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.
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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 0.0cm
3 and the final reading was 240cm
3. Find the
stoichiometry of the reaction. [RAM Mg=24, all volumes measured at room temperature,
1 mole of gas = 24dm
3]
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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
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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]
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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
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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
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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
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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
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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
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Q364-05 A 2.5 g sample of
limestone (calcium carbonate) reacted completely with exactly 50 cm
3
of 1.0M Hydrochloric acid. Find the stoichiometry of the reaction. [Ca=40, C=12,
O=16, H=1, Cl=35.5]
Answer
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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
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Q364-06 When 100cm
3
of a hydrocarbon was burned in excess oxygen 300cm
3 of carbon dioxide
and 300cm
3 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
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Q364-07 When 20cm
3
of a hydrocarbon was burned in 200cm
3 oxygen, the total volume decreased
by 60cm
3. When the gas mixture was passed through concentrated NaOH
solution the gas mixture decreased by a further 60cm
3. 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
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Q364-08 Calculate the volume
of 1.2M hydrochloric acid required to react with 8.0g of copper(II) oxide. [Cu=64,
O=16]
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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
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Q364-09 Calculate the volume
of 2M barium chloride solution needed to completely precipitate the sulfate
ions from 25cm
3 of 1.6M iron(III) sulfate solution.
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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
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Q364-10 Calculate the volume
of 0.1M silver nitrate solution required to completely precipitate the chloride
ions from 40cm
3 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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Now test yourself
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