Synthesis of iron iii acetylacetonate lab

Foundations of Chemistry Laboratory Manual EQUILIBRIUM and LE CHÂTELIER’S PRINCIPLE

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EXPERIMENT 4F

Equilibrium and

Le Châtelier’s Principle (This experiment is done in pairs. Note: you may wish to divide part 1 and 2 between partners.)

Useful background reading (this is not compulsory but may be helpful):

Tro, 4th and 5th Edition: Sections 15.3, 15.7, 15.8, 14.9 (Intro only) – Questions 1 and 2 Sections 12.1 and 12.6 – Question 3

What is the relevance of this prac…?

The prac brings together several concepts that underpin many areas of chemistry study. You will undertake your first laboratory synthesis in which you make a compound (much like cooking but you don’t get to lick the bowl!).

You will then analyse, using Le Châtelier’s Principle, how the reaction conditions may be optimised in order to maximise the amount of product you obtain. Le Châtelier’s Principle can be used to predict outcomes on a small scale such as your reaction vessel, on a miniscule scale such as in cells and on a planetary scale such as in Earth’s atmosphere.

Finally, you will examine how the charge of a species determines what solvents it can be

dissolved in. The type of possible intermolecular forces present between the solute and solvent

will dictate solubility and this is investigated during this practical. Intermolecular forces are

incredibly important and we take them for granted all the time. They are responsible for oxygen

being a gas at room temperature so we can breathe it in and water being a liquid at room

temperature so we can drink it.

Learning objectives (remember these are different to the scientific objectives):

On completion of this practical, you should have:

 Become familiar with the class of chemical compounds called “co-ordination complexes”

 Understand that a co-ordination complex consists of a metal cation at the centre

surrounded by ligands

 Recall the concept of equilibrium from lectures and consider how it relates to this

practical

A BIG Question

What is life?

Life is dependent on many things working

together in concert to give a cohesive whole.

One of the many things on which human life

is dependent is the process of equilibrium.

Equilibrium processes are involved in

controlling the acidity of our blood and the

transport of oxygen in our bodies, among

many other things.

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Fe3+ and 6 x O

H H

 Become familiar with Le Châtelier’s Principle and use it to predict and explain changes in

the equilibrium position based on a change of reaction conditions

Note: this practical has three parts and can be quite long. However, many of the questions

(including all of those in Part Three) do not rely on experimental results and can be answered

prior to the practical if you have an understanding of equilibrium, Le Châtelier’s Principle and

intermolecular forces. It is possible to thoroughly prepare for this practical before you step into

the lab and students who do this should not find any problems with its length.

Introduction (extra background)

CO-ORDINATION COMPLEXES

This experiment involves the synthesis and investigation of a co-ordination complex:

tris(acetylacetonato) iron (III). Co-ordination complexes are a class of compounds that most

commonly involve a central metal ion which is surrounded by a certain number of molecules or

ions called ligands. These ligands are said to “co-ordinate” to the metal centre and therefore

the bond formed between them is referred to as a co-ordinate bond.

The complex you begin with in solution at the start of your reaction in Part One is made up of

the following species:

Fe3+ is the central metal ion and it is surrounded by six water ligands.

The formula for this complex is displayed like this: [Fe(H2O)6] 3+. This

species is also referred to as the “aquated Fe3+ ion” because it is a

metal cation surrounded by only water ligands. Everything inside the

square brackets is part of the complex – the metal centre and ligands.

The charge outside is that of the overall complex. The complex is also

shown diagrammatically in two ways below:

In diagram (a) we can see the Fe3+ cation at the centre with six positions around it for six water

ligands. Diagram (b) is conventionally how complexes are displayed and corresponds to the

formula: [Fe(H2O)6] 3+. The charge on the overall complex is a result of adding all the charges on

the metal centre and ligands together. The metal in this case has a 3+ charge and all six water

Fe3+

H2O

H2O OH2

OH2

OH2

OH2

Fe

H2O

H2O OH2

OH2

OH2

OH2

3+

a b

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O

C H

O

H3C CH3

ligands are neutral and contribute no charge (water molecules are neutral). Therefore the

overall charge on the complex is simply 3+, as shown on the outside of the square brackets.

For a species to have the potential to act as a ligand in a complex, it must have a lone pair of

electrons in its structure that it can donate to the metal centre. The atom that has the lone

electron pair on it is known as the donor atom for that ligand. This is the oxygen atom in water

for the complex above. Notice that in the complex diagrams this is why the ligand water

formula is always written to show the oxygen bonded to the metal (ie on the right hand side it

is written OH2 rather than the usual H2O to show O as the donor atom).

All ligands will therefore contain at least one donor atom with a lone electron pair on it. Each

donor atom can form one bond to the metal centre and therefore occupy one position around

it. Some ligands contain more than one donor atom and these are called multidentate ligands.

The ligand present on your final product complex is one such example shown below:

acetylacetonate

acac¯ ion

This is the acetylacetonate ion (acac¯ in shorthand). The two oxygen atoms share the negative

charge and therefore both contain lone electron pairs and both act as donor atoms. The acac¯ is

known as a bidentate ligand because it has two donor atoms which will co-ordinate at two sites.

The chemistry of coordination complexes has been a fertile area of research for many decades.

Coordination complexes are involved in many biological and industrial processes and are the

crucial component in many biologically active compounds. Hæmoglobin, Vitamin B12 and

chlorophyll are coordination complexes of iron, cobalt and magnesium respectively.

GENERAL COMPLEX FORMATION

It is useful to assume that the formation of complexes takes place in a series of steps eg:

M?+(aq) + L?–(aq) [M(L)]?(aq)

[M(L)]?(aq) + L?–(aq) [M(L)2] ?(aq) etc...

We have already seen that the aquated metal ion M?+(aq) may itself be represented as a

coordination complex with water molecules as ligands; [M(H2O)n] ?+(aq). (Eg [Fe(H2O)6]

3+.)

The formation of [M(L)]?(aq) will occur when water molecules are replaced by the ligand, L. If L

is a mono-dentate ligand then it will displace one water molecule. If L is multidentate then it

will displace several water molecules depending on how many donor atoms it has.

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O

C H

O

H3C CH3

O

C

O

H3C CH3

HH

acidic hydrogens

base

acetylacteone acacH

acetylacetonate ion acac

FORMATION OF THE COMPLEX TRIS(ACETYLACETONATO) IRON(III)

Acetylacetonate, acac¯, has already been provided as an example of a multidentate ligand. It is

formed by treating the organic compound acetylacetone, acacH, with a base:

In Part One of this experiment you

will use the acacH molecule to synthesise the complex tris(acetylacetonato) iron(III) – known in

shorthand as “tris-acac iron”.

The first step in the formation of the tris-acac iron complex is the hydrolysis of ferric chloride,

FeCl3.6H2O. The FeCl3.6H2O species is simply the source of Fe 3+ ions that become surrounded by

water ligands in solution:

[FeCl 3 ] + 6H2O [Fe(H2O)6]

3+(aq) + 3Cl–(aq)

In many cases, water ligands co-ordinated to metal ions in complexes become more reactive

than “free” water molecules. They are able to break one of the O-H covalent bonds to lose a

proton, H+ and also form the hydroxide anion, OH¯. This is seen in the following equation:

[Fe(H2O)6] 3+(aq) [Fe(H2O)5OH]

2+(aq) + H+(aq) Equation 4.1

Notice how the charge on the overall complex changes from 3+ to 2+ because the metal ion

stays the same – Fe3+ – but one of the neutral water ligands is converted to a hydroxide ion,

OH¯. This equation may be useful when you are determining the effect of pH on the formation

of tris-acac iron in Part Two. Also note, again, that you cannot see the charges of the individual

species in the complex formula, only the overall charge on the outside.

The formation of tris-acac iron is a stepwise process in which the water ligands coordinated to

Fe3+ ions are replaced sequentially by acetylacetonate ions, acac¯. The first step in the process is

shown in Equation 4.2.

[Fe(H2O)6] 3+(aq) + acac¯(aq) [Fe(H2O)4(acac)]

2+(aq) + 2H2O(l) Equation 4.2

Notice how two water ligands are replaced by only one acac¯ because the acac¯ is bidentate

and can therefore co-ordinate at two positions with its two donor atoms.

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The next step is the replacement of a further two water molecules with acac¯, and so on until

[Fe(acac)3] is produced.

The overall balanced equation describing the formation of tris(acetylacetonato) iron(III) (tris-

acac iron) from acetylacetone and aquated ferric ions is shown below.

[Fe(H2O)6] 3+(aq) + 3acacH(aq) [Fe(acac)3](aq) + 6H2O(l) +3H

+(aq) Equation 4.3

It is important to realise that the balanced equation does not necessarily represent the way in

which the reaction occurs since only acac¯ can act as a ligand, as shown in Equation 4.2. The

balanced equation is useful, for example, when yield calculations are necessary. The yield tells a

chemist how much of the product they have managed to produce. Every chemist performs yield

calculations after synthesising a molecule and you may become familiar with this later in your

studies.

In Part Two of this experiment, you will examine how pH affects the formation of the tris-acac

iron complex. When acetate is added in the form of the sodium salt, the yield of tris-acac iron

complex is increased. This occurs since sodium acetate acts as a base, decreasing the

concentration of H+ in Equation 4.3. According to Le Châtelier’s principle, if the concentration

of H+ is reduced, the equilibrium will be pushed to the right, increasing the formation of

[Fe(acac)3].

CH3COO¯(aq) + H +(aq) CH3COOH(aq) Equation 4.4

Chemistry connections…

Remember, even though we label things as reactants or products and we write their formulae on the

left or right hand side of the reaction arrow, this is merely a convention. We already know that at

equilibrium the rate of the forward reaction equals the rate of the back reaction. Therefore in an

equilibrium reaction, by definition, both the directions are occurring at the same time and consequently

all the species are constantly acting as both products and reactants.

In a reaction vessel (such as a test tube or a flask) all the species (whether they’re labelled reactants or

products) are able to interact with each other. Species that are reactants in the written equation are not

separated from species that are products. Therefore the protons, H+, being used up in Equation 4.4 are

also the same protons being produced in Equation 4.3 – they all simply count as “protons in the reaction

vessel”. Many equilibria can therefore be connected to each other and if you make a change to one you

can affect many others.

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LE CHÂTELIER’S PRINCIPLE

So how can we make that prediction above, that adding a base will promote the production of

more of our desired product [Fe(acac)3]?

Le Châtelier’s Principle states that if an outside force is applied to a system at equilibrium, then

the system will adjust to oppose the force and re-establish equilibrium. The “outside force” is

something that the chemist has control over and includes changes in concentration, pressure

and temperature. For example, consider the equilibrium below:

A + B C + D

If product C is removed, then the reaction will favour the right in order to replace what has

been taken away (ie reacting to minimise the disturbance).

If more of product D is added to the reaction mixture, then the reaction will shift to favour the

left, to try to remove the excess D and restore the balance. A similar shift in reaction would

occur if some of reactant B was removed from the reaction mixture.

Chemistry connections…

If you were in industry, what would you do to the concentrations of A, B, C and/or D once equilibrium

was reached in order to get more product?

SOLUBILITY (INTERMOLECULAR FORCES)

In Part Three of this experiment, you will examine the solubility of charged metal complexes

relative to a metal complex with no charge (ie. neutral). Examples of charged metal complexes

are the two intermediates in this experiment, [Fe(H2O)4(acac)] 2+ and [Fe(H2O)2(acac)2]

+, shown

on page 4. An example of a metal complex with no charge is the product of this experiment,

tris-acac iron or [Fe(acac)3].

The following solubility properties are relevant to this experiment:

 charged metal complexes tend to be insoluble in non-polar organic solvents and more

soluble in water

 complexes with no charge tend to be very insoluble in water and soluble in non-polar

organic solvents

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This can be expressed by the “like dissolves like” statement which chemists use to predict

whether species will dissolve in each other or not. Charged complexes and water are similar in

that one has a full charge (the complex ion) and the other has permanent partial charges (polar

water molecules). Neutral complexes and organic solvents are similar because neither have

charges; the complex is neutral and organic solvents are non-polar (have no or only very small

permanent partial charges).

The organic solvent you will use in this experiment is dichloromethane. In general chlorinated

hydrocarbons are denser than water, therefore the bottom layer will be the organic layer in

your test tubes.

Chemistry connections…

Beyond “like dissolves like” we should be able to identify the type of intermolecular forces present

between any pair of species. This is what actually determines solubility. Fortunately, the name of each

type of force is derived from the two species it is acting between. So in order to identify the force we

just have to decide what each species is – a fully charged species will contribute “ion” to the

intermolecular force name, a polar molecule will contribute “dipole” to the intermolecular force name

and a non-polar molecule will contribute “induced dipole” to the intermolecular force name.

For example, why does salt – NaCl – dissolve in water in the sea? It breaks up into Na+ cations and Cl-

anions, both of which interact with the water molecules. The specific name of the forces acting between

these ions and the water is called “ion-dipole” because the charged Na+ or Cl - are ions and polar water

molecules are dipoles.

What would be the type of intermolecular force experienced between a charged metal complex and

water molecules?

Experimental (procedure)

Remember: drawing a diagrammatic representation of what you will do in your experiment can

help you to visualise what you’ll need to do during the session and also to not accidentally miss

crucial steps.

The experimental section of this practical can be divided as follows:

Part One preparation of tris(acetylacetonato) iron (III) - [Fe(acac)3]

Part Two investigation of the effect of pH on the formation of

acetylacetonato iron (III) complexes

Part Three investigation of the solubility of complexes

You are required to answer a number of questions relating to each of these sections. The

questions relating to Part Two follow immediately after the relevant section. Those relating to

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Part One have been placed at the end of the script because you may find them easier to answer

after first considering the issues raised in Part Two.

Hazardous substances

Ferric chloride•6H2O FeCl3•6H2O harmful

Acetylacetone CH3COCH2COCH3 harmful

Ferric nitrate•9H2O FeNO3•9H2O irritant

Dichloromethane CH2Cl2 harmful

PART ONE PREPARATION OF THE COMPLEX TRIS(ACETYLACETONATO) IRON (III)

CAUTION

 Place all residues and discarded solutions in the residue buckets provided. Wear gloves during this practical.

PROCEDURE

1 Weigh approximately 1.4 g of ferric chloride, FeCl3.6H2O (M=270.3 g mol–1), into a

100mL conical flask. Add 30mL of deionised water and warm on steam bath.

2 Weigh approximately 3.0 g of sodium acetate, CH3COONa.3H2O in a suitable beaker.

Add 20mL of deionised water and warm on a steam bath.

3 Dissolve 2.0mL of acetylacetone in 10mL of ethanol in a small beaker and add this solution to the ferric chloride solution. Record your observations in your report book.

4 Combine your two remaining solutions, note carefully any changes and return this solution to the steam bath for a further ten minutes. Record your observations in your report book.

5 Allow your solution to cool slightly in the fumehood. Then place in an ice bath for 5 minutes.

6 Take your solution back to your bench and collect the precipitate using a Büchner funnel. (You should watch the video 'Buchner Filtration' prior to using a Büchner funnel). Wash the residue twice with deionised water and dry it on a watch glass for ten minutes at 100°C in the oven provided. You may need to dry your product in the oven for more than ten minutes depending on how wet it was to begin with. Consult your demonstrator for advice.

7 Determine and record the weight of your product and describe its appearance. Show the product to your demonstrator for assessment.

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PART TWO INVESTIGATION OF THE EFFECT OF pH ON THE FORMATION OF

ACETYLACETONATO IRON COMPLEXES

Record your observations in Table One and answer the questions following.

PROCEDURE Use the pH meter provided for this section. Rinse the electrode with deionised water between readings. Instructions are located on the benchtop next to the pH meter.

Solution 1:

1 Dissolve approximately 0.4 g of ferric nitrate, Fe(NO3)3.9H2O, in 50mL of deionised water in a beaker. Measure the pH of this solution using a pH meter. Note the colour of the crystals of ferric nitrate and the colour of the resulting solution. Record your observations in your report book.

Solution 2:

2 Dissolve 5 drops (0.1 g) of acetylacetone in 1mL of ethanol and add to the ferric nitrate solution. Stir and record the pH of this solution in Table 4.1. Pour 4mL of this solution into a clean test tube labelled Solution 2.

Solution 3:

3 Add approximately 0.5 g sodium acetate to the solution in the beaker note any changes, stir and record the pH of this solution in Table 4.1. Pour 4mL of this solution into a clean test tube labelled Solution 3.

Chemistry connections…

The pH scale – most commonly ranging between 0 – 14 – is used to define how acidic a solution

is. The more acidic a solution is, the more protons (H+) are present and therefore the lower its

pH. Acidic solutions have pHs below 7, for example orange juice has a pH of around 3.5 and the

hydrochloric acid in gastric juice has a pH around 2.5. A neutral solution has a pH of exactly 7,

such as pure water. Basic solutions have pHs above 7, for example laundry cleaning products at

about pH 11. Notice that the more acidic a solution is, the lower its pH becomes.

The pH scale is a logarithmic scale which means that a change in one pH unit actually

represents a 10-fold change in acidity. This is just like the Richter Scale which measures the

destructive power of an earthquake. For example, an earthquake that is 6.0 on the Richter

Scale has ten times more shaking power than an earthquake that is 5.0 on the scale. A solution

that has a pH of 6.0 is ten times less acidic (has ten times less protons present) than a solution

of pH 5.0.

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Question 1

a) Write down the overall equation showing the formation of [Fe(acac)3] from

[Fe(H2O)6] 3+ and acetylacetone (equation 4.3 in the Introduction).

b) What reagent do you weigh out that ends up being the source of the [Fe(H2O)6] 3+?

Question 2

a) Circle the correct word in bold text to complete the following statements:

A pH of less than 7 means a solution is acidic/basic. A pH of greater than 7 means a

solution is acidic/basic.

b) The equation showing acetylacetone, acacH, forming the acac¯ ion is below:

acacH acac¯ + H+

What is it about this equation that results in a solution of acacH being slightly acidic?

c) Given that sodium acetate is a base (remember that sodium acetate can be

considered as sodium ions and acetate ions), complete the following equation

showing the reaction of acetate ions with H+:

CH3COO – + H+

Question 3

a) What is the connection between the two equations below?

acacH acac¯ + H+

[Fe(H2O)6] 3+ + acac¯ [Fe(H2O)4(acac)]

2+ + 2H2O

b) Using your answer to Question 3(a), explain why the pH of a mixture of [Fe(H2O)6] 3+

and acacH (solution 2, Table 4.1) is lower than a solution of either of the

components by themselves.

c) Explain why the pH increases on addition of sodium acetate (solution 3, Table 4.1)

d) Therefore, explain why more [Fe(acac)3] is produced after the addition of sodium

acetate. (Hint: use Equation 4.3 and Le Châtelier’s Principle.)

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PART THREE SOLUBILITY

In Part One, the complex [Fe(acac)3] was synthesised and then the equilibrium reactions that

produce this compound were investigated in Part Two. Finally in Part Three these equilibria will

be tested using knowledge of intermolecular forces.

Note: the following questions should be attempted after reading the PROCEDURE on page 13.

In your Laboratory Report Book there will also be “prediction” questions to answer before

completing the procedure for this part.

These diagrams of the test tubes in this part may be helpful:

Diagram 1 Diagram 2

Chemistry connections…

It has already been stated that dichloromethane is the denser of the two solvents (ie the bottom

layer). If, however, you did not know which layer was which, can you think of a simple test you

could do to ascertain this? Hint: think about the “like dissolves like” statement. What do you

have on your bench that you could add to the test tube to help with this problem but that will

not affect the experiment?

CH2Cl2 CH2Cl2

CH2Cl2

CH2Cl2

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Question 4

a) Steps 1 and 2 in the PROCEDURE describe test tubes A and B in Diagram 1. Test tube A

should contain the neutral [Fe(acac)3] complex synthesised in Part One and polar water,

test tube B should contain [Fe(acac)3] and non-polar dichloromethane. Using the idea of

“like dissolves like”, which test tube – A or B – do you predict will contain the dissolved

[Fe(acac)3]? (Hint: see the notes on solubility in the Introduction section for help.)

b) If the intensity of the colour of the solutions can be taken as a rough guide as to the

concentration, which solvent layer in test tube B, the top or the bottom, contains the

most [Fe(acac)3]?

c) Is the layer you chose in question 4 (b) the polar (water) layer or the non-polar

(dichloromethane) layer?

CAUTION

 Do not place dichloromethane solutions in sinks.

 Place all residues and discarded solutions into the appropriate residue container.

PROCEDURE

1 Place a small quantity (approximately a spatula full) of triacetylacetonato iron(III), which you prepared in Part One, into two new test tubes labelled A and B.

2 Attempt to dissolve these two samples in 2mL of water (test tube A) and 2mL of dichloromethane (test tube B). Record your observations in Table 4.2 in your report book.

3 Add 2mL of water now to test tube B.

4 Add 2mL of dichloromethane to the samples of solutions 2 and 3 that you set aside in Part Two.

5 Shake the three test tubes (test tube B from Part Three and the two test tubes from Part Two). Allow the solutions to settle for several minutes and record your observations in Table 4.3 in your report book.

Chemistry connections…

Imagine you are a chemist in a research lab and you have reached the end of another synthesis.

Unfortunately, tests show that your desired product is present in the reaction vessel but there are also

several unwanted by-products. If your desired product is non-polar and the unwanted by-products are

polar, can you think of a way you could separate them and isolate the compound you’re after?

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Question 5

a) Complete the equations that describe the stepwise formation of the tris-acac iron

complex, [Fe(acac)3]:

1) [Fe(H2O)6] 3+ + acac¯ [Fe(H2O)4(acac)]

2+ + 2H2O

2)

3) [Fe(acac)3] + 2H2O

b) In Question 5(a) you have written the stepwise equilibrium reactions that must occur in

order to produce the final complex, [Fe(acac)3]. All three equilibria can be occurring in

solution to differing extents depending on what species are present and how Le

Châtelier’s Principle might affect the reaction position. The three test tubes in Diagram 2

(steps 3, 4 and 5 in the PROCEDURE) show varying extents of reaction. Of the four

complexes present in the reaction scheme, which solvent would you expect each one to

be soluble in: polar water or non-polar dichloromethane? (First answer is provided.)

COMPLEX SOLUBILITY

WATER DICHLOROMETHANE

[Fe(H2O)6] 3+ YES NO

[Fe(H2O)4(acac)] 2+

[Fe(H2O)2(acac)2] 1+

[Fe(acac)3]

Question 6

If the colour intensity of the layers changes position on going from test tube 2 to test tube

3, what does this suggest about the complexes that are present and, therefore, what has

happened to the equilibrium position?

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Chemistry connections…

Test tube B in Diagram 2 contains [Fe(acac)3] and the two solvents water and dichloromethane (your Question 2a)

and b) answers should confirm which layer the complex is dissolved in). Solution 2 contains the starting ions, Fe 3+

(which become [Fe(H2O)6] 3+

in water) as well as the starting acacH molecule (which can become the acac¯ ligand in

water). Where does the equilibrium lie for this combination of starting species? Ie in Question 5(a) will reactions 1

and 2 dominate, or perhaps reactions 2 and 3? Which complexes do you think will be the predominant species in

Solution 2? Matching these complexes with the solvents they’ll dissolve in from Question 5(b), what solvent layer

do you think will contain the most dissolved species in test tube 2?

Solution 3 contains the starting complex, [Fe(H2O)6] 3+

, the acacH molecule that can turn into the acac¯ ligand AND

the base, CH3COO¯. From Question 3(d), how did the addition of base affect the equilibrium position? If the

equilibrium position has shifted, which complexes do you now think are the predominant species in Solution 3?

Matching these complexes with the solvents they’ll dissolve in from Question 2b), what solvent layer do you think

will contain the most dissolved species in test tube 3?