Formulas, Moles, and Chemical Composition

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 (N_A).

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.

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. The molar mass serves as the critical conversion factor between grams and moles.

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 × N_A ↔ 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.

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:

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).

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.

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:

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).

In balanced chemical equations, coefficients represent mole ratios between reactants and products. These ratios are the stoichiometric factors you use to convert between moles of different substances.

For the reaction:

The coefficients tell us that 1 mol N⁠₂ reacts with 3 mol H⁠₂ to produce 2 mol NH⁠₃. So the stoichiometric factors are:

• 3 mol H⁠₂ / 1 mol N⁠₂ (or its inverse)
• 2 mol NH⁠₃ / 3 mol H⁠₂ (or its inverse)
• 2 mol NH⁠₃ / 1 mol N⁠₂ (or its inverse)

To convert moles of one substance to moles of another, simply multiply by the appropriate ratio. For instance, 4.5 mol H⁠₂ × (2 mol NH⁠₃ / 3 mol H⁠₂) = 3.0 mol NH⁠₃.

Molar mass is the two-way bridge between grams and moles. Every mass-to-mole problem uses the same pattern:

When solving stoichiometry problems that start and end with mass, the workflow is:

1. Grams → moles (divide by molar mass of given substance)
2. Moles → moles (use stoichiometric ratio from balanced equation)
3. Moles → grams (multiply by molar mass of desired substance)

This three-step pattern — grams in, mole ratio, grams out — is the backbone of virtually every quantitative chemistry problem.

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 N_A. Example: 0.50 mol O⁠₂ × 6.022 × 10⁠²³ = 3.01 × 10⁠²³ molecules of O⁠₂.
• Particles → moles: Divide by N_A. 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.

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 = 0.4096 g, mass of H = 0.0458 g, and mass of O (by difference) = 0.545 g, leading to the empirical formula.

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.

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.

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.