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Percent Yield Definition and Formula

What is Percent Yield?

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Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
B.A., Physics and Mathematics, Hastings College

Percent yield is the percent ratio of actual yield to the theoretical yield. It is calculated to be the experimental yield divided by the theoretical yield multiplied by 100%. If the actual and theoretical yield are the same, percent yield is 100%. Usually, percent yield is lower than 100% because the actual yield is often less than the theoretical value. Reasons for this can include incomplete or competing reactions and loss of sample during recovery.

It's possible for percent yield to be over 100%, which means more sample was recovered from a reaction than predicted. This can happen when other reactions occurred that also formed the product. It can also be a source of error if the excess is due to incomplete removal of water or other impurities from the sample. Percent yield is always a positive value.

Also Known As: percentage yield

What is the Percent Yield Formula?

The equation for percent yield is:

percent yield = (actual yield/theoretical yield) x 100%

actual yield is the amount of product obtained from a chemical reaction
theoretical yield is the amount of product obtained from the stoichiometric or balanced equation , using the limiting reactant to determine product

Units for both actual and theoretical yield need to be the same (moles or grams).

Example Percent Yield Calculation

For example, the decomposition of magnesium carbonate forms 15 grams of magnesium oxide in an experiment. The theoretical yield is known to be 19 grams. What is the percent yield of magnesium oxide?

MgCO 3 → MgO + CO 2

The calculation is simple if you know the actual and theoretical yields. All you need to do is plug the values into the formula:

percent yield = actual yield / theoretical yield x 100%

percent yield = 15 g / 19 g x 100%

percent yield = 79%

Usually, you have to calculate the theoretical yield based on the balanced equation. In this equation, the reactant and the product have a 1:1 mole ratio , so if you know the amount of reactant, you know the theoretical yield is the same value in moles (not grams!). You take the number of grams of reactant you have, convert it to moles, and then use this number of moles to find out how many grams of product to expect.

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How to Calculate Theoretical Yield – Definition and Example

The theoretical yield of a chemical reaction is the amount of product you get if the reactants fully react. Here are the steps for calculating theoretical yield, along with a worked example problem.

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 (KClO 3 ) produces oxygen gas (O 2 ) and potassium chloride (KCl).

2 KClO 3 (s) → 3 O 2 (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 KClO 3 produce 3 moles of O 2 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 KClO 3 ?

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 KClO 3 . Do this by converting grams KClO 3 to moles KClO 3 . To make this easier, know the molecular mass of KClO 3 is 122.55 g/mol.

6 = x moles KClO 3

Use the chemical equation to relate moles KClO 3 to moles O 2 . This is the mole ratio between the two compounds. We see 2 moles of KClO 3 produces 3 moles of O 2 gas. Use the mole ratio and find the number of moles of oxygen formed by 6 moles of potassium chlorate.

x moles O 2 = 3 x 3 moles O 2 x moles O 2 = 9 moles O 2

6 moles of KClO 3 (735.3 grams of KClO 3 ) produce 9 moles of O 2 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 O 2 molecule.

x grams O 2 = (2)(16.00 grams O 2 /mole) x grams O 2 = 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 O 2 = (9 moles)(32 grams/mole) theoretical yield of O 2 = 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:

H 2 (g) + O 2 (g) → H 2 O(l)

Balancing the equation yields the mole ratios:

2 H 2 (g) + O 2 (g) → 2 H 2 O(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 H 2 O = (90 grams H 2 O)(1 mole H 2 O/18.00 grams H 2 O) moles H 2 O = 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 H 2 and H 2 O. So, 5 moles of water comes from reacting 5 moles of hydrogen.

However, there is a 1:2 ratio between the moles of O 2 and H 2 O. You need half the number of moles of oxygen gas compared to the number of moles of water.

moles O 2 = (mole ratio)(moles water) moles O 2 = (1 mol O 2 /2 mol H 2 O)(5 mol H 2 O) moles O 2 = 2.5 mol

Step 5: Convert moles to grams.

grams H 2 = (moles H 2 )(2 g H 2 /1 mol H 2 ) grams H 2 = (5 moles H 2 )(2 g H 2 /1 mol H 2 ) grams H 2 = (5 moles H 2 )(2 g H 2 /1 mol H 2 ) grams H 2 = 10 grams

grams O 2 = (moles O 2 )(32 g O 2 /1 mol O 2 ) grams O 2 = (2.5 mol O 2 )(32 g O 2 /1 mol O 2 ) grams O 2 = 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.