Percentage yield:

                Is the measure of conversion of reactants to products.  It is a measure of the waste of the reactants

1 Equilibria: The reaction may not go to completion.
2 Side reactions: This will produce 'by - products' reducing the amount of desirable product.
3 Reactant purity: The reactants may be impure meaning you have started with less than you thought you did.
4 Transfers: Every time you move your reactants / products from one place to another, you will leave some behind.
5 Separation / purification: This inevitable results n the loss of product.
  % Yield  = Actual amount of product (moles) x100
  Theoretical amount of product (moles)

The rules:

1 Write a balanced chemical equation
2 Identify the limiting reactant
3 Calculate the theoretical amount of moles of product starting from the limiting reactant
4 Calculate the actual amount of moles of product obtained
5 Calculate % yield using the formula

Examples:

A)  Preparation of ethanoic acid: 

A student reacted 9.20g of ethanol with and excess of sulphuric acid and sodium dichromate (the oxidising agent).  The student obtained 4.35g of ethanoic acid.

Calculate the % yield:

 

1)  Write a balanced chemical equation:

 

CH3CH2OH + 2[O] CH3COOH + H2O

2)  Identify the limiting reactant

3)  Calculate the theoretical amount of moles of product starting from the limiting reactant:

CH3CH2OH + 2[O] CH3COOH + H2O
9.20g            
i            
Moles = mass  /  Mr            
Moles = 9.20  /  46.0     1:1      
Moles = 0.200 moles     Theoretical Moles 0.200 moles    

4)  Calculate the actual amount of moles of product obtained:

CH3CH2OH + 2[O] CH3COOH + H2O
        4.35g    
        i    
          Moles = mass  /  Mr    
          Moles = 4.35  /  60.0    
          Actual Moles 0.0725 moles    

5)  Calculate % yield using the formula:

% Yield = Actual number of moles x 100
    Theoretical number of moles
         
  = 0.0725 x 100
    0.200
         
  = 36.25%    

 

 

B)  Preparation of propyl methanoate

A student prepared propyl methanoate from propan - 1 - ol and methanoic acid. 

The students reacted 3.00g of propan - 1 - ol witth 2.50g of methanoic acid in the presence of a sulphuric acid catalyst.  He was disappointed to obtain only 1.75g of propyl methanoate.

Calculate the % yield of propyl methanoate:

 

1)  Write a balanced chemical equation:

 

CH3CH2CH2OH + HCOOH HCOOCH2CH2CH3 + H2O

2)  Identify the limiting reactant

CH3CH2CH2OH + HCOOH HCOOCH2CH2CH3 + H2O
3.00g   2.50g        
i   i        
Moles = mass  /  Mr   Moles = mass  /  Mr        
Moles = 3.00  /  60.0   Moles = 2.50  /  46.0        
Moles = 0.0500     0.0543        

3)  Calculate the theoretical amount of moles of product starting from the limiting reactant:

CH3CH2CH2OH + HCOOH HCOOCH2CH2CH3 + H2O
3.00g            
i            
Moles = mass  /  Mr              
Moles = 3.00  /  60.0       1:1      
Moles = 0.0500       Theoretical Moles = 0.0500    

4)  Calculate the actual amount of moles of product obtained:

CH3CH2CH2OH + HCOOH HCOOCH2CH2CH3 + H2O
3.00g       1.75g    
i       i    
            Moles = mass  /  Mr    
            Moles = 1.75  /  88.0    
            Actual Moles = 0.0199    

5)  Calculate % yield using the formula:

% Yield = Actual number of moles x 100
    Theoretical number of moles
         
  = 0.0199 x 100
    0.050
         
  = 39.8%    

Questions  P163 1-2

Atom economy:

               Is the measure of the waste associated with the products

Atom economy = Mr of the desired product x 100
    Sum of Mr's of all products

Calculating atom economy:

A)  Bromination of propene:

Atom economy = Mr of the desired product x 100
    Sum of Mr's of all products
         
  = 201.8 x 100
    201.8
         
  = 100%    

B)  Preparation of butan - 1 - ol:

Atom economy = Mr of the desired product x 100
    Sum of Mr's of all products
         
  = 74.0 x 100
    176.9
         
  = 41.8%    

Atom economy and type of reaction:

Questions  P165  1-2

Infrared spectroscopy

Infrared radiation and molecules:

        1)  Bond strength

        2)  Bond length

        3)  Mass of atom at either end of the bond

 

How it works:

What does the spectrum look like:

Applications of IR spectroscopy:

- Paint fragments in hit and run offences

- Monitor unsaturation in polymerisation

- Drug analysis  P167

- Perfume quality control

Questions  P167  1-3

Infrared spectroscopy:  Functional groups

Identification of functional groups:

Bond Functional group Wavenumber
C=O Aldehydes, ketones, carboxylic acids 1640 - 1750
C- H Organic conpounds 2850 - 3100
O- H Carboxylic acids 2500 - 3300 (very broad)
O- H Alcohols (hydrogen bonded) 3200 - 3550 (broad)

Alcohols:

Aldehydes and ketones:

Aldehydes and ketones:

Questions  P169 1-2

Mass Spectrometry

Uses of mass spectroscopy:

Examples:

How a mass spectrometer works:

Explanation

Deflection Time of flight
  • The sample is vaporised then ionised = mass / charge.
  • Ionisation is done by electron impact, chemical ionisation, electrospray or lasers.
  • Electroplates repel and accelerate the ion into the chamber.
  • A strong magnetic field deflects the beam of ions.
  • This is used to determine mass / charge.
  • The ion with a large mass is deflected less and hits the outer edge.
  • The ion with a small mass is deflected most hits the inside edge.
  • Only if the ion has the correct mass / charge ratio will it reach the detector.
  • The magnetic field strength can be varied to get a range of mass / charge readings.
  • The sample is vaporised then ionised = mass / charge.
  • Ionisation is done by electron impact, chemical ionisation, electrospray or lasers.
  • Electroplates repel and accelerate the ion into the chamber.
  • A short time of flight = small mass / charge
  • A long time of flight = large mass / charge
  • The time taken to reach the detector determine the mass / charge of the ion.

 

Mass spectra of elements:

From mass spectra From table of data

RAM of silicon isotopes

% Abundance

28

92.2

29

4.7

30

3.1

Use the formula:-

 

RAM = (%  x  Ar)  +  (%  x  Ar)  +  ....etc

                              100

Use the formula:-

 

         RAM = (%  x  Ar)  +  (%  x  Ar)  +  ....etc

                                        100

RAM = (90.9  x  20) + (0.2  x  21) + (8.9  x  22)

                                      100

 

                               

         RAM = (92.2  x  28)  +  (4.7  x  29)  +   (3.1  x  30)

                                                   100

 

RAM = 20.8

RAM = 28.1

Questions 1-2  P171

Mass spectrometry in organic chemistry

Mass spectrometry and molecules:

C2H5OH

+

e-

C2H5OH+

+

2e-

Fragmentation:

C2H5OH+

CH3

+

CH2OH+

Fragmentation patterns:

Questions  1-2  P173

Mass spectrometry:  Fragmentation patterns

Identifying fragment ions:

m/z value Possible identity of the fragment ion
15 CH3+
29 C2H5+
43 C3H7+
57 C4H9+
17 OH+

Identification of organic structures:

Questions  1-2  P175  /  14  P179  /  3  P181