The theoretical mass is the most product you could possibly make
Definition
Theoretical mass
The maximum mass of a product that could be made if all the limiting reactant is converted and none is lost.
- The theoretical mass assumes every bit of the limiting reactant is converted and no product is lost.
- It is worked out from the balanced equation, so it is a prediction rather than a measured result.
Key Idea
The theoretical mass is the target the reaction is measured against, so it is the bottom of a percentage yield calculation.
Find it with the mole method
Definition
Mole
The amount of a substance, measured in mol, where one mole has a mass in grams numerically equal to its relative formula mass.
Definition
Relative formula mass
The sum of the relative atomic masses of all the atoms shown in a chemical formula.
- Change the mass of the reactant into moles using n=mMrn = \dfrac{m}{M_r}n=Mrm.
- Use the mole ratio from the balanced equation to find the moles of product.
- Change those moles into a mass using m=n×Mrm = n \times M_rm=n×Mr.
Exam technique
- Base the calculation on the limiting reactant, not on any reactant that is in excess.
- Show your moles and ratio clearly, so each stage scores even if the final figure is wrong.
Example
Calcium oxide from a carbonate
- Find the theoretical mass of calcium oxide from 50 g50\ \text{g}50 g of calcium carbonate in CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2CaCO3→CaO+CO2.
- Moles of calcium carbonate =50100=0.5 mol= \dfrac{50}{100} = 0.5\ \text{mol}=10050=0.5 mol.
- The ratio is 1:11:11:1, so 0.5 mol0.5\ \text{mol}0.5 mol of calcium oxide forms.
- Mass of calcium oxide =0.5×56=28 g= 0.5 \times 56 = 28\ \text{g}=0.5×56=28 g, which is the theoretical mass.
- If only 21 g21\ \text{g}21 g were actually collected, this 28 g28\ \text{g}28 g would be the value you divide by to find the percentage yield.
Self review
- What two assumptions does a theoretical mass make?
- List the three steps of the mole method for finding a theoretical mass.
- Find the theoretical mass of magnesium oxide from 6 g6\ \text{g}6 g of magnesium in 2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}2Mg+O2→2MgO.
- Why should the calculation be based on the limiting reactant?
