Calculation of quantities required for a chemical
reaction, % yield and atom economy

Suppose you want to make 50g of the blue hydrated copper(II) sulphate crystals
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. Your reactants are dilute sulphuric acid and the black solid copper(II) oxide.
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You can use copper(II) carbonate, but this is not a pure simple compound and the
predictive nature of the calculations will not be as good.
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copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water
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(i) CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l)
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on crystallisation you get the blue hydrated crystals of formula CuSO4.5H2O
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so strictly speaking, after evaporation-crystallisation the equation is
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(ii) CuO(s) + H2SO4(aq) + 4H2O(l) ==> CuSO4.5H2O(s)
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How much copper(II) oxide is needed?
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A 'non-moles' calculation first of all, involving a reacting mass calculation.
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The crucial change overall is CuO ==> CuSO4.5H2O
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(Note: In reacting mass calculations you can often ignore other reactant/product
masses)
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Atomic masses: Cu = 64, S = 32, H = 1, O = 16
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Formula masses are for: CuO = 64 + 16 = 80, CuSO4 = 64 + 32 + (4 x 16) = 160
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and CuSO4.5H2O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250
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The crucial reacting mass ratio is: 80 ==> 250 since formula ratio is 1:1 in the
equation.
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Therefore, theoretically, to make 50g of the crystals (1/5th of 250),
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you need 1/5th of 80g of copper(II) oxide,
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and 80/5 = 16g of copper(II) oxide is required.
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However, in reality, things are not so simple because the method involves adding
excess copper(II) oxide to the dilute sulphuric acid. (see salt preparation method
(b))
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So in practice you would need to take more of the CuO to get anything like 50g of
the salt crystals.
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There is another way to calculate the quantities required based on the acid.
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How much dilute sulphuric acid (of concentration 1 mol dm-3) is required?

Mol = mass in g / formula mass,

so moles of CuSO4.5H2O required = 50/250 = 0.2 mol (see basics of moles)

Therefore 0.2 mol of H2SO4 is required (1/5th mol), since the mole ratio CuO :
H2SO4 is 1 : 1 in the equation.

1dm3 (1000 cm3) of a 1 mol dm-3 solution of contains 1 mole (by definition see molarity page)

Therefore 1/5th of 1dm3 is required, so 200 cm3 of 1 mol dm-3 dilute
sulphuric acid is required,

or 100 cm3 of 2M dilute sulphuric acid. (2M is old fashioned notation for 2
mol dm-3 as seen on many laboratory bottles!)

You would then add copper(II) oxide in small amounts until no more
dissolves in the warm-hot acid and the excess black powder is filtered off.
There is no need to weigh out an exact amount of copper oxide.
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Suppose after carrying out the preparation you finally crystallise 34g of pure the
blue crystals of CuSO4.5H2O and weigh the product when dry.
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What is the 'atom economy' of the preparation? (you need to refer to equations
(i)/(ii) at the start of section 14.5

Atom economy = useful theoretical products x 100/mass of all reactants

based on equation (i) Atom economy = mass CuSO4 x 100 / (mass CuO +
H2SO4)

= 160 x 100 / (80 + 98) = 16000/178 = 89.9%

based on equation (ii) the atom economy is 100% if you include water as a
'reactant', can you see why?

What is the % yield? i.e. comparing what you actually get with the
maximum possible, i.e. a 'reality check'!

% yield = mass of product obtained x 100 / theoretical mass from
the equation

% yield = 34 x 100 / 50 = 68%