Percentage yield:

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

• When we think about reactions, we always think of them as going 100% to products.
• This is usually not the case due to:
 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.
• Percentage yield is like a score in a test. It is an indication of what you achieved out of what you could have got:
 % 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] à CH3COOH + H2O

2)  Identify the limiting reactant

• You are told in the question that you have an excess of sulphuric acid and sodium dichromate.
• This means that the limiting reactant is ethanol.

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

• Calculate the amount of moles of ethanoic acid you could have made:
 CH3CH2OH + 2[O] à CH3COOH + H2O 9.20g i Moles = mass  /  Mr
 Moles = 9.20  /  46.0 1:1
 Moles = 0.200 moles à Theoretical Moles 0.200 moles

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

• Calculate the number of moles you actually made:
 CH3CH2OH + 2[O] à 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 à HCOOCH2CH2CH3 + H2O

2)  Identify the limiting reactant

• You are given 2 starting amounts which means you have to work out which one is the limiting reactant:
 CH3CH2CH2OH + HCOOH à 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
• Propan - 1 - ol is the limiting reactant so the theoretical calculation must be made using this

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

• Calculate the amount of moles of ethanoic acid you could have made:
 CH3CH2CH2OH + HCOOH à HCOOCH2CH2CH3 + H2O 3.00g i Moles = mass  /  Mr Moles = 3.00  /  60.0 1:1 Moles = 0.0500 à Theoretical Moles = 0.0500

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

• Calculate the number of moles you actually made:
 CH3CH2CH2OH + HCOOH à 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

• % yield tells us how much of our product is made from our starting materials but it doesn't take into account any undesirable or side products.
• Atom economy takes into account any wasteful by products too
• By products are considered wasteful as they are usually disposed of.  This is costly and can cause environmental problems.
• A more efficient way of dealing with by products would be to sell them on to companies that would make use of them.
 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%
• Any reaction that gives only one product is very atom economic, addition reactions for example.

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%
• This means that most of the starting materials ended up as waste.

Atom economy and type of reaction:

• Reactions having only one product have a high atom economy.  The type of reactions giving only one product are addition reactions.
• Reactions giving more than one product have a low atom economy.  The type of reactions giving more than one product are substitution / elimination reactions.
• To improve the atom economy for substitution / elimination reactions, a use for the undesired product should be found.
• If the undesired product is toxic, we have even bigger problems -disposal.

Questions  P165  1-2

Infrared spectroscopy

• All molecules absorb IR light.
• The IR light makes the bonds in a molecule vibrate (like the engine of a bus making the windows vibrate).
• Vibrations occur in one of 2 ways, a stretching vibration or a bending vibration:

• Every bond vibrates at its own unique frequency depending on:

1)  Bond strength

2)  Bond length

3)  Mass of atom at either end of the bond

How it works:

• The full IR spectrum is passed through a sample.

• The frequencies are called wavenumbers - 300 - 4000cm-1

• Some frequencies make some of the bonds in a molecule vibrate

• When these bonds vibrate they absorb energy from the IR light source.

• This means less IR light gets through the sample to the detector.

• Each absorbance peak is characteristic of a particular bond / atoms vibrating.

• A trace which we call a spectrum is produced.

What does the spectrum look like:

• The spectrum gives us 'peaks' which are actually absorbance troughs.

• These troughs are caused by a frequency of IR light being absorbed from a bond vibrating bond.

• Each 'peak' is characteristic to a specific bond / atoms

Applications of IR spectroscopy:

• It is used widely in forensic science analysing:

- 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:

• We have just seen that the peak on an IR spectra are due to specific bonds (and atoms) vibrating or stretching.
• The frequency at which you find an absorbance peak is therefore unique to bonds and atoms at each end of the bond.
• This means that functional groups will give specific peaks.
• The groups you need to know are:
 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)
• Do not get the peaks for C - H bonds confused with O - H bonds.

Alcohols:

• The IR spectrum for methanol, CH3OH is shown below:

• The peak at 3230 - 3500 represents an O - H group in alcohols.

Aldehydes and ketones:

• The IR spectrum for propanal, CH3CHO is shown below:

• The peak at 1680 - 1750 represents a C=O group in aldehydes and ketones.

Aldehydes and ketones:

• The IR spectrum for propanoic acid, CH3CH2COOH is shown below:

• The peak at 2500 - 3300 represents an O - H group in a carboxylic acid.
• The peak at 1680 - 1750 represents a C=O group in a carboxylic acid.

Questions  P169 1-2

Mass Spectrometry

Uses of mass spectroscopy:

• First developed by JJ Thompson at the start of the 20th Century.
• It is used:
• To identify unknown compounds.
• To determine the abundance of isotopes
• To gain further information about the structure and chemical properties of molecules.

Examples:

• To examine patients breath while under anaesthetic.
• Detecting banned substances - steroids in athletes.
• Detecting traces of toxic chemicals in contaminated marine life

How a mass spectrometer works:

 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:

• One of the most important uses is to determine the isotopes present in a natural sample of an element.
• A mass spectrum shows the mass charge (the Ar) and the abundance as a %.
• This information can be used to determine the relative atomic mass:
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:

• Ionisation in a mass spectroscope is usually done by electron bombardment.
• Electron bombardment knocks another electron out of the molecule producing a positive molecular ion:
 C2H5OH + e- à C2H5OH+ + 2e-
• This is called the molecular ion, M+.
• The mass of the electron lost electron is negligible.
• The molecular ion has the same mass as the Mr of the molecule.
• As we have a mass and a charge we can use a mass spectrometer to determine the Mr (m/z).

Fragmentation:

• Excess energy from the ionisation process causes bonds in the organic molecule to vibrate and weaken.
• This causes the molecule to split or fragment into smaller pieces.
• Fragmentation gives a positively charged molecular fragment ion and a neutral molecule:
 C2H5OH+ à CH3 + CH2OH+
• The fragment ion, CH2OH+ has a mass and charge so we can use a mass spectrometer to determine the Mr (m/z) of that fragment.
• Fragment ions can be broken up further to give a range of m/z values.
• The m/z values correspond to the Mr's of the molecule and its fragments.
• The Mr of the molecule is always the highest m/z value - ie this molecule has not been fragmented so it must have the highest Mr.
• The one below is for ethanol.  It has a m/z of 46 which is also its Mr.

Fragmentation patterns:

• Mass spectroscopy is used to identify and determine the structures of unknown compounds.
• Although 2 isomers will have exactly the same M+ peak, the fragmentation patterns will be unique to that molecule, like a fingerprint.
• In practice mass spectrometers are linked to a database and the spectra is compared until an exact match is found:
• These are the mass spectra for pentane and a structural isomer of pentane, 2 methyl butane.
• The M+ peak is the same for each but the fragmentation patterns are different.

Questions  1-2  P173

Mass spectrometry:  Fragmentation patterns

Identifying fragment ions:

• When you look at a mass spectrum, other peaks seem to look more important than the M+ peak.
• These fragment peaks give clues to the structure of the compound.
• Even simple structures give common peaks that can be identified:
 m/z value Possible identity of the fragment ion 15 CH3+ 29 C2H5+ 43 C3H7+ 57 C4H9+ 17 OH+
• Functional groups are a good place to start, OH = m/z of 17
• Some fragments are more difficult to identify as these will have undergone molecular rearrangement.

Identification of organic structures:

• A mass spectrum will not only tell you the Mr (from the M+ peak), but it can also tell you some of the structural detail.
• These peaks have been labelled with a letter:

• The mass spectrum above has been produced from hexane.
• The following reactions show how the molecule could fragment to form the fragment ions 57 and 43:

Questions  1-2  P175  /  14  P179  /  3  P181