Theoretical Yield – amount of product that is predicted using stoichiometry Actual Yield –...
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Transcript of Theoretical Yield – amount of product that is predicted using stoichiometry Actual Yield –...
Theoretical Yield – amount of product that is predicted using stoichiometry
Actual Yield – amount of product that is obtained in an experiment
Percent Yield – compares the mass of product obtained by experiment with the mass of product determined by stoichiometric calculations
Reasons For Low Yield
1. Sources of error. (Experimental procedures)
2. Impurities in reagents used (different grades of chemicals)
3. Side reactions (other products formed)
4. Reactions are reversible
Reasons for High Yield
1. Sources of error (experimental procedures)
Solving Percent Yield Problems
• First, determine the theoretical yield (how much you should get) using stoichiometry
• Second using the experimental/ actual yield, use the formula.
% yield = Experimental yield x 100% theoretical yield
Example 1: In a particular experiment 10.0 g of sugar should be produced but only 0.664 g is produced. What is the percentage yield?
% yield = x 100
= x 100 = 6.64 %
Experimental yield theoretical yield
0.664 g 10.0g
Example 2: Aluminium reacts with oxygen to form aluminium oxide. If 635g of Aluminium
oxide is obtained from reacting 1150g of aluminium, what is the percentage yield?
• Al + O2 Al2O3
• 4Al + 3O2 2Al2O3
Step 1: Write out the balanced chemical equation
• Step 2: Fill in chart with information you know
Balanced
Equation4Al 3O2
2Al2O3
Mole Ratio
4 3 2
Mass (m) 1150gMolar Mass (M)
26.98g/mol
Moles (n)
Step 3: Convert given mass into moles (n=m/M)
= 42.59 mol of Al
Moles of Al = 1150g
27.0g/mol
Part 1: Find the theoretical yield through stoichmetry (how much you should have got)
Fill in chart with information you know
Balanced
Equation4Al 3O2
2Al2O3
Mole Ratio
4 3 2
Mass (m) 1150gMolar Mass (M)
26.98g/mol
Moles (n) 42.59 mol
Step 4: Use moles of given substance to find moles of required substance (mole to mole ratio)
4Al + 3O2 2Al2O3 x mols of Al2O3 = 2 mol of Al2O3 42.59 mol of Al 4mol of Al
42.59 mols of Al 2 mol of Al2O3
4 mol Al
or
X
= 21.31 mols of Al2O3
= 21.31 mols of Al2O3
Fill in chart with information you know
Balanced
Equation4Al 3O2
2Al2O3
Mole Ratio
4 3 2
Mass (m) 1150gMolar Mass (M)
26.98g/mol 102g/mol
Moles (n) 42.59 mol
21.31 mol
Step 5: Convert moles of required substance to the mass (m=n x M)
mass of Al2O3 = 21.31 mol of Al2O3 X 102g/mol
= 2173 g of Al2O3
This is your theoretical yield
Part 2: Calculate Percent Yield
% yield = x 100
= x 100 = 29.2 %
Experimental yield theoretical yield
635 g 2173g
Step 6: Using experimental and theoretical yield find percent yield
Iron (III) oxide reacts with carbon monoxide to produce carbon dioxide and iron. If 300g of iron is produced when 425 of iron ore is used. What is the percentage yield?
Step 1: Write out the balanced chemical equation
Fe2O3 + 3CO 2Fe + 3CO2
m= 425g m= ?
Example 3:
• Step 2: Fill in chart with information you know
Balanced
EquationFe2O3 3CO 2Fe 3CO2
Mole Ratio
1 3 2 3
Mass (m) 425g ? g
Molar Mass (M)
159.7 g/mol
Moles (n)
Step 3: Convert given mass into moles (n=m/M)
Moles of Fe2O3 = 425g
159.7g/mol= 2.66 mol of Fe2O3
Part 1: Find the theoretical yield through stoichmetry (how much you should have got)
• Fill in chart with information
Balanced
EquationFe2O3 3CO 2Fe 3CO2
Mole Ratio
1 3 2 3
Mass (m) 425g ? g
Molar Mass (M)
159.7 g/mol
Moles (n) 2.66mol
Step 4: Calculate the number of mols of the required substance (mol to mole ratio)
Fe2O3 + 3CO 2Fe + 3CO2
x mols of Fe = 2 mol of Fe 2.66 mols of Fe2O3 1mol of Fe2O3
2.66 mols of Fe2O3 2 mol of Fe
1 mols Fe2O3
or
X
= 5.32 mols of Fe
= 5.32 mols of Fe
• Fill in chart with information
Balanced
EquationFe2O3 3CO 2Fe 3CO2
Mole Ratio
1 3 2 3
Mass (m) 425g ? g
Molar Mass (M)
159.7 g/mol
55.85 g/mol
Moles (n) 2.66mol 5.32 mol
Step 5: Convert moles of required substance to the mass
mass of Fe = 5.32 mols of Fe X 55.85g/mol of Fe
= 297g of Fe
Part 2: use percent yield formula
% yield = x 100
= x 100 = 101 %
Experimental yield theoretical yield
300 g 297g
Step 6: Using experimental and theoretical yield find percent yield