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ATOM ECONOMY CALCULATIONS

See also % YIELD notes too

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

study examples carefully14. Other GCSE chemical calculations

ATOM ECONOMY CALCULATIONSstudy examples carefully

soluble salt preparation from insoluble base-acid neutralisationQuantitative 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 (91) chemistry science courses.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?


See also

14.1 % purity of a product and assay calculations

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling


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.

A chemical reaction may give, and often does, more than one product, but of the mixture of products, perhaps only one of them is the desired useful product.

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, percent yield, rate of reaction, 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.

 

   MASS of desired USEFUL PRODUCT

ATOM ECONOMY   =  100  x  

 -------------------------------------------------------------------------------
        TOTAL MASS of all REACTANTS or PRODUCTS

In atom economy calculations you can say REACTANTS or PRODUCTS because of the law of conservation of mass.

The greater the % atom economy of a reaction, the more 'efficient' or 'economic' it is likely to be, though this is a gross simplification when complex and costly chemical synthesis are looked at.

Many reactions give more than one product, and not all of them are useful, so it is useful to calculate what % of the products is theoretically useful, and we call this the atom economy of the reaction.

The reactions that only give one product, have the maximum atom economy of 100% and these are the most economic reactions e.g. synthesis of ammonia and reacting ethene with water to make ethanol.

N2 + 3H2 ==> 2NH3    and    CH2=CH2  +  H2O  ===>  CH3CH2OH

These are simple addition reactions where two reactants give one product e.g. synthesis of ammonia from nitrogen and hydrogen and the synthesis of ethanol from ethene and water.

100 minus the atom economy gives you the % waste, but reactions with only one product will always give the highest atom economy

If a reaction gives more than one product it may be possible to use and sell these other products, thereby making the process more economic overall.

Whatever, waste products have to be dealt with and disposed of in some safe way, and this costs money!

Reactions with a low atom economy are very wasteful and use up resources at faster rate than high atom economy reactions.

They are usually less sustainable in the long run and the cost of raw materials will increase over time.

e.g. the process may involve non-renewable raw materials which will increase in costs as reserves become depleted.

BUT take care on this 'low economy' point, e.g. many products can be recycled and, as already mentioned, the waste products may have some economic value after further separation and processing, known as useful by-product.

The less waste there is, the higher the atom economy, the less materials are wasted, less energy used, so making the process more economic, 'greener' and sustainable.

Quite simply, the larger the atom economy of a reaction, the less waste products are produced and hopeful the waste is easy and cheap to deal with, or even better, some use can be found for these 'by-products'.

It can be defined numerically in words in several ways, all of which amount to the same theoretical % number!

The chemical industry is continuously working on chemical synthetic routes that have the highest atom economy and give the highest % yield (see section 14.2a % reaction yield)

If a chemical reaction that has a low atom economy or gives a low yield of useful product research would be undertaken to find ways the reaction can been improved to increase the yield of useful product.

CALCULATING the ATOM ECONOMY of a chemical reaction

You can do the calculation in any mass units you want, or non at all by simply using the atomic/formula masses of the reactants and products as appropriate , and I suggest you just think like that.

The formula to calculate atom economy can be written in several different ways and they are ALL equivalent to each other because of the law of conservation of mass e.g.

     MASS of desired USEFUL PRODUCT

ATOM ECONOMY   =  100  x  

 ---------------------------------------------------------------
        TOTAL MASS of all REACTANTS

 

     MASS of desired USEFUL PRODUCT

ATOM ECONOMY   =  100  x  

 ---------------------------------------------------------------
        TOTAL MASS of all PRODUCTS

 

     TOTAL FORMULA MASSES of  USEFUL PRODUCT

ATOM ECONOMY   =  100  x  

 -----------------------------------------------------------------------------------
        TOTAL FORMULA MASSES of all REACTANTS

 

     TOTAL FORMULA MASSES of  USEFUL PRODUCT

ATOM ECONOMY   =  100  x  

 -----------------------------------------------------------------------------------
        TOTAL FORMULA MASSES of all PRODUCTS

THEY ALL GIVE THE SAME ANSWER!

but are theoretical values, you don't get these results in real life chemistry - see % purity of a product  & % reaction yield

 


 

Atom economy calculation Example 14.2b (1) See Extraction of Iron and Steel Making for detailed chemistry

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

(a) The fermentation of sugar to make ethanol ('alcohol') and (b) converting ethanol to ethene

(a) 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 in the equation = 2 x 46 = 92

Atom economy = 100 x 92/180 = 51.1%

(b) ethanol === heat/catalyst ===> ethene + water

It possible to use ethanol from fermentation to produce ethene for plastics (polymers) manufacture instead of relying on the cracking of crude oil fractions (e.g. a country like Brazil with a huge agriculture system but no oil deposits.

CH3CH2OH  ===>  CH2=CH2  +  H2O

formula masses: ethanol = 46, ethene = 28, water = 18

% atom economy = 100 x 28/46 = 60.9%

 


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 the ratios gives

16 + 18 ==> (3 x 2) + 28

34 mass units of reactants ==> 6 mass units of useful product

Atom economy = 100 x 6 / 34 = 17.6%

so the 82.4% of waste toxic carbon monoxide must be dealt with in some way!

It seems a very inefficient process BUT ...

(i) hydrogen has very small molecular mass and 75% of the molecules are the desired product

(ii) you can actually burn the carbon monoxide as a fuel to provide energy for power generation.

So, a low atom economy doesn't always mean the process is not economically viable or necessarily produces undue waste.

This reaction 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


Atom economy calculation Example 14.2b (4) See Ammonia Synthesis

You can use either (a) hydrogen or (b) a hydrocarbon gas like methane to reduce the oxides of metals of low reactivity to obtain the metal itself.

e.g. the reduction of copper(II) oxide.

Using the atomic masses: Cu = 63.5,  H = 1,  O =16,  C = 12

(a) CuO  + H2  ==>  Cu  +  H2O

Calculate the atom economy of the reaction.

Formula masses: CuO = 79.5,  H2 = 2,  Cu = 63.5,  H2O = 18

Total mass of reactants = total mass of products 79.5 + 2 = 63.5 + 18 = 81.5

atom economy of desired product Cu = 100 x 63.5/81.5 = 77.9%

(b) 4CuO  +  CH4  ==>  4Cu  +  2H2O  +  CO2

Calculate the atom economy of the reaction.

Formula masses: CuO = 79.5,  CH4 = 16,  Cu = 63.5,  H2O = 18,  CO2 = 44

Total mass of reactants = total mass of products

(4 x 79.5) + 16 = (4 x 63.5) + (2 x 18) + 44 = 334

atom economy of desired product Cu = 100 x (4 x 63.5)/334 = 76.0%

(c) Which reaction has the higher atom economy? BUT, would this be the preferential method used?

Reaction (a) has the higher atom economy, BUT, hydrogen is probably more costly to produce than cheap methane gas from crude oil. Therefore method (b) is probably more economic.


Rather than here, I've added more atom economy calculations to the Reacting mass ratio calculations of reactants and products from equations page.


See also

14.1 % purity of a product and assay calculations

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling


 

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


 

OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Calculating relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

  8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

  9. Moles and the molar volume of a gas, Avogadro's Law

  10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

  13. Electrolysis products calculations (negative cathode and positive anode products)

  14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials


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