A balanced equation predicts the largest mass of product you could get. A real experiment usually gives less, and the difference comes from the experiment, not from the equation.
This lesson is part of conservation reasoning across a reaction. If your school has not yet introduced percentage yield, the reasoning in the first two sections still applies.
Why does the measured mass come out smaller?
The equation describes every particle reacting as planned. A real reaction may stop early, form something else on the side, or lose product while you handle it.
In each case the atoms still exist somewhere, so conservation of mass is not broken. The product you wanted is simply less than the calculated amount.
Worked example: heating calcium carbonate
A student heats 5.00 g of calcium carbonate: CaCO₃ → CaO + CO₂. The calcium oxide left in the crucible weighs 2.38 g. (Ca = 40, C = 12, O = 16.)
- Mr of CaCO₃ = 100, so moles = 5.00 ÷ 100 = 0.0500 mol.
- The ratio CaCO₃ : CaO is 1 : 1, so the moles of CaO = 0.0500 mol.
- Calculated mass of CaO = 0.0500 × 56 = 2.80 g.
- Percentage yield = 2.38 ÷ 2.80 × 100 = 85.0%.
The missing 0.42 g of calcium oxide is explained by something practical. Perhaps some carbonate did not decompose, or a little powder was lost when the crucible was moved.
Which reasons earn the mark?
A good reason names a physical cause and links it to the product. Each of the following works for a yield below 100%:
- The reaction did not go to completion, so some reactant remained.
- A side reaction made a different product.
- Some product was lost during transfer, filtering or washing.
- Some product stayed on the apparatus.
- The reactant was impure, so less of it was actually present.
Two answers do not earn marks. “The equation is wrong” and “mass was destroyed” both contradict what the equation and the law of conservation of mass say.
The mistake that costs marks
Some students try to repair the gap by changing coefficients, for example writing 2CaCO₃ → CaO + CO₂ to force the numbers to fit 2.38 g. That breaks the atom count on both sides.
| Approach | What it does | Result |
|---|---|---|
| Change the coefficients to fit the data | Atoms no longer balance | Equation is false |
| Keep the equation, compare masses | Calculated 2.80 g against measured 2.38 g | Gap is a percentage yield |
| Name a cause of loss | Links the gap to the method | Marks for explanation |
Keep the balanced equation fixed. Put the difference into the percentage and then the explanation.
Check yourself
3.25 g of zinc reacts with excess copper(II) sulfate solution: Zn + CuSO₄ → ZnSO₄ + Cu. The dried copper weighs 2.56 g. Find the percentage yield and give one reason the yield is below 100%. (Zn = 65, Cu = 64.)
Answer
Moles of Zn = 3.25 ÷ 65 = 0.0500 mol. The ratio Zn : Cu is 1 : 1, so calculated Cu = 0.0500 × 64 = 3.20 g.
Percentage yield = 2.56 ÷ 3.20 × 100 = 80.0%.
One valid reason: some copper was lost when the solid was filtered and washed. Another: the zinc was coated and did not all react.
What to study next
The next lesson treats mass and charge as two separate checks in checking mass and charge independently in an ionic equation. For more calculation routes, see solving reacting-mass and gas-volume calculations.
If you would like a teacher to go through your own practical write-ups, see online one-to-one Chemistry tuition.