Introduction
Balanced chemical equations are more than just a tidy way of writing a reaction — they are a recipe. The numbers in front of each formula (the stoichiometric coefficients) tell you exactly what ratio of particles (and therefore moles) react together and what ratio of products forms. Today's lesson introduces how to turn that ratio information into real calculations: working out masses of reactants and products from a balanced equation (Talbot, p.418).
Why equations must be balanced
A chemical equation must always be balanced because mass is conserved in a chemical reaction — atoms are rearranged, not created or destroyed, so the same number of each type of atom must appear on both sides (Talbot, p.82). This is why we write:
N₂(g) + 3H₂(g) → 2NH₃(g)
This tells us that 1 molecule (or 1 mole) of N₂ reacts with 3 moles of H₂ to form 2 moles of NH₃. The coefficients give the mole ratio, not a mass ratio — this distinction is the key idea of the whole topic.
From moles to mass: the toolkit
To do calculations from equations, you need two tools working together:
- n = m / M (moles = mass ÷ molar mass)
- The mole ratio from the balanced equation
General method
- Write the balanced equation.
- Convert the given mass into moles using n = m / M.
- Use the mole ratio from the equation to find moles of the substance you want.
- Convert back to mass using m = n × M.
Worked Example 1
Question: What mass of magnesium oxide forms when 6.0 g of magnesium burns completely in oxygen?
2Mg(s) + O₂(g) → 2MgO(s)
- Step 1: M(Mg) = 24.3 g mol⁻¹ n(Mg) = m / M = 6.0 / 24.3 = 0.247 mol
- Step 2: Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so n(MgO) = 0.247 mol
- Step 3: M(MgO) = 24.3 + 16.0 = 40.3 g mol⁻¹ m(MgO) = n × M = 0.247 × 40.3 = 9.95 g
Answer: about 10.0 g of MgO forms.
Worked Example 2
Question: Calculate the mass of carbon dioxide produced when 5.00 g of calcium carbonate decomposes completely