5 Percentage yield and atom economy

% Yield Example 7: 25.0g of Fe2O3 was reacted and it produced 10.0g of Fe. What is the percentage yield? Fe2O3 + 3CO  2Fe + 3 CO2 % yield = (actual yield/theoretical yield) x 100 = (10/ 17.48) x 100 =57.2% First calculate maximum mass of Fe that could be produced Step 1: work out amount in mol of Iron oxide amount = mass / Mr =25 / 159.6 = 0.1566 mol Step 2: use balanced equation to give moles of Fe 1 moles Fe2O3 : 2 moles Fe So 0.1566 Fe2O3 : 0.313moles Fe Step 3: work out mass of Fe Mass = amount x Mr = 0.313 x 55.8 =17.48g % yield in a process can be lowered through incomplete reactions, side reactions, losses during transfers of substances, losses during purification stages. % Atom Economy percentage atom economy Mass of useful products Mass of all reactants = x 100 Example 8 : What is the % atom economy for the following reaction where Fe is the desired product assuming the reaction goes to completion? Fe2O3 + 3CO  2Fe + 3 CO2 % atom economy = (2 x 55.8) (2 x 55.8 + 3×16) + 3 x (12+16) x 100 =45.8% Do take into account balancing numbers when working out % atom economy. Reactions where there is only one product where all atoms are used making product are ideal and have 100% atom economy. e.g. CH2=CH2 + H2  CH3CH3 Sustainable chemistry requires chemists to design processes with high atom economy that minimise production of waste products. If a process does have a side, waste product the economics of the process can be improved by selling the bi-product for other uses. percentage yield = x 100 actual yield theoretical yield