Steps to Calculate Theoretical Yield
- Write the balanced chemical equation for the reaction.
- Identify the limiting reactant.
- Convert grams of limiting reactant to moles.
- Use the mole ratio between the limiting reactant and the product and find the theoretical number of moles of product.
- Convert the number of moles of product to grams.
Sometimes you’ll know some of these steps without having to figure them out. For example, you might know the balanced equation or be given the limiting reactant. For example, when one reactant is “in excess,” you know the other (if there are only two reactants) is the limiting reactant.
Theoretical Yield Example Problem
Let’s look at the following reaction where heating potassium chlorate (KClO3) produces oxygen gas (O2) and potassium chloride (KCl).
2 KClO3 (s) → 3 O2 (g) + 2 KCl (s)
This reaction is fairly common in school laboratories since it is a relatively inexpensive method of obtaining oxygen gas.
The balanced reaction shows that 2 moles of KClO3 produce 3 moles of O2 and 2 moles of KCl. To calculate the theoretical yield, you use these ratios as a conversion factor. Here is a typical example problem.
Question: How many moles of oxygen gas will be produced from heating 735.3 grams of KClO3?
The problem gives the balanced equation and identifies the limiting reactant (in this case, the only reactant), so now we need to know the number of moles of KClO3. Do this by converting grams KClO3 to moles KClO3. To make this easier, know the molecular mass of KClO3 is 122.55 g/mol.
6 = x moles KClO3
Use the chemical equation to relate moles KClO3 to moles O2. This is the mole ratio between the two compounds. We see 2 moles of KClO3 produces 3 moles of O2 gas. Use the mole ratio and find the number of moles of oxygen formed by 6 moles of potassium chlorate.
x moles O2 = 3 x 3 moles O2
x moles O2 = 9 moles O2
6 moles of KClO3 (735.3 grams of KClO3) produce 9 moles of O2 gas.
Technically, this is the theoretical yield, but the answer becomes more useful when you convert moles to grams. Use the atomic mass of oxygen and the molecular formula for the conversion. From the periodic table, the atomic mass of oxygen is 16.00. There are two oxygen atoms in each O2 molecule.
x grams O2 = (2)(16.00 grams O2/mole)
x grams O2 = 32 g/mol
Finally, the theoretical yield is the number of moles of oxygen gas multiplied by the moles-to-grams conversion factor:
theoretical yield of O2 = (9 moles)(32 grams/mole)
theoretical yield of O2 = 288 grams
Calculate Reactant Needed to Make Product
A variation of the theoretical yield calculation helps you find how much reactant you use when you want a predetermined amount of product. Here again, start with the balanced equation and use the mole ratio between reactant and product.
Question: How many grams of hydrogen gas and oxygen gas are needed to produce 90 grams of water?
Step 1: Write the balanced equation.
Start with the unbalanced equation. Hydrogen gas and oxygen gas react, producing water:
H2(g) + O2(g) → H2O(l)
Balancing the equation yields the mole ratios:
2 H2(g) + O2(g) → 2 H2O(l)
Step 2: Identify the limiting reactant.
Well, in this case, the amount of product (water) is your limit because you’re working the reaction backwards.
Step 3: Convert grams of limiting reactant to moles.
moles H2O = (90 grams H2O)(1 mole H2O/18.00 grams H2O)
moles H2O = 5 moles
Step 4: Use the mole ratio.
From the balanced equation, there is a 1:1 mole relationship between the number of moles of H2 and H2O. So, 5 moles of water comes from reacting 5 moles of hydrogen.
However, there is a 1:2 ratio between the moles of O2 and H2O. You need half the number of moles of oxygen gas compared to the number of moles of water.
moles O2 = (mole ratio)(moles water)
moles O2 = (1 mol O2/2 mol H2O)(5 mol H2O)
moles O2 = 2.5 mol
Step 5: Convert moles to grams.
grams H2 = (moles H2)(2 g H2/1 mol H2)
grams H2 = (5 moles H2)(2 g H2/1 mol H2)
grams H2 = (5 moles H2)(2 g H2/1 mol H2)
grams H2 = 10 grams
grams O2 = (moles O2)(32 g O2/1 mol O2)
grams O2 = (2.5 mol O2)(32 g O2/1 mol O2)
grams O2 = 80 grams
So, you need 10 grams of hydrogen gas and 80 grams of oxygen gas to make 90 grams of water.
- Petrucci, R.H., Harwood, W.S.; Herring, F.G. (2002) General Chemistry (8th ed.). Prentice Hall. ISBN 0130143294.
- Vogel, A. I.; Tatchell, A. R.; Furnis, B. S.; Hannaford, A. J.; Smith, P. W. G. (1996) Vogel’s Textbook of Practical Organic Chemistry (5th ed.). Pearson. ISBN 978-0582462366.
- Whitten, K.W., Gailey, K.D; Davis, R.E. (1992) General Chemistry (4th ed.). Saunders College Publishing. ISBN 0030723736.