Higher Tier, Chemistry-only: derive the theoretical yield first
| English | 中文 | Pinyin |
|---|---|---|
| theoretical yield/θɪəˈretɪkl jiːld/ | 理论产量 | lǐ lùn chǎn liàng |
| complete conversion | 完全转化 | wán quán zhuǎn huà |
What would explain this observation?
- Higher Tier: If a question gives only a reactant mass and recovered product, there is one extra step before finding yield: determine the theoretical product mass from the equation.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Convert the stated limiting-reactant mass into moles, use the balanced coefficient ratio and multiply by product molar mass. The result is theoretical yield 理论产量 under complete conversion 完全转化 with sufficient other reactants. Then use actual/theoretical×100 for percentage yield. This mass derivation is Higher-only, while dividing two supplied product masses is common-tier Chemistry-only work.
- theoretical yield: The maximum desired product amount predicted by the stated reaction model; complete conversion: The stated model in which the relevant limiting reactant is fully converted as written.
For a 1:1 mole ratio, what must be done after finding product moles?
For CaCO₃ → CaO + CO₂, the amount ratio CaCO₃:CaO is 1:1. Supplied molar masses are 100 and 56 g per mol. A 10.0 g CaCO₃ sample contains 0.100 mol and can give 0.100 mol CaO, mass 5.60 g. The other 4.40 g is theoretical carbon dioxide, so the product-residue mass is lower than the original carbonate mass even at complete conversion.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- For CaCO₃ → CaO + CO₂, the amount ratio CaCO₃:CaO is 1:1. Supplied molar masses are 100 and 56 $\dfrac{\text{g}}{\text{mol}}$. A 10.0 g CaCO₃ sample contains 0.100 mol and can give 0.100 mol CaO, mass 5.60 g. The other 4.40 g is theoretical carbon dioxide, so the product-residue mass is lower than the original carbonate mass even at complete conversion.
- Label theoretical and actual masses clearly. State sample purity and which reactant limits if relevant. Use unrounded intermediate values when multiplying a ratio. Check the total product masses against reactant mass for a decomposition with no added reactant. For an actual school experiment, follow approved heating and handling procedures; calculation alone does not establish purity or completion.
Which two habits make the investigation or model in this case more defensible?
Label theoretical and actual masses clearly. State sample purity and which reactant limits if relevant. Use unrounded intermediate values when multiplying a ratio. Check the total product masses against reactant mass for a decomposition with no added reactant. For an actual school experiment, follow approved heating and handling procedures; calculation alone does not establish purity or completion.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: 10.0 g pure CaCO₃ theoretically gives 5.60 g CaO. If 4.48 g dry CaO is recovered, yield=4.48/5.60×100=80.0%. Dividing 4.48 by 10.0 instead would compare different substances and incorrectly give 44.8%. Recovery must be compared with the possible mass of the same desired product.
CaCO₃ → CaO + CO₂ has molar masses 100 and 56 $\dfrac{\text{g}}{\text{mol}}$ for CaCO₃ and CaO. Find theoretical CaO mass from 25 g pure CaCO₃. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
CaCO₃ → CaO + CO₂ has molar masses 100 and 56 g per mol for CaCO₃ and CaO. Find theoretical CaO mass from 25 g pure CaCO₃.
The result is 14 g. Known: 10.0 g pure CaCO₃ theoretically gives 5.60 g CaO. If 4.48 g dry CaO is recovered, yield=4.48/5.60×100=80.0%. Dividing 4.48 by 10.0 instead would compare different substances and incorrectly give 44.8%. Recovery must be compared with the possible mass of the same desired product.
Check the conclusion and its limits
- A smaller solid residue is not itself low yield when gas is an expected product. Impure starting material changes the available reactant mass. Do not use the actual product mass to calculate the theoretical amount and then claim an independent yield result.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Yield is always actual product mass divided by the initial mass of any reactant. This claim is false: A smaller solid residue is not itself low yield when gas is an expected product. Impure starting material changes the available reactant mass. Do not use the actual product mass to calculate the theoretical amount and then claim an independent yield result.
Higher Tier, Chemistry-only: derive the theoretical yield first: For CaCO₃ → CaO + CO₂, the amount ratio CaCO₃:CaO is 1:1. Supplied molar masses are 100 and 56 $\dfrac{\text{g}}{\text{mol}}$. A 10.0 g CaCO₃ sample contains 0.100 mol and can give 0.100 mol CaO, mass 5.60 g. The other 4.40 g is theoretical carbon dioxide, so the product-residue mass is lower than the original carbonate mass even at complete conversion.
Yield is always actual product mass divided by the initial mass of any reactant.
A smaller solid residue is not itself low yield when gas is an expected product. Impure starting material changes the available reactant mass. Do not use the actual product mass to calculate the theoretical amount and then claim an independent yield result.
The maximum desired product amount predicted by the stated reaction model: write the technical term.
theoretical yield means The maximum desired product amount predicted by the stated reaction model.