Introduction
By this point you already know how to find the number of moles of a substance using n = m / M. Today we combine that skill with balanced chemical equations to answer the question chemists ask constantly: if I use this much of one substance, how much of another substance do I get or need? This is the heart of stoichiometry (Talbot, p.418).
Why balanced equations matter
A balanced symbol equation tells you the ratio in which reactants combine and products form. For example:
N₂(g) + 3H₂(g) → 2NH₃(g)
This means 1 molecule of N₂ reacts with 3 molecules of H₂ to give 2 molecules of NH₃ — and because moles are just "chemical counting units," it also means 1 mole of N₂ reacts with 3 moles of H₂ to give 2 moles of NH₃ (Talbot, p.82). The equation must always be balanced, because atoms cannot be created or destroyed in a chemical reaction — mass is conserved (Talbot, p.82).
The general method
To go from "how much of one thing" to "how much of another," follow these steps:
- Write a balanced equation for the reaction.
- Convert the given mass (or volume, or concentration) into moles using n = m / M.
- Use the mole ratio from the balanced equation to find the moles of the substance you want.
- Convert back from moles into the units the question asks for (mass, volume, etc.).
A simple mole ratio is just the ratio of the coefficients in the equation.
Worked Example 1: Mass to mass
Question: How many grams of magnesium oxide (MgO) form when 6.0 g of magnesium burns completely in oxygen?
2Mg(s) + O₂(g) → 2MgO(s)
- Step 1: Equation already balanced.
- Step 2: M(Mg) = 24.3 g mol⁻¹ n(Mg) = 6.0 / 24.3 = 0.247 mol
- Step 3: Mole ratio Mg : MgO = 2 : 2 = 1 : 1 n(MgO) = 0.247 mol
- Step 4: M(MgO) = 24.3 + 16.0 = 40.3 g mol⁻¹ m(MgO) = 0.247 × 40.3 = 10.0 g
Answer: 10.0 g of MgO is formed.
Worked Example 2: Mass to moles of another product
Question: How many moles of CO₂ are produced when 4.4 g of propane (C₃H₈) burns completely?
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
- M(C₃H₈) = 3(12.0) + 8(1.0) = 44.0 g mol⁻¹
- n(C₃H₈) = 4.4 / 44.0 = 0.100 mol
- Ratio C₃H₈ : CO₂ = 1 : 3
- n(CO₂) = 0.100 × 3 = 0.300 mol
Answer: 0.300 mol of CO₂ is formed.
Worked Example 3: Using ratios that aren't 1:1
Question: How many grams of oxygen are needed to react completely with 5.00