2.1.3 (h) Percentage yields and atom economy

eg.

A student makes 2.4g of Calcium Hydroxide. She had predicted that she would make 3.6g

What is the percentage yield?

(2.4 / 3.6) x 100 = 66.7%

You can safely assume an A level examiner will expect more of you.

eg

A student completely reacts 4.6 g of Sodium with Sulphuric acid. 9.23 g of Sodium Sulphate is made.

What is the percentage yield?

2 Na_(s) + H₂SO_{4 (aq)} → Na₂SO_4(aq) + H₂O_(l)

Moles of Na = Mass/ A_r = 4.6/23 = 0.2

Ratio Na: Na₂SO₄ = 2:1

Moles of Na₂SO₄ expected = 0.2/2 = 0.1

M_r(Na₂SO₄) = 23+23+32+64 =142 g/mol

Mass of Na₂SO₄ expected = Moles x M_r = 0.1 x 142 = 14.2

% yield = (Actual/Predicted) x 100 = (9.23/14.2) x 100 = 65%

NOTE: The % yield calculation can be carried out in moles or in grams.

In reality, this means that you work out the M_r of the desired product and the M_r of the other products and slot them into the equation above.

So, for our example above: 2 Na_(s) + H₂SO_{4 (aq)} → Na₂SO_4(aq) + H₂O_(l)

Assuming that the desired product is Sodium Sulphate rather than water!

M_r(Desired product) = 23+23+32+64 =142 g/mol

M_r(Water) = 2+16 =18 g/mol

M_r(All products) = 142 + 18 = 160 g/mol

Atom Economy = (142/160) x 100 = 88.75%

If there is a coefficient then you must take it into account

Or: 2Na_(s) + 2 HCl_(aq) → 2NaCl_(aq) + H_2(g)

If the desired product is Sodium Chloride rather than Hydrogen!

M_r(Desired product) = 23+35.5=58.5g/mol

M_r(Hydrogen) = 2 g/mol

M_r(All products) = (2 x 58.5) + 2 = 119 g/mol

Atom Economy = (2 x 58.5 /119) x 100 = 98.3 %

If the desired product is Hydrogen rather than Sodium Chloride!

M_r(Desired Product) = 2 g/mol

M_r(Sodium Chloride) = 23+35.5=58.5g/mol

M_r(All products) = (2 x 58.5) + 2 = 119 g/mol

Atom Economy = (2 /119) x 100 = 1.7 %

Likely follow-ups include:

1) Why would a company producing Hydrogen from the following process be unconcerned by a poor atom economy?

Ca + H₂SO_4(aq) → CaSO_4(aq) + H_2(g)

The answer is generally because the other product is also useful and can be sold. So it isn't actually waste.

At which point atom economy is irrelevant & the company would be much more interested in percentage yield.

2) How does developing chemical processes with high atom economy benefit sustainability?

Sustainability is an environmental idea.

Using a high percentage of your raw material to make "dumpable" waste is less sustainable than either

a) Coming up with a process that only produces one product - 100% atom economy

b) Makes several useful products - 100% effective atom economy.

3) Given a process that can be carried out one way with a mediocre atom economy or a second way with a better atom economy why might a company choose the first?

Either because

a) Method 1 produces largely harmless waste, like water, when Method 2 produces less but toxic waste

Or

b) Method 2 may have a terrible percentage yield or a bad rate of reaction or both.

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