Mass-to-mass stoichiometry

≈ 48 min

Mass-to-mass stoichiometry

Mass-to-mass problems are mole-ratio problems with two conversions. First convert the given mass to moles using n = m/M. Then use the coefficient ratio from a balanced equation. Finally convert the moles of the required substance back to mass using m = nM. Do not compare masses directly with coefficients because coefficients represent mole ratios, not gram ratios. Show a balanced equation even when you know it from memory; it is the evidence for the ratio and protects against a common exam error.

Work it through

For CaCO₃(s) → CaO(s) + CO₂(g), the mole ratio CaCO₃:CaO is 1:1. If 100 g CaCO₃ reacts completely, n(CaCO₃) = 100/100 = 1.0 mol using rounded molar masses. Therefore 1.0 mol CaO forms, with mass 1.0 × 56 = 56 g. The missing 44 g is CO₂, so the total mass is conserved across all products.

Mastery target

Complete a mass-to-mass calculation using moles and explain how the balanced equation and conservation of mass check the result.

In CaCO₃ → CaO + CO₂, what is the mole ratio CaCO₃:CO₂?

1:1

2:1

1:2

44:56

Name the key chemistry term from Mass-to-mass stoichiometry that best fits the explanation and visual model.

Using M(CaCO₃) = 100 g mol⁻¹ and M(CaO) = 56 g mol⁻¹, what mass of CaO forms from 100 g CaCO₃ if the reaction is complete?

Which statement corrects a common misunderstanding in Mass-to-mass stoichiometry?

Correct: Conservation of mass applies to all reactants and products together. It does not mean each product has the same mass as the reactant.

A chemical label is enough; evidence, quantities and conditions never matter.

A model can be applied without checking its assumptions or limits.

A calculation is reliable even when its units and equation are not checked.