Before performing chemical reactions, it is helpful to know how much product will be produced with given quantities of reactants. This is known as the **theoretical yield**. This is a strategy to use when calculating the theoretical yield of a chemical reaction. The same strategy can be applied to determine the amount of each reagent needed to produce a desired amount of product.

## Theoretical Yield Sample Calculation

10 grams of hydrogen gas are burned in the presence of excess oxygen gas to produce water. How much water is produced?

The reaction where hydrogen gas combines with oxygen gas to produce water is:

H_{2}(g) + O_{2}(g) → H_{2}O(l)

**Step 1: Make sure your chemical equations are balanced equations.**

The equation above is not balanced. After balancing, the equation becomes:

2 H_{2}(g) + O_{2}(g) → 2 H_{2}O(l)

**Step 2: Determine the mole ratios between the reactants and the product.**

This value is the bridge between the reactant and the product.

The mole ratio is the stoichiometric ratio between the amount of one compound and the amount of another compound in a reaction. For this reaction, for every two moles of hydrogen gas used, two moles of water are produced. The mole ratio between H_{2} and H_{2}O is 1 mol H_{2}/1 mol H_{2}O.

**Step 3: Calculate the theoretical yield of the reaction.**

There is now enough information to determine the theoretical yield. Use the strategy:

- Use molar mass of reactant to convert grams of reactant to moles of reactant
- Use the mole ratio between reactant and product to convert moles reactant to moles product
- Use the molar mass of the product to convert moles product to grams of product.

In equation form:

grams product = grams reactant x (1 mol reactant/molar mass of reactant) x (mole ratio product/reactant) x (molar mass of product/1 mol product)

The theoretical yield of our reaction is calculated using:

- molar mass of H
_{2}gas = 2 grams - molar mass of H
_{2}O = 18 grams

grams H_{2}O = grams H_{2}x (1 mol H_{2}/2 grams H_{2}) x (1 mol H_{2}O/1 mol H_{2}) x (18 grams H_{2}O/1 mol H_{2}O)

We had 10 grams of H_{2} gas, so:

grams H_{2}O = 10 g H_{2}x (1 mol H_{2}/2 g H_{2}) x (1 mol H_{2}O/1 mol H_{2}) x (18 g H_{2}O/1 mol H_{2}O)

All units except grams H_{2}O cancel out, leaving:

grams H_{2}O = (10 x 1/2 x 1 x 18) grams H_{2}O

grams H_{2}O = 90 grams H_{2}O

Ten grams of hydrogen gas with excess oxygen will theoretically produce 90 grams of water.

## Calculate Reactant Needed to Make a Set Amount of Product

This strategy can be slightly modified to calculate the amount of reactants needed to produce a set amount of product. Let's change our example slightly: How many grams of hydrogen gas and oxygen gas are needed to produce 90 grams of water?

We know the amount of hydrogen needed by the first example, but to do the calculation:

grams reactant = grams product x (1 mol product/molar mass product) x (mole ratio reactant/product) x (grams reactant/molar mass reactant)

For hydrogen gas:

grams H_{2}= 90 grams H_{2}O x (1 mol H_{2}O/18 g) x (1 mol H_{2}/1 mol H_{2}O) x (2 g H_{2}/1 mol H_{2})

grams H_{2}= (90 x 1/18 x 1 x 2) grams H_{2}grams H_{2}= 10 grams H_{2}

This agrees with the first example. To determine the amount of oxygen needed, the mole ratio of oxygen to water is needed. For every mole of oxygen gas used, 2 moles of water are produced. The mole ratio between oxygen gas and water is 1 mol O^{2}/2 mol H_{2}O.

The equation for grams O_{2} becomes:

grams O_{2}= 90 grams H_{2}O x (1 mol H_{2}O/18 g) x (1 mol O_{2}/2 mol H_{2}O) x (32 g O_{2}/1 mol H_{2})

grams O_{2}= (90 x 1/18 x 1/2 x 32) grams O_{2}

grams O_{2}= 80 grams O_{2}

To produce 90 grams of water, 10 grams of hydrogen gas and 80 grams of oxygen gas are needed.

Theoretical yield calculations are straightforward as long as you have balanced equations to find the mole ratios needed to bridge the reactants and the product.

## Theoretical Yield Quick Review

- Balance your equations.
- Find the mole ratio between the reactant and the product.
- Calculate using the following strategy: Convert grams to moles, use the mole ratio to bridge products and reactants, and then convert moles back to grams. In other words, work with moles and then convert them to grams. Don't work with grams and assume you'll get the right answer.

For more examples, examine the theoretical yield worked problem and aqueous solution chemical reaction example problems.

## Sources

- Petrucci, R.H., Harwood, W.S. and Herring, F.G. (2002)
*General Chemistry*, 8th Edition. 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. and Davis, R.E. (1992)
*General Chemistry*, 4th Edition. Saunders College Publishing. ISBN 0030723736.