Practical Guide EDEXCEL

N Goalby chemrevise.org 1

Potential errors in using a gas syringe

•gas escapes before bung inserted

•syringe sticks

• some gases like carbon dioxide or sulphur dioxide are

soluble in water so the true amount of gas is not

measured.

Using a gas syringe

If drawing a gas syringe make sure

you draw it with some

measurement markings on the

barrel to show measurements can

be made.

The volume of a gas depends on pressure

and temperature so when recording volume

it is important to note down the

temperature and pressure of the room.

Make sure you don’t leave gaps in

your diagram where gas could

escape

Gas syringes can be used for a variety of

experiments where the volume of a gas is

measured, possibly to work out moles of

gas or to follow reaction rates.

Moles of gas can be calculated from gas

volume (and temperature and pressure)

using ideal gas equation PV = nRT or using

the molar gas volume (1mol gas =24dm

room temperature and pressure

Measuring gas volumes

Alternatively gas volumes can be

measured ‘over water’ with an up-turned

measuring cylinder in a trough of water

Practical Guide EDEXCEL

This guide includes details about the core practicals for A-level chemistry. It also contains information about

other experiments that often occur in A-level examinations. You may be asked to describe these experiments in

details or be asked about reasons for doing individual steps.

You may be asked about other unfamiliar experiments but these will be using the skills and techniques that are

described in the following experiments.

Irritant - dilute acid and alkalis- wear googles

Corrosive- stronger acids and alkalis wear goggles

Flammable – keep away from naked flames

Toxic – wear gloves- avoid skin contact- wash hands after use

Oxidising- Keep away from flammable / easily oxidised materials

Hazardous substances in low

concentrations or amounts will

not pose the same risks as the

pure substance.

Safety and hazards

N Goalby chemrevise.org 2

Detailed method

1. Measure 30 cm

of 1 mol dm⁻

ethanoic acid and transfer

to a conical flask.

2. Attach conical flask to gas syringe or use collection over

water method (see previous page)

3. Measure the mass of a weighing bottle with approximately

0.05 g of calcium carbonate

4. Add the calcium carbonate to the conical flask- quickly

resealing the bung so no gas escapes

5. Measure the final total volume of gas

6. Reweigh the empty weighing bottle test tube from step 3

7. Repeat the experiment several more times, increasing the

mass of calcium carbonate by about 0.05 g each time.

Method for using a gas syringe to calculate the Mr of propanone

1. Extract 0.20 cm

of propanone into a hypodermic syringe and then measure the mass of this syringe

2. using hand protection, remove a gas syringe from the oven and note the volume of air already in the

barrel – about 5 cm

3. inject the propanone through the self-seal cap into the barrel. The plunger will move straight away.

4. Put the gas syringe back into the oven.

5. Measure the mass of the empty hypodermic syringe immediately.

6. After a few minutes measure the volume of the gas in the gas syringe, record the temperature of the oven

shelf and the pressure of the room.

Example 1 : 0.150g of a volatile liquid was injected into a sealed gas syringe. The gas syringe was placed in an oven at

C at a pressure of 100kPa and a volume of 80cm

was measured. What is the Mr of the volatile liquid ? (R = 8.31)

moles = PV/RT

= 100 000 x 0.00008 / (8.31 x 343)

= 0.00281 mol

100 kPa = 100 000 Pa

80 cm

= 0.00008 m

Mr = mass/moles

= 0.15 / 0.00281

= 53.4 g mol

-1

Core Practical 1:Measure the molar volume of a gas

Mass of CaCO

in g

Volume of CO

in cm

Analysis

From the graph read the volume of CO

given off with 0.25 g CaCO

Work out the moles of CaCO

in 0.25g = 0.25/100.1 = 2.5 x 10

-3

Assume the moles of CO

= moles of CaCO

Work out molar volume of CO

= volume of CO

/ moles of CO

The water of crystallisation in calcium sulphate crystals can be

removed as water vapour by heating as shown in the following

equation.

CaSO

.xH

O(s) → CaSO

(s) + xH

O(g)

Method.

•Weigh an empty clean dry crucible and lid .

•Add 2g of hydrated calcium sulphate to the crucible and weigh

again

•Heat strongly with a Bunsen for a couple of minutes

•Allow to cool

•Weigh the crucible and contents again

•Heat crucible again and reweigh until you reach a constant mass (

do this to ensure reaction is complete).

Small amounts of the solid , such as

0.100 g, should not be used in this

experiment as the percentage

uncertainties in weighing will be too

high.

Large amounts of hydrated calcium sulphate, such as 50g,

should not be used in this experiment as the decomposition is

likely to be incomplete.

The lid improves the accuracy of the

experiment as it prevents loss of solid

from the crucible but should be loose

fitting to allow gas to escape.

The crucible needs to be dry otherwise a wet crucible would

give an inaccurate result. It would cause mass loss to be too

large as the water would be lost when heating.

Heating in a crucible

This method could be used for measuring mass loss in various

thermal decomposition reactions and also for mass gain when

reacting magnesium in oxygen.

N Goalby chemrevise.org

Example 2. 3.51 g of hydrated zinc sulphate were heated and 1.97 g of anhydrous zinc sulphate were

obtained. Use these data to calculate the value of the integer x in ZnSO

.xH

Calculate the mass of H

O = 3.51 – 1.97 = 1.54g

Calculate moles of

ZnSO

Calculate moles of

1.97

161.5

1.54

= 0.0122

=0.085

Calculate ratio of mole of

ZnSO

to H

0.0122

= 0.085

0.0122

X = 7

N Goalby chemrevise.org 4

• Weigh the sample bottle containing the required mass of solid

on a 2 dp balance

• Transfer to beaker

• Reweigh empty sample bottle

• Record the difference in mass

• Add 100cm

of distilled water to the beaker. Use a glass rod to

stir to help dissolve the solid.

•Sometimes the substance may not dissolve well in cold water so

the beaker and its contents could be heated gently until all the

solid had dissolved.

• Pour solution into a 250cm

graduated flask via a funnel.

• Rinse beaker and funnel and add washings from the beaker

and glass rod to the volumetric flask.

• make up to the mark with distilled water using a dropping

pipette for last few drops.

• Invert flask several times to ensure uniform solution.

Making a solution

Alternatively the known mass of solid

in the weighing bottle could be

transferred to beaker, washed and

washings added to the beaker.

Remember to fill so the bottom of the

meniscus sits on the line on the neck of the

flask. With dark liquids like potassium

manganate it can be difficult to see the

meniscus.

Shake the volumetric flask thoroughly to

ensure a uniform concentration

Graduated/volumetric flask

A graduated flask has one mark on the neck which the level to

fill to get the accurate volume. Do not heat or put hot

solutions in the volumetric flask because the heat would cause

the flask to expand and the volume would then be incorrect.

Use a teat pipette to make up to the mark in

volumetric flask to ensure volume of solution

accurately measured and one doesn’t go over

the line

Diluting a solution

•Pipette 25cm

of original solution into a 250cm

volumetric

flask

•make up to the mark with distilled water using a dropping

pipette for last few drops.

• Invert flask several times to ensure uniform solution.

Using a volumetric pipette is more accurate

than a measuring cylinder because it has a

smaller uncertainty

Dilutions

Core practical 2. Make up a volumetric solution and carry out a simple acid–base titration

Measuring mass accurately:

In many experiments the best method for measuring mass is

1. Measure mass on 2 or 3d.p. balance of a weighing bottle

with the required quantity of solid in it

2. Empty mass into reaction vessel/flask

3. Reweigh the now empty weighing bottle

4. Subtract the mass of the empty weighing bottle from the

first reading to give exact of mass actually added.

N Goalby chemrevise.org 5

Titrations are done often to find out the concentration of one

substance by reacting it with another substance of known

concentration.

They are often done with neutralisation reactions, but can be

done with redox reactions.

However, the standard phrase: titrate solution A

with solution B means that A should be in the

conical flask and B should be in the burette.

One substance (generally the one we don’t know the

concentration) is put in the conical flask. It is measured

using a volumetric pipette.

The other substance is placed in the burette

burette

pipette

conical

flask

Using the pipette

Detailed Method for Titration

•rinse pipette with substance to go in it (often alkali).

•pipette 25 cm

of solution A into conical flask. The

volumetric pipette will have a mark on its neck to show

the level to fill to. The bottom of the meniscus should

sit on this line.

•touch surface of solution with pipette ( to ensure

correct amount is added). A small amount of solution

will be left in the pipette at this stage. The calibration

of the pipette will take into account this effect. It

should not be forced out.

Make sure bottom of

meniscus is on line on

neck of pipette

Titrations

A conical flask is used in preference to a beaker because

it is easier to swirl the mixture in a conical flask without

spilling the contents.

General Method

•rinse equipment (burette with acid, pipette with alkali, conical flask with distilled water)

•pipette 25 cm

of alkali into conical flask

•touch surface of alkali with pipette ( to ensure correct amount is added)

•adds acid solution from burette

•make sure the jet space in the burette is filled with acid

•add a few drops of indicator and refer to colour change at end point

•phenolphthalein [pink (alkali) to colourless (acid): end point pink colour just disappears] [use if NaOH is used]

•methyl orange [yellow (alkali) to red (acid): end point orange] [use if HCl is used]

•use a white tile underneath the flask to help observe the colour change

•add acid to alkali whilst swirling the mixture and add acid drop wise at end point

•note burette reading before and after addition of acid

•repeats titration until at least 2 concordant results are obtained- two readings within 0.1 of each other

Core Practical 3: Make up a volumetric solution and carry out a simple acid–base titration

N Goalby chemrevise.org

Using the burette

The burette should be rinsed out with substance that will be

put in it. If it is not rinsed out the acid or alkali added may be

diluted by residual water in the burette or may react with

substances left from a previous titration. This would lead to

the concentration of the substance being lowered and a larger

titre being delivered.

Don’t leave the funnel in the burette because small drops of

liquid may fall from the funnel during the titration leading to a

false burette reading (would give a lower titre volume)

make sure the jet space in the burette is filled with the

solution and air bubbles are removed.

If the jet space in the burette is not filled properly prior to commencing the

titration it will lead to errors if it then fills during the titration, leading to a

larger than expected titre reading.

Read the bottom of the meniscus on the burette

This is reading 9.00cm

Even though a burette has marking reading to 0.1cm

the burette readings

should always be given to 2dp either ending in 0.00 or 0.05. 0.05cm

is the

volume of 1 drop of solution delivered from a burette and so this is the

smallest difference in readings that can be measured. If the bottom of the

meniscus sits on a line it should end with a 0.00 as in the above example

9.00cm

If the meniscus sits between two lines it should end 0.05. e.g. if the

bottom of the meniscus sits between the lines marked 9.1 and 9.2, you should

record 9.15

Add a few drops of indicator and refer to colour change

at end point

Adding indicator

phenolphthalein

If acid is added from the burette the colour change would

be pink (alkali) to colourless (acid): end point pink colour

just disappears [use with titrations using strong alkalis e.g.

NaOH ]

phenolphthalein

Alkali colour

phenolphthalein acid

colour

Methyl orange

Use a white tile underneath the flask to help

observe the colour change

Methyl orange is a suitable indicator for neutralisation

reactions where strong acids are used.

It is red in acid and yellow in alkali. It is orange at the end

point.

Methyl orange

Alkali colour

Methyl orange

acid colour

Methyl orange

end point

Indicators are generally weak acids so only add a few

drops of them. If too much is added they will affect

the titration result

N Goalby chemrevise.org 7

Add solution from burette whilst swirling the mixture and add drop-wise at end

point

note burette reading before and after addition of solution

repeats titration until at least 2 concordant results are

obtained- two readings within 0.1 of each other

Distilled water can be added to the conical flask during a titration to wash the

sides of the flask so that all the acid on the side is washed into the reaction

mixture to react with the alkali.

It does not affect the titration reading as water does not react with the reagents

or change the number of moles of acid added.

Only distilled water should be used to

wash out conical flasks between titrations

because it does not add any extra moles

of reagents

Recording results

•Results should be clearly recorded in a table

•Result should be recorded in full (i.e. both initial and final readings)

•Record titre volumes to 2dp (0.05 cm

)

Working out average titre results

Only make an average of the

concordant titre results

lf 2 or 3 values are within 0.10cm

and

therefore concordant or close then we

can say results are accurate and

repeatable and the titration technique

is good and consistent

Titration number 1 2 3

Initial burette reading (cm

) 0.50 2.50 1.55

Final burette reading (cm

) 24.50 27.00 25.95

Titre (cm

) 24.00 24.50 24.40

Average titre = (24.50+ 24.40)/2 =

24.45

Safety precautions

Acids and alkalis are corrosive

(at low concentrations acids are irritants)

Wear eye protection and gloves

If spilled immediately wash affected parts after spillage

If substance is unknown treat it as potentially toxic and wear

gloves.

A single titration could be flawed. Repeating allows for

anomalous titres to be spotted and discounted

Testing batches

In quality control it will be necessary to do titrations/testing

on several samples as the amount/concentration of the

chemical being tested may vary between samples.

Titrating mixtures

If titrating a mixture to work out the concentration of

an active ingredient it is necessary to consider if the

mixture contains other substances that have acid

base properties.

If they don’t have acid base properties we can titrate

with confidence.

Common Titration Equations

H + NaOH  CH

+ H

+ 2NaOH  Na

+2H

HCl + NaOH  NaCl +H

NaHCO

+ HCl  NaCl + CO

+ H

+ 2HCl 2NaCl + CO

+ H

N Goalby chemrevise.org 8

Calculating Apparatus Uncertainties

Each type of apparatus has a sensitivity uncertainty

•balance  0.001 g (if using a 3 d.p. balance)

•volumetric flask  0.1 cm

•25 cm

pipette  0.1 cm

•burette (start & end readings and end point )  0.15 cm

Calculate the percentage error for each piece of equipment used by

% uncertainty =  uncertainty x 100

Measurement made on apparatus

e.g. for burette

% uncertainty = 0.15/average titre result x100

To calculate the maximum total percentage apparatus uncertainty in the

final result add all the individual equipment uncertainties together.

Reducing uncertainties in a titration

Replacing measuring cylinders with pipettes or burettes which have

lower apparatus uncertainty will lower the error.

To reduce the uncertainty in a burette reading it is necessary to make

the titre a larger volume. This could be done by: increasing the volume

and concentration of the substance in the conical flask or by decreasing

the concentration of the substance in the burette.

To decrease the apparatus uncertainties

you can either decrease the sensitivity

uncertainty by using apparatus with a

greater resolution (finer scale divisions ) or

you can increase the size of the

measurement made.

If looking at a series of measurements in

an investigation, the experiments with

the smallest readings will have the

highest experimental uncertainties.

Reducing uncertainties in measuring mass

Using a more accurate balance or a larger mass will reduce the

uncertainty in weighing a solid

Weighing sample before and after addition and then

calculating difference will ensure a more accurate

measurement of the mass added.

Calculating the percentage difference between the actual

value and the calculated value

If we calculated an Mr of 203 and the real value is 214, then

the calculation is as follows:

Calculate difference 214-203 = 11

% = 11/214 x100

=5.41%

If the %uncertainty due to the apparatus < percentage

difference between the actual value and the calculated

value then there is a discrepancy in the result due to

other errors.

If the %uncertainty due to the apparatus > percentage

difference between the actual value and the calculated

value then there is no discrepancy and all errors in the

results can be explained by the sensitivity of the

equipment.

Uncertainty

In general, if uncertainty is not indicated

on apparatus, the following assumptions

are made:

For an analogue scale-

The uncertainty of a reading (one

judgement) is at least ±0.5 of the smallest

scale reading.

The uncertainty of a measurement (two

judgements) is at least ±1 of the smallest

scale reading.

- If the apparatus has a digital scale, the

uncertainty is  the resolution of the

apparatus in each measurement

Readings and Measurements

Readings

the values found from a single

judgement when using a piece

of equipment

Measurements

the values taken as the

difference between the

judgements of two values

(e.g. using a burette in a

titration)

Uncertainty of a measurement using

a burette. If the burette used in the

titration had an uncertainty for each

reading of +/– 0.05 cm

then during a

titration two readings would be taken

so the uncertainty on the titre volume

would be +/– 0.10 cm

N Goalby chemrevise.org 9

Testing for halogenoalkanes method

• Arrange three test tubes in a row and add three drops of halogenoalkane in the sequence 1-chlorobutane, 1-

bromobutane, 1-iodobutane.

• Add 4 cm

of 0.02 M silver nitrate to each halogenoalkane.

• Without delay, put all three test tubes simultaneously in a hot water bath.

• Note the order in which precipitates appear

Comparing the rate of hydrolysis of halogenoalkanes reaction

Water is a poor nucleophile but it can react

slowly with haloalkanes in a substitution

reaction

Hydrolysis is defined as the splitting of a molecule ( in this case a

haloalkane) by a reaction with water

X + H

O  CH

OH + X

+ H

Aqueous silver nitrate is added to a haloalkane and the halide

leaving group combines with a silver ion to form a SILVER HALIDE

PRECIPITATE.

The precipitate only forms when the halide ion has left the

haloalkane and so the rate of formation of the precipitate can be

used to compare the reactivity of the different haloalkanes.

I + H

O  CH

OH + I

+ H

(aq)

+ I

(aq)

 AgI

(s)

- yellow precipitate

The iodoalkane forms a precipitate with the

silver nitrate first as the C-I bond is weakest

and so it hydrolyses the quickest

The quicker the precipitate is formed, the faster the substitution

reaction and the more reactive the haloalkane

AgI

(s)

- yellow precipitate

AgBr

(s)

– cream precipitate

AgCl

(s)

– white precipitate

forms faster

The rate of these substitution reactions depends on the strength of the

C-X bond . The weaker the bond, the easier it is to break and the faster

the reaction.

Core practical 4: Investigation of the rates of hydrolysis of some halogenoalkanes

N Goalby chemrevise.org 10

Detailed Method: The partial oxidation of propan-1-ol

This experiment uses a limited quantity of oxidising agent (0.01 mol) and the product is distilled from the reaction

mixture immediately it is formed. In this way we hope to achieve a partial oxidation of propan-1-ol.

 Place about 10 cm

of dilute sulphuric acid in a flask and add about 3g of potassium dichromate(VI) and 2 or 3

anti-bumping granules. Shake the contents of the flask until solution is complete (do not warm).

 Add 1.5 cm

of propan-1-ol in drops from a dropping pipette, shaking the flask so as to mix the contents, and

then assemble distillation apparatus as shown below

 Gently heat and slowly distil 2 cm

of liquid into a test tube, taking care that none of the reaction mixture

splashes over.

SAFETY

You must wear gloves when handling

solid potassium dichromate(Vl) since it is

highly toxic and a category 2 carcinogen;

it is also an irritant. Avoid inhaling any

dust.

Concentrated sulphuric acid is corrosive.

Partial Oxidation of Primary Alcohols

Reaction: primary alcohol  aldehyde

Reagent: potassium dichromate (VI) solution and dilute sulphuric acid.

Conditions: (use a limited amount of dichromate) warm gently and distil out

the aldehyde as it forms:

Ethanal

Observation: the orange

dichromate ion (Cr

) reduces

to the green Cr

ion

propan-1-ol

propanal

+ [O]

+ H

OH + [O]  CH

CHO + H

+ [O]

+ H

C O

Distillation

In general used as separation technique to separate an

organic product from its reacting mixture. Need to

collect the distillate of the approximate boiling point

range of the desired liquid.

Water in

Water

out

Liebig condenser

thermometer

Heat

Note the bulb of the thermometer should be

at the T junction connecting to the

condenser to measure the correct boiling

point

Note the water goes in the bottom of the

condenser to go against gravity. This allows more

efficient cooling and prevents back flow of water.

It’s important to be able to

draw and label this apparatus

accurately. Don’t draw lines

between flask, adaptor and

condenser.

Round

bottomed

flask

Electric heaters are often used to heat organic

chemicals. This is because organic chemicals are

normally highly flammable and could set on fire

with a naked flame.

Core practical 5. Oxidation of an alcohol

N Goalby chemrevise.org 11

Detailed method

 Measure 5 cm

of water into a boiling tube. Add 6 g of sodium dichromate(VI), shake and set aside to dissolve.

 Put about 1.5 cm

propan-1-ol into a 50 cm

round bottomed flask and add about 5 cm

of water and two or

three anti-bumping granules. Put a condenser on the flask for reflux, as shown in figure below.

 Add 2 cm

of concentrated sulphuric acid down the condenser in drops from a dropping pipette. While the

mixture is still warm, start to add your sodium dichromate(VI) solution down the condenser in drops from a

dropping pipette. The energy released from the reaction should make the mixture boil. Add the solution a drop

at a time so that the mixture continues to boil without any external heating.

 When all the sodium dichromate(VI) solution has been added, use a low Bunsen burner flame to keep the

mixture boiling for 10 minutes, not allowing any vapour to escape.

 At the end of that time remove the Bunsen burner and arrange the apparatus for distillation. Gently distil 2-3

of liquid into a test tube.

Reflux: Full Oxidation of Primary Alcohols

Reaction: primary alcohol  carboxylic acid

Reagent: potassium/sodium dichromate(VI) solution and

sulphuric acid

Conditions: use an excess of dichromate, and heat under

reflux: (distil off product after the reaction has

finished)

C C

O H

Propanoic acid

propan-1-ol

Propanoic acid

Observation: the

orange dichromate

ion (Cr

)

reduces to the

green Cr

ion

OH + 2[O]  CH

COOH + H

+ 2[O]

+ H

Reflux

Reflux is used when heating organic reaction mixtures for long periods. The

condenser prevents organic vapours from escaping by condensing them back

to liquids. The reactant vapours of volatile compound are condensed and

returned to the reaction mixture.

Never seal the end of the condenser as the build up of gas

pressure could cause the apparatus to explode. This is true of any

apparatus where volatile liquids are heated including the

distillation set up

Water in

Water out

Heat

Anti-bumping granules are added to the flask in both distillation

and reflux to prevent vigorous, uneven boiling by making small

bubbles form instead of large bubbles

It’s important to be able to draw and label this apparatus

accurately.

• Don’t draw lines between flask and condenser.

• Don’t have top of condenser sealed

• Condenser must have outer tube for water that is sealed

at top and bottom

• Condenser must have two openings for water in and out

that are open

Round

bottomed

flask

condenser

N Goalby chemrevise.org 12

Fractional Distillation: In the laboratory

• Heat the flask, with a Bunsen burner or electric mantle

• This causes vapours of all the components in the mixture

to be produced.

• Vapours pass up the fractionating column.

• The vapour of the substance with the lower boiling point

reaches the top of the fractionating column first.

• The thermometer should be at or below the boiling point

of the most volatile substance.

• The vapours with higher boiling points condense back

into the flask.

• Only the most volatile vapour passes into the condenser.

• The condenser cools the vapours and condenses to a

liquid and is collected.

Fractional distillation is used to

separate liquids with different

boiling points

flask

condenser

fractionating column

N Goalby chemrevise.org 13

The drying agent should

•be insoluble in the organic liquid

• not react with the organic liquid

• Put the distillate of impure product into a separating funnel

• wash product by adding either

• sodium hydrogencarbonate solution , shaking and

releasing the pressure from CO

produced.

• Saturated sodium chloride solution

•Allow the layers to separate in the funnel, and then run and

discard the aqueous layer.

•Run the organic layer into a clean, dry conical flask and add three

spatula loads of drying agent (e.g. anhydrous sodium sulphate,

calcium chloride) to dry the organic liquid. When dry the organic

liquid should appear clear.

• Carefully decant the liquid into the distillation flask

•Distill to collect pure product

Sodium hydrogencarbonate will neutralise

any remaining reactant acid.

Sodium chloride will help separate the

organic layer from the aqueous layer

Purifying an organic liquid

Separating funnel

General method

The layer with lower density will be the

upper layer. This is usually the organic layer

Decant means carefully pour off organic

liquid leaving the drying agent in the

conical flask

Distillation

In general used as separation technique to separate an

organic product from its reacting mixture. Need to

collect the distillate of the approximate boiling point

range of the desired liquid.

Water in

Water

out

Liebig condenser

thermometer

Heat

Note the bulb of the thermometer should be

at the T junction connecting to the

condenser to measure the correct boiling

point

Note the water goes in the bottom of the

condenser to go against gravity. This allows more

efficient cooling and prevents back flow of water.

It’s important to be able to

draw and label this apparatus

accurately. Don’t draw lines

between flask, adaptor and

condenser.

Round

bottomed

flask

Electric heaters are often used to heat organic

chemicals. This is because organic chemicals are

normally highly flammable and could set on fire

with a naked flame.

N Goalby chemrevise.org 14

1. Measure 8 cm

of 2-methylpropan-2-ol in a measuring cylinder and measure its mass.

2. Pour the 2-methylpropan-2-ol into a separating funnel, and reweigh the measuring cylinder to find the

mass of the 2-methylpropan-2-ol used.

3. In a fume cupboard, add 20 cm

of concentrated hydrochloric acid to the separating funnel, in portions

of 3cm

. After each portion, stopper the flask and invert it several times . Open the tap when doing this

to release the pressure.

4. Allow the separating funnel to stand in the fume cupboard for about 20 minutes. Gently shake it at

intervals.

5. After 20 minutes, allow the layers to separate in the funnel. Open the tap and remove the lower

aqueous layer. Dispose of this layer.

6. Add sodium hydrogencarbonate solution in 2 cm

portions to the separating funnel. This neutralises any

remaining acid. Shake the funnel after each addition, and release the pressure. Continue until no more

bubbles of CO

are seen.

7. Allow the layers to separate in the funnel. Again remove and pour away the lower aqueous layer. Run

off the organic layer into a clean conical flask and add two spatulas of anhydrous sodium sulfate.

Stopper the flask, shake the contents and allow this to stand until the liquid becomes clear. This step

dries the organic liquid.

8. Decant the liquid into a weighed clean distillation flask.

9. Distil the liquid by holding a 250ml beaker half-full of boiled water around the flask using standard

distillation set up. Collect the liquid that distils in the range 47-53

10. Measure the mass of the 2-chloro-2-methylpropane collected.

Detailed method for preparing and purifying a halogenoalkane from an alcohol

a) Pour about 20 cm

of cyclohexanol into a weighed 50 cm

pear-shaped flask. Reweigh the flask and record

the mass of cyclohexanol.

b) Using a plastic graduated dropping pipette, carefully and with frequent shaking, add to the flask

approximately 8 cm

of concentrated phosphoric acid.

c ) Add a few anti-bumping granules to the flask and assemble the distillation apparatus, so that the contents

of the flask may be distilled. Heat the flask gently, distilling over any liquid which boils below 100 °C.

d) Pour the distillate into a separating funnel and add 50 cm

of saturated sodium chloride solution. Shake

the mixture and allow the two layers to separate.

e) run off the lower layer into a beaker and then transfer the upper layer, which contains the crude

cyclohexene, into a small conical flask.

f) Add a few lumps of anhydrous calcium chloride or anhydrous sodium sulfate(VI) or anhydrous magnesium

sulfate to the crude cyclohexene to remove water. Stopper the flask, shake the contents and allow this to

stand until the liquid becomes clear.

g) Decant the liquid into a clean, dry, weighed sample container.

h) Reweigh the container, calculate the mass of dry cyclohexene produced

Detailed method for preparing and purifying Cyclohexene from cyclohexanol

Conc

CORE PRACTICAL 6: Chlorination of 2-methylpropan-2-ol using concentrated hydrochloric acid

N Goalby chemrevise.org 15

Detailed Method for Preparing and Purifying an Ester

Propyl ethanoate can be made in the laboratory from propan-

1-ol and ethanoic acid.

The equation for the reaction is

COOH + CH

OH  CH

COOCH

+ H

Procedure

1. Propan-1-ol (50 cm

) and ethanoic acid (50 cm

) are mixed

thoroughly in a 250 cm

round-bottomed flask.

2. Concentrated sulfuric acid (10 cm

) is added drop by drop

to the mixture, keeping the contents of the flask well-shaken

and cooled in an ice-water bath.

3. When the acid has all been added, a reflux condenser is

fitted to the flask and the mixture gently boiled over an

electric heating mantle for about 30 minutes.

4. The mixture is cooled, and the apparatus rearranged for

distillation. The crude ester (about 60 cm

) is distilled off.

5. The distillate is placed in a separating funnel and shaken

with about half its volume of 30% sodium carbonate solution,

with the pressure being released at intervals. The lower

aqueous layer is then discarded.

6. The crude ester is shaken in a separating funnel with about

half its volume of 50% calcium chloride solution, which

removes unreacted alcohol. The lower layer is discarded.

7. The ester is run into a clean, dry flask containing some

anhydrous calcium chloride and swirled.

8. The ester is filtered into a clean, dry flask, with a few anti-

bumping granules, and distilled. The fraction boiling between

100°C and 103°C is collected.

Sulfuric acid is a catalyst

Adding conc H

is an exothermic reaction- to

prevent uncontrolled boiling over add drop by drop

and cool

In reflux the reactant vapours of volatile compound

are condensed and returned to the reaction mixture.

The reaction is slow so it is heated for 30 minutes

The electric heating mantle allows for controlled

heating and stops flammable vapour lighting

Sodium carbonate reacts with unreacted acid and

remaining catalyst still present after distillation.

The reaction produces CO

so the pressure of gas

needs to be released.

The upper layer is organic because it has a lower

density than water

Calcium chloride is a drying agent. The liquid will

appear clear when dry.

Anti-bumping granules are added to the prevent

vigorous, uneven boiling by making small bubbles

form instead of large bubbles

Measuring boiling point

Purity of liquid can be determined by measuring a boiling point. This can be done

in a distillation set up or by simply boiling a tube of the sample in an heating oil

bath. If the liquid is pure it will have the boiling point referred to in data books. If

impure the boiling point tends to be higher than the pure liquid

To get a correct measure of

boiling point the

thermometer should be

above the level of the surface

of the boiling liquid and be

measuring the temperature

of the saturated vapour.

Pressure should be noted as changing pressure can change the boiling point of

a liquid

Measuring boiling point is not the most accurate method of identifying a

substance as several substances may have the same boiling point.

N Goalby chemrevise.org 16

Testing for Ammonium ions (NH

)

a) Place about 10 drops of 0.1 mol dm

–3

ammonium chloride in a test tube.

b) Add about 10 drops of 0.4 mol dm

–3

sodium hydroxide solution. Shake the mixture.

c) Warm the mixture in the test tube gently using a water bath.

d) Test the fumes released from the mixture by holding a piece of damp red litmus

paper in the mouth of the test tube.

Method: adding dilute sodium hydroxide

a) Place about 10 drops of 0.1 mol dm

–3

metal ion solution in a test tube.

b) Add about 10 drops of 0.6 mol dm

–3

sodium hydroxide solution, mixing well.

c) Continue to add sodium hydroxide solution, dropwise with gentle shaking, until in excess

This test can be used

on group 2 metal

ions and transition

metal ions.

Results for Group 2

Magnesium hydroxide is classed as insoluble in water and

will appear as a white precipitate.

Simplest Ionic Equation for formation of Mg(OH)

(s)

(aq) + 2OH

(aq)  Mg(OH)

(s).

A suspension of magnesium hydroxide in water

will appear slightly alkaline (pH 9) so some

hydroxide ions must therefore have been

produced by a very slight dissolving.

Calcium hydroxide is classed as partially soluble in water and

will appear as a white precipitate (it may need more sodium

hydroxide to be added before it appears compared to a

magnesium solution.)

Simplest Ionic Equation for formation of Ca(OH)

(s)

(aq) + 2OH

(aq)  Ca(OH)

(s).

A suspension of calcium hydroxide in water will

appear more alkaline (pH 11) than magnesium

hydroxide as it is more soluble so there will be

more hydroxide ions present in solution.

The results in this test are an application of the trend that group II hydroxides become more soluble down the group.

Strontium and barium salts will not form a hydroxide precipitate on addition of

sodium hydroxide due to their high solubility. The solutions will be highly alkaline

Results for transition metals

Copper(II) solutions form a blue ppt,

Cobalt(II) solutions form a blue ppt,

iron (II) solutions form a green ppt

iron (III) solutions form a brown ppt

Chromium (III) solutions form a green

ppt which dissolves in excess to form a

green solution

[Cu(H

]

(aq)

+ 2OH

(aq)

 Cu(H

(OH)

2 (s)

+ 2H

(l)

[Fe(H

]

(aq)

+ 2OH

(aq)

 Fe(H

(OH)

2 (s)

+ 2H

(l)

[Fe(H

]

(aq)

+ 3OH

(aq)

 Fe(H

(OH)

3 (s)

+ 3H

(l)

Results: alkaline

ammonia gas is released

which turns the red

litmus paper blue

CORE PRACTICAL 7 + 15: Analysis of some inorganic unknowns

Testing for cations

Lithium : Scarlet red

Sodium : Yellow

Potassium : lilac

Rubidium : red

Caesium: blue

Magnesium: no flame colour (energy emitted of a

wavelength outside visible spectrum)

Calcium: brick red

Strontium: red

Barium: apple green

Method

Use a nichrome wire ( nichrome is an unreactive metal and

will not give out any flame colour)

Clean the wire by dipping in concentrated hydrochloric acid

and then heating in Bunsen flame

If the sample is not powdered then grind it up.

Dip wire in solid and put in Bunsen flame and observe flame

Flame tests

[Co(H

]

(aq)

+ 2OH

(aq)

 Co(H

(OH)

2 (s)

+ 2H

(l)

[Cr(H

]

(aq)

+ 3OH

(aq)

 Cr(H

(OH)

3 (s)

+ 3H

(l)

Cr(H

(OH)

3 (s)

+ 3OH

(aq )

 [Cr(OH)

]

(aq)

+ 3H

(l)

N Goalby chemrevise.org 17

Testing for presence of a sulfate ion

BaCl

solution acidified with hydrochloric acid is used as a reagent to test for

sulphate ions.

If acidified Barium Chloride is added to a solution that contains sulfate ions a white

precipitate of Barium Sulfate forms.

Simplest ionic equation

(aq) + SO

(aq)  BaSO

(s).

2HCl + Na

 2NaCl + H

O + CO

Fizzing due to CO

would be observed if a carbonate was present.

Other anions should give a negative

result which is no precipitate

forming.

The hydrochloric acid is needed to react with carbonate impurities that are often found in salts which

would form a white Barium carbonate precipitate and so give a false result. You could not used

sulphuric acid because it contains sulphate ions and so would give a false positive result.

Testing for presence of halide ions with silver nitrate.

This reaction is used as a test to identify which halide ion is

present. The test solution is made acidic with nitric acid, and

then Silver nitrate solution is added dropwise.

The role of nitric acid is to react with any

carbonates present to prevent formation of the

precipitate Ag

. This would mask the desired

observations

2 HNO

+ Na

 2 NaNO

+ H

O + CO

Fluorides produce no precipitate

Chlorides produce a white precipitate

(aq) + Cl

(aq)  AgCl(s)

Bromides produce a cream precipitate

(aq) + Br

(aq)  AgBr(s)

Iodides produce a pale yellow precipitate

(aq) + I

(aq)  AgI(s)

The silver halide precipitates can be treated with ammonia solution to help differentiate between them

if the colours look similar:

Silver chloride dissolves in dilute ammonia to form a complex ion

AgCl(s) + 2NH

(aq) [Ag(NH

)

]

(aq) + Cl

(aq)

Colourless solution

Silver bromide dissolves in concentrated ammonia to form a complex ion

AgBr(s) + 2NH

(aq) [Ag(NH

)

]

(aq) + Br

(aq)

Colourless solution

Silver iodide does not react with ammonia – it is too insoluble.

Fizzing due to CO

would be

observed if a carbonate was

present

Testing for presence of carbonate ions

Add any dilute acid and observe effervescence.

Bubble gas through limewater to test for CO

– will turn limewater cloudy

2HCl + Na

 2NaCl + H

O + CO

Testing for presence of a hydroxide ions

Alkaline hydroxide ions will turn red litmus paper blue

CORE PRACTICAL 7+15: Analysis of some inorganic unknowns

Testing for anions: – Group 7 (halide ions), OH

–

, CO

2–

, SO

2–

Reactions of halide salts with concentrated sulphuric acid.

Explanation of differing reducing power of halides

A reducing agent donates electrons.

The reducing power of the halides increases down group 7

They have a greater tendency to donate electrons.

This is because as the ions get bigger it is easier for the outer

electrons to be given away as the pull from the nucleus on them

becomes smaller.

The Halides show increasing power as reducing

agents as one goes down the group. This can be

clearly demonstrated in the various reactions of

the solid halides with concentrated sulphuric

acid.

Know the equations and observations of these

reactions very well.

and Cl

ions are not strong enough reducing agents to reduce the S in H

. No redox

reactions occur. Only acid-base reactions occur.

Fluoride and Chloride

NaF(s) + H

(l) NaHSO

(s) + HF(g)

Observations: White steamy fumes of HF are evolved.

NaCl(s) + H

(l)  NaHSO

(s) + HCl(g)

Observations: White steamy fumes of HCl are evolved.

These are acid –base reactions and not

redox reactions. H

plays the role of

an acid (proton donor).

Br- ions are stronger reducing agents than Cl- and F- and after the initial acid-base

reaction reduce the Sulphur in H

from +6 to + 4 in SO

Bromide

Acid- base step: NaBr(s) + H

(l)  NaHSO

(s) + HBr(g)

Redox step: 2HBr + H

 Br

(g) + SO

(g) + 2H

O(l)

Observations: White steamy fumes of

HBr are evolved.

Red fumes of Bromine are also evolved

and a colourless, acidic gas SO

Ox ½ equation 2Br

 Br

+ 2e

Re ½ equation H

+ 2 H

+ 2 e

 SO

+ 2 H

Iodide

I- ions are the strongest halide reducing agents. They can reduce the Sulphur from +6 in

to + 4 in SO

, to 0 in S and -2 in H

NaI(s) + H

(l)  NaHSO

(s) + HI(g)

2HI + H

 I

(s) + SO

(g) + 2H

O(l)

6HI + H

 3 I

+ S (s) + 4 H

O (l)

8HI + H

4I

(s) + H

S(g) + 4H

O(l)

Observations:

White steamy fumes of HI are evolved.

Black solid and purple fumes of Iodine are also

evolved

A colourless, acidic gas SO

A yellow solid of Sulphur

S (Hydrogen Sulphide), a gas with a bad egg

smell,

Ox ½ equation 2I

 I

+ 2e

Re ½ equation H

+ 2 H

+ 2 e

 SO

+ 2 H

Re ½ equation H

+ 6 H

+ 6 e

 S + 4 H

Re ½ equation H

+ 8 H

+ 8 e

 H

S + 4 H

Often in exam questions these redox reactions are

worked out after first making the half-equations

Reduction product = sulphur dioxide

Note the H

plays the role of acid in the first step producing HBr and then

acts as an oxidising agent in the second redox step.

Note the H

plays the role of acid in the first step producing HI and then

acts as an oxidising agent in the three redox steps

Reduction products = sulphur dioxide, sulphur

and hydrogen sulphide

N Goalby chemrevise.org

N Goalby chemrevise.org 19

More on Insoluble salts and Precipitation reactions

Insoluble salts can be made by mixing appropriate solutions of ions so that a precipitate is formed

Barium nitrate (aq) + sodium sulfate (aq)  Barium Sulfate (s) + sodium nitrate (aq)

These are called precipitation reactions. A precipitate is a solid

When making an insoluble salt, normally the salt would be removed by filtration, washed with distilled

water to remove soluble impurities and then dried on filter paper

Writing Ionic equations for precipitation reactions

We usually write ionic equations to show precipitation

reactions. Ionic equations only show the ions that are

reacting and leave out spectator ions.

Spectator ions are ions that are

• Not changing state

• Not changing oxidation number

Ba(NO

)

(aq) + Na

(aq)  BaSO

(s) + 2 NaNO

(aq)

Take full equation

Separate (aq) solutions into

ions

(aq)

+ 2NO

(aq)

+ 2Na

(aq)

+ SO

(aq)

 BaSO

4(s)

+ 2 Na

(aq)

+ 2NO

(aq)

Cancel out spectator ions leaving

the simplest ionic equation

(aq) + SO

(aq)  BaSO

(s).

There are some common rules for solubility of salts. No syllabus requires these to be learnt but a good

chemist does know them.

Soluble salts Insoluble salts

All sodium, potassium and ammonium salts

All nitrates

Most chlorides, bromides, iodides

Silver, lead chlorides, bromides iodides

Most

sulfates

Lead

sulfate strontium and barium sulfate

Sodium, potassium and ammonium

carbonates

Most other carbonates

Sodium, potassium and ammonium

hydroxides

Most other hydroxides

Filter

funnel

Filter

paper

residue

filtrate

This is gravitational filtration. Use

if small amounts of solid are

formed.

Buchner flask (has

thicker glass walls than

a normal flask to cope

with the vacuum )

Filter paper

This is vacuum filtration. The apparatus is

connected to a water pump which will

produce a vacuum. Use if larger amounts of

solid are formed.

Air outlet to

water pump

Buchner

funnel

For both types of filtration apparatus AQA expect filter paper to be drawn on the diagram

Filtration

N Goalby chemrevise.org 20

Tollen's reagent method

Place 1 cm

of silver nitrate solution in each of two clean boiling tubes.

Then add one drop of sodium hydroxide solution to form a precipitate of

silver oxide. Add ammonia solution dropwise until a clear, colourless

solution is formed. Add a few drops of the unknown and leave in the

water bath for a few minutes.

Functional group tests for an Aldehyde

Tollen’s Reagent

Reagent: Tollen’s Reagent formed by mixing aqueous ammonia and silver

nitrate. The active substance is the complex ion of [Ag(NH

)

]

Conditions: heat gently

Reaction: aldehydes only are oxidised by Tollen’s reagent into a carboxylic

acid and the silver(I) ions are reduced to silver atoms

Observation: with aldehydes, a silver mirror forms coating the inside of the

test tube. Ketones result in no change.

CHO + 2Ag

+ H

O  CH

COOH + 2Ag + 2H

Fehling's solution method

Place 1 cm

of Fehling's A into each of

two boiling tubes, and then add Fehling's

B until the blue precipitate redissolves.

Add a few drops of the unknown and

leave in the water bath for a few

minutes.

Reagent: Fehling’s Solution containing blue Cu

ions.

Conditions: heat gently

Reaction: aldehydes only are oxidised by Fehling’s Solution

into a carboxylic acid and the copper (II) ions are

reduced to copper(I) oxide .

Observation: Aldehydes :Blue Cu

ions in solution change

to a red precipitate of Cu

O. Ketones do not react

Fehling’s solution

CHO + 2Cu

+ 2H

O  CH

COOH + Cu

O + 4H

Functional group test for an Alkene

To 0.5 cm

of bromine water in a test tube add a few drops of the unknown and shake.

Observation: alkenes should decolourise bromine water

Tests for alcohol, aldehyde, alkene and carboxylic acid

CORE PRACTICAL 7+15: Analysis of some organic unknowns

The melting point of the crystal formed can be used to help

identify which carbonyl was used. Take the melting point of

orange crystals product from 2,4-DNP. Compare melting point

with known values in database

Reaction with 2,4-dinitro phenylhydrazine

2,4-DNP reacts with both aldehydes and ketones. The product

is an orange precipitate, It can be used as a test for a carbonyl

group in a compound.

Use 2,4-DNP to identify if the compound is a

carbonyl. Then to differentiate an aldehyde

from a ketone use Tollen’s reagent.

N Goalby chemrevise.org 21

Summary of Identification of Functional Groups by test-tube reactions

Functional group test for a Carboxylic acid

To 0.5 cm

of your unknown solution in a test tube add a

small amount of sodium carbonate solid and observe.

Result carboxylic acids will fizz with sodium carbonate due

to CO

produced

The presence of a carboxylic acid can be tested by

addition of sodium carbonate. It will fizz and produce

carbon dioxide

2CH

H + Na

 2CH

+ H

O + CO

Reaction of carbonyls with iodine in presence of alkali

Reagents: Iodine and sodium hydroxide

Conditions: warm very gently

Only carbonyls with a methyl group next to the

C=O bond will do this reaction. Ethanal is the

only aldehyde that reacts. More commonly is

methyl ketones.

The product CHI

is a yellow crystalline

precipitate with an antiseptic smell

This reaction is called the Iodoform test

Functional group Reagent Result

Alkene Bromine water Orange colour

decolourises

Alcohols + carboxylic acids PCl

Misty fumes of HCl

produced

Alcohols, phenols,

carboxylic acids

Sodium metal Efferevesence due to H

gas

Carbonyls 2,4,DNP Orange/red crystals

produced

Aldehyde Fehlings solution Blue solution to red

precipitate

Aldehyde Tollens Reagent Silver mirror formed

Carboxylic acid Sodium carbonate Effervescence of CO

evolved

alcohol and

aldehyde

Sodium dichromate and

sulphuric acid

Orange to green colour

change

chloroalkane Warm with silver nitrate Slow formation of white

precipitate of AgCl

Acyl chloride Silver nitrate Vigorous reaction- steamy

fumes of HCl- rapid white

precipitate of AgCl

N Goalby chemrevise.org 22

Measuring the Enthalpy Change for a Reaction Experimentally

Calorimetric method

If the reaction is slow then the exact temperature rise can be

difficult to obtain as cooling occurs simultaneously with the

reaction

To counteract this we take readings at regular time intervals and

extrapolate the temperature curve/line back to the time the

reactants were added together.

We also take the temperature of the reactants for a few minutes

before they are added together to get a better average

temperature. If the two reactants are solutions then the

temperature of both solutions need to be measured before

addition and an average temperature is used.

For a reaction in solution we use the following equation

energy change = mass of solution x heat capacity x temperature change

Q (J) = m (g) x c

(J g

-1

) x T ( K)

This equation will only give the

energy for the actual quantities

used. Normally this value is

converted into the energy

change per mole of one of the

reactants. (The enthalpy change

of reaction, H)

General method

 washes the equipment (cup and pipettes etc) with the solutions to be used

 dry the cup after washing

 put polystyrene cup in a beaker for insulation and support

 Measure out desired volumes of solutions with volumetric pipettes and transfer to

insulated cup

 clamp thermometer into place making sure the thermometer bulb is immersed in

solution

 measure the initial temperatures of the solution or both solutions if 2 are used. Do this

every minute for 2-3 minutes

 At minute 3 transfer second reagent to cup. If a solid reagent is used then add the

solution to the cup first and then add the solid weighed out on a balance.

 If using a solid reagent then use ‘before and after’ weighing method

 stirs mixture (ensures that all of the solution is at the same temperature)

 Record temperature every minute after addition for several minutes

One type of experiment is one in which substances are mixed

in an insulated container and the temperature rise measured.

This could be a solid dissolving or reacting in a

solution or it could be two solutions reacting

together

Errors in this method

• energy transfer from surroundings (usually loss)

• approximation in specific heat capacity of solution. The method assumes all

solutions have the heat capacity of water.

• neglecting the specific heat capacity of the calorimeter- we ignore any energy

absorbed by the apparatus.

• reaction or dissolving may be incomplete or slow.

• Density of solution is taken to be the same as water.

Calorimetric method

Read question carefully. It

may be necessary to

describe:

• Method

• Drawing of graph with

extrapolation

• Description of the

calculation

CORE PRACTICAL 8: To determine the enthalpy change of a reaction using Hess’s Law

Calculating the enthalpy change of reaction, H

from experimental data

General method

1. Using q= m x c

x T calculate energy change for quantities used

2. Work out the moles of the reactants used

3. Divide q by the number of moles of the reactant not in excess to give H

4. Add a sign and unit (divide by a thousand to convert Jmol

-1

to kJmol

-1

The heat capacity of water is 4.18

J g

-1

. In any reaction where the

reactants are dissolved in water

we assume that the heat capacity

is the same as pure water.

Also assume that the solutions

have the density of water, which is

1g cm

-3

. Eg 25cm

will weigh 25 g

Example 3. Calculate the enthalpy change of reaction for the reaction where 25.0cm

of 0.20M

copper sulphate was reacted with 0.01mol (excess of zinc). The temperature increased 7.0

C .

Step 1: Calculate the energy change for the amount of reactants in the test tube.

Q = m x c

x T

Q = 25 x 4.18 x 7

Q = 731.5 J

Step 2 : calculate the number of moles of the reactant not in excess.

moles of CuSO

= conc x vol

= 0.2 x 25/1000

= 0.005 mol

If you are not told what is in excess, then you need to work

out the moles of both reactants and work out using the

balanced equation which one is in excess.

Step 3 : calculate the enthalpy change per mole which is often called H (the enthalpy change of reaction)

H = Q/ no of moles

= 731.5/0.005

= 146300 J mol

-1

= 146 kJ mol

-1

to 3 sf

Finally add in the sign to represent the energy change: if temp increases the

reaction is exothermic and is given a minus sign e.g. –146 kJ mol

-1

Remember in these

questions: sign, unit

Example 4. 25.0cm

of 2.0M HCl was neutralised by 25.0cm

of 2.0M NaOH. The Temperature increased 13.5

What was the energy change per mole of HCl?

Step 1: Calculate the energy change for the amount of reactants in the test tube.

Q = m x c

x T

Q = 50 x 4.18 x13.5

Q = 2821.5 J

Step 2 : calculate the number of moles of the HCl.

moles of HCl = conc x vol

= 2 x 25/1000

= 0. 05 mol

Step 3 : calculate H the enthalpy change per mole which might be called the enthalpy change of neutralisation

H = Q/ no of moles

= 2821.5/0.05

= 564300 J mol

-1

= -56.4 kJ mol

-1

to 3 sf

Exothermic and so is given a minus sign

Remember in these

questions: sign, unit,

Note the mass equals the mass of acid + the

mass of alkali, as they are both solutions.

Note the mass is the mass of the copper sulphate

solution only. Do not include mass of zinc powder.

N Goalby chemrevise.org

N Goalby chemrevise.org 24

CuSO

(aq)

CuSO

4 (s)

+ 5H

(l)

CuSO

.5H

(s)

+ 11kJmol

-1

= -66.1 kJ mol

-1

H reaction

+ aq

H reaction

+11kJ mol

-1

-66.1 kJ mol

-1

H reaction

= -66.1

- 11

= -77.1 kJ mol

-1

This Hess’s law is used to work out the

enthalpy change to form a hydrated salt from

an anhydrous salt.

This cannot be done experimentally because it

is impossible to add the exact amount of water

without the solid dissolving and it is not easy to

measure the temperature change of a solid.

Often Hess’s law cycles are used to measure the enthalpy change for a reaction that cannot be measured directly by

experiments. Instead alternative reactions are carried out that can be measured experimentally.

Instead both salts are dissolved in excess water

to form a solution of copper sulphate. The

temperature changes can be measured for

these reactions.

Detailed method for measuring enthalpy change of solution of anhydrous copper(II) sulfate

1. Weigh out between 3.90 g and 4.10 g of anhydrous copper(II) sulfate in a dry weighing bottle. The precise mass

should be recorded.

2. Using a volumetric pipette, place 25 cm

of deionised water into a polystyrene cup and record its temperature at the

beginning (t=0), start the timer and then record the temperature again every minute, stirring the liquid continuously.

3. At the fourth minute, add the powdered anhydrous copper(II) sulfate rapidly to the water in the polystyrene cup and

continue to stir, but do not record the temperature.

4. Reweigh the empty weighing bottle

5. At the fifth minute and for every minute up to 15 minutes, stir and record the temperature of the solution in the

polystyrene cup.

6. Plot a graph of temperature (on the y-axis) against time. Draw two separate best fit lines; one, which joins the points

before the addition, and one, which joins the points after the addition, extrapolating both lines to the fourth minute.

7. Use your graph to determine the temperature change at the fourth minute, which theoretically should have

occurred immediately on addition of the solid.

8. Using q= m x c

x T calculate energy change

= 20 x 4.18 x T

9. Calculate H

solution

by dividing q by number of moles of anhydrous copper(II) sulfate in mass added

The above method is then repeated using hydrated copper sulfate. The two H

solution

can then be used to calculate

the H for the enthalpy change of forming a hydrated salt as in the example above

Example 5. Calculate the enthalpy change of combustion for the reaction where 0.65g of propan-1-ol was

completely combusted and used to heat up 150g of water from 20.1 to 45.5

Step 1: Calculate the energy change used to heat up the water.

Q = m x c

x T

Q = 150 x 4.18 x 25.4

Q = 15925.8 J

Step 2 : calculate the number of moles of alcohol combusted.

moles of propan-1-ol = mass/ Mr

= 0.65 / 60

= 0.01083 mol

Step 3 : calculate the enthalpy change per mole which is called Hc (the enthalpy change of combustion)

H = Q/ no of moles

= 15925.8/0.01083

= 1470073 J mol

-1

= 1470 kJ mol

-1

to 3 sf

Finally add in the sign to represent the energy change: if temp increases the

reaction is exothermic and is given a minus sign eg –1470 kJ mol

-1

Remember in these

questions: sign, unit

Note the mass in this equation is the mass of water in

the calorimeter and not the alcohol

Enthalpies of combustion can be calculated by using calorimetry.

Generally the fuel is burnt and the flame is used to heat up water in a

metal cup.

Need to measure

• mass of spirit burner before and after

• Temperature change of water

• Volume of water in cup

N Goalby chemrevise.org

Measuring Enthalpies of Combustion using Flame Calorimetry

• energy losses from calorimeter

• Incomplete combustion of fuel

• Incomplete transfer of energy

• Evaporation of fuel after weighing

• Heat capacity of calorimeter not included

• Measurements not carried out under standard conditions as H

O is

gas, not liquid, in this experiment

Errors in this method

CORE PRACTICAL 9: Finding the Ka value for a weak acid – titration curves

Strong acid – Strong base

of base

Long steep part from

around 3 to 9

pH at equivalence

point = 7

The equivalence point lies

at the mid point of the

extrapolated vertical

portion of the curve.

e.g. HCl and NaOH

N Goalby chemrevise.org

Weak acid – Strong base

e.g. CH

H and NaOH

of base

At the start the pH rises quickly and then levels off. The

flattened part is called the buffer region and is formed because

a buffer solution is made

Steep part of curve >7 (around

7 to 9)

Equivalence point >7

pH starts

near 3

Half neutralisation volume

(aq)

][A

(aq)

]

[HA (aq)]

Ka=

For weak acids

At ½ the neutralisation volume

the [HA] = [A

]

So Ka= [H

] and pKa = pH

If we know the Ka we can then work out the pH

at ½ V or vice versa.

If a pH curve is plotted then the pH of a weak

acid at half neutralisation (½ V) will equal the pKa

Constructing a pH curve

Calibrate meter first by measuring known pH of a

buffer solution. This is necessary because pH meters

can lose accuracy on storage.

Most pH probes are calibrated by putting probe in a

set buffer (often pH 4) and pressing a calibration

button/setting for that pH. Sometimes this is

repeated with a second buffer at a different pH

Can also improve accuracy by maintaining

constant temperature

1. Transfer 25cm

of acid to a conical flask with a volumetric

pipette

2. Measure initial pH of the acid with a pH meter

3. Add alkali in small amounts (2cm

) noting the volume

added

4. Stir mixture to equalise the pH

5. Measure and record the pH to 1 d.p.

6. Repeat steps 3-5 but when approaching endpoint add in

smaller volumes of alkali

7. Add until alkali in excess

½ V

of base

Strong acid – Weak base

Equivalence point < 7

Steep part of curve <7

(around 4 to 7)

Weak acid – Weak base

e.g. CH

H and NH

e.g. HCl and NH

of base

No Steep part of the curve

CORE PRACTICAL 10: Investigating some electrochemical cells

Salt Bridge

The salt bridge is used to connect up the circuit. The free moving ions conduct the charge.

A salt bridge is usually made from a piece of filter paper (or material) soaked in a salt solution, usually Potassium Nitrate.

It can also be a glass U tube containing a salt solution plugged with cotton wool

The salt should be unreactive with the electrodes and electrode solutions.. E.g. potassium chloride would not be suitable

for copper systems because chloride ions can form complexes with copper ions.

A wire is not used because the metal wire would set up its own electrode system with the solutions.

N Goalby chemrevise.org

Zinc

electrode

copper

electrode

1M zinc

sulphate

solution

copper

sulphate

solution

Salt bridge

Electron

flow

Method

• Clean the zinc and copper foils with emery before use.

Degrease the metal using some cotton wool and

propanone.

• Place the copper strip into a 100 cm

beaker with

about 50 cm

of 1 mol dm

–3

CuSO

solution.

• Place the zinc strip into a 100 cm

beaker with about

50 cm

of 1 mol dm

–3

ZnSO

solution.

• Use a strip of filter paper soaked in saturated

potassium nitrate solution for the salt bridge

• Connect the Cu(s)|Cu

(aq) and Zn(s)|Zn

(aq) half-

cells by connecting the metals using the crocodile clips

and leads provided to the voltmeter

Method

If one or both of the half cells do not contain a

conducting metal, we must use an inert platinum

electrode.

Set up a copper half cell using a similar

arrangement to the previous one. Combine it

with a Fe

/Fe

half-cell with a platinum

electrode.

The half cell should have a mixture of acidified

1.0M iron(II) sulphate solution and an equal

volume of 0.5M iron(III) sulphate solution as the

electrolyte. Use a fresh salt bridge.

1M FeSO

and 0.5 M Fe

(SO

)

Salt bridge

KNO

(aq)

Note: in the electrode system containing two solutions it is

necessary to use a platinum electrode and both ion solutions must

be of a 1M concentration so [Fe

] = 1M and [Fe

] = 1M .

copper

electrode

copper

sulphate

solution

Pt electrode

A platinum electrode is used because it is

unreactive and can conduct electricity

N Goalby chemrevise.org 28

. Detailed Procedure : how much iron in iron tablets

• Weigh accurately two 'ferrous sulphate' tablets.

• Grind up the tablets with a little 1M sulphuric acid, using a pestle and mortar.

• Through a funnel, transfer the resulting paste into a 100cm

volumetric flask. Use further small volumes of 1 M

sulphuric acid to rinse the ground-up tablets into the flask.

• Then add sufficient 1M sulphuric acid to make up the solution to exactly 100cm

. Stopper the flask and shake it to

make sure that all the contents are thoroughly mixed. They will not all be in solution although the Fe

ions which

were present in the tablets will be dissolved.

• Titrate 10.0 cm

portions of the solution with 0.0050 M potassium manganate(VII). The end-point is marked by the

first permanent purple colour.

Manganate Redox Titrations

The redox titration between Fe

with MnO

–

(purple) is a very common

exercise. This titration is self indicating because of the significant colour

change from reactant to product.

MnO

(aq) + 8H

(aq) + 5Fe

(aq)  Mn

(aq) + 4H

O (l) + 5Fe

(aq)

Purple colourless

The purple colour of manganate can make it

difficult to see the bottom of meniscus in

the burette.

If the manganate is in the burette then the end

point of the titration will be the first permanent

pink colour.

Colourless  purple

Choosing correct acid for manganate titrations.

The acid is needed to supply the 8H

ions. Some acids are not suitable as they set up alternative redox reactions and

hence make the titration readings inaccurate.

Only use dilute sulphuric acid for manganate titrations.

Insufficient volumes of sulphuric acid will mean the solution is not acidic enough and MnO

will be produced instead

of Mn

MnO

(aq) + 4H

(aq) + 3e-  MnO

(s) + 2H

The brown MnO

will mask the colour change and lead to a greater (inaccurate) volume of Manganate being used in

the titration.

Using a weak acid like ethanoic acid would have the same effect as it cannot supply the large amount of hydrogen

ions needed (8H

It cannot be conc HCl as the Cl

ions would be oxidised to Cl

by MnO

as the E

MnO

/Mn

> E

/Cl

MnO

(aq) + 8H

(aq) + 5e

–

 Mn

(aq) + 4H

O(l) E+1.51V

(aq) +2e

–

 2Cl

–

(aq) E +1.36V

This would lead to a greater volume of manganate being used and poisonous Cl

being produced.

It cannot be nitric acid as it is an oxidising agent. It oxidises Fe

to Fe

as E

/HNO

> E

/Fe

(aq) + 3H

(aq) + 2e

–

 HNO

(aq) + H

O(l) E

+0.94V

(aq)+e

–

 Fe

(aq) E

+0.77 V

This would lead to a smaller volume of manganate being used.

A 2.41g nail made from an alloy containing iron is

dissolved in 100cm

acid. The solution formed contains

Fe(II) ions.

10cm

portions of this solution are titrated with

potassium manganate (VII) solution of 0.02M. 9.80cm

of KMnO

were needed to react with the solution

containing the iron.

What is the percentage of Iron by mass in the nail?

Example 6 Manganate titration

MnO

(aq)

+ 8H

(aq)

+ 5Fe

 Mn

(aq)

+ 4H

O + 5Fe

Step1 : find moles of KMnO

moles = conc x vol

0.02 x 9.8/1000

= 1.96x10

-4

mol

Step 2 : using balanced equation find moles Fe

in 10cm

= moles of KMnO

x 5

= 9.8x10

-4

mol

Step 3 : find moles Fe

in 100cm

= 9.8x10

-4

mol x 10

= 9.8x10

-3

mol

Step 4 : find mass of Fe in 9.8x10

-3

mol

mass= moles x RAM = 9.8x10

-3

x 55.8 = 0.547g

Step 5 : find % mass

%mass = 0.547/2.41 x100

= 22.6%

CORE PRACTICAL 11: Redox titration

Other useful manganate titrations

With hydrogen peroxide

Ox H

 O

+ 2H

+ 2e

Red MnO

(aq) + 8H

(aq) + 5e

 Mn

(aq) + 4H

Overall 2MnO

(aq) + 6H

(aq) + 5H

 5O

+ 2Mn

(aq) + 8H

Ox C

 2CO

+ 2e

Red MnO

(aq) + 8H

(aq) + 5e

 Mn

(aq) + 4H

Overall 2MnO

(aq) + 16H

(aq) + 5C

(aq)  10CO

(g) + 2Mn

(aq) + 8H

O(l)

With ethanedioate

With Iron (II) ethanedioate both the Fe

and the C

react with the MnO

1MnO

reacts with 5Fe

and 2 MnO

reacts with 5C

MnO

(aq) + 8H

(aq) + 5Fe

 Mn

(aq) + 4H

O + 5Fe

2MnO

(aq) + 16H

(aq) + 5C

 10CO

+ 2Mn

(aq) + 8H

So overall

3MnO

(aq) + 24H

(aq) + 5FeC

 10CO

+ 3Mn

(aq) + 5Fe

+ 12H

So overall the ratio is 3 MnO

to 5 FeC

The reaction between MnO

and

is slow to begin with (as the

reaction is between two negative

ions). To do as a titration the conical

flask can be heated to 60

C to

speed up the initial reaction.

N Goalby chemrevise.org

Example 7

A 1.412 g sample of impure FeC

.2H

O was

dissolved in an excess of dilute sulphuric acid

and made up to 250 cm

of solution. 25.0 cm

of this solution decolourised 23.45 cm

of a

0.0189 mol dm

–3

solution of potassium

manganate(VII).

What is the percentage by mass of

FeC

.2H

O in the original sample?

Step1 : find moles of KMnO

moles = conc x vol

0.0189 x 23.45/1000

= 4.43x10

-4

mol

Step 2 : using balanced equation find moles FeC

.2H

O in 25cm

= moles of KMnO

x 5/3 (see above for ratio)

= 7.39x10

-4

mol

Step 3 : find moles FeC

.2H

O in 250 cm

= 7.39x10

-4

mol x 10

= 7.39x10

-3

mol

Step 4 : find mass of FeC

.2H

O in 7.39x10

-3

mol

mass= moles x Mr = 7.39x10

-3

x 179.8 = 1.33g

Step 5 ; find % mass

%mass = 1.33/1.412 x100

= 94.1%

EDTA titrations

The formation of the stable EDTA complex with metal ions can with the choice of suitable indicator be done in a

quantitative titration.

[Cu(H

]

+ EDTA

 [Cu(EDTA)]

+ 6H

Always the same 1:1 ratio with any metal ion

Example 8

A river was polluted with copper(II)

ions. 25.0 cm

sample of the river

water was titrated with a 0.0150

mol dm

–3

solution of EDTA

4–

, 6.45

were required for complete

reaction.

Calculate the concentration, in mol

–3

, of copper(II) ions in the river

water.

Step1 : find moles of EDTA

moles = conc x vol = 0.0150 x 6.45/1000

= 9.68x10

-5

mol

Step 2 : using balanced equation find moles Cu

1:1 ratio

= 9.68x10

-5

mol

Step 3 : find conc Cu

in 25cm

= 9.68x10

-5

/0.025

= 0.00387 moldm

-3

N Goalby chemrevise.org 30

1. Add 1.5 g of hydrated copper(II) sulfate to weighing

bottle and measure the combined mass.

2. Transfer the copper (II) sulfate to a test tube and

measure the mass of the empty weighing bottle.

3. Add 4 cm

of water to the test tube

4. Place the test tube in the water bath (a beaker with

freshly boiled water).

5. Stir gently to dissolve the copper(II) sulfate.

6. Remove the test tube containing copper(II) sulfate

solution from the water bath

7. In a fume cupboard, add 2 cm

of concentrated

ammonia solution to the copper(II) sulfate solution

8. Pour the contents of the test tube into a beaker

containing 6 cm

of ethanol. Stir well and cool the

mixture in an ice bath.

9. Using a Buchner funnel and flask, filter the crystals.

Wash the test tube with some cold ethanol and add

the washings to the Buchner funnel. Finally, rinse the

crystals with a little cold ethanol.

10. Scrape the crystals onto a fresh piece of filter paper

and cover with a second piece of filter paper and

press to dry the crystals.

11. Measure the mass of the dry crystals

The mass of the copper(II) sulfate is the

difference between the two masses.

Buchner flask (has

thicker glass walls than

a normal flask to cope

with the vacuum )

Filter paper

This is vacuum filtration. The apparatus is

connected to a water pump which will

produce a vacuum. Use if larger amounts of

solid are formed.

Air outlet to

water pump

Buchner

funnel

Concentrated ammonia is corrosive.

Use in a fume cupboard and wear

gloves

Core practical 12: Prepare a transition metal complex

Equation for reaction

CuSO

•5H

O + 4NH

→ Cu(NH

)

•H

O + 4H

Loss of yield in this process

• Crystals lost when filtering or washing

• Some product stays in solution after recrystallisation

• other side reactions occurring

If the crystals are not dried properly the mass will be larger

than expected which can lead to a percentage yield >100%

N Goalby chemrevise.org

Techniques to investigate rates of reaction

measurement of the change in volume of a gas

Titrating samples of reaction mixture with acid, alkali, sodium thiosulphate etc

Colorimetry.

Measurement of optical activity.

Measurement of change of mass

Measuring change in electrical conductivity

(aq) + 2I

(aq) + 2H

(aq)  2H

O(l) + I

(aq)

HCOOCH

(aq) + NaOH(aq)  HCOONa(aq) + CH

OH(aq)

(CH

)

C=CH

(g) + HI(g) (CH

)

CI(g)

BrO

–

(aq) + 5Br

–

(aq) + 6H

(aq)  3Br

(aq) + 3H

O(l)

HCOOH(aq) + Br

(aq)  2H

(aq) + 2Br

(aq) + CO

(g)

HCOOH(aq) + Br

(aq)  2H

(aq) + 2Br

(aq) + CO

(g)

There are several different methods for measuring reactions rates. Some reactions can be measured

in several ways

This works if there is a change in the number of moles of

gas in the reaction. Using a gas syringe is a common

way of following this.

This works if there is a gas produced which is allowed to

escape. Works better with heavy gases such as CO

HCOOH(aq) + Br

(aq)  2H

(aq) + 2Br

(aq) + CO

(g)

COCH

(aq) + I

(aq) → CH

COCH

I(aq) + H

(aq) + I

–

(aq)

Small samples are removed from the reaction mixture, quenched (which

stops the reaction) and the titrated with a suitable reagent.

The NaOH could be titrated with an acid

The H

could be titrated with an alkali

The I

could be titrated with sodium

thiosulphate

If one of the reactants or products is coloured

then colorimetry can be used to measure the

change in colour of the reacting mixtures

The I

produced is a brown solution

Can be used if there is a change in the number

of ions in the reaction mixture

If there is a change in the optical activity through

the reaction this could be followed in a

polarimeter

CHBrCH

(l) + OH

−

(aq)  CH

CH(OH)CH

(l) + Br

−

(aq)

Investigating rates of reaction

32N Goalby chemrevise.org

When we follow one experiment over time recording the change in

concentration it is the continuous rate method.

The gradient represents the rate of reaction. The reaction is

fastest at the start where the gradient is steepest. The rate

drops as the reactants start to get used up and their

concentration drops. The graph will eventual become

horizontal and the gradient becomes zero which represents

the reaction having stopped.

time

concentration

Measurement of the change in volume of a gas

Mg + HCl  MgCl

This works if there is a change in the number of moles of gas

in the reaction. Using a gas syringe is a common way of

following this. It works quite well for measuring continuous

rate but a typical gas syringe only measures 100ml of gas so

you don’t want at reaction to produce more than this volume.

Quantities of reactants need to be calculated carefully.

gas syringe

Typical Method

• Measure 50 cm

of the 1.0 mol dm

–3

hydrochloric acid

and add to conical flask.

• Set up the gas syringe in the stand

• Weigh 0.20 g of magnesium.

• Add the magnesium ribbon to the conical flask, place the

bung firmly into the top of the flask and start the timer.

• Record the volume of hydrogen gas collected every 15

seconds for 3 minutes.

The initial rate is the rate at the start of the

reaction, where it is fastest. It is obtained by taking

the gradient of a continuous monitoring conc vs

time graph at time = zero. A measure of initial rate

is preferable as we know the concentrations at the

start of the reaction

Large Excess of reactants

In reactions where there are several reactants, if the

concentration of one of the reactant is kept in a large excess

then that reactant will appear not to affect rate and will be

pseudo-zero order . This is because its concentration stays

virtually constant and does not affect rate.

General: Measuring the rate of reaction: by an continuous monitoring method

Continuous rate data

This is data from one experiment where the concentration of a

substance is followed throughout the experiment.

Continuous rate experiments

This data is processed by plotting the data and calculating

successive half-lives.

If half-lives are constant

then the order is 1

order

The half-life of a first-order reaction

is independent of the concentration and is constant

If half-lives rapidly increase then the order is

2nd order

0.060

0.030

0.015

0.0075

t ½

Time (min)

[A]

Propanone reacts with iodine in acidic solution (the acid is a catalyst) as shown in the equation

below.

COCH

(aq) + I

(aq) → CH

COCH

I(aq) + H

(aq) + I

–

(aq)

The rate equation for the reaction is

Rate = k[CH

COCH

(aq)][H

(aq)]

This reaction can be followed by removing small samples from the reaction mixture with a

volumetric pipette. The sample is then quenched by adding excess sodium hydrogencarbonate to

neutralize acid catalyst which stops the reaction. Then the sample can be titrated with sodium

thiosulphate using a starch catalyst

(aq) + I

(aq)  2I

(aq) + S

(aq)

yellow/brown sol colourless sol

Time (min)

]

This reaction is zero order with respect to I

but 1

order

with respect to the propanone and acid catalyst

If there is a zero order reactant there must

be at least two steps in the mechanism

because the rate determining step will not

involve the zero order reactant

The rate determining step of this reaction must therefore contain one propanone molecule and one H

ion

forming an intermediate. The iodine will be involved in a subsequent faster step.

Detailed Method

1. To a beaker add 25 cm

of 1 mol dm

−3

aqueous propanone and 25 cm

of 1 mol dm

−3

sulfuric acid.

2. Add 50 cm

of 0.02 mol dm

− 3

iodine solution. Start the clock swirl the beaker well to mix.

3. Using a 10cm

pipette, withdraw a sample of the mixture and transfer it to a conical flask.

4. Add a spatula measure of sodium hydrogencarbonate (This stops the reaction). Record the time at which the sodium

hydrogencarbonate is added.

5. Titrate the iodine present in the conical flask with 0.01 mol dm

− 3

sodium thiosulfate solution. When the colour

turns a pale yellow add the starch indicator. The end point is then when the mixture goes from blue to colourless.

6 Every 5 minutes withdraw another 10 cm

sample and repeat steps 4 and 5

33N Goalby chemrevise.org

This method allows the order with respect to Iodine to be calculated because the propanone and

acid are in large excess so their concentrations do not change during the reaction

CORE PRACTICAL 13a: Rates of reaction: Following the rate of the iodine propanone

reaction by a titrimetric method

N Goalby chemrevise.org

Detailed method

• Put each of the chemicals in the table in separate burettes.

•

In each experiment, measure out required volumes of the potassium iodide, sodium thiosulphate, starch and

water into a small conical flask from the burettes

• Measure the hydrogen peroxide into a test tube

• Pour the hydrogen peroxide from the test tube into the conical flaks and immediately start the timer. Stir the

mixture.

• Time until the first hint of blue/ black colour appears

CORE PRACTICAL 13b: Rates of reaction: clock Reaction

The initial rate can be calculated from taking the

gradient of a continuous monitoring conc vs time

graph at time = zero

Initial rate can also be calculated from clock reactions

where the time taken to reach a fixed concentration is

measured.

A Common Clock Reaction (no need to learn details)

Hydrogen peroxide reacts with iodide ions to form iodine. The thiosulfate

ion then immediately reacts with iodine formed in the second reaction as

shown below.

(aq) + 2H

(aq) + 2I

–

(aq) → I

(aq) + 2H

O(l)

2–

(aq) + I

(aq) → 2I

–

(aq) + S

2–

(aq)

When the I

produced has reacted with all of the limited amount of

thiosulfate ions present, excess I

remains in solution. Reaction with the

starch then suddenly forms a dark blue-black colour.

A series of experiments is carried out, in which the concentration of iodide

ions is varied, while keeping the concentrations of all of the other reagents

the same. In each experiment the time taken (t) for the reaction mixture to

turn blue is measured.

In clock reactions there are often two

successive reactions and an end point is

achieved when one limited reactant runs

out, resulting in a sudden colour change

By repeating the experiment several times,

varying the concentration of a reactant e.g.

–

, ( keeping the other reactants at constant

concentration )you can determine the

order of reaction with respect to that

reactant

The initial rate of the reaction can be

represented as (1/t )

Experiment

Sulfuric

acid (H

) ml Starch ml Water ml

Potassium

iodide(I

) ml

Sodium

Thiosulfate S

1 25 1 20 5 5

2 25 1 15 10 5

3 25 1 10 15 5

4 25 1 5 20 5

5 25 1 0 25 5

Hydrogen

peroxide ml

Working out rate order graphically

In an experiment where the concentration of one of

the reagents is changed and the reaction rate measured

it is possible to calculate the order graphically

Rate = k [I

]

Log both sides of equation

Log rate = log k + n log [Y]

A graph of log rate vs log [I

] will yield a straight line where the

gradient is equal to the order n

Taking rate equation

Y = c + m x

In this experiment high concentrations with quick

times will have the biggest percentage errors.

Normally to work out the rate equation we do a series of experiments where the initial concentrations of reactants are

changed (one at a time) and measure the initial rate each time.

log (rate)

log [I

]

y intercept = log K

gradient = n

= change in y

change in x

N Goalby chemrevise.org

Experiment Determine the activation energy for the reaction between bromide ions and bromate(V) ions

The Arrhenius equation can be rearranged

ln k = constant – Ea/(RT)

k is proportional to the rate of reaction so ln k can be

replaced by ln(rate)

From plotting a graph of ln(rate) or ln k against 1/T the

activation energy can be calculated from measuring the

gradient of the line

ln (Rate)

1/T

Gradient = - Ea/ R

Ea = - gradient x R

-4.1

-3.6

-3.1

-2.6

-2.1

-1.6

0.0029 0.003 0.0031 0.0032 0.0033 0.0034

ln (Rate)

1/T

use a line of best fit

use all graph paper

choose points far apart on the graph to

calculate the gradient

Temperature

T (K) 1/T

time t

(s) 1/t Ln (1/t)

297.3 0.003364 53 0.018868 -3.9703

310.6 0.00322 24 0.041667 -3.1781

317.2 0.003153 16 0.0625 -2.7726

323.9 0.003087 12 0.083333 -2.4849

335.6 0.00298 6 0.166667 -1.7918

gradient = y

-y

-x

In above example gradient =-5680

The gradient should

always be -ve

Ea = - gradient x R (8.31)

= - -5680 x8.31

= 47200 J mol

-1

The unit of Ea using this equation will be J mol

-1

. Convert

into kJ mol

-1

by dividing 1000

Ea = +47.2 kJ mol

-1

Example 9

Analysis of results to calculate Activation Energy

CORE PRACTICAL 14: Finding the activation energy of a reaction

Detailed Method

1. Pipette 10 cm

of phenol solution and 10 cm

of bromide/bromate solution into a boiling tube.

2. Add a few drops of methyl red indicator to the mixture.

3. Into a second boiling tube pipette 5 cm

of sulfuric acid.

4. Place the two boiling tubes in a water bath at temperature 20

5. When the substances have reached the water temperature, mix the contents into one of the boiling tubes and

swirl. Start the stop clock.

6. Place the boiling tube containing the reaction mixture in the water bath.

7. Stop the clock when the methyl red indicator disappears.

8. Repeat the experiment at 30,40 50 60

In this experiment rate is 1/time where the time is the time taken for the indicator to change colour.

This is an approximation for initial rate of reaction as it does not include the change in concentration term. We can use

this because we can assume the amount of phenol used in each experiment is the same and constant. The change in

concentration is therefore the same for each experiment so only the time taken to reach this concentration is relevant.

OH + 3Br

→ C

OH + 3HBr

BrO

–

+ 5Br

–

+ 6H

 3Br

+ 3H

The Bromine produced in the first reaction reacts with the phenol. When

the phenol is used up, the bromine is no longer removed. The bromine

then bleaches the methyl red indicator at the ‘end of the reaction’

Step Reason

1. Dissolve the impure compound in a minimum volume

of hot (near boiling) solvent.

An appropriate solvent is one which will dissolve both

compound and impurities when hot and one in which the

compound itself does not dissolve well when cold.

The minimum volume is used to obtain saturated

solution and to enable crystallisation on cooling

2. Hot filter solution through (fluted) filter paper quickly. This step will remove any insoluble impurities and heat

will prevent crystals reforming during filtration

3. Cool the filtered solution by inserting beaker in ice Crystals will reform but soluble impurities will remain in

solution form because they are present in small quantities

so solution is not saturated. Ice will increase the yield of

crystals

4. Suction filtrate with a Buchner flask to separate out

crystals

The water pump connected to the Buchner flask reduces

the pressure and speeds up the filtration.

5 Wash the crystals with distilled water To remove soluble impurities

6. Dry the crystals between absorbent paper

Purifying an organic solid: Recrystallisation

Loss of yield in this process

• Crystals lost when filtering or washing

• Some product stays in solution after recrystallisation

• other side reactions occurring

buchner flask

Used for purifying aspirin

N Goalby chemrevise.org

If the crystals are not dried properly the mass will be larger

than expected which can lead to a percentage yield >100%

Preparation of a pure organic solid and test of its purity

N Goalby chemrevise.org 37

16. Core activity: The preparation of aspirin

Add to a 50 cm

pear-shaped flask 2.0 g of 2-hydroxybenzoic acid and

4 cm

of ethanoic anhydride.

To this mixture add 5 drops of 85% phosphoric(v) acid and swirl to

mix, Fit the flask with a reflux condenser and heat the mixture on a

boiling water bath for about 5 minutes. Without cooling the mixture,

carefully add 2 cm

of water in one portion down the condenser.

When the vigorous reaction has ended, pour the mixture into 40 cm

of cold water in a 100 cm

beaker, stir and rub the sides of the beaker

with a stirring rod necessary to induce crystallisation and, finally,

allow the mixture to stand in ice bath to complete crystallisation.

Collect the product by suction filtration and wash it with a little water.

Purification stage: recrystallisation

Using a measuring cylinder, measure out 15 cm

of ethanol into a

boiling tube.

Prepare a beaker half-filled with hot water from a kettle at a

temperature of approximately 75 °C.

Use a spatula to add the crude aspirin to the boiling tube with ethanol

and place the tube in the beaker of hot water.

Stir the contents of the boiling tube until all of the aspirin dissolves

into the ethanol.

Pour the hot solution containing dissolved aspirin through a warmed

filter funnel and fluted filter paper to hot filter

Then pour filtrate into 40 cm

of water in a conical flask.

Allow the conical flask to cool slowly and white needles of aspirin

should separate.

Cool the whole mixture in an ice bath.

Filter off the purified solid under reduced pressure and allow it to dry

on filter paper.

Record the mass of the dry purified solid

Detailed method for Nitration Procedure

Measure 2.5 cm

of methyl benzoate into a small conical flask and then

dissolve it in 5 cm

of concentrated sulfuric acid. When the liquid has

dissolved, cool the mixture in ice.

Prepare the nitrating mixture by adding drop by drop 2 cm

concentrated sulfuric acid to 2 cm

of concentrated nitric acid. Cool this

mixture in ice as well.

Now add the nitrating mixture drop by drop from a dropping pipette to

the solution of methyl benzoate. Stir the mixture with a thermometer and

keep the temperature below 10 °C. When the addition is complete, allow

the mixture to stand at room temperature for another 15 minutes.

After this time, pour the reaction mixture on to about 25 g of crushed ice

and stir until all the ice has melted and crystalline methyl 3-nitrobenzoate

has formed.

Then use same purification method as in aspirin above

Conc acids are corrosive- wear

gloves

The acids react together to make

the NO

ion

This reaction is exothermic so acids

are kept cool and acid is added

dropwise

The excess ethanoic anhydride will

hydrolyse and the contents of the flask

will boil.

The temperature is kept low at this

stage to prevent multiple

substitution of nitro groups on the

benzene ring

Avoid naked flames due to

flammability of ethanol

This step will remove any insoluble

impurities and heat will prevent crystals

reforming during filtration

Soluble impurities will remain in solution form

because they are present in small quantities

so solution is not saturated. Ice will increase

the yield of crystals

CCH

O C

+ CH



Aspirin is made from 2-hydroxybenzoic

acid which contains a phenol group. In the

reaction the phenol group is turned into an

ester by reacting it with the reactive

ethanoic anhydride

Ethanoic anhydride is used instead of

acid chlorides because it is cheaper, less

corrosive, less vulnerable to hydrolysis,

and less dangerous to use.

Aspirin

Detailed method for Preparation of Aspirin

N Goalby chemrevise.org

If the sample is very pure then the melting point will be a

sharp one, at the same value as quoted in data books.

One way of testing for the degree of purity is to determine the melting

“point”, or melting range, of the sample.

Heat

Heating oil- needs to

have boiling point

higher than samples

melting point and low

flammability

Thermometer with

capillary tube

strapped to it

containing sample

Measuring melting point

Comparing an experimentally determined melting point value

with one quoted in a data source will verify the degree of

purity.

Sometimes an error may occur if the temperature

on the thermometer is not the same as the

temperature in the actual sample tube.

Melting point can be measured in an electronic melting point machine or

by using a practical set up where the capillary tube is strapped to a

thermometer immersed in some heating oil.

In both cases a small amount of the sample is put into a capillary tube. The

tube is heated up and is heated slowly near the melting point

If impurities are present (and this can include solvent from the

recrystallisation process) the melting point will be lowered and the

sample will melt over a range of several degrees Celsius

In steam distillation steam is passed into

the mixture and the product vapour is

distilled off with the water and condensed

Steam distillation

Water

out

steam

Advantage of steam distillation:

The product distils at a lower temperature

which can prevents decomposition of the

product if it has a high boiling point

Solvent extraction

Mix organic solvent and oil-water mixture in a

separating funnel then separate the oil layer.

Distil to separate oil from organic solvent

Add anhydrous CaCl

to clove oil to dry oil

Decant to remove CaCl

Separating funnel

N Goalby chemrevise.org 39

Separation of species by thin-layer chromatography

Method: Thin-layer chromatography

a) Wearing gloves, draw a pencil line 1 cm above the bottom of a

TLC plate and mark spots for each sample, equally spaced along

line.

b) Use a capillary tube to add a tiny drop of each solution to a

different spot and allow the plate to air dry.

c) Add solvent to a chamber or large beaker with a lid so that is no

more than 1cm in depth

d) Place the TLC plate into the chamber, making sure that the level

of the solvent is below the pencil line. Replace the lid to get a

tight seal.

e) When the level of the solvent reaches about 1 cm from the top

of the plate, remove the plate and mark the solvent level with a

pencil. Allow the plate to dry in the fume cupboard.

f) Place the plate under a UV lamp in order to see the spots. Draw

around them lightly in pencil.

g) Calculate the Rf values of the observed spots.

value = distance moved by amino acid

distance moved by the solvent

Wear plastic gloves to prevent contamination

from the hands to the plate

pencil line –will not dissolve in the solvent

tiny drop – too big a drop will cause different

spots to merge

Depth of solvent– if the solvent is too deep it will

dissolve the sample spots from the plate

Will get more accurate results if the solvent is

allowed to rise to near the top of the plate but the

Rf value can be calculated if the solvent front does

not reach the top of the plate

lid– to prevent evaporation of toxic solvent

dry in a fume cupboard as the solvent is toxic

UV lamp used if the spots are colourless and not

visible

A solid stationary phase separates by adsorption,

A liquid stationary phase separates by relative solubility

If the stationary phase was polar and the moving

phase was non- polar e.g. Hexane. Then non-polar

compounds would pass through the plate more quickly

than polar compounds as they would have a greater

solubility in the non-polar moving phase.

(Think about intermolecular forces)

Separation by chromatography depends on the

balance between solubility in the moving phase and

retention in the stationary phase.

Rf values are used to identify different substances.

N Goalby chemrevise.org

Method

Part 1 Preparing the equilibrium mixture

1 Use burettes to prepare a mixture in boiling tube of carboxylic acid, alcohol, and dilute sulfuric acid.

2 Swirl and bung tube. Leave the mixture to reach equilibrium for one week

Part 2 Titrating the equilibrium mixture

1 Rinse a 250 cm

volumetric flask with distilled water.

Use a funnel to transfer the contents of the boiling tube into the flask. Rinse the boiling tube with water and add

the washings to the volumetric flask.

2 Use distilled water to make up the solution in the volumetric flask to exactly 250 cm

Stopper the flask, then invert and shake the contents thoroughly.

3 Use the pipette to transfer 25.0 cm

of the diluted equilibrium mixture to a 250 cm

conical flask.

4 Add 3 or 4 drops of phenolphthalein indicator to the conical flask.

5 Set up the burette with sodium hydroxide solution..

6 Add the sodium hydroxide solution from the burette until the mixture in the conical flask just turns pink. Record

this burette reading in your table.

7 Repeat the titration until you obtain a minimum of two concordant titres.

Practical: Working out equilibrium constant Kc

The sodium hydroxide will react with the sulphuric acid catalyst and any unreacted carboxylic acid in the

equilibrium mixture

H + CH

OH CH

+ H

Ethanoic acid Ethanol Ethyl Ethanoate

A common experiment is working out the equilibrium constant for an esterification reaction. Ethanol and

ethanoic acid are mixed together with a sulphuric acid catalyst.

The pink colour of the phenolphthalein in the titration can fade after the end-point of the titration has been reached

because the addition of sodium hydroxide may make the equilibrium shift towards the reactants

There are many different calculations that can be based on this experiment. Let’s

look at general stages. Not all calculations will use all the stages.

Working out initial amount of moles of reactants

The amount of moles of alcohol and carboxylic acid can be calculated from the densities

and volumes of liquids added

Mass = density x volume

then

Moles = mass X Mr

The initial amount of moles of acid catalyst used is usually determined by titrating a

separate sample of catalyst with sodium hydroxide

Working out equilibrium amount of moles of acid present from the titre results

39.0 cm

of 0.400 mol dm

-3

sodium hydroxide was used in the above titration. The initial moles of sulphuric acid

was 5x10

-4

mol. Calculate the moles of ethanoic acid present at equilibrium

Amount of NaOH = vol X conc

= 0.039 x 0.400

= 0.0156 mol

So total amount of H

present in 25cm

= 0.0156 mol

So total amount of H

present in 250cm

= 0.156 mol

Total mol acid present = moles of carboxylic acid + moles of acid catalyst

Amount of carboxylic acid at equilibrium = 0.156 – (5x10

-4

x 2)

= 0.155 mol

X 2 because H

has 2 H

N Goalby chemrevise.org 41

Calculating the equilibrium constant

Finally calculate the equilibrium constant.

To work out equilibrium concentrations divide the equilibrium amounts by the total volume. Then put in Kc

expression

Kc = [CH

] [H

[CH

H] [CH

OH]

In order to confirm that one week was sufficient time for equilibrium to be established in the mixture

from Part 1, several mixtures could be made and left for different amount of time. If the resulting Kc is

the same value then it can be concluded the time is sufficient

Absorption of visible light is used in spectrometry to

determine the concentration of coloured ions.

method

•Add an appropriate ligand to intensify colour

•Make up solutions of known concentration

•Measure absorption or transmission

•Plot graph of absorption vs concentration

•Measure absorption of unknown and compare

If visible light of increasing frequency is passed through a

sample of a coloured complex ion, some of the light is

absorbed.

The amount of light absorbed is proportional to the

concentration of the absorbing species (and to the distance

travelled through the solution).

Some complexes have only pale colours and do not absorb

light strongly. In these cases a suitable ligand is added to

intensify the colour.

Spectrophotometry

Spectrometers contain a coloured filter. The colour of the filter is chosen to only allow the wavelengths of light

through that would be most strongly absorbed by the coloured solution.

Detailed method- measuring absorption of copper solutions

• Take nine 100cm

graduated flasks and pipette 20cm

of 2M ammonia solution into each one.

• Use the 0.05M solution of aqueous copper sulphate to make up solutions which are 0.005 to 0.04M

[Cu(NH

)

]

• Mix each solution thoroughly.

• Insert the red filter into the colorimeter.

• Use a cuvette with distilled water to zero the colorimeter

• Then put each prepared solution in cuvette and measure the absorbance of each solution.

• Plot graph of absorption vs concentration

• Measure absorption of unknown solution and determine its concentration from the calibration curve

Working out equilibrium amount of moles of other substances

Calculate the equilibrium amount of ethanol, ethyl ethanoate and water if there were initially 0.400 mol of

ethanol and 0.500 mol of ethanoic acid and at equilibrium there were 0.155 mol of ethanoic acid.

Amount of ethanoic acid that reacted = initial amount – equilibrium amount

= 0.5 – 0.155

= 0.344mol

Amount of ethanol at equilibrium = initial amount - amount that reacted

= 0.400 – 0.344

= 0.056 mol

Amount of ethyl ethanoate at equilibrium = initial amount + amount that formed

= 0 + 0.344

= 0.344 mol

Amount of water at equilibrium = initial amount + amount that formed

= 0 + 0.344

= 0.344 mol

The amount of water at

equilibrium would not really be

0 as there would be water

present in the acid catalyst

N Goalby chemrevise.org 42

Potassium manganate(VII) will oxidise ethanedioic acid (oxalic acid) to carbon

dioxide and water, in the presence of an excess of acid:

2MnO

(aq) + 6H

(aq) + 5(CO

(aq) 2Mn

(aq) +10CO

(g) + 8H

O (I)

Detailed method

1. Prepare a reaction mixture according to the table, using

measuring cylinders.

2. Some members of your group should use Mixture 1 and some

Mixture 2. The results should then be shared.

3. Add 50 cm

of 0.02 M potassium manganate(VII) and start timing.

Shake the mixture for about half a minute to mix it well.

4. After about a minute use a pipette to withdraw a 10.0 cm

portion of the reaction mixture and empty into a conical flask.

5. Note the time and add about 10cm

of 0.1 M potassium iodide

solution. This stops the reaction and releases iodine equivalent to

the remaining manganate.(VII) ions.

6. Titrate the liberated iodine with 0.01 M sodium thiosulphate,

adding a little starch solution near the end-point Record the titre

of sodium thiosulphate.

7. Remove further portions every 3 or 4 minutes and titrate them in

the same way. Continue until the titre is less than 3 cm

Solution

Mixture 1

Mixture 2

0.2 M ethanedioic acid

100cm

0.2 M manganese(ll)

sulphate

-- 15cm

1 M sulphuric acid

10 cm

10cm

Water

90 cm

75 cm

The autocatalysis by Mn

in titrations of C

with MnO

overall 2 MnO

+ 5 C

+ 16 H

 2Mn

+ 10 CO

+ 8 H

Catalysed alternative route

Step 1 4Mn

+ MnO

+ 8 H

 5Mn

+ + 4 H

Step 2 2Mn

+ C

 2Mn

+ 2 CO

The initial uncatalysed reaction is slow because the reaction is a

collision between two negative ions which repel each other

leading to a high activation energy.

The Mn

ions produced act as an autocatalyst and therefore the

reaction starts to speed up because they bring about the

alternative reaction route with lower activation energy.

The reaction eventually slows as the MnO

concentration drops.

This is an example of autocatalysis where one of

the products of the reaction can catalyse the

reaction.

Mixture with only

reactants showing

effect of autocatalysis

Mixture with

added catalyst

at start

Alternative method for following the reaction rate

This experiment can be done by removing samples at set times and titrating to work out the concentration of MnO

It could also be done by use of a spectrometer measuring the intensity of the purple colour. This method has the

advantage that it does not disrupt the reaction mixture, using up the reactants and it leads to a much quicker

determination of concentration.

Autocatalytic Reaction between Ethanedioate and Manganate ions

Explanation of results