Converting Between Moles, Molecules and Atoms

The most common mistake newcomers make is treating the conversion from moles to molecules as a separate, special calculation. It is not. The mole exists to count particles. One mole always contains exactly 6.02214076 × 10²³ particles, regardless of whether those particles are atoms, molecules, ions, or formula units. That number is Avogadro's constant, fixed by the 2019 SI redefinition and published by BIPM and CODATA. Converting between moles, molecules, and atoms is a single multiplication or division by that constant, nothing more.

Three conversions matter in first-year chemistry and lab work: amount of substance (moles) to number of molecules, molecules to atoms (using the atoms per formula unit), and the full chain from a mass on a balance to the number of elementary entities in the sample. Every instruction here assumes you have the periodic table or a reliable molar mass value for the substance you are working with.

The Exact Number Since 2019

Avogadro's constant (N_A) is 6.02214076 × 10²³ mol⁻¹. That number is exact. Since 20 May 2019, the SI defines one mole as exactly that many elementary entities. There is no uncertainty in the constant itself. A sample of 2.0 mol of any substance contains 2.0 × 6.02214076 × 10²³ particles. A sample of 0.50 mol contains half that number. The constant is the same for water, iron, glucose, or sodium chloride.

Rounded Value For Routine Work

The value you see in older textbooks, 6.022 × 10²³, is a rounding of the exact constant. For most homework problems and routine lab work, the rounded value produces an error of about 0.0023%, which is negligible. For high-precision work where you need the correct number of significant figures, use the full 6.02214076 × 10²³. The BIPM SI Brochure 9th edition is the authoritative source for this value.

Moles To Molecules

To convert moles to molecules, multiply the number of moles by Avogadro's constant.

Worked example: Convert 0.75 mol of carbon dioxide (CO₂) to molecules.

Molecules = 0.75 mol × 6.02214076 × 10²³ mol⁻¹ = 4.51660557 × 10²³ molecules. Rounded to two significant figures (matching the input), the answer is 4.5 × 10²³ molecules.

Molecules To Moles

To convert molecules to moles, divide the number of molecules by Avogadro's constant.

Worked example: A sample contains 1.5 × 10²⁴ molecules of water (H₂O). How many moles is that?

Moles = (1.5 × 10²⁴) ÷ (6.02214076 × 10²³ mol⁻¹) = 2.49 mol. Rounded to two significant figures, 2.5 mol.

The failure case: forgetting that Avogadro's constant has units of mol⁻¹. If you write the division backwards, you get a number that is 6 × 10²³ times too large or too small. Write the units with every factor in your work to catch this error.

The conversion from molecules to moles uses the same equation, solved for n instead of N. If you are given the number of molecules and need the amount of substance, divide by Avogadro's constant. The formula n = N ÷ N_A is direct and unambiguous.

The confusion pair that trips up most students is mole versus molecule. A mole is a counting unit, 6.02214076 × 10²³ of something. A molecule is a single particle. When a problem asks 'how many molecules are in 2.5 mol of ammonia (NH₃)?', the answer is 2.5 × 6.02214076 × 10²³ molecules, not 2.5 molecules.

Worked example: A reaction produces 3.0 × 10²¹ molecules of oxygen gas (O₂). Convert this to moles.

Moles = (3.0 × 10²¹) ÷ (6.02214076 × 10²³ mol⁻¹) = 4.98 × 10⁻³ mol. Rounded to two significant figures, 5.0 × 10⁻³ mol (5.0 mmol).

When a problem asks for the number of atoms in a sample, you first find the number of molecules, then multiply by the number of atoms per molecule. This is the step where the formula mass of the compound matters, specifically, the number of atoms in one formula unit.

Worked example: How many atoms are in 0.20 mol of water (H₂O)?

First, find the number of molecules: 0.20 mol × 6.02214076 × 10²³ mol⁻¹ = 1.204428152 × 10²³ molecules. Each water molecule contains three atoms (two hydrogen, one oxygen). Multiply molecules by three: 1.204428152 × 10²³ × 3 = 3.613284456 × 10²³ atoms. Rounded to two significant figures, 3.6 × 10²³ atoms.

Worked example: How many oxygen atoms are in 1.5 mol of carbon dioxide (CO₂)?

Molecules = 1.5 × 6.02214076 × 10²³ = 9.03321114 × 10²³ molecules CO₂. Each CO₂ molecule contains two oxygen atoms. Oxygen atoms = 9.03321114 × 10²³ × 2 = 1.806642228 × 10²⁴ atoms. Rounded to two significant figures, 1.8 × 10²⁴ oxygen atoms.

The failure case: using the atomic mass of an element instead of the molecular mass when the substance is diatomic or polyatomic. For oxygen gas (O₂), the molar mass is 31.998 g/mol, not 16.00 g/mol. The number of atoms per molecule is two, not one.

The conversion from grams to atoms or molecules requires two steps: grams to moles, then moles to particles. The first step uses the formula n = m ÷ M, where m is the mass in grams and M is the molar mass in grams per mole. The second step multiplies moles by Avogadro's constant.

Worked example: A lab technician has 5.00 g of sodium chloride (NaCl, molar mass 58.44 g/mol). How many formula units of NaCl are in the sample?

Step 1: Moles = 5.00 g ÷ 58.44 g/mol = 0.0856 mol.

Step 2: Formula units = 0.0856 mol × 6.02214076 × 10²³ mol⁻¹ = 5.15 × 10²² formula units.

Worked example: How many atoms are in 12.0 g of carbon (C, molar mass 12.011 g/mol)?

Step 1: Moles = 12.0 g ÷ 12.011 g/mol = 0.999 mol.

Step 2: Atoms = 0.999 mol × 6.02214076 × 10²³ mol⁻¹ = 6.02 × 10²³ atoms.

Grams ↔ Moles ↔ Particles flow chart:

Given mass (g) → divide by molar mass (g/mol) → moles (mol) → multiply by Avogadro's constant (mol⁻¹) → number of particles.

Reverse direction: Given number of particles → divide by Avogadro's constant → moles → multiply by molar mass → mass in grams.

The failure case: using mass in kilograms instead of grams. The molar mass is defined in grams per mole. If you have 0.050 kg of a substance, convert to 50 g before dividing by the molar mass.

Hydrated compounds add a complication to the grams-to-atoms conversion. The molar mass must include the water of hydration. For example, copper(II) sulfate pentahydrate (CuSO₄·5H₂O) has five water molecules per formula unit. The molar mass is the sum of CuSO₄ (159.609 g/mol) plus 5 × 18.015 g/mol (water) = 249.684 g/mol. Forgetting the water of hydration underestimates the molar mass by about 36%.

Worked example: A sample contains 10.00 g of CuSO₄·5H₂O. How many copper atoms are in the sample?

Moles of CuSO₄·5H₂O = 10.00 g ÷ 249.684 g/mol = 0.04005 mol. Each formula unit contains one copper atom. Copper atoms = 0.04005 mol × 6.02214076 × 10²³ mol⁻¹ = 2.412 × 10²² atoms.

If you used the anhydrous molar mass (159.609 g/mol), you would calculate 0.0626 mol and 3.77 × 10²² atoms, an error of over 56%.

Substance	Formula	Molar Mass (g/mol)	Common Error
Water	H₂O	18.015	Using 18.0; acceptable for homework but imprecise for lab work
Sodium chloride	NaCl	58.44	Using 58.5; the IUPAC value is 58.44 g/mol
Carbon dioxide	CO₂	44.01	Using 44.0; atomic weights of C (12.011) + 2×O (15.999) = 44.009, rounded to 44.01
Glucose	C₆H₁₂O₆	180.156	Forgetting the six water molecules in a hydrate that is actually the anhydrous form
Sodium hydroxide	NaOH	39.997	Using 40.0; close but NaOH absorbs water and CO₂, changing effective molar mass in practice
Calcium carbonate	CaCO₃	100.09	Using 100.0; Ca (40.078) + C (12.011) + 3×O (15.999) = 100.086, rounded to 100.09
Sulfuric acid	H₂SO₄	98.079	Using 98.0; the precise value matters for molarity calculations
Ethanol	C₂H₅OH	46.07	Using 46.0; common in high school texts
Ammonia	NH₃	17.031	Using 17.0; N (14.007) + 3×H (1.008) = 17.031
Oxygen gas	O₂	31.998	Using 16.00 g/mol for O instead of 32.00 for O₂; a very common failure

Avogadro's constant is a large number, and your calculator will display results in scientific notation. Enter the constant as 6.02214076E23 on most calculators. For the reverse conversion, enter the number of molecules, then divide by 6.02214076E23.

When multiplying 0.75 mol by 6.02214076 × 10²³, your calculator shows 4.51660557E23. Write this as 4.51660557 × 10²³ and round to the same number of significant figures as the least precise input. The input 0.75 has two significant figures, so the answer is 4.5 × 10²³.

The failure case: misreading the exponent. A number like 6.02 × 10²¹ is 100 times smaller than 6.02 × 10²³. Check the exponent every time. A difference of two in the exponent is a factor of 100 in the answer.

In reaction stoichiometry, the conversion chain is: mass of reactant A → moles of A (via molar mass) → mole ratio (from balanced equation) → moles of product B → mass of product B (via molar mass of B). The mole concept connects every step. OpenStax Chemistry 2e, sections 3.1 and 4.3, covers this chain in detail.

Worked example: How many grams of carbon dioxide are produced from 10.00 g of glucose (C₆H₁₂O₆, 180.156 g/mol) in complete combustion? The balanced equation is C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

Moles of glucose = 10.00 g ÷ 180.156 g/mol = 0.05551 mol. The mole ratio is 1 mol glucose : 6 mol CO₂. Moles of CO₂ = 0.05551 mol × 6 = 0.3331 mol. Mass of CO₂ = 0.3331 mol × 44.01 g/mol = 14.66 g.

If you use the wrong mole ratio, for example, 1:1 instead of 1:6, the mass of CO₂ calculated would be 2.44 g, missing by a factor of six.

The single thing that most often goes wrong is using the wrong molar mass. The failure cases are predictable: using the atomic mass of an element instead of the molecular mass of its diatomic form, using the anhydrous molar mass for a hydrated compound, or using the mass of the solute without accounting for the water of hydration. Every conversion starts with the correct molar mass from the periodic table or the IUPAC CIAAW standard atomic weights. Verify the formula of the substance before you multiply or divide.