DSE Chemistry | Alkanes Tutor Prep Guide | Formula, Naming and Combustion

2026-11-01

Tutor lesson-prep guides · Secondary Chemistry

Students can recite the general formula CnH2n+2 yet freeze at a question that gives 5.8 g of an alkane and asks for the oxygen needed. They treat the formula as a string to memorise instead of an equation with n in it. This lesson starts by solving for n from a relative molecular mass.

Objectives — Students can: (1) move between the general formula and the relative molecular mass; (2) name alkanes with up to ten carbons by the IUPAC rules; (3) write complete and incomplete combustion equations and compute the amounts; (4) distinguish substitution from addition and explain why alkanes only substitute.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Mole calculations	n = m / Mr	Mr from atomic masses
Balancing equations	carbon, then hydrogen, then oxygen	fixed order for combustion
Covalent bonding	single and double bonds	alkanes are all single bonds

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Formula and homologues	each differs by CH2	Mr of C4H10 is 58
8-20 min	Finding n from Mr	set up the equation	5.8 g is 0.1 mol
20-32 min	Naming	longest chain then numbering	butane, 2-methylbutane
32-44 min	Complete combustion	balance C then H	needs 0.65 mol O2
44-54 min	Substitution	UV light, mono first	CH4 + Cl2 gives CH3Cl + HCl
54-60 min	Exit check	collect and mark	independent work

03 Teaching script (read it out)

Scene 1: First move: turn the general formula into an equation, never a memorised number

For an alkane CnH2n+2, Mr = 12n + (2n + 2) = 14n + 2. If a sample has Mr = 58, solve 14n + 2 = 58 to get n = 4, so the formula is C4H10, butane. In reverse, 5.8 g of butane is 5.8 / 58 = 0.1 mol.
The key idea is that each step up the series adds one carbon and two hydrogens, so the mass step is always 14. Ask: why does the series step by CH2 and not CH3? (Each new carbon must be saturated, so two hydrogens come with it.)

Task: an alkane has Mr = 72, how many carbons? (Solve 14n + 2 = 72.)

Scene 2: Naming: longest chain first, then number from the end nearest a branch

Three steps: longest chain, lowest locants, substituents in alphabetical order. In CH3CH(CH3)CH2CH3 the longest chain has four carbons, so the parent is butane and the methyl sits on carbon 2, giving 2-methylbutane.
The usual trap is numbering left to right instead of from the end that gives the smaller locant. Two identical substituents take the prefix di and both numbers, as in 2,2-dimethylpropane.

Task: draw 2,2-dimethylpropane and count the hydrogens to check that H = 2n + 2. Ask: how many isomers does C5H12 have? (Three: pentane, 2-methylbutane and 2,2-dimethylpropane.)

Scene 3: Combustion: balance first, then use the mole ratio

Complete combustion of butane: C4H10 + 6.5O2 gives 4CO2 + 5H2O. The oxygen coefficient is n + (2n + 2)/4 = 4 + 2.50 = 6.5, so 0.1 mol of butane needs 0.1 x 6.5 = 0.65 mol of oxygen, produces 0.4 mol of carbon dioxide which occupies 0.4 x 24 = 9.6 dm3 at room conditions, and 9.0 g of water.
With limited oxygen the fuel burns incompletely to carbon monoxide or soot, which is why gas stoves need ventilation.

Task: write and balance the equation for methane burning to CO. Ask: why does an alkane not decolourise bromine water? (No double bond, so no addition.)

04 Question types and worked steps

Type	Example	Steps	Answer
Find n from Mr	given the relative molecular mass	14n + 2 = Mr	n = 4
Naming	given a structural formula	longest chain plus lowest number	2-methylbutane
Combustion	given a mass	balance then mole ratio	0.65 mol O2
Reaction test	will it decolourise bromine water	look for a double bond	alkanes do not

Sanity check (teach this): An alkane always has an even number of hydrogens (10); an odd count means n was misused. The H to C ratio starts near 4 for methane and approaches 2, but it is always above 2 for an alkane. Oxygen needed always exceeds the moles of fuel itself (0.65 against 0.1). Moles of CO2 equal moles of fuel times carbons and do not depend on the oxygen figure, so it is a good cross-check. Gas volumes may only be compared in moles at the same temperature and pressure.

05 Independent practice (3 questions)

Question	Answer	What to watch
Formula of an alkane with Mr = 58	C4H10	14n + 2 = Mr
Oxygen needed to burn 5.8 g of butane	0.65 mol	moles x 6.5
Volume of CO2 from 0.1 mol of butane at room conditions	9.6 dm3	x 24

06 Common mistakes and what to say

Mistake	Student writes	Say this
Writing CnH2n	answers C4H8 and calls it butane	Alkanes are 2n + 2; 2n is the alkene family.
Numbering left to right	answers 3-methylbutane	Number from the end that gives the smaller locant.
Converting before balancing	answers 0.1 mol O2	The ratio 1 to 6.5 only appears after balancing.
Expecting bromine water to fade	answers that it fades	Alkanes have only single bonds, so they substitute rather than add.

07 Homework (paper, section and question numbers)

Secondary Chemistry chapter index — questions 6 and 9 are the same type
Prep guide contents — other chapters in this subject
DSE Chemistry practice papers — organic set, questions 1 to 8

08 Exit check (last 5 minutes)

Question	Answer	Decision
Formula of an alkane with Mr = 58	C4H10	three right means move on to combustion maths
Alkane versus alkene reactions	substitution versus addition	one wrong means reteach next lesson
Oxygen for 0.1 mol of butane	0.65 mol	two wrong means return to act one

09 Tutor FAQ

Why is the hydrogen count always even?

Each carbon forms four bonds and the two chain ends each carry an extra hydrogen, so the count is 2n + 2 and therefore even.

Do isomers change the oxygen demand?

No. The molecular formula fixes the moles of oxygen for complete combustion; only the structure differs.

Why does substitution need ultraviolet light?

UV supplies the energy to break the Cl to Cl bond and start the radical chain.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.