teaching-notes — Chemistry (THE MOLE CONCEPT)

Topic: THE MOLE CONCEPT

Subtopic: The mole

MEANING OF THE MOLE

Definition: A mole is the amount of substance that contains 6.02 × 10²³ particles (atoms, molecules or ions). This number is called Avogadro's constant and is written as N_A = 6.02 × 10²³.

Example: 1 mole of carbon atoms = 6.02 × 10²³ carbon atoms. 2 moles of H₂ molecules = 2 × 6.02 × 10²³ H₂ molecules.

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RELATION BETWEEN MOLES, MASS AND VOLUME OF A GAS

We use two main formulae for the mole:

Formula 1 (from mass):

n = mM_r

Where: n = number of moles, m = mass in grams, M_r = relative (molar) mass in g mol^-1.

Formula 2 (for gases at r.t.p or s.t.p):

n = vV_m

Where: v = gas volume in dm³, V_m = molar volume of gas. At room temperature and pressure (r.t.p) V_m ≈ 24 dm³ mol^-1. At standard temperature and pressure (s.t.p) V_m = 22.4 dm³ mol^-1.

Worked Examples on USING n = m / M_r and n = v / V_m

A sample exam-style question and worked solution using the mass formula:

Question: Calculate the mass of 0.250 mol of calcium carbonate, CaCO₃. (Relative atomic masses: Ca = 40.1, C = 12.0, O = 16.0)

Worked Example: Finding mass from moles

Worked example using gas volumes and V_m at r.t.p:

Question: A sample of hydrogen gas occupies 48.0 dm³ at r.t.p. Calculate the number of moles of H₂. (Use V_m = 24 dm³ mol^-1 at r.t.p.)

Worked Example: Moles from gas volume at r.t.p

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PHYSICAL MASSES (M) OF ANY SUBSTANCE USING THE MOLAR MASS (MR) AND THE PHYSICAL VOLUME (V) OF ANY GAS AT R.T.P AND VICE VERSA

Use the two formulae above to move between mass, moles and gas volume. Steps to follow:

• To find moles from mass: Calculate M_r, then use n = m/M_r.
• To find mass from moles: Use m = n × M_r.
• To find moles from gas volume at r.t.p or s.t.p: Use n = v/V_m with V_m = 24 dm³ (r.t.p) or 22.4 dm³ (s.t.p).
• To find gas volume from moles: Use v = n × V_m.

Example: How many grams of O₂ are in 3.00 dm³ of O₂ at r.t.p? (M_r(O₂) = 32.0)

Working:

Step 1: n = v / V_m = 3.00 / 24 = 0.125 mol

Step 2: m = n × M_r = 0.125 × 32.0 = 4.00 g

Answer: 4.00 g O₂.

RELATIONSHIP OF AVOGADRO'S LAW TO REACTING MOLES AND VOLUMES OF GASES AT R.T.P AND S.T.P

Avogadro's law: Equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules (or moles). Therefore, under the same conditions, gas volumes are directly proportional to the number of moles.

Consequences for reacting gases:

• For a gaseous reaction, the ratio of volumes of gases that react (or form) equals the ratio of moles from the balanced equation, provided all volumes are measured at the same T and P (r.t.p or s.t.p).
• Example using the balanced equation: N₂(g) + 3H₂(g) → 2NH₃(g). Volume ratio = 1 : 3 : 2 for N₂ : H₂ : NH₃.

Worked example involving reacting gas volumes:

Question: If 4.00 dm³ of N₂ at r.t.p reacts completely with hydrogen according to N₂ + 3H₂ → 2NH₃, what volume of H₂ at r.t.p is required?

Worked Example: Using Avogadro's law and mole ratios for gas volumes

COMMON MISTAKES TO AVOID

SUMMARY

• n = number of moles; N_A = 6.02 × 10²³ particles per mole.
• n = m / M_r allows conversion between mass and moles.
• n = v / V_m allows conversion between gas volume and moles when conditions are r.t.p or s.t.p.
• At r.t.p V_m ≈ 24 dm³ mol^-1; at s.t.p V_m = 22.4 dm³ mol^-1.
• Avogadro's law: equal volumes of gases at same T and P contain equal moles → use mole ratios as volume ratios for gases.

KEY TERMS

REVISION QUESTIONS

PRACTICE EXERCISE

ANSWERS TO PRACTICE EXERCISE (showing working)