Chemistry • Year 11 • Module 2 • Lesson 12
Mass–Mass Stoichiometry
Lock in the 4-step method, key vocabulary and the principle that all mass–mass conversions must pass through moles before tackling harder problems.
1. Term–definition match
The definitions below are shuffled. Write the matching term from this list in the right-hand column: mass–mass stoichiometry, given substance, wanted substance, molar mass, mole ratio, theoretical mass, balanced equation, reverse problem. 8 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A calculation that converts the mass of one substance to the mass of another using moles as an intermediary. | |
| 1.2 | The substance whose mass is provided in the problem and used to start the calculation. | |
| 1.3 | The substance whose mass you are asked to find at the end of the calculation. | |
| 1.4 | The mass of one mole of a substance in g mol−1, found by summing atomic masses of all atoms in the formula. | |
| 1.5 | The ratio of coefficients in the balanced equation used to convert moles of the given substance to moles of the wanted substance. | |
| 1.6 | The maximum mass of product predicted by stoichiometry, assuming complete reaction with no losses. | |
| 1.7 | The equation that must be written and verified before any mass–mass calculation can begin. | |
| 1.8 | A stoichiometry problem where the mass of a product is given and the mass of a reactant must be found, using the same 4-step method. |
2. Sequence the 4 steps
The four steps of mass–mass stoichiometry are listed below in the wrong order. Number them 1–4 to show the correct sequence. Then write the mathematical formula for each step. 8 marks (1 number + 1 formula each)
| Order (1–4) | Step description | Mathematical formula |
|---|---|---|
| Convert moles of given substance to moles of wanted substance using the mole ratio. | ||
| Write and balance the chemical equation; extract the mole ratio. | n/a | |
| Convert the given mass to moles. | ||
| Convert moles of wanted substance to mass. |
3. True or false — with correction
Circle T or F. If false, write the corrected statement on the line below. 10 marks (1 T/F + 1 correction each)
3.1 The equation coefficients compare masses, so you can multiply the given mass by the coefficient ratio to get the product mass directly. T / F
3.2 A mole ratio of 1:1 means the given and wanted substances have the same mass per mole. T / F
3.3 For the diatomic molecule O2, the molar mass is approximately 32.00 g mol−1. T / F
3.4 In a reverse stoichiometry problem (given product, find reactant), you need a different method from the standard 4-step approach. T / F
3.5 In the reaction 2Al + 3Cl2 → 2AlCl3, the mole ratio of Al to AlCl3 is 1:1. T / F
4. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
balanced · coefficients · molar mass · mole ratio · moles · ratio · wanted · mass
In a mass–mass stoichiometry problem, you are given the ___________ of one substance and asked to find the mass of a ___________ substance. The calculation must always pass through ___________ — you can never multiply masses directly by equation ___________. Step 1 requires that the chemical equation is ___________ so that the correct ___________ between species can be extracted. In Step 3, the ___________ converts moles of the given substance to moles of the wanted substance. Finally, in Step 4, you multiply moles by the ___________ of the wanted substance to obtain its mass.
5. Short recall
Answer each question in 1–2 sentences. 6 marks (2 each)
5.1 Explain why 12 g of carbon (C) does NOT produce 12 g of carbon dioxide (CO2) even though the mole ratio in C + O2 → CO2 is 1:1.
5.2 What is the most common error students make when they encounter a mole ratio that is NOT 1:1? How is it avoided?
5.3 Why must diatomic molecules (H2, O2, N2, Cl2) be treated carefully when calculating molar mass in Step 4?
Q1 — Term–definition match
1.1 mass–mass stoichiometry • 1.2 given substance • 1.3 wanted substance • 1.4 molar mass • 1.5 mole ratio • 1.6 theoretical mass • 1.7 balanced equation • 1.8 reverse problem
Q2 — 4-step sequence
Correct order:
- Step 1 (order = 1): Write and balance the equation; extract mole ratio. Formula: n/a — this is a prerequisite step.
- Step 2 (order = 2): Convert given mass to moles. Formula: n(given) = m(given) ÷ MM(given)
- Step 3 (order = 3): Convert moles of given to moles of wanted using mole ratio. Formula: n(wanted) = n(given) × coeff(wanted) ÷ coeff(given)
- Step 4 (order = 4): Convert moles of wanted to mass. Formula: m(wanted) = n(wanted) × MM(wanted)
Q3 — True / false
3.1 False. Equation coefficients compare moles, not grams. You must convert mass to moles, apply the mole ratio, then convert back to mass. Multiplying masses directly by the coefficient ratio gives the wrong answer because different substances have different molar masses.
3.2 False. A 1:1 mole ratio means equal moles, not equal masses. The substances can still have very different molar masses (e.g. C = 12 g mol−1, CO2 = 44 g mol−1) so equal moles weigh different amounts.
3.3 True. MM(O2) = 2 × 15.999 = 31.998 ≈ 32.00 g mol−1.
3.4 False. The same 4-step method is used. In a reverse problem, the product is the “given” substance and the reactant is the “wanted” substance — the steps are identical.
3.5 True. In 2Al + 3Cl2 → 2AlCl3, the ratio of Al to AlCl3 is 2:2, which simplifies to 1:1.
Q4 — Cloze paragraph
In order: mass / wanted / moles / coefficients / balanced / ratio / mole ratio / molar mass
Q5.1 — Carbon → CO2 mass change
Although the mole ratio is 1:1 (one mole of C produces one mole of CO2), the molar masses differ: MM(C) = 12.011 g mol−1, MM(CO2) = 44.009 g mol−1. So 12 g of C (1 mol) produces 44 g of CO2 (also 1 mol). Equal moles does not mean equal mass.
Q5.2 — Most common non-1:1 ratio error
Students assume a 1:1 ratio or forget to apply the ratio, giving n(wanted) = n(given) instead of the correct n(wanted) = n(given) × coeff(wanted) ÷ coeff(given). It is avoided by writing the balanced equation first, circling the two species in the question, and explicitly writing out the ratio (e.g. “Fe2O3 : Fe = 1 : 2”) before calculating.
Q5.3 — Diatomic molar mass
Diatomic molecules (H2, O2, N2, Cl2, Br2, I2, F2) consist of two atoms per formula unit, so their molar mass is double the atomic mass listed in the periodic table. Using the atomic mass instead of the molecular mass (e.g. 16 instead of 32 for O2) will halve or double the answer. Always use the formula as written in the balanced equation.