1.1 Moles & Molar Mass

Mole Map

Type any formula, then enter one quantity and watch particles, mass, and gas volume solve through the mole hub. Every conversion is a factor-label step with units cancelling, and each rail lights the divide or multiply lane your own arithmetic travels.

Avogadro's NumberMolar MassFactor Labeln = m/M
Topic 1.1

Moles and Molar Mass

Calculate quantities of a substance or its relative number of particles using dimensional analysis and the mole concept.

You cannot count atoms in a laboratory. You can weigh things. The mole is the bridge the CED builds between those two facts: a connection between the masses of substances that react and the number of particles actually undergoing chemical change.

One mole is Avogadro's number of particles, NA = 6.022 × 10²³ mol⁻¹. The particles can be atoms, molecules, ions, electrons, or formula units — the mole never specifies which, so you must.

Molar mass (M) is the mass of one mole in grams. The reason the number on the periodic table works for both scales is the point of EK 1.1.A.3: the average mass in amu of one particle of a substance is always numerically equal to the molar mass of that substance in grams. Carbon: one atom averages 12.01 amu, one mole weighs 12.01 g.

Three conversions cover essentially every Unit 1 calculation, and they all pass through moles:

  • mass ↔ moles: n = m / M
  • moles ↔ particles: N = n × NA
  • moles of gas ↔ volume at STP: V = n × 22.4 L/mol

Chain them with dimensional analysis and cancel units as you go. If the units do not cancel to the units you want, the setup is wrong regardless of the arithmetic.

Key points

  • The mole exists to connect a measurable mass to an uncountable number of particles.
  • Molar mass in g/mol is numerically identical to average particle mass in amu — that is not a coincidence, it is how the amu is defined.
  • 22.4 L/mol applies only to an ideal gas at STP (273.15 K, 1.0 atm). It is not a universal conversion.
  • Subscripts multiply through: 1 mol Ca(NO₃)₂ contains 2 mol N and 6 mol O.

Equations

  • on the exam sheetPrinted on the AP equation sheet under Gases, Liquids, and Solutions.
    • moles
    • mass (g)
    • molar mass (g/mol)
  • not on the sheetNA = 6.022 × 10²³ mol⁻¹ is given as a constant on the sheet, but this relationship is not written out for you.
    • number of particles
    • Avogadro’s number

Common mistakes

  • Particles vs. atoms. 1 mol of CO₂ is 6.022 × 10²³ molecules but 1.807 × 10²⁴ atoms. Read which one the question wants.
  • Formula unit is not a molecule. Ionic compounds have no discrete molecules; 1 mol NaCl means 1 mol Na⁺ and 1 mol Cl⁻.
  • STP is 273.15 K and 1.0 atm, not 25 °C. 22.4 L/mol at any other conditions is wrong — use PV = nRT instead.
  • Molar mass has units. AP graders take off for a bare number where g/mol was required.

Worked example

How many oxygen atoms are in 4.50 g of glucose, C₆H₁₂O₆ (M = 180.16 g/mol)?

Step 1 — mass to moles of glucose.
n = 4.50 g ÷ 180.16 g/mol = 0.02498 mol C₆H₁₂O₆

Step 2 — moles of glucose to moles of O (6 O atoms per formula):
0.02498 mol × 6 = 0.1499 mol O

Step 3 — moles to atoms.
0.1499 mol × 6.022 × 10²³ mol⁻¹ = 9.03 × 10²² O atoms

Sanity check: less than a tenth of a mole of glucose, so an answer near 10²² (not 10²³) is right.

Full notes for topic 1.1 →