Some mixed questions give you a mass change and an equation, and expect you to find an amount of a substance. The method is the same each time: turn mass into moles, use the equation ratio, then turn moles back into mass.
This lesson belongs to chemical reasoning in mixed tasks. The data below is invented for teaching and is labelled as such. Only the calculation is taught here, and no experimental procedure.
What is the method?
Take the equation for heating copper(II) carbonate, which breaks down into copper(II) oxide and carbon dioxide:
CuCO₃ → CuO + CO₂
The relative formula masses are CuCO₃ = 64 + 12 + 48 = 124, CuO = 64 + 16 = 80 and CO₂ = 12 + 32 = 44.
- Identify the mass change. The solid loses mass because CO₂ escapes. Mass lost = mass of CO₂.
- Convert to moles: moles = mass ÷ relative formula mass.
- Use the ratio. Here every 1 mole of CuCO₃ gives 1 mole of CO₂ and 1 mole of CuO.
- Convert back to mass: mass = moles × relative formula mass.
Worked example (invented data)
A 6.20 g sample of copper(II) carbonate is heated until the reaction is complete. The mass of the solid left is measured.
Question: What mass of carbon dioxide is released, and what is the mass of the solid left?
Step 1, moles of CuCO₃: 6.20 ÷ 124 = 0.0500 mol.
Step 2, ratio: 1 : 1 : 1, so 0.0500 mol of CO₂ and 0.0500 mol of CuO form.
Step 3, mass of CO₂: 0.0500 × 44 = 2.20 g.
Step 4, mass of solid left: this is CuO, so 0.0500 × 80 = 4.00 g.
Check: the starting mass should equal the products. 4.00 + 2.20 = 6.20 g, which matches. The mass lost is 6.20 − 4.00 = 2.20 g, which matches the CO₂.
The mistake to watch for
Mistaken answer: “Moles of CO₂ = 2.20 ÷ 124 = 0.0177 mol.”
The student divided the mass of CO₂ by the formula mass of CuCO₃. Each mass must be divided by the formula mass of its own substance: CO₂ is 44, so 2.20 ÷ 44 = 0.0500 mol.
The correction is to write the formula mass beside each substance before dividing. A second slip is forgetting that the mass lost is the gas, not the solid that remains.
Check yourself
1. A 3.10 g sample of CuCO₃ is heated until complete. What mass of CO₂ is released?
Show answer
Moles of CuCO₃ = 3.10 ÷ 124 = 0.0250 mol. Ratio 1 : 1, so 0.0250 mol of CO₂. Mass = 0.0250 × 44 = 1.10 g.
2. The mass of solid falls by 0.88 g when CuCO₃ is heated until complete. What mass of CuCO₃ was in the sample?
Show answer
The mass lost is CO₂, so moles of CO₂ = 0.88 ÷ 44 = 0.020 mol. Ratio 1 : 1, so 0.020 mol of CuCO₃. Mass = 0.020 × 124 = 2.48 g.
3. A 6.20 g sample loses only 1.76 g in total. What percentage of the CuCO₃ has decomposed?
Show answer
Moles of CO₂ = 1.76 ÷ 44 = 0.0400 mol, so 0.0400 mol of CuCO₃ reacted. Mass reacted = 0.0400 × 124 = 4.96 g. Percentage = 4.96 ÷ 6.20 × 100 = 80.0%.
Where this leads next
Next, interpret a separation sequence in an environmental case to practise reading a method as a series of decisions. You can go back to using a particle model to explain a state observation at any time, and the mixed practice set includes more mass-change questions.
If the step from mass change to moles is where your marks go, our teachers can drill it with fresh numbers in online one-to-one Combined Science tuition.