Calculations of Product | Edexcel International A Level Chemistry Revision Notes 2017

Calculating Yield & Atom Economy

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
- Products are lost during separation and purification
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:

$p e r c e n t a g e y i e l d = \frac{a c t u a l y i e l d}{t h e o r e t i c a l y i e l d} \times 100$

- The actual yield is the number of moles or mass of product obtained experimentally
- The theoretical yield is the number of moles or mass obtained by a reacting mass calculation

Worked example

In an experiment to displace copper from copper(II) sulfate, 6.5 g of zinc was added to an excess of copper(II) sulfate solution. The resulting copper was filtered off, washed and dried. The mass of copper obtained was 4.8 g.

Calculate the percentage yield of copper.

Answer:

Step 1: The balanced symbol equation is:

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

Step 2: Calculate the amount of zinc reacted in moles

n u m b e r o f m o l e s = \frac{6.5 g}{65.4 g m o l^{- 1}} = 0.10 m o l

Step 3: Calculate the maximum amount of copper that could be formed from the molar ratio:

Since the ratio of Zn(s) to Cu(s) is 1:1 a maximum of 0.10 moles can be produced

Step 4: Calculate the maximum mass of copper that could be formed (theoretical yield)

mass = mol x M

mass = 0.10 mol x 63.55 g mol-1

mass = 6.4 g (2 sig figs)

Step 5: Calculate the percentage yield of copper

p e r c e n t a g e y i e l d = \frac{4.8 g}{6.4 g} \times 100 = 75 %

Examiner Tip

It is possible to calculate a percentage yield that is greater than 100%. This would be an error which could occur when preparing crystals due to:

The crystals may still be dry when weighed
There are impurities in the crystals
The mass of the filter paper or container could mistakingly be included in the total mass of the product

Atom economy

The atom economy of a reaction shows how many of the atoms used in the reaction become the desired product
- The rest of the atoms or mass is wasted
It is found directly from the balanced equation by calculating the M_r of the desired product

A t o m e c o n o m y = \frac{m o l e c u l a r m a s s o f d e s i r e d p r o d u c t}{s u m o f m o l e c u l a r m a s s e s o f A L L r e a c t a n t s} \times 100

In addition reactions, the atom economy will always be 100% because all of the atoms are used to make the desired product
- Whenever there is only one product, the atom economy will always be 100%
- For example, in the reaction between ethene and bromine:

CH₂=CH₂ + Br₂ → CH₂BrCH2Br

The atom economy could also be calculated using mass, instead or M_r
- In this case, you would divide the mass of the desired product formed by the total mass of all reactants, and then multiply by 100
Questions about atom economy often ask in qualitative or quantitative terms

Worked example

Qualitative atom economy
Ethanol can be produced by various reactions, such as:

Hydration of ethene: C₂H₄ + H₂O → C₂H₅OH

Substitution of bromoethane: C₂H₅Br + NaOH → C₂H₅OH + NaBr

Explain which reaction has a higher atom economy.

Answer

Hydration of ethene has a higher atom economy (of 100%) because all of the reactants are converted into products, whereas the substitution of bromoethane produces NaBr as a waste product

Worked example

Quantitative atom economy

The blast furnace uses carbon monoxide to reduce iron(III) oxide to iron.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Calculate the atom economy for this reaction, assuming that iron is the desired product.

(A_r / M_r data: Fe₂O₃ = 159.6, CO = 28.0, Fe = 55.8, CO₂ = 44.0)

Answer

Step 1: Write the equation:

$A t o m e c o n o m y = \frac{m o l e c u l a r m a s s o f d e s i r e d p r o d u c t}{s u m o f m o l e c u l a r m a s s e s o f A L L r e a c t a n t s} \times 100$

Step 2: Substitute values and evaluate:

A t o m e c o n o m y = \frac{2 \times 55.8}{159.6 + (3 \times 28.0)} \times 100 = 45.9 %

Examiner Tip

Careful: Sometimes a question may ask you to show your working when calculating atom economy.

In this case, even if it is an addition reaction and it is obvious that the atom economy is 100%, you will still need to show your working.

Calculations of Product (Edexcel International A Level Chemistry)

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1. Structure, Bonding & Introduction to Organic Chemistry

1.1 Formulae & Equations

1.1.1 Formulae & Mass

1.1.2 Avogadro & the Mole

1.1.3 Full & Ionic Equations

1.1.4 Calculating Formulae

1.2 Amount of Substance

1.2.1 Concentration Calculations

1.2.2 Reacting Mass Calculations

1.2.3 Reacting Volume Calculations

1.2.4 Calculations of Product

1.3 Atomic Structure

1.3.1 Sub-Atomic Particles

1.3.2 Isotopes & Mass Spectra

1.4 Electrons & Ions

1.4.1 Endothermic Ionisation Energy

1.4.2 Electronic Structures

1.4.3 Shapes of Orbitals

1.4.4 Electronic Configurations & Chemical Properties

1.5 The Periodic Table

1.5.1 Periodicity - Ionisation Energy

1.5.2 Periodicity - Thermal Trends

1.6 Ionic & Metallic Bonding & Structure

1.6.1 Evidence of Ions

1.6.2 Ionic Bonding & Structures

1.6.3 Ionic Bond Strength

1.6.4 Polarisation

1.6.5 Metallic Bonding & Lattices

1.7 Covalent Bonding & Structure

1.7.1 Evidence of Covalent Bonding

1.7.2 Covalent Bonding

1.7.3 Carbon Allotropes

1.7.4 Electronegativity

1.7.5 VSEPR Theory

1.7.6 Predicting Shapes & Angles

1.8 Introductory Organic Chemistry

1.8.1 Hazards & Risks

1.8.2 Functional Groups & Homologous Series

1.8.3 Nomenclature & Classification

1.8.4 Isomers - Structural

1.9 Alkanes

1.9.1 Alkanes - Introduction

1.9.2 Alkanes as Fuels

1.9.3 Combustion of Alkanes

1.9.4 Alternative Fuels

1.9.5 Free Radicals & Fission

1.9.6 The Free Radical Substitution Mechanism

1.10 Alkenes

1.10.1 Alkenes - Introduction

1.10.2 Isomers - Geometric

1.10.3 Reactions of Alkenes

1.10.4 Saturation Test

1.10.5 Electrophilic Addition - Mechanism

1.10.6 Addition Polymerisation

1.10.7 Polymer Disposal

2. Energetics, Group Chemistry, Halogenoalkanes & Alcohols

2.1 Energetics

2.1.1 Enthalpy Level Diagrams

2.1.2 Enthalpy Change - Definitions

2.1.3 Using Calorimetry

2.1.4 Using Hess Cycles

2.1.5 Bond Enthalpies

2.2 Intermolecular Forces

2.2.1 Intermolecular Forces - Introduction

2.2.2 Hydrogen Bonding

2.2.3 Physical Properties

2.2.4 Solvent Choice

2.3 Redox Chemistry & Acid-Base Titrations

2.3.1 Oxidation Numbers - Introduction

2.3.2 Oxidation & Reduction

2.3.3 Disproportionation Reactions

2.3.4 Ionic Equations

2.3.5 Acid-Base Titrations with Indicators