0.0845 = 0.0566 0.0566/0.0845 = 67.1 % ( (Mr of all reactants or Mr of all products is 134.0 (Mr of desired product) is 116.0  Atom economy = 100 × 116.0/134.0 = 86.6%.

  22. Process 2 is a two-step process. CH3COOH + SOCl2 → CH3COCl + SO2 + HCl CH3CH2CH2CH2OH + CH3COCl → CH3COOCH2CH2CH2CH3 + HCl 5.450 grms of ethanoic acid produces 9.806 g of Butyl ethanoate The overall percentage yield for process 2 is The overall atom economy for process 2 is CH3COOH → CH3COOCH2CH2CH2CH3 5.450 g 9.806 g 60 116 5.45 / 60 = 0.0908 9.806 / 116 = 0.0845 0.0845 / 0.0908 = 93.1 % Total mass of products = 64 + 2 x 36.5 + 116 = 253 Total mass of reactants = 60 + 74 + 119 = 253 atom economy is 116 / 253 = 45.8 % 93.1%. 45.8%.

  23. Explain why process 2 has a high percentage yield but a low atom economy. ........................................................................................................................ [2] Link between yield AND explanation required: (high percentage) yield shows a high % conversion (of reactants into products)  Link between atom economy AND explanation required: (low) atom economy shows a lot of waste (product) OR (low) atom economy shows not much desired product 

  24. Suggest two reasons why butyl ethanoate is manufactured by process 1 rather than by process 2. ..............................................................................................................................................[2] ANY TWO FROM - Comparison essential throughout Less waste (products) OR higher atom economy  Less toxic reactants OR less toxic (waste) products OR less corrosive reactants OR less corrosive (waste) products OR less harmful reactants OR less harmful (waste) products OR less hazardous reactants OR less hazardous (waste) products  Cheaper starting materials OR more readily available starting materials  Fewer steps OR one step rather than two steps 

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