Worked solutions to textbook questions
Chapter 8 Compounds of carbon
E1.
From the molecular formulas shown on page 140 for ethyne, propyne and butyne,
deduce the general formula for the alkyne homologous series.
AE1.
CnH2n–2
E2.
What is the molecular formula for an alkyne with five carbon atoms?
AE2.
C5H8
E3.
Explain why there isn’t a branched-chain isomer of molecular formula C4H6.
AE3.
The carbon atoms at each end of the C–C triple bond are free to join only with 1 other
atom—either H or C—not the 2 C atoms required if a branched-chain isomer is to
occur.
Q1.
Draw a diagram that includes some of the carbon-based substances that are important
in your life. Your diagram should show where the carbon in these substances comes
from and how it moves on to other roles in the environment when you have finished
with these substances.
A1.
A possible answer is shown.
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Q2.
Why is the study of carbon compounds sometimes called organic chemistry?
A2.
Carbon compounds are the basis of many living systems and the term ‘organic
chemistry’ is used to signify this. Berzelius coined the name ‘organic chemistry’, as
he believed that only carbon compounds existed in living systems.
It was originally believed that all organic compounds had a biological origin. It was
thought that organic compounds contained a ‘vital force’ that led to life.
In the nineteenth century, this idea was shown to be false when urea, CO(NH2)2 (a
component of urine), was made in a laboratory from inorganic materials.
Q3.
Methane is the smallest hydrocarbon molecule.
a What is the molecular formula of methane?
b Why is methane a hydrocarbon?
c Why does the carbon atom in a methane molecule bond to four, rather than two,
three, five or any other number of hydrogen atoms?
d Draw the structural formula of methane. Why does methane have this particular
arrangement of hydrogen atoms around each carbon atom?
A3.
a
b
c
CH4
Methane is a hydrocarbon because it is a compound of carbon and hydrogen.
Carbon has the electronic configuration of 2,4. Each carbon atom needs four
electrons to complete its outer shell. Hydrogen has an electronic configuration of
1. Each hydrogen needs one electron to complete its outer shell. Because both
atoms need electrons, they will share electrons, that is, form covalent bonds. It
will take four hydrogen atoms to provide the four electrons required by each
carbon atom.
d
This arrangement gives minimum electrostatic repulsion between the four pairs of
bonding electrons. The hydrogen atoms are arranged around the central carbon
atom in a tetrahedral configuration.
Q4.
Ethane and ethene are both hydrocarbons. Do these hydrocarbons belong to the same
homologous series? Explain your answer.
A4.
Ethane (C2H6) and ethene (C2H4) do not belong to the same homologous series, as
their formulas do not differ by a –CH2– group (or a multiple of this). Also, ethane
contains only single bonds (i.e. it is saturated) whereas ethene has a double bond (i.e.
it is unsaturated). Members of the same homologous series have the same functional
group.
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Q5.
The following molecular formulas all represent hydrocarbons:
CH4, C5H12, C3H4, C6H12, C3H6, C5H10, C25H52, C6H6
a Which of these hydrocarbons belong to the same homologous series as ethane?
b Which of these hydrocarbons belong to the same homologous series as ethene?
A5.
a
b
CH4, C5H12, C25H52
C6H12, C5H10
Q6.
Give:
i the structural formula, and
ii the semistructural formula
for the hydrocarbon with molecular formula C3H8.
A6.
a
b
CH3CH2CH3
E4.
Give the structural formula of:
a a member of the alkanol series that has two carbon atoms
b an alkanoic acid with a total of three carbon atoms
AE4.
a
ethanol
b
C2H5COOH
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E5.
A cyclic hydrocarbon has the molecular formula C4H8. Give the structural formula of
this hydrocarbon.
AE5
E6.
Under appropriate conditions, it is possible to add hydrogen gas (H2) to benzene and
convert it to cyclohexane. How many molecules of hydrogen would need to be added
to each molecule of benzene in such a conversion?
AE6.
Benzene C6H6; cyclohexane C6H12. So 6 extra hydrogen atoms are required per
molecule of benzene, which is 3 molecules of H2.
Q7.
Draw structural formulas for each of the following hydrocarbons:
a propane
b propene
c methane
d but-1-ene
e 2-methylpropane
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A7.
Q8.
Name the following hydrocarbons:
a CH3CH2CH3
b CH3CH2CH2CH2CH2CH2CH3
c CH3CH=CHCH3
d CH3CH(CH3)CH2CH3
e CH3CH(CH3)CH(CH3)CH3
A8.
a
b
c
d
e
propane
hexane
but-2-ene
2 methylbutane
2,3 dimethylbutane
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Q9.
Explain why methane is a gas at room temperature whereas octane, one of the
constituents of petrol, is a liquid.
A9.
Octane (C8H18) (b.p. 125.7°C) and methane (CH4) (b.p. –161.5°C) are both non-polar
and so are held together only by dispersion forces. Octane is the larger of these two
molecules and has more electrons, so the dispersion forces between octane molecules
will be greater than between the methane molecules. Because there are stronger
dispersion forces between octane molecules, octane has a higher boiling temperature
than methane.
Q10.
Write a balanced chemical equation for the combustion reaction of:
a pentane
b ethane
c ethene
(Assume there is sufficient oxygen to react fully with the hydrocarbon.)
A10.
a
b
c
C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(g)
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
C2H4(g) + 4O2 (g)  2CO2 (g) + 2H2O(g)
Q11.
Write a balanced equation for the addition of chlorine gas (Cl2) to propene.
A11.
C3H6(g) + Cl2(g)  CH2Cl.CHCl.CH3(g)
Q12.
Refer to Table 8.9 (page 156) and draw diagrams to represent the formation of:
a polyvinyl chloride
b polypropene
c Teflon
A12.
a
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b
c
Q13.
The polymerisations in Question 12 are addition reactions. Why are they classified in
this way?
A13.
Each of the polymers is formed from monomers that contain double bonds. The
double bond in each of the monomers is broken and the monomers add onto each
other to form a polymer.
Q14.
What atoms form the backbone in a polymer chain of:
a rubber?
b Teflon?
A14.
Carbon forms the backbone of all polymers. Carbon is a unique element because of its
ability to bond to itself. Most modern construction materials and pharmaceuticals
have carbon backbones.
a carbon
b carbon
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Q15.
a
b
In terms of their structures, explain the difference in properties between HDPE
and LDPE.
Which of these two forms would be a suitable material for:
i a soft, flexible plastic wrap?
ii a 2-litre drink container?
iii wrapping material for frozen food?
A15.
a
b
High-density polyethene (HDPE) is made of relatively unbranched chains of
polyethene, which can pack more closely together than the low-density
polyethene (LDPE). LDPE contains branched chains of polyethene that cannot
pack together as closely. HDPE is therefore stronger and slightly less flexible
than LDPE. Both HDPE and LDPE are chemically unreactive, waterproof, nonconductors and only slightly permeable to gases.
i LDPE
ii HDPE
iii LDPE
Q16.
How do each of the following pairs of terms differ. Where possible, give examples to
support your answer.
a thermoplastic and thermosetting polymer
b atactic and isotactic polypropene
c crystalline and amorphous regions
d branched and cross-linked polymers
A16.
a
b
c
Thermoplastic polymers can be heated and reshaped. Thermosetting polymers
cannot be heated and reshaped. For example, LDPE is a thermoplastic and can be
reshaped on heating; a vulcanised rubber article like a shoe sole cannot be
reshaped when heated, but decomposes.
In atactic polypropylene, the side groups are distributed randomly and, as a result,
the polymer chains do not stack well. It is used as a grease. In isotactic
polypropylene, the side groups are all on the same side. The polymer chains can
stack well, forming many crystalline regions and a substance that is stiff and
strong. Isotactic polypropylene is frequently used to make ropes.
Crystalline regions are found in plastic materials in which many of the polymer
chains line up in parallel. Such regions maximise the intermolecular forces,
increasing the strength of the plastic material and scattering the light that passes
through them. Amorphous regions are found in materials in which the polymer
chains do not line up. Instead, the polymer chains are arranged randomly. An
example is the two different forms of polyethene. HDPE contains more regions of
crystallinity and is stronger and more opaque due to light scattering. LDPE
contains few or no regions of crystallinity, is very soft and flexible, and
transparent.
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d
9
Branched polymers consist of a main carbon backbone with further carbon chains
bonded onto one or more carbons in the backbone, creating a branch. The
branched polymer chains are weakly bonded to each other and so exist as separate
entities. Cross-linked polymers contain covalent bonds linking one polymer chain
to another.
Chapter review
Q17.
Why can carbon form so many compounds?
A17.
Carbon can form a large number of compounds for several reasons. Carbon atoms can
use two electrons each to bond to form very long chains. This leaves the other two
valence electrons able to bond other non-metal atoms onto the chain. In addition, there
can be double and triple bonds between carbon atoms as well as ring structures. All
these possible variations mean that an enormous number of different compounds
containing carbon can be formed.
Q18.
Classify each of the following hydrocarbons as alkanes, alkenes or neither:
a C2H6
b C3H6
c C20H42
d C2H2
e C5H10
f C5H12
g C8H14
A18.
a
b
c
d
e
f
g
alkane
alkene
alkane
neither
alkene
alkane
neither
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Q19.
Draw the structural formulas and give the systematic names of:
a C3H8
b C2H4
c the isomers of C4H10
A19.
c
Q20.
Select appropriate examples to help explain the following terms:
a homologous series
b polar bond
c polar molecule
A20.
a
b
c
A homologous series is a family of compounds in which the molecular formula of
each member differs from the previous member by –CH2–, and each compound in
the series has the same functional group(s); for example, the alkanes, CH4, C2H6,
C3H8; carboxylic acids, HCOOH, CH3COOH, C2H5COOH.
A polar bond is one in which the bonding electrons are shared unequally because
one atom has a higher electronegativity than the other; for example, O–H, N–H,
C–Cl.
A polar molecule is one which has one or more polar bonds and the molecule has
no overall symmetry of charge. For example, H2O is a polar molecule because it
is angular, whereas CO2 (linear) and CF4 (tetrahedral) have a symmetrical
distribution of charge and are therefore non-polar.
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Q21.
The formula of a hydrocarbon is C16H34.
a To which homologous series does it belong?
b What is the formula of the next hydrocarbon in the same homologous series?
c What is the formula of the previous hydrocarbon in the same homologous series?
d What mass of carbon is present in 275 g of the hydrocarbon C16H34?
A21.
a
b
c
d
alkane
C17H36
C15H32
Percentage by mass of an element
=
mass of element in 1 mol of compound
× 100%
mass of 1 mol of the compound
M(C16H34) = 226 g mol–1.
 %(C)
=
16  12.0  100
226
= 85.0%
85.0  275 g
100
= 234 g
So, M(C) in 275 g of compound =
Q22.
A sample of hexane has a mass of 63.8 g.
a What amount, in mol, of hexane is in the sample?
b How many hexane molecules are there in the sample?
A22.
a
n=
m
M
n(C6H14) =
63.8
86.0
= 0.742 mol (to 3 significant figures)
b
Number of particles = n × NA
Number of C6H14 molecules = 0.742 × 6.02  1023
= 4.47 × 1023 molecules
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Q23.
Give:
i a structural formula, and
ii a semi-structural formula
for each of the following compounds.
a propene
b 2-methylpentane
c hex-2-ene
d 2,2-dimethylbutane
A23.
a
CH2CHCH3
b
c
d
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CH3CH(CH3)CH2CH2CH3
CH3CHCHCH2CH2CH3
CH3C(CH3) 2CH2CH3
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Q24.
The names for the following carbon compounds are incorrect. Give the correct names.
a
methylethane
b
1,2-dimethylethene
c
trimethylmethane
d
2,2-diethylpropane
A24.
a
b
c
d
propane
but-2-ene
2 methyl propane
3,3 dimethyl pentane
Q25.
Table 8.7 (page 147) gives the boiling temperatures of the first ten alkanes.
a What happens to the boiling temperatures of these alkanes as molecular size
increases?
b Which alkanes would you expect to be gases at 25°C?
c Explain why there is a change in boiling temperature with molecular size.
A25.
a
b
c
The boiling temperature increases.
methane, ethane, propane and butane
All of the hydrocarbons are alkanes. They are all non-polar and are held together
only by dispersion forces. The larger the alkane, the more electrons it has and the
greater the dispersion forces between the alkane molecules. The stronger the
dispersion forces, the more energy is needed to vaporise the alkane and, hence,
the higher the boiling temperature of the alkane.
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Q26.
Write a balanced chemical equation for each of the following reactions. Remember to
include the states of the reactants and products.
a Gaseous methane and oxygen in the air react to form carbon dioxide gas and
water vapour.
b Gaseous hexane and oxygen react to produce carbon dioxide gas and water
vapour.
c Gaseous propene and oxygen react to produce carbon dioxide gas and water
vapour.
d Gaseous butene and hydrogen react in the presence of a catalyst at 130°C to form
a single product.
A26.
a
b
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
19
C6H14(g) + 2 O2(g)  6CO2(g) + 7H2O(g)
or
2C6H14(g) + 19O2(g)  12CO2(g) + 14H2O(g)
c
C3H6(g) + 2 O2(g)  3CO2(g) + 3H2O(g)
or
2C3H6(g) + 9O2(g)  6CO2(g) + 6H2O(g)
C4H8(g) + H2(g)  C4H10(g)
d
9
Q27.
A hydrocarbon contains 85.7% carbon and its molar mass is between 40 and 50 g mol–1.
a Calculate the empirical formula of the hydrocarbon.
b Determine its molecular formula.
c To which homologous series does the compound belong?
d Write a chemical equation for one reaction typical of this compound.
A27.
a
The molecular formula is always a whole-number multiple of the empirical
formula. The empirical formula provides the simplest whole-number ratio of
m
atoms in a compound. The amount of each atom is found by using n =
,
M
where m is the mass in grams and M is the molar mass in g mol–1.
Mass
Molar mass
m
M
Divide all by smallest
amount
Amount, using n =
C
85.7 g
12 g mol–1
85.7
n=
= 7.14 mol
12.0
7.14
=1
7.14
 empirical formula is CH2
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H
14.3 g
1 g mol–1
14.3
n=
= 14.3 mol
1.0
14.3
= 2.0
7.14
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b
c
d
15
Molar mass of a CH2 unit (empirical formula) = 12 + 2 = 14 g mol–1
Molar mass of the compound (molecular formula) = between 40 and 50 g mol–1
 Number of CH2 units in one molecule = 3
 Molecular formula is C3H6
alkene
A typical reaction is addition; for example, C3H6(g) + H2(g)  C3H8(g)
Q28.
A sample of polyethene has a mass of 175 kg. How many ethene monomers have
joined together to form this sample of polyethene?
A28.
Each ethene-recurring unit (C2H4) has a molar mass of 2 × 12 + 4 = 28 g mol–1.
175 000
 Number of C2H4 units in the sample of polyethene =
28
= 6250
Q29.
Ethene (C2H4) is the smallest alkene.
a Why is it described as unsaturated?
b Draw the structural formula of ethene. Why does ethene have this particular
arrangement of atoms?
c Could ethane (C2H6) act as a monomer? Explain.
A29.
a
b
c
The ethene molecule has a carbon-to-carbon double bond.
Ethene has this structural formula because it ensures each atom has a stable outer
electron shell, and this arrangement of atoms gives the minimum repulsion
between electron pairs.
Ethane cannot undergo addition polymerisation because it is a saturated
compound.
Q30.
Use polyethene as an example to explain the following terms:
a monomer
b polymer
c addition polymerisation
d unsaturated and saturated
e empirical formula
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A30.
a
b
Ethene, an alkene (CH2=CH2), can combine with other ethene molecules to form
a long chain polymer, polyethene. Ethene contains a double carbon–carbon bond
and is able to react with itself. This makes it an ideal monomer.
When molecules of ethene join together they form a long chain molecule, a
polymer, with the following structure:
–CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH2–.
Each ethene, CH2=CH2, becomes a –CH2–CH2– in the polymer.
c
d
e
Ethene (CH2=CH2) is unsaturated because it contains a double bond. Polyethene
is saturated because it contains only single bonds between carbon atoms.
The empirical formula is the simplest whole-number ratio of elements in a
compound. The empirical formula of ethene (CH2=CH2) is CH2.
Q31.
Look at the figure below. Draw a section of the polymer made from each of the
monomers in an addition polymerisation process.
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A31.
Q32.
PVA (or white) adhesive contains polyvinyl acetate in water. A section of the polymer
is shown below.
Give the structural formula for its monomer, vinyl acetate.
A32.
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Q33.
The polymer chains of a certain material can be cross-linked to varying extents.
a Sketch the polymer chains when there is:
i no cross-linking
ii a little cross-linking
iii a large degree of cross-linking
b Use diagrams to show the effect of stretching each of these materials.
A33.
Q34.
Would a thermoplastic or thermosetting polymer be the most suitable material for the
following purposes?
a the handle of a kettle
b a ‘squeeze’ container for shampoo
c the knob of a saucepan lid
d a shopping bag
e a rope
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A34.
Distinguishing between thermoplastic and thermosetting polymers is important. The
former get soft when heated, whereas the latter remain rigid and will char if heating
continues.
a thermosetting
b thermoplastic
c thermosetting
d thermoplastic
e thermoplastic
Q35.
Elastic bands, golf balls and saucepan handles are made from polymers with some
cross-linking.
a Which material has the greatest degree of cross-linking?
b Describe the properties of the material you chose in part a to support your
answer.
A35.
a
b
The polymer used in saucepan handles.
Saucepan handles are much harder and more resistant to the effect of heat than
elastic bands. The polymer on the outside of the golf ball is also very hard but,
unlike the polymer in saucepan handles, is quite elastic.
Q36.
Plasticisers can be added to polymers to change their properties. Plasticisers are small
molecules that hold polymer chains further apart. What effect does the addition of a
plasticiser have on the hardness and softening temperature of a polymer?
A36.
As the polymer chains are held further apart by the plasticisers, it makes it easier for
the polymer chains to slide over each other. As a result, these materials have a lower
softening temperature, and are more malleable and soft.
Q37.
Silicon, the element below carbon in the periodic table, forms a small number of
compounds called silanes. The smallest of these silane molecules has the molecular
formula SiH4.
a Draw a structural formula for SiH4.
b Predict the molecular formula and structural formula of the next member in the
series of silanes.
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A37.
Q38.
Explain the following:
a The first member of the alkene homologous series is ethene, not methene.
b Carbon compounds usually have four covalent bonds around each carbon atom.
c Compounds containing a quadruple carbon–carbon bond do not exist.
A38.
a
b
c
An alkene contains one double carbon–carbon bond, which requires two carbon
atoms. The first alkene is therefore ethene.
The carbon atom has four electrons in the outer shell, which are available for
sharing with other atoms to produce four covalent bonds.
A quadruple carbon–carbon bond of eight electrons would not allow room for
other bonds to either carbon atom.
Q39.
Labels on margarine and oil often include one of the following terms:
polyunsaturated, mono-unsaturated, saturated
a What do you think each of these terms means?
b Check some food containers at home or in a supermarket and note the details on
any labels that include these terms.
A39.
a
b
Polyunsaturated: contains many double bonds; mono-unsaturated: contains one
double bond; saturated: contains only single bonds between carbon atoms.
Individual students’ response required.
Q40.
Explain whether you agree or disagree with the following statements:
a If one structural isomer of a compound of carbon and hydrogen is saturated, then
all of the other structural isomers of that compound must be saturated.
b A compound that contains three carbon atoms can have only one structural
isomer.
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A40.
a
b
Where isomers are possible, this is true for all saturated linear or branched
hydrocarbons containing four or more carbon atoms. Note: If the alkane was
formed into a ring, it would have the same structural formula as an alkene—
cyclobutane (C4H8) is an isomer of but-1-ene. It would be possible to have an
alkene that is an isomer of a saturated hydrocarbon.
This is true only for alkanes and alkenes, but not for three-carbon molecules with
a functional group. For example, propan-1-ol (CH3CH2CH2OH) and propan-2-ol
(CH3CH(OH)CH3) are isomers.
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