3.3.2Quantitative Chemistry
Theoretical Masses of Products
5 min read 5 parts Lesson 12 of 16 Unit quiz
Section 1 of 5Balanced Equations and Mole Ratios
A balanced equation gives the ratio in which substances react and form, and this ratio drives every theoretical mass calculation. Get the equation right before doing any arithmetic.
Balanced equation
- Balanced equation Shows the number of moles of each reactant and product, so it sets the mole ratio used in the calculation.
Relative formula mass
- Relative formula mass The sum of the relative atomic masses of all atoms in a formula, with no units, used to convert between mass and moles.
Mole ratio
- Mole ratio The numbers in front of formulae;
- for CaCO₃ → CaO + CO₂ the ratio is 1 : 1 : 1, so equal moles react and form.
Limiting reactant
- Limiting reactant The reactant that is used up first, which fixes the maximum mass of product that can form.
Section 2 of 5Method for Theoretical Mass
Work through these steps in order for any reaction where the predicted mass of a product is needed.
1
Write and balance the equation, then find the relative formula masses of the substance you know and the product you want.
2
Convert the known mass into moles by dividing it by the relative formula mass.
3
Use the mole ratio from the balanced equation to find the moles of the product.
4
Convert the moles of product back into a mass by multiplying by its relative formula mass.
Section 3 of 5Theoretical and Actual Yield
The theoretical mass is a prediction, and real experiments almost always give less than it.
Theoretical and Actual Yield
Theoretical yield
- Theoretical yield The maximum mass predicted from the equation, assuming complete reaction and no losses during collection.
Actual yield
- Actual yield The mass really obtained in an experiment, which is usually lower because of losses and incomplete reactions.
Conservation of mass
- Conservation of mass The total mass of products equals the total mass of reactants in a closed system, so the prediction is based on it.
| Quantity | Unit | Calculated from |
|---|---|---|
| Moles | mol | Mass ÷ relative formula mass |
| Mass | g | Moles × relative formula mass |
| Relative formula mass | no units | Sum of relative atomic masses |
Section 4 of 5Worked Example
Question
- Calculate the maximum mass of calcium oxide that can be made by heating 50 g of calcium carbonate.
- (Relative atomic masses: Ca = 40, C = 12, O = 16.)
- Relative formula mass of CaCO₃ = 40 + 12 + (3 × 16) = 100.
- Moles of CaCO₃ = 50 ÷ 100 = 0.5 mol.
- EquationCaCO₃ → CaO + CO₂, so the mole ratio CaCO₃ : CaO is 1 : 1, giving 0.5 mol of CaO.
- Relative formula mass of CaO = 40 + 16 = 56, so mass of CaO = 0.5 × 56 = 28 g.
Answer 28 g of calcium oxide (theoretical yield).
Section 5 of 5Exam Checklist & Tips
3.3.2 – Key Points for the Exam
- Relative formula mass Sum of the relative atomic masses of all atoms in a formula, with no units.
- Mole ratio Ratio of amounts in moles of substances, shown by the numbers in a balanced equation.
- Theoretical yield The maximum mass of product calculated from the balanced equation.
- Limiting reactant The reactant that is used up first and so limits how much product can form.
- Conservation of mass Mass is neither created nor destroyed in a chemical reaction.
Common Mistakes
- Using the mass of the reactant directly in the mole ratio instead of converting it into moles first.
- Forgetting to balance the equation before reading the mole ratio from it.
- Using the relative formula mass of the reactant when the product mass is needed.
- Giving the answer without units, or quoting the actual yield as if it were the theoretical yield.