To trace a carbon-containing process, follow the carbon atoms from the reactants to the products and check that none are lost or created. The same habit works for combustion, respiration, photosynthesis and the reaction of a carbonate with acid. Exam questions often pair this with a mass or mole calculation.
This lesson builds on reading pollutant data and relies on the ideas in formulas and equations and relative masses and amounts.
How do you trace carbon step by step?
- Write the word equation and then the formulas.
- Count carbon first. Carbon is usually in one compound on each side, so it is the easiest atom to track.
- Balance the other elements, leaving oxygen until last if it appears in several compounds.
- Check the total atoms for every element on each side.
- Use the ratio of coefficients for any amount question.
Four processes to know:
| Process | Balanced equation | Carbon moves from… to… |
|---|---|---|
| Complete combustion of methane | CH₄ + 2O₂ → CO₂ + 2H₂O | fuel to atmosphere |
| Photosynthesis | 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ | atmosphere to glucose in plants |
| Respiration | C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O | glucose to atmosphere |
| Carbonate with acid | CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ | rock to gas |
Worked example
Question: Methane burns completely in air. Calculate the mass of carbon dioxide formed from 8.0 g of methane. (Relative atomic masses: C = 12, H = 1, O = 16.)
Step 1, equation: CH₄ + 2O₂ → CO₂ + 2H₂O. Check carbon: 1 left, 1 right. Hydrogen: 4 left, 4 right. Oxygen: 4 left, 2 + 2 = 4 right. Balanced.
Step 2, relative formula masses: CH₄ = 12 + 4 = 16. CO₂ = 12 + 32 = 44.
Step 3, moles of methane: 8.0 ÷ 16 = 0.5 mol.
Step 4, use the ratio: the equation shows 1 mol CH₄ gives 1 mol CO₂, so 0.5 mol CO₂ forms.
Step 5, mass of carbon dioxide: 0.5 × 44 = 22 g.
Check by conservation of mass: oxygen used is 2 × 0.5 = 1.0 mol O₂ = 32 g. Reactants: 8.0 + 32 = 40 g. Products: 22 g CO₂ and 2 × 0.5 = 1.0 mol H₂O = 18 g, so 22 + 18 = 40 g. The masses agree.
This is an educational calculation only. It describes the amounts in a balanced equation, not how to carry out burning safely.
The mistake to watch for
Mistaken answer: “8.0 g of methane gives 8.0 g of carbon dioxide, because the carbon is just moving.”
The carbon atoms are conserved, but the mass of the compound is not. Carbon dioxide contains oxygen from the air as well as carbon. The mass of CO₂ is therefore larger than the mass of the carbon-containing fuel.
The correction is to convert mass to moles, use the coefficient ratio, then convert back to mass. The moles of carbon stay the same: 0.5 mol of carbon in the methane, and 0.5 mol in the carbon dioxide.
Check yourself
1. Balance: C₃H₈ + O₂ → CO₂ + H₂O
Show answer
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Check: carbon 3 and 3, hydrogen 8 and 8, oxygen 10 on the left and 6 + 4 = 10 on the right.
2. Calcium carbonate decomposes: CaCO₃ → CaO + CO₂. Calculate the mass of carbon dioxide from 10.0 g of CaCO₃. (Ar: Ca = 40, C = 12, O = 16.)
Show answer
Mr of CaCO₃ = 40 + 12 + 48 = 100. Moles = 10.0 ÷ 100 = 0.100 mol. The ratio is 1:1, so 0.100 mol CO₂. Mass = 0.100 × 44 = 4.4 g.
3. In which of the four processes in the table does carbon move from the atmosphere into living things?
Show answer
Photosynthesis. Carbon dioxide from the air is built into glucose.
Where this leads next
Next, use the same reasoning to weigh up a statement about emissions in comparing an environmental claim with its evidence and limits. The mole and equation-ratio tutor lets you check the amount relationships, and the equation balance reasoning trainer gives you atom counts to compare with yours. Then try the module practice set.
A common pattern is that the balancing is secure but the mass step slips. A teacher in online one-to-one Chemistry tuition can watch you work a whole calculation and stop you at the moment the reasoning changes.