Atom Economy
Contents
Key Stage 4
Meaning
Atom Economy is the percentage of atoms in the reactants which ends up in the desired products.
About Atom Economy
- During a chemical reaction there is often more than one product. When there is more than one product often one is desired and one is a waste product that must be disposed of.
- To reduce costs and increase the efficiency of a process chemical engineers design systems that produce the maximum amount of a desired product and a minimum amount of waste products.
Atom Economy Equation
Percentage Atom Economy = [(Relative Formula Mass of Desired Product)/(Sum of Relative Formula Masses of Reactants)] x 100%
% Atom Economy = \(\frac{M_desiredproduct}{M_allreactants}\)
Where:
Mdesiredproduct = Relative Formula Mass of Desired Product
Mallreactants = Sum of Relative Formula Masses of Reactants
Example Calculations
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Calculate the atom economy for producing Ammonia in the following reaction: N2(g) + 3H2(g) ⇌ 2NH3(g) |
Calculate the atom economy for producing Calcium Oxide in the following reaction: CaCO3(s) → CaO(s) + CO2(g) |
Calculate the atom economy for producing Sodium Chloride in the following reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) |
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Find the Relative Formula Mass of the relevant reactants and product. Mr of N2 = 28g Mr of H2 = 2g Mr of NH3 = 17g |
Find the Relative Formula Mass of the relevant reactants and product. Mr of CaCO3 = 100g Mr of CaO = 56g |
Find the Relative Formula Mass of the relevant reactants and product. Mr of HCl = 36.5g Mr of NaOH = 40g Mr of NaCl = 58.5g |
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Find the total mass of reactants needed and the total mass of the desired product. Mass of Reactants = 28 + 3 x 2 Mass of Reactants = 34g Mass of desired Product = 2 x 17 Mass of desired Product = 34g |
Find the total mass of reactants needed and the total mass of the desired product. Mass of Reactants = 100g Mass of desired Product = 56g |
Find the total mass of reactants needed and the total mass of the desired product. Mass of Reactants = 36.5 + 40 Mass of Reactants = 76.5g Mass of desired Product = 58.5g |
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Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{48}{24}\) No. Moles = 2 mol Therefore 2 moles of MgO is produced. 2 moles of MgO = 40x2 Yield = 80g |
Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{320}{16}\) No. Moles = 20 mol Therefore 40 moles of H2O are produced. 40 moles of CO2 = 18x40 Yield = 720g |
Find the number of moles supplied of the known mass. No. Moles = \(\frac{Mass}{M_r}\) No. Moles = \(\frac{7.3}{36.5}\) No. Moles = 0.2 mol Therefore 0.2 mole of Sodium Chloride is produced. 0.2 mole of NaCl = 11.7g Yield = 11.7g |