Moles, Avogadro's number and percentage yield

A mole is the amount of substance that contains 6.022 × 10²³ particles (Avogadro's number), and moles = mass ÷ molar mass. Chemists can't count atoms and molecules one by one, so they count them in this very large, fixed bundle. Once you can convert between grams and moles, you can work out how much of one chemical reacts with another, how much product a reaction should give, and how much it actually gave. This guide covers the mole, Avogadro's number, how to calculate moles, theoretical and percentage yield, and atom economy.

Terms are defined in the glossary.

What is a mole in chemistry?

A mole is a unit for an amount of substance, like a dozen is a unit for eggs, only far larger. One mole of anything contains 6.022 × 10²³ particles: atoms, molecules or ions, depending on what you are counting. Its symbol is mol.

The mole is useful because it links the number of particles to a mass you can weigh. One mole of a substance weighs its molar mass in grams:

Each of these contains the same number of molecules or formula units, 6.022 × 10²³. A mole of water and a mole of salt weigh different amounts because their particles have different masses. For how molar mass is worked out from a formula, see molar mass, molecular weight and monoisotopic mass.

Avogadro's number

Avogadro's number is the number of particles in one mole: 6.022 140 76 × 10²³. Strictly, the Avogadro constant is that number per mole (6.022 140 76 × 10²³ mol⁻¹), and Avogadro's number is the same figure without the unit. In everyday use the two names mean the same thing.

Since the SI units were redefined in 2019, the mole has been defined by this number. One mole is exactly 6.022 140 76 × 10²³ particles (BIPM, SI Brochure). Before 2019 it was defined as the number of atoms in 12 g of carbon-12, which gave almost exactly the same value.

To get a sense of the scale: one molecule of water weighs about 3 × 10⁻²³ g. You need 6 × 10²³ of them to make up 18 g.

How to calculate moles

The central equation links mass, moles and molar mass:

moles = mass (g) ÷ molar mass (g/mol)

It rearranges to give the other two:

• mass (g) = moles × molar mass
• molar mass = mass ÷ moles

Worked examples

From mass to moles. How many moles are in 5.844 g of sodium chloride?

5.844 g ÷ 58.44 g/mol = 0.1000 mol

From moles to mass. What mass of acetic acid is 0.25 mol?

0.25 mol × 60.05 g/mol = 15.0 g

From moles to particles. How many molecules are in 0.1000 mol?

0.1000 × 6.022 × 10²³ = 6.022 × 10²²

Moles in solution

In a solution, the amount is usually given as a concentration in moles per litre, or molarity:

moles = concentration (mol/L) × volume (L)

So 250 mL of a 0.1 M solution contains 0.1 × 0.250 = 0.025 mol. For making up solutions and diluting them, see molarity, mg/mL and dilutions and the molarity calculator.

Reacting amounts: why moles matter

Chemical equations describe reactions in moles, not grams. Take acetic acid reacting with sodium hydroxide to make sodium acetate and water:

CH₃COOH + NaOH → CH₃COONa + H₂O

The equation says one mole of acetic acid reacts with one mole of sodium hydroxide. It doesn't say one gram reacts with one gram, because the two have different molar masses (60.05 and 40.00 g/mol). Converting masses to moles is the only way to know whether you have the right proportions.

The limiting reagent

When reactants aren't present in exactly the right ratio, one runs out first. That one is the limiting reagent. It sets the most product the reaction can make. The others are in excess, and some of each is left over.

To find the limiting reagent, convert each reactant to moles and divide by its number in the equation. The smallest result is the limiting reagent.

Percentage yield

The theoretical yield is the amount of product you would get if every molecule of the limiting reagent reacted perfectly and nothing was lost. The actual yield is what you weigh at the end. Percentage yield compares the two.

Percentage yield formula

percentage yield = (actual yield ÷ theoretical yield) × 100

Both yields must be in the same units, usually grams. The theoretical yield comes from the balanced equation, as shown in the next section.

Worked example

6.005 g of acetic acid is reacted with excess sodium hydroxide, and the sodium acetate is dried and weighed.

1. Moles of the limiting reagent: 6.005 g ÷ 60.05 g/mol = 0.1000 mol of acetic acid.
2. Moles of product expected: the equation is one to one, so 0.1000 mol of sodium acetate.
3. Theoretical yield: 0.1000 mol × 82.03 g/mol = 8.203 g.
4. Actual yield: 7.38 g is recovered.
5. Percentage yield: 7.38 ÷ 8.203 × 100 = 90.0%.

How to calculate theoretical yield

1. Write the balanced equation. See balancing chemical equations.
2. Convert the limiting reagent to moles: mass ÷ molar mass.
3. Use the equation's ratio to find the moles of product.
4. Convert back to grams: moles × the product's molar mass.

Sodium carbonate reacting with excess hydrochloric acid shows why step 3 matters:

Na₂CO₃ + 2 HCl → 2 NaCl + H₂O + CO₂

• 5.30 g of sodium carbonate ÷ 105.99 g/mol = 0.0500 mol (PubChem).
• The equation gives two moles of sodium chloride for each mole of sodium carbonate, so 0.1000 mol.
• 0.1000 mol × 58.44 g/mol = a theoretical yield of 5.84 g of sodium chloride (PubChem).

Forgetting the 2 in the equation would halve the answer.

Why yields fall short of 100%

• Losses in handling: product left on glassware, filter paper or in the solution.
• Incomplete reactions: some reactant never reacts.
• Side reactions: some reactant forms something else.
• Purification: each step that removes impurities also removes some product.

When a yield is over 100%

A percentage yield above 100% doesn't mean the reaction made matter from nothing. It means the weighed product contains something else: usually water or solvent that wasn't dried off, or an impurity. It is a sign to check the product's purity and water content.

Atom economy

Atom economy is the share of the total mass of the reactants that ends up in the product you want (Cognito, atom economy). The idea was put forward by Barry Trost as a measure of how efficient a synthesis is (Trost, Science, 1991):

atom economy = (molar mass of desired product ÷ total molar mass of all reactants) × 100

Use the numbers in front of each formula in the balanced equation.

Worked example. In CH₃COOH + NaOH → CH₃COONa + H₂O, the desired product is sodium acetate:

• Reactants: 60.05 + 40.00 = 100.05 g/mol.
• Desired product: 82.03 g/mol.
• Atom economy: 82.03 ÷ 100.05 × 100 = 82.0%. The other 18% leaves as water.

Atom economy and percentage yield measure different things. Atom economy is worked out from the equation alone and shows how much of the starting material could end up in the product. Percentage yield is measured after the reaction and shows how much actually did. A reaction can have a high yield and a poor atom economy, or the reverse. A reaction with only one product has an atom economy of 100%.

Yield in peptide synthesis

Peptides are built one amino acid at a time by solid-phase peptide synthesis. Each amino acid needs its own coupling step, and no step is perfect. The small losses multiply.

Each coupling adds one peptide bond to a growing polypeptide chain. If each coupling works on a fraction p of the chains, the share of chains that are complete after n couplings is pⁿ. A peptide of a given length needs one fewer coupling than it has residues:

These figures are a simplified illustration. They ignore losses in deprotection, cleavage and purification, which lower the final yield much further. But they show why a long sequence is harder and dearer to make than a short one. The chains that fail aren't lost: they become impurities, such as sequences with one residue missing, that purification then has to remove. See reading impurity peaks for how they show up.

Related guides

For turning moles into solution concentrations, see molarity, mg/mL and dilutions. For working out molar mass from a formula or sequence, see molar mass, molecular weight and monoisotopic mass and the molecular weight calculator. For rounding results sensibly, see accuracy, precision and significant figures. For working out a formula from percentage composition, see empirical and molecular formulas. For writing and balancing the equation before working out amounts, see balancing chemical equations.