CHEMISTRY · LESSON 07 OF 12
Moles, molar mass and chemical formulas
The mole links the particles in a formula to the grams on a balance: n = m/M and N = nNA. Formulas give atom ratios, percentage composition and empirical formulas.
What this lesson explains
The mole connects the particles in an atomic model to the mass measured on a balance. Start by identifying what is being counted: atoms, molecules, ions or formula units. The same number of moles can have different masses because the particles have different masses.
Reaction stoichiometry, solution concentrations and gas amounts are all expressed in moles. This lesson is the bridge from formulas to every later calculation.
Before you begin
Reading a formula
Subscripts count atoms in one formula unit; parentheses multiply a group (see Writing formulas and naming compounds).
Measurement and units
Carry units through every step (g ÷ g/mol = mol) and round only the final answer to the least precise measurement. Mass is measured on a balance in grams; volumes in mL or L.
Average atomic mass
Periodic-table atomic masses are isotope-weighted averages (see Atomic structure); they are the values used for molar masses.
The idea, made visible
One mole contains exactly 6.02214076 × 1023 specified entities. This number is the Avogadro constant NA in mol−1. An entity can be an atom, a molecule, an ion or a formula unit; always name it. One mole of O₂ molecules contains two moles of O atoms. One mole of NaCl formula units represents one mole each of Na⁺ and Cl⁻ ions in the ionic lattice.
A subscript belongs to the symbol immediately before it; a subscript outside parentheses multiplies everything inside. Ca(NO₃)₂ contains one Ca, two N and six O atoms per formula unit. A coefficient multiplies the whole formula: 3 Ca(NO₃)₂ represents three formula units, so the atom counts become 3, 6 and 18. Changing a coefficient changes the amount; changing a subscript changes the substance.
Molar mass M is mass per mole, usually expressed in g/mol. Add the listed atomic molar masses, with each multiplied by its atom count. Formula mass in u describes one formula unit; molar mass in g/mol describes a mole. Their numerical values agree to the precision used in introductory chemistry, but their units and meanings differ. Use your course’s periodic-table values consistently.
An empirical formula is the simplest whole-number ratio of atoms. A molecular formula gives the actual atom counts in one molecule. C₆H₁₂O₆ has empirical formula CH₂O. Ionic formulas describe neutral ratios in a lattice, not separate molecules. A measured percentage composition gives an atom ratio only after masses have been divided by the corresponding atomic molar masses.
| Formula | Ca atoms | N atoms | O atoms | Meaning |
|---|---|---|---|---|
| Ca(NO₃)₂ | 1 | 2 | 6 | One formula unit |
| 2 Ca(NO₃)₂ | 2 | 4 | 12 | Two formula units |
| 1 mol Ca(NO₃)₂ | 1 mol | 2 mol | 6 mol | Amounts of each atom type |
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| m | sample mass | g |
| namount | amount of substance; often written n | mol |
| M | molar mass | g/mol |
| N | number of specified entities | count |
| NA | Avogadro constant, exactly 6.02214076 × 1023 | mol−1 |
Mass to amount
namount = m / M; m = namountM
Conditions and limits: m and M must use the same mass unit. Here n means amount, not the neutron count in Atomic structure.
Amount to entities
N = namountNA; namount = N / NA
Conditions and limits: Identify the entity. Formula subscripts give the number of atoms of each element per molecule or formula unit.
Mass percentage
mass % of element = mass of that element in one mole of compound / molar mass of compound × 100%
Conditions and limits: Use all atoms of the element in the formula. Element percentages should sum to approximately 100%.
Molecular formula multiplier
k = measured molecular molar mass / empirical-formula molar mass
Conditions and limits: k should be close to a positive whole number within the measurement precision. Multiply every empirical subscript by k.
A first worked example
Mass, moles and particles
- Name the entity (atoms, molecules, formula units, ions).
- Find the molar mass from the formula: sum of (atom count × atomic molar mass).
- Convert grams → moles with n = m/M, moles → entities with N = nNA; reverse as needed.
- For composition: mass of element in one mole ÷ molar mass × 100%. For an empirical formula: grams → moles for each element, divide by the smallest, clear fractions.
Mass to moles to molecules
Problem. Find the amount and number of molecules in 9.008 g of water. Use H = 1.008 and O = 16.00 g/mol.
M(H₂O) = 2(1.008) + 16.00 = 18.016 g/mol.
The subscript 2 applies to H only.
n = 9.008/18.016 = 0.5000 mol.
Grams divided by grams per mole leaves moles.
N = 0.5000 × 6.02214076 × 1023 = 3.011 × 1023 molecules.
Multiply by the entity count per mole.
O atoms = N; H atoms = 2N = 6.022 × 1023.
Every water molecule contains one O and two H atoms.
Result: 0.5000 mol H₂O; 3.011 × 1023 molecules.
What it means: The number of hydrogen atoms is twice the number of water molecules, not the same number.
A different case
Parentheses and ionic formula units
Problem. Find the molar mass of calcium nitrate, Ca(NO₃)₂, and the amount of oxygen atoms in 0.250 mol of the compound. Use Ca = 40.08, N = 14.01 and O = 16.00 g/mol.
Counts: Ca = 1, N = 2, O = 6.
The outside 2 multiplies the complete nitrate group.
M = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol.
Add every atom’s contribution.
Amount of oxygen atoms = 6 × 0.250 = 1.50 mol.
Each formula unit contains six O atoms.
Result: M = 164.10 g/mol; 1.50 mol of oxygen atoms.
What it means: This counts oxygen atoms chemically present in nitrate; it does not mean 1.50 mol of free O₂ gas is produced.
More worked cases
Each case below uses a different skill. Every step and result is shown.
An empirical formula from percentage composition
Problem. A compound contains 40.00% C, 6.71% H and 53.29% O by mass. Find its empirical formula using C = 12.01, H = 1.008 and O = 16.00 g/mol.
Assume a 100.00 g sample: 40.00 g C, 6.71 g H and 53.29 g O.
Percentages then become convenient masses without changing the ratio.
Amounts: C = 40.00/12.01 = 3.331 mol; H = 6.71/1.008 = 6.657 mol; O = 53.29/16.00 = 3.331 mol.
Divide each mass by its own atomic molar mass.
Divide by the smallest amount: C:H:O ≈ 1:2:1.
Mole ratios equal atom-number ratios. The small mismatch comes from rounded percentages.
Result: Empirical formula CH₂O.
What it means: Do not use the mass ratio 40:6.71:53.29 as subscripts. For a ratio near 1:1.5, multiply all ratios by 2; do not round 1.5 to 2.
From empirical formula to molecular formula
Problem. A molecular compound has empirical formula CH₂O and measured molar mass about 180.16 g/mol. Find its molecular formula using the same atomic masses.
Empirical molar mass = 12.01 + 2(1.008) + 16.00 = 30.026 g/mol.
Use one empirical formula unit.
k = 180.16/30.026 ≈ 6.000.
The molecular formula is a whole-number multiple of the empirical formula.
Multiply every subscript by 6: C₆H₁₂O₆.
Preserve the 1:2:1 ratio.
Result: Molecular formula C₆H₁₂O₆.
What it means: Composition and molar mass identify a molecular formula, not a unique structure. Different compounds can share that formula.
Mass fraction from a formula
Problem. Find the oxygen mass percentage in water using the atomic masses above.
One mole of water contains 16.00 g oxygen in 18.016 g total.
There is one O atom per water molecule.
Oxygen mass percentage = (16.00/18.016) × 100% = 88.81%.
Compare masses, not the 1:2 oxygen-to-hydrogen atom ratio.
Result: Water is about 88.81% oxygen by mass with these atomic masses.
What it means: Hydrogen contributes two thirds of the atoms but only about 11.19% of the mass.
Common misunderstandings
Misunderstanding: Using an average atomic mass as a neutron count.
Correct idea: Average atomic mass is used for molar-mass calculations; neutron counts require the mass number of a specific isotope.
Misunderstanding: Calling a mole a mass.
Correct idea: The mole measures amount. The molar mass connects that amount to grams.
Misunderstanding: Rounding every empirical ratio immediately.
Correct idea: Keep several digits, identify a common multiplier, and check against the original composition.
Misunderstanding: Changing subscripts to balance a reaction.
Correct idea: Change coefficients; the formula of each substance stays fixed.
Keep in mind
- namount = m / M; m = namountMMass to amount
- N = namountNA; namount = N / NAAmount to entities
- mass % of element = mass of that element in one mole of compound / molar mass of compound × 100%Mass percentage
- k = measured molecular molar mass / empirical-formula molar massMolecular formula multiplier
Scope of this lesson
- Combustion analysis and hydrate formulas are not treated. Stoichiometry in reactions continues in Reactions in solution.
Next: Reactions in solution. In water, many compounds exist as ions. Reactions in solution — precipitation, acid–base and redox — are written as net ionic equations, and their amounts are calculated through moles using balanced equations.
Further reading: OpenStax Chemistry 2e — Formula mass and the mole · OpenStax Chemistry 2e — Empirical and molecular formulas. Sources and credits.