Lesson 8

The Mole and Chemical Formulas

Understand the mole concept, molar mass, percent composition, empirical formulas, and molecular formulas.

7 learning objectivesquantitative
Chemistry reference tables

The Mole and Avogadro's Number

Atoms and molecules are far too small to count individually, so chemists use a counting unit called the mole (mol). One mole of anything contains exactly 6.022 × 10⁠23 items — this is Avogadro's number (NA).

To put that in perspective, one mole of pennies stacked up would reach the sun and back more than a billion times. Yet one mole of water molecules is just 18 grams — roughly a tablespoon.

The mole bridges the gap between the atomic scale (where mass is measured in amu) and the laboratory scale (grams). This makes it the chemist's most essential conversion unit: every quantitative calculation in chemistry passes through moles at some point.

Go deeperHow anyone counted to Avogadro’s number

Nobody tallied a mole of atoms one by one; the number was pinned down by measuring in bulk and dividing. One modern route: grow a nearly perfect silicon sphere, measure its volume, then use X-ray diffraction to measure the exact spacing of atoms in the crystal, which tells you how many atoms fit inside. The counting became so precise that in 2019 the logic was inverted: the mole is now defined as exactly 6.02214076 × 10⁠23 entities, a fixed constant of the SI system rather than a measured property of carbon-12.

Go deeperWhy this matters: a sip of water outnumbers the ocean

A single glass of water holds more molecules than the entire ocean holds glasses of water. The arithmetic is quick: 250 mL of water is about 14 moles, roughly 8 × 10⁠24 molecules, while all of Earth’s oceans (about 1.3 × 10⁠21 L) contain only around 5 × 10⁠21 glass-sized portions. That gap of more than a thousandfold is why chemistry needs the mole: the objects are so absurdly numerous that only a bulk counting unit makes the numbers speakable.

Calculating Molar Mass

The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol). Numerically, it equals the formula mass in amu.

To calculate the molar mass of a compound, sum the atomic masses of every atom in the formula:

  1. Write the chemical formula (e.g., H⁠2O)
  2. Multiply each element's atomic mass by its subscript
  3. Add the contributions: H⁠2O = 2(1.008) + 1(16.00) = 18.02 g/mol

For ionic compounds, the same procedure applies using the formula unit. For example, NaCl has a molar mass of 22.99 + 35.45 = 58.44 g/mol.

Compounds with parenthesized polyatomic ions. When a polyatomic ion has an outer subscript, multiply the masses of every atom inside the parentheses by that subscript. For Fe(NO3)3: 1 Fe + 3 N + 9 O = 55.85 + 3(14.01) + 9(16.00) = 241.88 g/mol.

Hydrates include water of crystallization; the centered dot indicates that many H2O per formula unit. For CuSO4·5H2O: 63.55 + 32.07 + 4(16.00) + 5(18.02) = 249.72 g/mol.

Doubly-parenthesized double salts work the same way: each parenthesis distributes its subscript over its contents. For Fe(NH4)2(SO4)2: 1 Fe + 2 N + 8 H + 2 S + 8 O = 284.07 g/mol.

Go deeperCommon mistake: forgetting the parentheses multiply everything

In Ca(NO⁠3)⁠2, the subscript 2 multiplies the entire nitrate group: two N atoms and six O atoms, not two N and three O. The molar mass is 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol. Dropping the distribution (counting only three oxygens) gives 116.10, an error of nearly 30% that then poisons every downstream calculation.

Reliable habit: before summing, expand the formula into a full atom inventory (Ca: 1, N: 2, O: 6) and only then reach for the periodic table.

Converting Between Moles, Mass, and Particles

Three quantities sit at the heart of chemical calculations: mass (grams), moles, and number of particles. Two conversion factors connect them:

  • Molar mass (g/mol) converts between grams and moles
  • Avogadro's number (6.022 × 10⁠23/mol) converts between moles and particles

The conversion map looks like this:

Mass (g) ↔ ÷ or × molar mass ↔ Moles ↔ ÷ or × NA ↔ Particles

For example, to find how many molecules are in 36.0 g of water: 36.0 g ÷ 18.02 g/mol = 2.00 mol, then 2.00 mol × 6.022 × 10⁠23 = 1.20 × 10⁠24 molecules. Always pass through moles as the central hub when converting between mass and particle counts.

Go deeperCommon mistake: multiplying when you should divide

The fastest error-catcher in mole work is a sense of scale. Moles of a lab-sized sample should be a modest number (a few grams of most substances is well under one mole). Particle counts should be astronomical: anything visible contains upwards of 10⁠20 particles. So if your “number of molecules” comes out as 10⁠-23, you divided by Avogadro’s number where you should have multiplied; if your “moles” from a 5-gram sample come out in the hundreds, the molar mass went in upside down.

Sanity-check every answer against those two magnitudes before moving on; the check takes two seconds and catches the two most common inversions.

Percent Composition

Percent composition tells you the mass percentage of each element in a compound. It answers the question: what fraction of the compound's total mass comes from each element?

To calculate from a formula:

% element = (atoms of element × atomic mass) / molar mass of compound × 100%

For example, in water (H⁠2O, molar mass 18.02 g/mol):

  • % H = (2 × 1.008) / 18.02 × 100% = 11.19%
  • % O = 16.00 / 18.02 × 100% = 88.81%

Percent composition is useful for comparing how much of an active element different compounds provide. Farmers choose nitrogen fertilizers partly based on the mass percent of nitrogen: NH⁠3 (82.2% N) is much richer in nitrogen than NH⁠4NO⁠3 (35.0% N).

Go deeperWhy this matters: the numbers on a fertilizer bag

Farmers buy nitrogen by percent composition. Urea, CO(NH⁠2)⁠2, delivers 28.02 / 60.06 = 46.7% nitrogen by mass, the highest of any common solid fertilizer, while ammonium nitrate carries 35.0%. Those percentages, computed exactly the way this section teaches, determine shipping costs, application rates, and prices per acre.

The same calculation runs through ore refining (which iron ore is richest?) and nutrition labels. Percent composition is the answer to a very commercial question: how much of what I paid for is the element I actually wanted?

Empirical Formulas from Percent Composition

The empirical formula gives the simplest whole-number ratio of atoms in a compound. Determining it from experimental data follows a consistent procedure:

  1. Start with percent composition (or mass data). If given percentages, assume a 100-g sample so that percentages become grams directly.
  2. Convert grams of each element to moles by dividing by atomic mass.
  3. Divide every mole value by the smallest mole value to get mole ratios.
  4. If the ratios are not whole numbers, multiply all by the smallest integer that makes them whole (e.g., multiply by 2 if you see a 0.5, by 3 if you see a 0.33).

For example, a compound that is 40.0% C, 6.7% H, and 53.3% O yields a mole ratio of C : H : O = 1 : 2 : 1, giving the empirical formula CH⁠2O.

Go deeperCommon mistake: rounding 1.33 to 1

After dividing by the smallest mole value, ratios like 1.33, 1.5, or 1.25 are not rounding errors to squash into whole numbers; they are exact small fractions announcing a multiplier. A ratio of 1.33 is 4/3 (multiply everything by 3); 1.5 is 3/2 (multiply by 2); 1.25 is 5/4 (multiply by 4). Rounding 1.33 down to 1 quietly changes the compound.

Rule of thumb: treat a value within about 0.05 of a whole number as that whole number (experimental data are never perfect), but treat anything near .25, .33, .5, .67, or .75 as a fraction demanding its multiplier.

Molecular Formulas from Empirical Formulas

The molecular formula shows the actual number of atoms per molecule. It is always a whole-number multiple of the empirical formula.

To determine the molecular formula, you need two things:

  1. The empirical formula (and its mass)
  2. The compound's molar mass (from experiment — often measured via gas density, mass spectrometry, or other techniques)

Calculate the multiplier:

n = molar mass / empirical formula mass

Then multiply every subscript in the empirical formula by n. For example, if the empirical formula is CH⁠2O (mass 30.03 g/mol) and the molar mass is 180.18 g/mol, then n = 180.18 / 30.03 = 6, and the molecular formula is C⁠6H⁠12O⁠6 (glucose).

Go deeperWhy this matters: three compounds, one empirical formula

Formaldehyde (CH⁠2O), a preservative you would never eat; acetic acid (C⁠2H⁠4O⁠2), the tang in vinegar; and glucose (C⁠6H⁠12O⁠6), the sugar your cells burn: all three reduce to the same empirical formula, CH⁠2O. Composition analysis alone literally cannot tell a poison from a sugar.

That is why the molar mass is the essential second measurement: 30 g/mol picks out formaldehyde, 60 picks acetic acid, 180 picks glucose. The empirical formula narrows the suspects; the molar mass makes the identification.

Using Avogadro's Number in Calculations

Avogadro's number (6.022 × 10⁠23 mol−1) converts between moles and individual particles — atoms, molecules, ions, or formula units, depending on the substance.

Common calculation patterns:

  • Moles → particles: Multiply by NA. Example: 0.50 mol O⁠2 × 6.022 × 10⁠23 = 3.01 × 10⁠23 molecules of O⁠2.
  • Particles → moles: Divide by NA. Example: 1.81 × 10⁠24 atoms Fe ÷ 6.022 × 10⁠23 = 3.00 mol Fe.

Remember that Avogadro's number always relates moles to whatever particle is specified by the formula. One mole of H⁠2O gives 6.022 × 10⁠23 molecules, but since each molecule contains 3 atoms, it contains 3 × 6.022 × 10⁠23 = 1.81 × 10⁠24 total atoms.

Empirical Formulas from Combustion Analysis

Combustion analysis is a common experimental technique for determining the empirical formula of organic compounds containing C, H, and sometimes O, N, or S.

In a combustion analysis:

  1. A known mass of the compound is burned completely in excess O⁠2
  2. All carbon is converted to CO⁠2 (collected and weighed)
  3. All hydrogen is converted to H⁠2O (collected and weighed)
  4. Oxygen (if present) is found by difference: mass O = sample mass − mass C − mass H

From the masses of CO⁠2 and H⁠2O produced, you can calculate grams of C and H, convert to moles, and determine the empirical formula using the standard procedure. For example, if burning 1.000 g of a compound yields 1.500 g CO⁠2 and 0.409 g H⁠2O, then the mass of C = 0.4093 g, mass of H = 0.0458 g, and mass of O (by difference) = 0.545 g, leading to the empirical formula.

Go deeperWhy this matters: the instrument that identified organic chemistry

For over 150 years, burning a sample and weighing the CO⁠2 and H⁠2O was the way to determine what an organic compound was made of; Justus von Liebig industrialized the technique in the 1830s with a five-bulb glass apparatus so effective it became the emblem of chemistry itself. Modern CHN analyzers automate the same combustion into a benchtop instrument, and pharmaceutical labs still run elemental analysis to confirm that a newly synthesized drug has exactly the composition its formula claims. The problems you solve here are that instrument’s arithmetic.

Counting Atoms in a Given Mass

To determine how many atoms are in a sample of a substance, chain together two conversions: mass → moles → atoms.

  1. Convert mass to moles: divide by the molar mass
  2. Convert moles to atoms: multiply by Avogadro's number
  3. If the substance is molecular, account for how many atoms of the target element exist per molecule

For example, how many oxygen atoms are in 50.0 g of CaCO⁠3 (molar mass 100.09 g/mol)?

  • 50.0 g ÷ 100.09 g/mol = 0.4996 mol CaCO⁠3
  • Each formula unit has 3 oxygen atoms
  • 0.4996 mol × 3 × 6.022 × 10⁠23 = 9.02 × 10⁠23 oxygen atoms

The key is to always identify how many atoms of the element of interest appear in each formula unit or molecule.

Go deeperTry it: count the atoms in a flake of gold

A 1.0 g flake of gold: how many atoms is that? Work it, then check below.

Answer: 1.0 g ÷ 196.97 g/mol = 5.1 × 10⁠-3 mol, then 5.1 × 10⁠-3 mol × 6.022 × 10⁠23 atoms/mol = 3.1 × 10⁠21 atoms (2 sig figs, from the 1.0 g). Both scale checks pass: the mole count is small, the atom count is astronomical. Three thousand billion billion atoms in a flake lighter than a paperclip: the two-step chain (mass → moles → atoms) is short, but it crosses twenty orders of magnitude.

Formula Subscripts as Mole Ratios

The subscripts in a chemical formula do double duty: they tell you both the number of atoms per molecule and the number of moles of each element per mole of compound.

For example, in glucose (C⁠6H⁠12O⁠6):

  • One molecule contains 6 C atoms, 12 H atoms, and 6 O atoms
  • One mole contains 6 mol C, 12 mol H, and 6 mol O

This equivalence makes molar-level calculations straightforward. Need the moles of hydrogen in 3.0 mol of glucose? Simply multiply: 3.0 mol C⁠6H⁠12O⁠6 × 12 mol H / 1 mol C⁠6H⁠12O⁠6 = 36 mol H.

Subscript ratios also directly define the empirical formula. If analysis shows a compound contains elements in a 1:2:1 mole ratio, the empirical formula is XY⁠2Z (where X, Y, Z are the elements). The subscripts are the mole ratio.

A Practical Problem Map for Mole and Formula Calculations

Most quantitative composition problems become simple once you identify the target and route:

  1. Target = particles? Use moles → particles with Avogadro’s number.
  2. Target = mass? Use moles → mass with molar mass.
  3. Target = empirical formula? Convert each element to moles, divide by the smallest, then scale to whole numbers.
  4. Target = molecular formula? Find empirical-formula mass, compute multiplier = (molar mass)/(empirical mass), then multiply subscripts.

Common pitfalls: rounding subscripts too early, skipping unit tracking, and forgetting that subscripts represent mole ratios. Keep at least one guardrail in every solution: each conversion factor must cancel units cleanly before moving to the next step.

Key Equations

Moles from Mass
moles = mass (g) / molar mass (g/mol)
Number of Particles
particles = moles x 6.022 × 1023
Avogadro's number (NA)
Percent Composition
% element = (atoms of element x atomic mass) / molar mass of compound x 100%
Molecular Formula Multiplier
n = molar mass / empirical formula mass
Multiply all empirical subscripts by n

Learning Objectives

After studying this topic, you should be able to:

  1. Define the mole and use Avogadro's number to convert between moles and number of particles
  2. Calculate molar mass of elements and compounds
  3. Interconvert among moles, mass, and number of particles using molar mass and Avogadro's number
  4. Calculate percent composition by mass from a chemical formula
  5. Determine an empirical formula from percent composition or mass data
  6. Determine a molecular formula from the empirical formula and molar mass
  7. Determine an empirical formula from combustion analysis data

How-To Procedure

How to Determine an Empirical Formula from Percent Composition

  1. Assume a 100.0 g sample so that each percentage converts directly to grams.
  2. Convert grams of each element to moles by dividing by that element's atomic mass from the periodic table.
  3. Divide every mole value by the smallest mole value to obtain mole ratios.
  4. If any ratio is not close to a whole number, multiply all ratios by the smallest integer that converts them to whole numbers (multiply by 2 for halves, by 3 for thirds, etc.).
  5. Write the empirical formula using the whole-number ratios as subscripts.
  6. If the molar mass is known, divide it by the empirical formula mass to find the multiplier n, then multiply all subscripts by n to get the molecular formula.

Worked Example

Determining Molecular Formula from Percent Composition and Molar Mass

Problem

A compound is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen by mass. Its molar mass is 180.16 g/mol. Determine the empirical and molecular formulas.

Solution
  1. Assume a 100.00 g sample: 40.00 g C, 6.71 g H, 53.29 g O
  2. Convert to moles: C = 40.00 / 12.01 = 3.331 mol; H = 6.71 / 1.008 = 6.657 mol; O = 53.29 / 16.00 = 3.331 mol
  3. Divide by the smallest (3.331): C = 1.000, H = 1.998 ≈ 2, O = 1.000
  4. Empirical formula = CH⁠2O with empirical formula mass = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol
  5. Find the multiplier: n = 180.16 / 30.03 = 6.00
  6. Multiply all subscripts by 6: molecular formula = C⁠6H⁠12O⁠6
Answer

Empirical formula: CH⁠2O. Molecular formula: C⁠6H⁠12O⁠6 (glucose).

Test Your Understanding

Two compounds have the same empirical formula, CH2O. Compound A has a molar mass of 30 g/mol and Compound B has a molar mass of 180 g/mol. Are these the same substance? What are their molecular formulas?

Practice Problems

calculation

A compound contains 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen by mass. Determine its empirical formula.

calculation

How many individual oxygen atoms are present in 25.0 g of calcium carbonate (CaCO3, molar mass 100.09 g/mol)?

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Self-Study Questions

What is the mole and what is Avogadro’s number?

What is molar mass and how does it relate to the periodic table?

How do you convert between grams, moles, and number of particles?

Hint: Two conversion factors are involved — one uses molar mass, the other uses Avogadro’s number.

What is percent composition by mass and how is it calculated from a formula?

What is the difference between an empirical formula and a molecular formula?

How do you determine an empirical formula from percent composition data?

How do you find a molecular formula from an empirical formula?

What is combustion analysis and what information does it provide?

How are mole ratios used as conversion factors?

Why is the mole concept central to all quantitative chemistry?

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Units & MeasurementSolutions & MolarityStoichiometry

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Content Sources

Concept sections adapted from open educational resources under Creative Commons licensing:

  • OpenStax Chemistry 2e, Ch 3.1: Formula Mass and the Mole Concept (CC BY 4.0)
  • OpenStax Chemistry 2e, Ch 3.2: Determining Empirical and Molecular Formulas (CC BY 4.0)
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