Worked solutions to textbook questions 21
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.
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 CH4
b Methane is a hydrocarbon because it is a compound of carbon and hydrogen.
c 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.
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 CH4, C5H12, C25H52
b 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
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
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 propane
b hexane
c but-2-ene
d 2 methylbutane
e 2,3 dimethylbutane
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 C5H12(g) + 8O2(g) ® 5CO2(g) + 6H2O(g)
b 2C2H6(g) + 7O2(g) ® 4CO2(g) + 6H2O(g)
c 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
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
Q15.
a In terms of their structures, explain the difference in properties between HDPE and LDPE.
b 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 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, non-conductors and only slightly permeable to gases.
b 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 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.
b 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.
c 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.
d 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 alkane
b alkene
c alkane
d neither
e alkene
f alkane
g neither
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 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.
b 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.
c 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.
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 alkane
b C17H36
c C15H32
d Percentage by mass of an element
=×100%
M(C16H34) = 226 g mol–1.
\ %(C) =
= 85.0%
So, M(C) in 275 g of compound =
= 234 g
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 =
n(C6H14) =
= 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
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 CH3CH(CH3)CH2CH2CH3
c CH3CHCHCH2CH2CH3
d CH3C(CH3) 2CH2CH3
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 propane
b but-2-ene
c 2 methyl propane
d 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 The boiling temperature increases.
b methane, ethane, propane and butane
c 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.
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 CH4(g) + 2O2(g) ® CO2(g) + 2H2O(g)
b C6H14(g) + O2(g) ® 6CO2(g) + 7H2O(g)
or
2C6H14(g) + 19O2(g) ® 12CO2(g) + 14H2O(g)
c C3H6(g) + O2(g) ® 3CO2(g) + 3H2O(g)
or
2C3H6(g) + 9O2(g) ® 6CO2(g) + 6H2O(g)
d C4H8(g) + H2(g) ® C4H10(g)
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 atoms in a compound. The amount of each atom is found by using n = , where m is the mass in grams and M is the molar mass in g mol–1.