ATOM ECONOMY CALCULATIONS
Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level Online Chemical Calculations
14. Other GCSE chemical calculations - ATOM ECONOMY CALCULATIONS
Quantitative chemistry calculations What is atom economy? Help on how to do atom economy calculations. What is the formula to calculate the 'atom economy' of a chemical reaction? Calculating the atom economy of a chemical reaction is explained on this page - fully worked out examples of atom economy calculations. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses. These revision notes and practice questions on how to do atom economy chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.
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14. Other GCSE chemical calculations
14.2b The Atom economy of a chemical reaction
The atom economy (a measure of atom utilisation or efficiency) is a measure of the amount of starting materials that end up as useful products. It is important for sustainable development and for good economic reasons to use reactions with high atom economy. The percentage atom economy of a reaction is readily calculated using the balanced equation for the reaction expressed in reacting masses. You need to be able to calculate the atom economy of a reaction to form a desired product from the balanced equation and perhaps explain why a particular reaction pathway is chosen to produce a specified product given appropriate information such as atom economy, yield, rate, equilibrium position and usefulness of by-products.
The atom economy of a reaction is a theoretical percentage measure of the amount of starting materials that ends up as the 'desired' useful reaction products. Its sometimes referred to as atom utilisation.
THEY ALL GIVE THE SAME ANSWER!
Atom economy calculation Example 14.2b (1) See Extraction of Iron and Steel Making
This is illustrated by using the blast furnace reaction from example 14.2a.3 above.
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]
so 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 160 + 84 = 112 + 132 = 244.
Therefore the atom economy = 100 x 112 / 244 = 45.9%
Note: It doesn't matter whether you use the total mass of reactants or the total mass products in the calculations, they are the same from the law of conservation of mass
Atom economy calculation Example 14.2b (2) See Ethanol Chemistry
The fermentation of sugar to make ethanol ('alcohol')
e.g. glucose (sugar) == enzyme ==> ethanol + carbon dioxide
C6H12O6(aq) ==> 2C2H5OH(aq) + 2CO2(g)
atomic masses: C = 12, H = 1 and O = 16
formula mass of glucose reactant = 180 (6x12 + 12x1 + 6x16)
formula mass of ethanol product = 46 (2x12 + 5x1 + 1x16 + 1x1)
relative mass of desired useful product = 2 x 46 = 92
Atom economy = 100 x 92/180 = 51.1%
Atom economy calculation Example 14.2b (3) See Ammonia Synthesis
Hydrogen is used in synthesising ammonia and is made on a large scale from reacting methane with water
methane + water ==> hydrogen + carbon monoxide
CH4 + H2O ==> 3H2 + CO
atomic masses: C = 12, H = 1, O = 16
using formula masses gives 16 + 18 ==> (3 x 2) + 28, 34 reactants ==> 6 of useful product
Atom economy = 100 x 6 / 34 = 17.6%
so the 82.4% of waste carbon monoxide must be dealt with in some way
this contrasts with 100% atom economy for ammonia production, because its an addition reaction with no extra waste products
N2 + 3H2 ==> 2NH3
mass of reactants = mass of useful products = 100% atom economy
Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses
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