Calculations using n = N/NA, n = m/M, n = V/Vm (II)

Introduction

By now you know three key "amount of substance" relationships: n = N/N_A (particles), n = m/M (mass), and n = V/V_m (gas volume at STP). Today we combine them with balanced equations to solve real stoichiometry problems — working out unknown masses, volumes, or particle numbers from a given quantity of reactant or product (Talbot, p.424).

The general strategy

Every stoichiometry calculation follows the same three-step logic, no matter which of the three "n =" formulas you start with:

1. Convert the given quantity (mass, volume, or number of particles) into moles using the appropriate formula.
2. Scale using the mole ratio from the balanced chemical equation to find moles of the substance you want.
3. Convert back from moles of the required substance into whatever unit the question asks for (mass, volume, or number of particles).

This works because a balanced equation tells us the ratio of moles reacting and forming — not masses or volumes directly (Talbot, p.418).

Worked Example 1 – mass to mass

Question: What mass of mercury forms when 0.00125 kg of mercury(II) oxide decomposes?

2HgO(s) → 2Hg(l) + O₂(g)

Steps:

• n(HgO) = m / M = 1.25 g / 216.6 g mol⁻¹ = 5.77 × 10⁻³ mol
• Mole ratio HgO : Hg = 2 : 2 = 1 : 1, so n(Hg) = 5.77 × 10⁻³ mol
• m(Hg) = n × M = 5.77 × 10⁻³ × 200.6 = 1.16 g

Worked Example 2 – mass to gas volume

Question: What volume of oxygen gas (at STP, V_m = 22.7 dm³ mol⁻¹) is produced from the same reaction above?

• n(HgO) = 5.77 × 10⁻³ mol (from before)
• Mole ratio HgO : O₂ = 2 : 1, so n(O₂) = 2.89 × 10⁻³ mol
• V(O₂)