Chem U1 - Moles

Exam outcomes (what I must be able to do)

• Explain the concept of Moles
• Explain what avogadro’s number is and do practical examples

Core idea (1–2 lines)

A mole is a counting unit: 1 mol = 6.022×10^23 particles.
We use molar mass to convert mass ↔ moles, then use Avogadro’s constant to convert moles ↔ particles.

Must-know facts / rules / Equations

• Avogadro’s Number $N_A$ $=6.02214076\ast 10^{22}$
• $N_A=6.022\ast 10^{23}$
• Mole = A counting unit, you have a specific amount of atoms. Aka an Avogadro’s amount of atoms/particles/ions in that unit.
• Mass to moles equation

$$n=m/M$$

Where:

• $n$ = Moles in a substance
• $m$ = Mass (usually in grams) (if the question stats g/mol, then use grams. Else convert to the required)
• $M$ = Molar Mass

Typical exam questions

• I have 15.4mg of Ge, how many atoms of Germanium am i dealing with.

  $Ge:72.63g/mol$

• Step 1: Convert to grams of germanium since it is g/mol

$$15.4\mathrm{mg}\times \frac{1\mathrm{g}}{1000\mathrm{mg}}=0.0154\mathrm{g}$$

• Step 2: Convert the grams to moles using molar mass. The g Ge cancels out.

$$0.0154\mathrm{g}\mathrm{Ge}\times \frac{1\mathrm{mol}\mathrm{Ge}}{72.63\mathrm{g}\mathrm{Ge}}=2.12\times 10^{-4}\mathrm{mol}\mathrm{Ge}$$

• Step 3: Cancel out the mol Ge, multiply by converting the moles to atoms using Avogadro’s Number.

$$2.12\times 10^{-4}\mathrm{mol}\mathrm{Ge}\times \frac{6.022\times 10^{23}\mathrm{atoms}\mathrm{Ge}}{1\mathrm{mol}\mathrm{Ge}}=1.28\times 10^{20}\mathrm{atoms}\mathrm{Ge}$$

Common traps

• Forgetting unit conversion: mg → g, or cm^3 → dm^3, etc.
• Mixing up molar mass vs Avogadro:
  • Use molar mass for g ↔ mol
  • Use Avogadro for mol ↔ particles
• Significant figures: final answer should match the given data (15.4 mg = 3 s.f.).

Quick self-test (5 mins)

• What are moles
• Explain Avogadro’s number
• How do you get the amount of moles when given values.

Links

Chem U1 - Average Atomic Mass