Shapes of Molecules

Shapes of Molecules
Ionic bonds formed by electrostatic attraction between ions are non-directional in space. When covalent bonds are formed between atoms they occur by the overlapping of atomic orbitals. Some of these orbitals point in particular directions in space (p, d and f). It therefore follows that the bonds formed will have preferred directions in space and the molecules will have definite shapes.

Explanation in terms of electron pair repulsion theory for the shapes of molecules containing up to four pairs of electrons around the central atom such as BeCl2, BF3, CH4, NH3, H2O, and CO2. (Questions will not be set on hybridisation of orbitals). The departure of the bond angles in NH3 and H2O from the predicted tetrahedral, explained in terms of the increasing repulsion between bonding pair-bonding pair, lone pair-bonding pair and lone pair-lone pair electrons.
When you have finished this section you should be able to:

· · Apply the electron-pair-repulsion theory to molecules involving two, three and four pairs of bonding electrons.
· · Extend the electron-pair-repulsion theory to include non-bonding pairs (lone pairs) of electrons.
· · Use the following terms correctly to describe the shapes of molecules - linear, bent or V-shaped, trigonal planar, tetrahedral, trigonal pyramidal, trigonal bipyramidal, octahedral.


The shapes of covalent molecules
A covalent molecule will have a shape which is determined by the angles between the bonds joining the atoms together. A simple theory to account for the shapes of molecules was put forward by Sidgwick and Powell in 1940. It is called the VSEPR (Valence Shell Electron Pair Repulsion) theory. Note that the term valence shell is often used for what we call the outer-shell electrons.
The basic idea of the theory is that charge clouds (i.e. electron pairs) in the outer –shell of an atom in a molecule will arrange themselves to stay as far away from each other as possible. This will minimise the total energy of electrostatic repulsion between them.
With the help of this theory it is possible to predict
(i) The general geometric shape of a molecule.
(ii) The departure of interbond angles from the values for regular
geometric shapes.


Linear molecules
Exercise 1
(a) (a) How many pairs of bonding electrons are there in a molecule of beryllium hydride, BeH2?

(b) If the electron pairs repel each other so that they occupy regions of space as far apart as possible, what shape must the molecules have?


Molecules formed from the bonding of two atoms such as HCl, Cl2, HI and Br2 must be linear as it will always be possible to connect the nuclei of the two atoms by a straight line.
Molecules of the type
A B A

having just two electron pairs around the central atom (an electron deficient compound) will also be linear. A linear arrangement of atoms ( with a bond angle of 180o ) puts the two electron clouds as far apart as possible. e.g. Cl-Be-Cl

You can apply the same principle to molecules containing multiple bonds. Electron pairs in multiple bonds are assumed to occupy the position of one electron pair in a single bond.


Exercise 2
What shape would you expect for the following molecules?
Explain your answer.
(a) CO2 (b) HCN

Trigonal (triangular) planar molecules

Exercise 3
(a) How many pairs of bonding electrons are there in a molecule of boron trifluoride, BF3?

(b) By considering repulsion of electron pairs, what shape do you expect for BF3?



When there are three pairs of electrons around the central
Atom (or their equivalent), the bonds lie in the same plane at an angle of 120o. The arrangement is described as trigonal planar.
A


B A


A

Exercise 4
The molecule of methanal (formaldehyde) is trigonal planar, and the molecule of ethene is based on a trigonal planar arrangement around each carbon atom, as shown below.

H H H
120o C=O C=C
H H H
methanal ethene

Explain why the bonds around each carbon atom are in a trigonal planar arrangement.


Tetrahedral molecules
Both the shapes we have considered so far are planar and are therefore easy to draw on paper. For this reason, dot-and-cross diagrams for linear and trigonal planar molecules can correspond also to their actual shapes. This is not the case for tetrahedral molecules, which are three-dimensional, as you can see in the next exercise.


Exercise 5
(a) How many pairs of bonding electrons are there in a molecule of methane, CH4?

(b) (b) By considering repulsion of electron pairs, what shape do you expect for the CH4 molecule?

(c) By considering bond angles, show that electron pair repulsion would not give the square planar shape of the dot-and-cross diagram below.
(d) State whether you expect the following molecules and ions to have an identical shape or a very similar shape to the CH4 molecule. Give reasons.

(i) NH4+ (ii) SiCl4


When there is a complete octet in the valence shell with four electron pairs around the central atom the molecule adopts a tetrahedral structure with bond angles of 109.5o.
The 3-dimensional tetrahedral shape can be represented as shown :
B


109½ o
A
B
B
B
It is most important for you to be able to visualise tetrahedral bonding and to be able to draw it adequately, because it forms the basis of so many molecular shapes, especially in organic chemistry.
Molecules with lone pairs
So far we have considered only molecules where all the outer shell electrons are used in bonding. However, many molecules have some electrons , called non-bonding pairs or lone pairs, which are not shared between two atoms. These electrons also affect the shape of the molecule by electron pair repulsion.



Exercise 6
(a) (a) Draw a dot-and-cross diagram of the ammonia molecule, and count the number of pairs of electrons surrounding the nitrogen atom.
(b) (b) In which directions (from the N atom) would you expect to find the electron pairs (or regions of greatest charge density)?
(c) (c) What shape is outlined by the four atoms in the molecule?



The tetrahedron is not quite regular, because the four electron pairs are not identical. The lone pair, since it is not shared, remains closer to the N atom than the bonding pairs. The repulsion between non-bonding pairs and bonding pairs is greater, which pushes the hydrogen atoms closer together. This gives a slightly smaller H-N-H bond angle than the tetrahedral angle of 109½ o.


The shape of the ammonia molecule is described as trigonal pyramidal.
When describing the shape only the positions of the bonded atoms are considered (even thought the non-bonded pairs help to determine the overall shape).



The strength of the repulsion between electron pairs decreases in the order:
lone-pair / lone-pair > bonded-pair / lone-pair > bonded-pair / bonded-pair
strongest strong least strong


Two non-bonding pairs

Exercise 7
(a) (a) Draw a dot-and-cross diagram of the water molecule.
(b) (b) Draw the shape of a water molecule, showing how it fits into a tetrahedron.
(c) (c) Use your answer to the previous exercise to predict approximately the H-O-H bond angle.

The shape of the water molecule is described as bent or V-shaped.

Isomerism in Organic Compounds

Isomerism in Organic Compounds

Structural isomerism for aliphatic compounds containing up to six carbon atoms, to include branched structures and position of the C=C double bonds in alkenes. (Cyclic compounds excluded).

Isomerism (greek isos, equal; meros, parts) occurs when there are two or more compounds (called isomers) with the same molecular formula but different arrangements of their atoms. Isomers have different physical and chemical properties but the differences may be great or small depending on the type of isomerism.
There are two main classes of isomerism;
(i) (i) structural isomerism
(ii) (ii) stereoisomerism
which are themselves sub-divided.

ISOMERISM


structural isomerism stereoisomerism


chain positional functional group geometric optical
isomerism isomerism isomerism isomerism isomerism

Structural isomerism
Structural isomers are molecules with the same molecular formula but with different structural arrangements of the atoms.

· · Chain isomers
This occurs where isomers have different arrangements of the carbon chain. There is only one alkane corresponding to each of the formulae CH4, C2H6 and C3H8.



For butane C4H10 two arrangements are possible.
H H H H H H H
│ │ │ │ │ │ │
H—C—C—C—C—H --- H—C C C—H
│ │ │ │ │ │ │
H H H H H H—C—H H
│
H
butane methylpropane
b.pt. 273K b.pt. 261K

As the number of carbon atoms increases the number of possible isomers increases rapidly.

Number of structural isomers
C5H12
3
C6H14
5
C7H16
9
C10H22
75
C20H42
366,319
C30H62
Over 4000 million

Positional isomers

This occurs when isomers have the same carbon skeleton but the functional group is in different positions in the molecule.
H H H H H H
│ │ │ │ │ │
H—C—C—C—OH-- H—C C C—H
│ │ │ │ │ │
H H H H OH H
propan-1-ol propan-2-ol


Exercise 1
(a) Draw and name the five structural isomers of C6H14.

(b) (b) Write the structural formulae of all the alcohols of molecular formula C4H10O. Name the isomers.


Stereoisomerism : cis-trans isomers for compounds containing one C=C bond, the energy barrier to rotation in these compounds.

Stereoisomerism
Stereoisomers have identical molecular formulae, and the atoms are linked together in the same order, but have different relative positions in space.
The two types of stereoisomerism are
(i) (i) Geometric (or cis-trans ) isomerism
(ii) (ii) Optical isomerism [See Module 4.5]

·Exercise 2
1. Draw and name all the structural isomers, including geometric isomers, of C4H8.

Bonding

Whenever atoms, ions or molecules approach each other, there are electrostatic forces acting between them. When the net forces are forces of attraction, and they are strong enough to bind the particles together, we refer to them as chemical bonds. When particles are bound together by chemical bonds, the resulting arrangement is known as the structure of the substance concerned.

All the noble gases except He have completely full inner shells and an outer octet of electrons. As these gases are monatomic and rarely enter into chemical combination it is assumed that the outer octet of electrons is a very stable arrangement and therefore when atoms combine they will try to obtain the noble gas configuration.
The three main types of bonds are:

1. 1. Ionic or electrovalent
2. 2. Covalent
3. 3. Metallic


Ionic bonding restricted to elements in groups I, II, VI and VII, the ions of which have a noble gas structure. Dot and cross diagrams. Characteristic properties of ionic compounds. Sodium chloride as a typical ionic crystal.

When you have finished this section you should be able to:

· · Understand that ionic bonding involves attraction between oppositely charged ions formed by electron transfer
· · Describe the various electron configurations of simple stable ions.
· · List properties typical of ionic compounds and indicate how useful each one is in deciding whether a substance is ionic.


Ionic Bonding

Atoms can achieve noble gas electronic configuration by loss or gain of electrons to form ions. Metals (with low electronegativities) lose electrons to form positive ions and non-metals gain electrons to form negative ions.

Sodium can attain the stable electron configuration of neon by losing one electron
Na (1s22s22p63s1) Na+ (1s22s22p6) + e-

With ten electrons and eleven protons he species formed has a positive charge.

Fluorine is one electron short of the neon electronic configuration. If it obtains one electron (from sodium) it can achieve a full outer shell of eight electrons.

F (1s22s22p5) + e- F- (1s22s22p6)

The species formed has ten electrons and nine protons; it is a negatively charged fluoride ion.

You will soon be aware of the limitations of these diagrams, but they are nevertheless quite useful even in A-level work for describing electron transfer and for checking that all electrons are accounted for. However, electron shell diagrams are rather tedious to draw, and give more information than we usually need when only the outer shell electrons are involved in reactions. We can simplify the above diagram to draw what we call a 'dot-and-cross' diagram showing only the outer shell electrons:


Electrons shown by dots and crosses are of course indistinguishable - we use different symbols only to show where they come from. Note also the use of brackets to separate the charge symbol from the symbols for electrons.

In the first exercise you draw some similar dot-and-cross diagrams yourself
Exercise 1

Draw dot-and-cross diagrams for the formation of:
(a) (a) sodium chloride from sodium atoms and chlorine atoms;
(b) (b) magnesium oxide from magnesium atoms and oxygen atoms;

(c ) calcium fluoride from calcium atoms and fluorine atoms.


Properties of Ionic Compounds

1. 1. Crystalline solids : Ionic compounds are crystalline solids. Ionic crystals are quite hard due to the strong electrostatic forces between the ions
2. 2. High melting and boiling points : All ionic compounds have high melting and boiling points as a large amount of heat energy is required to break the strong electrostatic forces of attraction between the ions.
3. 3. Soluble in water : Many ionic compounds are soluble in water. The water molecules attract the ions and pull the structure apart. They also stabalise the ions once in solution
4. 4. Conduct electricity when molten or dissolved in water : All ionic compounds conduct electricity when molten and are decomposed in the process, which is called electrolysis. Most ionic compounds dissolve in water, which frees the ions so that they can move and carry electrical charge.

Structure of sodium chloride

An ion may be regarded as an electrically charged sphere. A charged sphere is surrounded by a uniform electric field and therefore attracts oppositely charged spheres in all directions.

No particular orientation is favoured, so we often say that ionic bonding is non-directional. What we really mean by this is that the forces of attraction are non-directional.
However, when large numbers of oppositely-charged spherical ions are attracted to each other, the repulsion of the like-charges also comes into play. Mutual repulsion of similarly-charged ions limits the number which can come into contact with an oppositely-charged ion, and effectively fixes their relative positions. Ions, therefore, tend to cling together in large clusters known as ionic lattices in which attractive and repulsive forces are balanced. The particular arrangement of ions depends on their relative charges and sizes.



The ionic lattice of sodium chloride consists of a regular arrangement of alternating sodium and chloride ions extending in three dimensions:

Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl-
Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na+
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl-
Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na

The lattice can be imagined as consisting of two inter-penetrating face centred cubic structures of sodium ions and chloride ions.
Na+ Na+ Cl- Cl-
Na+ Cl-
Na+ Na+ Cl- Cl-

Each sodium ion is surrounded by six chloride ions as nearest neighbours in the lattice and each chloride ion by six sodium ions. The lattice is said to have 6:6 coordination.

Stereoscopic picture of sodium chloride. The large circles represent Cl- ions and the small circles Na+ ions.

You can see from the above figure that it is impossible to say that a particular sodium ion ‘belongs’ to a particular chloride ion, as they are an equal distance from six. This means that ‘molecules’ of an ionic compound are not formed. The formula used is the atoms in their combining ratio.


Formation of covalent bonds in terms of the sharing of electron pairs between atoms. Dot and cross diagrams. Multiple bonds exemplified by C2H4, N2 and CO2. The octet rule and its limitations, eg BeCl2, BF3. The coordinate bond as a special case of the covalent bond eg NH4+. Dot and cross diagrams are required to show only the outermost electrons but, where appropriate, charges should be included.

When you have finished this section you should be able to:

· · Describe covalent bonding in terms of shared pairs of electrons.
· · Draw dot-and-cross diagrams for a variety of covalent molecules.
· · Explain the term dative covalent bond.
· · Draw dot-and-cross diagrams and structural formulae including dative bonds.
· · For polyatomic ions, distinguish between the covalent bonding within the ion and the ionic bonding between ions.
· · Draw dot-and-cross diagrams for at least two covalent compounds in which an atom other than hydrogen has fewer than eight outer shell electrons.
· · Draw dot-and-cross diagrams for at least two covalent compounds in which an atom has more than eight outer shell electrons.


1.2 Covalent Bonding

Covalent bonds are formed by the sharing of electrons between atoms. The two atoms have to approach sufficiently close to each other for their atomic orbitals to overlap. The shared pair of electrons constitutes a single covalent bond and occupy the same orbital with opposing spins.

Dot-and cross Diagrams
Dot-and cross diagrams are simplified versions of the diagram for the chlorine molecule on the left below, showing only the outer shell electrons

Note that the use of two (or even more) different symbols for the electrons does not mean that the electrons are different - it simply identifies the ‘parent’ atoms.

You may also encounter dot-and-cross diagrams with bond lines added, and also versions showing only the bonding electrons,


Cleave crystal of calcite to show cleavage planes and hardness.

Show model of NaCl structure from booklet and from model building kit.

(Halogens : restricted to chlorine, bromine and iodine)
Valid deductions may be expected about other elements in the group.

Physical properties of halogens, limited to colour and physical state at room temperature.
The halogens are a group of reactive non-metals, which are essentially similar to each other with only gradual changes as the atomic number increases

They are all p-block elements with a simple molecular structure consisting of covalently bonded diatomic molecules, X2.

o o o o
o o o
o X o X o X X
o o o o
There are only weak Van der Waals forces between the molecules. The strength of the forces increases as the number of electrons (Mr) in the molecule increases.
F2< Cl2< Br2< I2

In the case of iodine, the forces are sufficiently strong to bind the iodine molecules together in a 3-D crystal lattice. The X-X bond strength decreases down the group
Cl2> Br2 > I2 as the atoms get larger and the attraction of the nucleus for the shared electrons decreases (electronegativity decreases).
There is a slight tendency to metallic character with increasing atomic number. The halogens complete their octet by gaining one electron forming a halide ion, X- (see electron affinity values) or by sharing one electron. Fluorine is restricted to an oxidation state of -1 but the remaining elements have empty d orbitals and can promote electrons to give oxidation states of +1, +3, +5 and +7.
They are all oxidising agents and combine readily with metals and hydrogen.

Chlorine is a greenish-yellow gas.
Bromine is a red-brown volatile liquid.
Iodine is a black shiny solid, which sublimes on heating to produce a purple vapour.



Reactions of chlorine gas with elements (see Section 7.5.1), water, alkalis (under different conditions to form ClO- and ClO3-), other halides in aqueous solution, iron(II) ions in solution, hydrocarbons (see Sections 7.3.2, 7.3.3 and 7.3.4).

Reaction of halides with elements
Metals
The halogens combine readily with most metals forming the metal halides.
The vigour of the reaction decreases from chlorine to iodine.
Group I and II halides are ionic.

2Na (s) + Cl2 (g) 2Na+Cl- (s)
Mg (s) + Cl2 (g) Mg2+2Cl- (s)

The halides of Group III are predominantly covalent.

2Al (s) + 3Cl2 (g) 2AlCl3 (s)

Non-metals
The elements react directly with many non-metals the oxidising power decreasing from chlorine to iodine.
The elements combine directly with phosphorus, the oxidation state of the product depending on the oxidising power of the halogen.

2P (s) + 5Cl2 (g) 2PCl5 (s)
2P (s) + 3Br2 (l) 2PBr3 (l)
Solubility of the elements
All three elements are only slightly soluble in water because of the relatively strong hydrogen-bonding between the water molecules, which does not exist between the halogen molecules
i.e. solvent-solvent attractions > solute-solvent attractions > solute-solute
attractions.
Cl2> Br2>I2
solubility decreasing

They are soluble in non-polar organic solvents such as toluene and TCE.
(Why?)


Chlorine reacts slowly with water forming hydrochloric acid and chloric(I) acid. This reaction involves disproportionation:- a change in which one particular molecule, atom or ion is simultaneously both oxidised and reduced.

reduction

Cl2 (g) + H2O (l) HCl (aq) + HClO (aq) (chlorine water)
o.n. 0 -1 +1
oxidation

Exercise 1


Write balanced equations for each of the above reactions.
(Use the symbol X for a general halogen reaction)
The identification of halide ions in solution by the use of silver ions and aqueous ammonia; the [Ag(NH3)2]+ ion

Reaction of the halide ions in solution, X-(aq)
Most metal halides are soluble except lead and silver halide. Therefore solutions of lead and silver ions are used to test for the presence of halide ions in solution.







Bromine and iodine disproportionate in a similar way but to a lesser extent.

Reaction of chlorine with aqueous sodium hydroxide.
Chlorine reacts faster with dilute sodium hydroxide than with water.
When chlorine is added to cold dilute alkali it disproportionates to chloride and chlorate(l).
(i)
reduction

Cl2 (g) + 2NaOH (aq) NaCl (aq) + NaOCl (aq) + H2O
o.n. 0 -1 +1
oxidation

( 2OH- + Cl2 Cl- + OCl- + H2O )

(ii) In hot concentrated alkali, if the solution is warmed to 70oC, the chlorate(I) disproportionates further to chlorate(V).

reduction

3NaOCl (aq) 2NaCl (aq) + NaClO3 (aq)
o.n. +1 -1 +5
oxidation

If chlorine is bubbled directly into hot conc. alkali then

(iii) reduction

3Cl2 (g) + 6NaOH(aq) 5NaCl (aq) + NaClO3 (aq)
o.n. 0 -1 +5
oxidation

( 6OH- + 3Cl2 5Cl- + ClO3- + 3H2O )

For bromine, both reactions (i) and (ii) are fast at 15oC.
For iodine, decomposition of IO- occurs rapidly at 0oC so it is difficult to prepare NaIO free from NaIO3.
NaClO is a mild antiseptic (Milton).
NaClO3 powerful weed killer.


Relative oxidising ability of the halogens linked to redox potentials.

Displacement reactions of the halogens
Since they are very electronegative, all the halogens are oxidising agents. Their standard electrode potentials, Eq, become less positive on descending the group, showing that their oxidising power decreases.
X2 + 2e- 2X-
s.e.p. Eq /volts
Cl2 (g) /2Cl- + 1.36
Br2 (g) / 2Br- + 1.09
I2 (g) / 2I- + 0.54

Therefore chlorine oxidises bromide ions to bromine and iodide ions to iodine.
These are displacement reactions.

Cl2 (g) + 2Br- (aq) ----- Br2 (l)+ 2Cl- (aq)
(colourless) (yellow/orange)

Cl2 (g) + 2I- (aq) -- I2 (s) + 2Cl- (aq)
(colourless) (red/brown)

Bromine oxidises iodide to iodine

Br2 (g) + 2I- (aq) ---- I2 (s) + 2Br- (aq)

Iodine does not oxidise any of the others.


Exercise 2
Write an equation for the reaction of sodium chloride solution with
(a) (a) lead nitrate solution and


(b) silver nitrate solution followed by the addition of ammonia.



The reaction of solid halides with concentrated sulphuric acid to illustrate the relative reducing ability of halide ions and the hydrogen halides. Presence of halide ions in sea water.

Reaction of the metal halides with conc. sulphuric acid H2SO4

When concentrated sulphuric acid is added to a sodium halide the first product is fumes of the hydrogen halide HX, because each of these compounds is more volatile than sulphuric acid.
NaCl (s) + H2SO4 (1) HCl (g) + NaHSO4 (s)

NaBr (s) + H2SO4 (1) HBr(g) + NaHSO4 (s)

However conc. sulphuric acid is also a strong oxidising agent and will oxidise
HBr Br2 and HI I2, but not HF and HCl.

oxidised

2HBr(g) + H2SO4 Br2 + SO2 (g) + 2H2O

reduced


Similarly 2HI(g) + H2SO4 I2 + SO2(g) + 2H2O

Therefore conc. sulphuric acid cannot be used for the preparation of
HBr(g) and HI(g).
However conc. phosphoric(V) acid, H3PO4, can be used for the preparation since it is relatively non volatile and a poor oxidising agent.

NaBr (s) + H3PO4 (1) HBr (g) + NaH2PO4
NaI (s) + H3PO4 (1) HI (g) + NaH2PO4

Exercise 3
1. 1. Research: Look up the presence of halide ions in sea water.
2. 2. Make notes on the uses of the halogens and their compounds.

The Periodic Table- work sheet

The Periodic Table
Section A
For each of the following questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the response sheet.

1.B Which one of the following equations does not represent a reaction of chlorine under suitable conditions?
A Cl2 + H2O HCl + HOCl
B 2Cl2 + CH4 CCl4 + 2H2
C 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O
D Cl2 + 2NaOH NaCl + NaOCl + H2O


2.A When the hydrogen halides dissolve in water acidic solutions are formed. Which has the highest pH (assuming all are equal concentrations)?
A Hydrogen fluoride
B Hydrogen chloride
C Hydrogen iodide
D Hydrogen bromide


3. In which one of the following changes has chlorine been oxidised?
A 3Cl2 + 2Fe 2FeCl3
B Cl2 + I2 2IC1
C Cl2 + 2KBr Br2 + 2KCl
D Cl2 + 2KOH H2O + KClO + KCl


4. 4. In which one of the following sequences are the oxides of the elements classified as basic, amphoteric and acidic respectively?
A Na, Mg, Al
B Na, K, S
C K, Al, P
D S, P, Mg


5. Which one of the following is true of the halogens as the Group is descended from chlorine to iodine?
A The atomic radius decreases.
B The colour of the element lightens.
C The melting point of the element increases.
D The oxidising power of the element increases.
5. 5. Which one of the following equations represents a reaction of chlorine?
A Cl2 + NaOH NaOCl + HCl
B Cl2 + Fe FeCl2
C Cl2 + H2O HCl + HOCl
D Cl2 + CH4 CH2Cl2 + H2

Determining the enthalpy change of a reaction

Determining the enthalpy change of a reaction
Aims
The purpose of this experiment is to determine the enthalpy change for the displacement reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
By adding an excess of zinc powder to a known amount of copper(II) sulphate solution, and measuring the temperature change over a period of time, you can calculate the enthalpy change for the reaction.
Apparatus

Goggles
Bench mat
25cm3 pipette
Pipette filler
Polystyrene cup with lid
Weighing bottle
Spatula
Balance
Thermometer
Stop clock
Zinc powder
1.0M copper(II) sulphate solution

Methods
1. Pipette 25.0cm3 of the copper(II) sulphate solution into the polystyrene cup.
2. Weigh about 6g of zinc powder in the weighing bottle – as this is an excess, there is no need to be accurate.
3. Put the thermometer through the hole in the lid, stir, and record the temperature every half minute for 2½ minutes in the table below.
4. At precisely 3 minutes, add the zinc powder to the cup.
5. Continue stirring, and record the temperature for an additional 6 minutes in the table below.
makaeResults table with time AND temperature



Plot the temperature (vertical axis) against time (horizontal axis).
2. Extrapolate the curve back to 3.0 minutes to establish the maximum temperature rise as shown in the example below: ΔT time (minutes)temperature (°C) 0 246810
3. Calculate the enthalpy change, ΔH, for the quantities used, using the formula:
ΔH = m c ΔT where m = mass of solution (g)
c = specific heat capacity of water = 4.18 J g–1 K–1
ΔT = rise in temperature (K)
4. Calculate the enthalpy change for one mole of Zn and CuSO4(aq).
5. Calculate the maximum error for each piece of apparatus, and then the total overall apparatus error.
6. The accepted value for this reaction is –217 kJ mol–1.
Compare your result with this value by calculating the percentage error in your answer:
error = experimental value - accepted valueaccepted value x 100%
Evaluation
1. Compare your total apparatus error with your answer to part 6 above – is the apparatus error enough to account for any difference between the accepted value and your experimental value?
2. List some possible reasons for any difference between your value and the accepted value (these should not be the apparatus errors mentioned above).
3. Why do you think the temperature increases for a few readings after adding the zinc?
(Hint: the temperature does not go even higher if more zinc is used, or if the powder is finely divided).

Simple calorimetry to find the enthalpy of combustion of alcohols

Simple calorimetry to find the enthalpy of combustion of alcohols
Aims
You will use simple calorimetry estimate the enthalpy of combustion of an alcohol.
Apparatus

Goggles
Bench mat
Stand, boss, clamp
Thermometer
100cm3 measuring cylinder
Steel can
Digital balance
Access to spirit burners containing:
methanol, ethanol,
propanol or butanol


Method
1. Draw up a suitable table or tables to record your results.
2. Measure 100cm3 of water in the measuring cylinder.
Pour the water into the steel can and record its temperature.
3. Choose a spirit burner.
Record the name of the fuel, and the mass of the whole burner (including the lid and fuel inside).
4. Clamp the steel can, and set it up so that the spirit burner will fit comfortably under it.
5. Light the wick of the spirit burner, and put it under the steel can.
6. Stir the water gently with the thermometer, and watch the temperature.
When it has increased by 20°C, put the lid on the spirit burner to put the flame out.
Record the new mass of the whole burner (including the lid and fuel inside).
7. Using fresh water each time, repeat the experiment at least twice with the same fuel.

Analysis

1. Calculate the energy transferred to the water using the equation q = mcΔT
Assume that 1cm3 of water has a mass of 1g and c = 4.18 J g–1 K–1.
2. For each replicate experiment, perform the calculations described below:
a) Calculate the mass of fuel burnt.
b) Calculate the Mr of the fuel used. Use your answer to part a) to work out the amount of fuel burnt.
c) Work out the energy transferred to the water in kJ mol–1, and so the enthalpy of combustion.
3. Estimate the maximum errors in the using each piece of apparatus, and the total apparatus error.