2.2.8
Percentage yield
Atom Economy
Perform calculations to determine the percentage yield
of a reaction
Perform calculations to determine the Atom Economy
of a reaction
Page 162-163
Calculating Percentage (%) Yield
2.3g of sodium reacts with an excess
of chlorine to produce 4.0g of sodium chloride.
What is the percentage yield?
2Na(s)
+
Cl2(g)

2NaCl(s)
(Ar reactants: Na=23 Cl=35.5
2.3g Na = 2.3 mol Na
23
Mr product: NaCl= 58.5)
= 0.1 mol Na
Theoretically 0.1 mol Na should yield 0.1 mol NaCl
Theoretical yield of NaCl = 58.5 x 0.1 = 5.85g
% Yield = Actual yield x 100%
Theoretical yield
% Yield = 4.0g x 100% = 68%
5.85g
Calculating Percentage (%) Yield
If 1.2g of magnesium reacts with an excess
of oxygen to produce 0.8g of magnesium
oxide…
What is the percentage yield?
2Mg(s) +
O2(g)

2MgO(s)
(Ar reactants: Mg=24 O=16 Mr product: MgO= 40)
1.2g Mg = 1.2 mol Mg = 0.05 mol Mg
24
Theoretically 0.05 mol Mg should yield 0.05 mol MgO
Theoretical yield of MgO = 40 x 0.05 = 2g
% Yield = Actual yield x 100%
Theoretical yield
% Yield = 0.8g x 100% = 40%
2g
Calculating Percentage (%) Yield
If 2g of calcium carbonate reacts with an excess of hydrochloric
acid to produce 1.11 g of calcium chloride….
What is the percentage yield?
2HCl(aq)+ CaCO3(s)  H2O(l) + CO2(g) + CaCl2(s)
(Mr values are: CaCO3 = 100 CaCl2 = 111)
2g CaCO3 = 2 mol CaCO3 = 0.02 mol CaCO3
100
Theoretically 0.02 mol CaCO3 should yield 0.02 mol CaCl2
Theoretical Yield of CaCl2 = 111 x 0.02 = 2.22g
% Yield = Actual yield x 100%
Theoretical yield
% Yield = 1.11 x 100 = 50%
2.22
2.2 8 Percentage Yield
Page 162-163
Questions 1 and 2
Key Definition : A Limiting Reagent is .
.
Question 1
CH3CH(Cl) CH3 + NaOH → CH3CH(OH) CH3 + NaCl
3.295 g
Molar mass
78.5 g
Moles
3.925 / 78.5
0.05
2.955 g
60 g
2.955 / 60
0.04925
So moles we should have got is = 0.05
or mass should have been = 0.05 x 60 = 3.00 g
so yield is
0.04925/0.050 x100 = 98.5 %
or using grams
2.955/3.00 x 100 = 98.5 %
Question 2
C2H5OH + CH3COOH → CH3COOC2H5 + H2O
4.0 g
4.5 g
Molar mass
46 g
60 g
Moles
4.0 / 46 g
0.087
4.5 / 60 g
0.075
5.5 g
88 g
5.5 / 88 g
0.0625
So moles we should have got is = 0.075
or mass should have been = 0.075 x 88 = 6.6g
so yield is
0.0625/0.075 x 100 = 83.3 %
or using grams 5.5/6.6 x100 = 83.3 %
Atom Economy
A sample of magnetite iron ore contains 76% of the iron
oxide compound Fe3O4 and 24% of waste silicate minerals.
(a) What is the maximum theoretical mass of iron that can
be extracted from each tonne (1000 kg) of magnetite ore
by carbon reduction?
[ Atomic masses: Fe = 55.8, C = 12 and O = 16 ]
o
The reduction equation is:
Fe3O4 + 2C ==> 3Fe + 2CO2
(b) What is the atom economy of the carbon
reduction reaction?
(c) Will the atom economy be smaller, the same, or
greater, if the reduction involves carbon monoxide (CO)
rather than carbon (C)? explain?
Fe3O4 + 2C ==> 3Fe + 2CO2
Fe3O4 + 4CO ==> 3Fe + 4CO2