Reaction Yields | AQA A Level Chemistry Revision Notes 2017

Percentage Yield

In a lot of reactions, not all reactants react to form products which can be due to several factors:
- Other reactions take place simultaneously
- The reaction does not go to completion
- Reactants or products are lost to the atmosphere
The percentage yield shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get:

percentage yield = $\frac{actual yield}{theoretical yield}$

Where actual yield is the number of moles or mass of product obtained experimentally
The predicted yield is the number of moles or mass obtained by calculation

You will often have to use the following equation to work out the reacting masses, to calculate the predicted yield

number of mol = $\frac{mass of a substance in grams (g)}{molar mass (g {mol}^{- 1})}$

It is important to be clear about the type of particle you are referring to when dealing with moles
- Eg. 1 mole of CaF₂ contains one mole of CaF₂ formula units, but one mole of Ca²⁺and two moles of F^-ions

Calculate % yield using moles

In an experiment to displace copper from copper(II) sulfate, 6.54 g of zinc was added to an excess of copper(II) sulfate solution.

The copper was filtered off, washed and dried.

The mass of copper obtained was 4.80 g.

Calculate the percentage yield of copper.

Answer:

Write the balanced symbol equation:
- Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)
Calculate the number of moles of zinc:
- n(Zn) = $\frac{6.54 g}{65.4 g {mol}^{- 1}}$ = 0.10 moles
Deduce the number of moles of copper, using the balanced chemical equation:
- 1 mole of zinc forms 1 mole of copper
  - The ratio is 1 : 1
- Therefore, n(Cu) = 0.10 moles
Calculate the maximum mass (theoretical yield) of copper:
- Mass = mol x M_r
- Mass = 0.10 mol x 63.5 g mol^-1
- Mass = 6.35 g
Calculate the percentage yield of copper:
- Percentage yield = $\frac{4.80 g}{6.35 g}$ x 100 = 75.6 %

Sometimes, there is an excess of one or more of the reactants (excess reagent)
The reactant which is not in excess is called the limiting reagent
To determine which reactant is limiting:
- The number of moles of each reactant should be calculated
- The ratio of the reactants shown in the equation should be taken into account e.g.

2Na + S → Na₂S

Here, the ratio of Na : S is 2 : 1, and this should be taken into account when doing calculations
Once all of one reactant has been used up, the reaction will stop, even if there are moles of the other reactant(s) leftover
- The reactant leftover is in excess, the reactant which causes the reaction to stop once it is used up is the limiting reagent

Excess & limiting reagent

9.2 g of sodium is reacted with 8.0 g of sulfur to produce sodium sulfide, Na₂S.

Which reactant is in excess and which is the limiting reactant?

Answer

Step 1: Calculate the moles of each reactant
- mol(Na) = $\frac{9.2 g}{23 g {mol}^{- 1}}$ = 0.40 mol
- mol(S) = $\frac{8.0}{32 g {mol}^{- 1}}$ = 0.25 mol
Step 2: Write the balanced equation and determine the molar ratio

2Na + S → Na₂S

- The molar ratio of Na: Na₂S is 2:1
Step 3: Compare the moles and determine the limiting reagent
- So, to react completely 0.40 moles of Na require 0.20 moles of S
- Since there are 0.25 moles of S, then S is in excess
- Na is therefore the limiting reactant.
- Once all of the Na has been used up, the reaction will stop, even though there is S left