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Atom economy - "Green Chemistry Project"
- 1. Definition Green chemistry is“the design of chemical products and processes that reduce or eliminate the use and generation of hazardous substances”
- 2. 12 Principles ofGreen Chemistry ...... ….. 7. Maximize atom economy What’s atom economy ?
- 3. ATOM ECONOMY Barry Trost, Stanford University “Because an Atom is a Terrible Thing to Waste” A Measure of the Efficiency of a Reaction How many of the atoms of the reactant are incorporated into the final product and how many are wasted?
- 4. All the designersof chemical processes want to make the maximum amount of product they can from a given raw material. It is possible to calculate how successful one of these processes is by using the idea of yield.
- 5. Mass of productactually made % Yield = __________________________ X 100 Maximum mass of product that could be made(theoretical yield)
- 6. In a chemicalsynthesis of calcium oxide, calcium carbonate is roasted in an oven. The equation for the reaction is: CaCO3(s)→ CaO(s) + CO2 (g) MMr: 100 56 44 The maximum mass of CaO that could be made from 1mol of CaCO3 (=100g) is 56 g
- 7. If 50 kgcalcium carbonate is used and 21 kg calcium oxide is made, what is the percentage yield of the reaction? theoretictal yeld = 28g calcium oxide 21kg _____________ % yeld = x 100 = 75% 28 kg
- 8. ATOM ECONOMY Theidea of yield is useful, but from a Green Chemistry and sustainable development perspective, it is not the full picture. This is because yield is calculated by considering only one product. One of the key principles of Green Chemistry is that processes should be designed so that the maximum amount of all the raw materials ends up in the product and a minimum amount of waste is produced
- 9. ATOM ECONOMY A reactioncan have a high percentage yield but also make a lot of waste product. This kind of reaction has a low atom economy. Both the yield and the atom economy have to be taken into account when designing a green chemical process.
- 10. Look again atthe reaction you considered CaCO3(s)→ CaO(s) + CO2 (g) MMr: 100 56 44 If we split up the formulae, we can look at what happens to each atom in the reaction. The atoms shown below in bold end up in the product we want, the rest do not: Waste Ca C O O O → Ca O COO box 1C 2O
- 11. From the originalatoms, one C atom and two O atoms are wasted – they are not in the final,useful product. Green chemists define atom economy as FW of atoms utilized % Atom Economy = ____________________ x 100 FW of all reactant So for this example, 56 (FW CaO) _______________ % Atom Economy = x 100 %A.E= 56% 100 (FW CaCO3)
- 12. ATOM ECONOMY INA SUBSTITUTION REACTION This kind of reaction has a low atom economy. CH3CH2CH2CH2OH + NaBr +H2SO4→CH3CH2CH2CH2Br + NaHSO4+ H2O 1 2 3 4 5 6 0.0108mole 0.0129 0.0200 0.0108 mole 0.08g 1.33 2.0 1.48 g (theoretical yield) limiting reagent Suppose the actual yield is 1.20 g of compound 4. %yeld= (1,20g/1,48g) x100 = 81%
- 13. CH3CH2CH2CH2OH + NaBr+H2SO4→CH3CH2CH2CH2Br + NaHSO4+ H2O 1 2 3 4 5 6 ATOM ECONOMY TABLE % Atom Economy = (FW of atoms utilized/FW of all reactants) X 100 = (137/275) X 100 = 50% FW = formula weight
- 14. Step-by-step: How tocalculate atom economy Step 1. Write out the balanced equation Step 2. Calculate the FW of each of the reactants (remember to account for stoichiometric coefficients) Step 3. Calculate the FW of the product (remember to account for stoichiometric coefficients) Step 4. Apply the formula FW of all atoms utilized ______________________ % atom economy = x 100 Total FW of all reactants
- 15. Thanks for yourattention
Editor's Notes
- #13 If one considers the above reaction where a total of 4.13 g of reactants (0.8 g of 1-butanol, 1.33 g of NaBr and 2.0 g of H2SO4) was used, and that at best this reaction will only yield 1.48 g of the desired product, the question might be asked "what happens to the bulk (4.13 g -1.48 g = 2.7 g) of the mass of reactants?". The answer is they end up in side products (NaHSO4 and H2O) that may be unwanted, unused, toxic and/or not recycled/reused. The side products are oftentimes treated as wastes and must be disposed of or otherwise treated. At best only 36% (1.48 g/4.13 g X 100) of the mass of the reactants end up in the desired product. If the actual yield is 81% then only 29% (.81 X .36 X 100) of the mass of the reactants actually ends up in the desired product!
- #14 In the equation we have illustrated the atom economy of this reaction by showing all of the reactant atoms that are incorporated into the desired product in green,while those that are wasted are shown in red. Likewise the atoms of the desired product are in green and the atoms composing the unwanted products are in red. Table 3 provides another view of the atom economy of this reaction. In columns 1 and 2 of this table, the formulas and formula weights (FW) of the reactants are listed. Shown in green (columns 3 and 4) are the atoms and weights of the atoms of the reactants that are incorporated into the desired product (4), and shown in brown (columns 5 and 6) are the atoms and weights of atoms of the reactants that end up in unwanted side products. Focusing on the last row of this table it can be seen that of all the atoms of the reactants (4C, 12H, 5O, 1Br, 1Na and 1S) only 4C, 9H, and 1Br are utilized in the desired product and the bulk (3H, 5O, 1Na, 1S) are wasted as components of unwanted products. This is an example of poor atom economy! continua nella prossima slide
- #15 The percentage atom economy can be calculate by taking the ratio of the mass of the utilized atoms (137) to the total mass of the atoms of all the reactants (275) and multiplying by 100. As shown below this reaction has only 50% atom economy. % Atom Economy = (FW of atoms utilized/FW of all reactants) X 100 = (137/275) X 100 = 50% 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.