Standard level
Atom economy is a concept in green chemistry that measures the efficiency of a chemical reaction in terms of how well the atoms of the reactants are incorporated into the desired products. A higher atom economy indicates a more efficient reaction with less waste generated.
Syllabus ref: R2.1.5Reactivity 2.1.5 - The atom economy is a measure of efficiency in green chemistry.
- Calculate the atom economy from the stoichiometry of a reaction.
Guidance
- Include discussion of the inverse relationship between atom economy and wastage in industrial processes.
- The equation for calculation of the atom economy is given in the data booklet.
Tools and links
- Structure 2.4, Reactivity 2.2 - The atom economy and the percentage yield both give important information about the “efficiency” of a chemical process. What other factors should be considered in this assessment?
Definition of atom economy
It is defined as the molecular weight of the desired product divided by the total molecular weight of all reactants, multiplied by 100 to give a percentage.
In the synthesis of 1-bromopropane from propane via free-radical bromination, the reaction is as follows:
C3H8 + Br2 → C3H7Br + HBr
Atom Economy Calculation:
- Molecular mass of propane (C3H8): 44 g/mol
- Molecular weight of bromine (Br2): 160 g/mol
- Molecular weight of 1-bromopropane (C3H7Br): 123 g/mol
- Total molecular weight of reactants (propane + bromine): 44 g/mol + 160 g/mol = 204 g/mol
The atom economy = 100 x molecular mass of desired product/total mass of reactants
= 100 x 123/204 = 60.3%
When chemical syntheses are carried out to prepare compounds required in industry, there are often by-products that have no useful purpose. This means that some of the atoms from the original reactants are "wasted".
The ultimate aim of green chemistry is to ensure that there is no wastage in any production process and that the atom economy is as close to 100% as possible.
In the synthesis methanol from carbon monoxide and hydrogen the reaction is:
CO + 2H2 → CH3OH
You can see that all of the atoms of the reactants end up in the (only) product.
The atom economy = 100%
The higher the value of the atom economy the lower the wastage in any process.
Worked examples
Q364-01 When a 0.24g sample of magnesium was dissolved in excess 2M HCl the volume of hydrogen gas produced was measured by collection in a gas syringe. The initial reading on the syringe was 0.0cm3 and the final reading was 240cm3. Find the stoichiometry of the reaction. [RAM Mg=24, all volumes measured at room temperature, 1 mole of gas = 24dm3]
Moles of magnesium reacting = 0.24/24 = 0.01 moles of hydrogen produced = 240/24000 cm3 = 0.01 therefore ratio of magnesium used to hydrogen produced = 1:1 the equation must be: Mg + 2HCl → MgCl2 + H2 |
Q364-02 When a sample of a copper oxide is heated in a stream of hydrogen gas the following results were obtained:
- Mass of copper oxide used = 0.8g
- Mass of copper remaining after heating 0.64g
[RAM Cu=64, O=16]
Moles of copper formed = 0.64/64 = 0.01 Mass of oxgen lost = 0.80 - 0.64 = 0.16 Therefore moles of oxygen lost = 0.16/16 = 0.01 Ratio of moles copper to moles oxygen = 0.01 : 0.01 = 1:1 Therefore formula of copper oxide = CuO |
Q364-03 In an experiment to determine the formula of a transition metal chloride, dry chlorine gas was passed over heated iron and the iron chloride collected and weighed. The following results were obtained:
- Mass of iron before heating = 2.8g
- Mass of iron chloride collected = 8.1g
Find the formula of the transition metal chloride. [RAM Fe=56, Cl=35.5]
Answer
Moles of iron used = 2.8/56 = 0.05 moles Mass of chlorine reacting with the iron = 8.1 - 2.8 = 5.3g Moles of chlorine reacting with the iron = 5.3/35.5 = 0.149 moles (approximately 0.15 moles) Therefore ratio of iron to chlorine = 0.05:0.15 = 1:3 Formula of the iron chloride = FeCl3 |
Q364-04 In an experiment to determine the formula of tin iodide the following results were obtained:
- Mass of tin used = 0.595g
- Mass of tin iodide obtained = 3.164g
Find the formula of the tin iodide. [RAM Sn=119, I=127]
Answer
Moles of tin used = 0.595/119 = 0.005 moles Mass of iodine reacting with the tin = 3.164 - 0.595 = 2.540g Moles of iodine reacting with the tin = 2.540/127 = 0.02 Therefore ratio of tin to iodine = 0.005 : 0.02 = 1:4 Therefore formula of tin iodide = SnI4 |
Q364-05 A 2.5 g sample of limestone (calcium carbonate) reacted completely with exactly 50 cm3 of 1.0M Hydrochloric acid. Find the stoichiometry of the reaction. [Ca=40, C=12, O=16, H=1, Cl=35.5]
Answer
Calcium carbonate has the formula CaCO3 Mr of calcium carbonate = 40 + 12 + 48 = 100 Moles of calcium carbonate = 2.5/100 = 0.025 moles Moles of hydrochloric acid = Molarity x volume (litres) = 1.0 x 0.05 = 0.05 moles Therefore ratio of reacting moles calcium carbonate to hydrochloric acid = 0.025 : 0.05 = 1 : 2 Therefore the reaction stoichiometry is: CaCO3 + 2HCl → CaCl2 + CO2 + H2O |
Q364-06 When 100cm3 of a hydrocarbon was burned in excess oxygen 300cm3 of carbon dioxide and 300cm3 of steam were formed. Calculate the formula of the hydrocarbon.
The number of moles of gas is directly proportional to the volume at any given pressure and temperature conditions. It may be seen that 1 volume of the hydrocarbon produces 3 volumes of CO2 and 3 volumes of H2O Hence 1 mole of the hydrocarbon gives 3 moles of CO2 and 3 moles of H2O The formula of any hydrocarbon is CxHy Therefore: CxHy + O2 → 3CO2 + 3H2O from this you can see that x = 3 and y = 6 So the formula of the hydrocarbon = C3H6 |
Q364-07 When 20cm3 of a hydrocarbon was burned in 200cm3 oxygen, the total volume decreased by 60cm3. When the gas mixture was passed through concentrated NaOH solution the gas mixture decreased by a further 60cm3. Calculate the formula of the hydrocarbon if all values were measured at STP.
The formula of a hydrocarbon is CxHy The volumes of a gas are proportional to the number of moles NaOH concentrated reacts with carbon dioxide gas according to the equation: 2NaOH + CO2 → Na2CO3 + H2O So the decrease in volume is due to CO2 gas. Hence 20 volumes of hydrocarbon produces 60 volumes of CO2 Hence 1 mole of CxHy produces 3 moles of CO2 therefore X=3 If the total volume decreased by 60cm3 at first (and 60cm3 CO2 were also formed), then 20cm3 CxHy must have reacted with 100cm3 oxygen. Therefore 1 mole of C3Hy reacts with 5 moles of oxygen Equation: C3Hy + 5O2 → 3CO2 + y/2H2O From this y/2= 4 , Y=8 Formula of the hydrocarbon = C3H8 |
Q364-08 Calculate the volume of 1.2M hydrochloric acid required to react with 8.0g of copper(II) oxide. [Cu=64, O=16]
The equation for the reaction is: copper(II) oxide + hydrochloric acid → copper(II) chloride + water CuO + 2HCl → CuCl2 + H2O Mr of copper(II) oxide (CuO) = 64 + 16 = 80 Therefore 8.0 g of copper(II) oxide = 8.0/80 moles = 0.1 moles From the equation 1 mole of copper(II) oxide requires 2 moles of hydrochloric acid Therefore 0.1 moles of of copper(II) oxide requires 0.2 moles of hydrochloric acid Molarity of the acid = 1.2M Therefore to obtain 0.2 moles of acid requires 0.2/1.2 litres = 167cm3 |
Q364-09 Calculate the volume of 2M barium chloride solution needed to completely precipitate the sulfate ions from 25cm3 of 1.6M iron(III) sulfate solution.
Barium 2+ ions precipitate sulfate SO42- ions according to the equation: Ba2+(aq) + SO42-(aq) → BaSO4(s) moles of sulfate ions in 25cm3 of 1.6M iron(III) sulfate Fe2(SO4)3 solution = 0.025 x 1.6 x 3 = 0.12 moles This will require the same number of moles of barium 2+ ions Moles of barium chloride = molarity x volume (dm3) Therefore volume = moles/molarity = 0.12/2 = 0.06dm3 Volume of barium chloride required = 60cm3 |
Q364-10 Calculate the volume of 0.1M silver nitrate solution required to completely precipitate the chloride ions from 40cm3 of 0.02M sodium chloride solution.
Silver ions react with chloride ions according to the equation: Ag+(aq) + Cl-(aq) → AgCl(s) moles of chloride ions = molarity x volume(dm3) = 0.02 x 0.04 = 8.0 x 10-4 moles This is equivalent to the moles of silver ions Volume of silver nitrate required = moles/molarity = 8.0 x 10-4 /0.1 = 8.0 x 10-3dm3 Volume of silver nitrate required = 8cm3 |
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