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 &quot;what happens to the bulk (4.13 g -1.48 g = 2.7 g) of the mass of reactants?&quot;. 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!
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
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.
Atom economy - "Green Chemistry Project"
DefinitionGreen chemistry is “the design ofchemical products and processesthat reduce or eliminate the use andgeneration of hazardous substances”
12 Principles of Green Chemistry ...... ….. 7. Maximize atom economy What’s atom economy ?
ATOM ECONOMY Barry Trost, Stanford University“Because an Atom is a Terrible Thing toWaste”A Measure of the Efficiency of a ReactionHow many of the atoms of the reactant areincorporated into the final product and howmany are wasted?
All the designers of chemical processeswant to make the maximum amount ofproduct they can from a given rawmaterial.It is possible to calculate how successfulone of these processes is by using theidea of yield.
Mass of product actually made% Yield = __________________________ X 100 Maximum mass of product that could bemade(theoretical yield)
In a chemical synthesis of calcium oxide, calciumcarbonate is roasted in an oven. The equation for thereaction 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
If 50 kg calcium carbonate is used and 21 kgcalcium oxide is made, what is the percentageyield of the reaction? theoretictal yeld = 28g calcium oxide 21kg _____________% yeld = x 100 = 75% 28 kg
ATOM ECONOMY The idea 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 thatprocesses should be designed so that the maximumamount of all the raw materials ends up in theproduct and a minimum amount of waste isproduced
ATOM ECONOMYA reaction can have a high percentageyield but also make a lot of waste product.This kind of reaction has a low atomeconomy.Both the yield and the atom economy haveto be taken into account when designing agreen chemical process.
Look again at the reaction you considered CaCO3(s)→ CaO(s) + CO2 (g)MMr: 100 56 44If we split up the formulae, we can look at whathappens to each atom in the reaction. The atomsshown below in bold end up in the product wewant, the rest do not: Waste Ca C O O O → Ca O COO box 1C 2O
From the original atoms, 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 reactantSo for this example, 56 (FW CaO) _______________% Atom Economy = x 100 %A.E= 56% 100 (FW CaCO3)
ATOM ECONOMY IN A SUBSTITUTION REACTIONThis 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%
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
Step-by-step: How to calculate atom economyStep 1. Write out the balanced equationStep 2. Calculate the FW of each of the reactants(remember to account for stoichiometric coefficients)Step 3. Calculate the FW of the product (remember toaccount for stoichiometric coefficients)Step 4. Apply the formula FW of all atoms utilized ______________________ % atom economy = x 100 Total FW of all reactants