SACE Board SACE 2019-style practice: The complete combustion of octane is 2C_8H_18 + 25O_2 → 16CO_2 + 18H_2O. Calculate the mass of CO_2 produced when 1.00\ kg of octane is burned completely. (M(C_8H_18) = 114.0, M(CO_2) = 44.0\ g mol^-1.)

Step 1: n(C_8H_18) = 1000114.0 = 8.77\ mol. (1 mark) Step 2: from the 2:16 (i.e. 1:8) ratio, n(CO_2) = 8 × 8.77 = 70.2\ mol. (1 mark) Step 3: m(CO_2) = nM = 70.2 × 44.0 = 3.09 × 10^3\ g = 3.09\ kg. (2 marks)

§-Syllabus dot point

SAChemistrySyllabus dot point

How does molecular structure explain the greenhouse effect?

Explain the greenhouse effect in terms of the absorption of infrared radiation by greenhouse gas molecules.

Why greenhouse gases absorb infrared while N2 and O2 do not, the role of bond vibrations and dipole change, the enhanced greenhouse effect, and worked SACE-style emission and energy calculations.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Lead worked calculation
Why some gases absorb infrared and others do not
The enhanced greenhouse effect
Monitoring greenhouse gases
Why it matters for monitoring

What this dot point is asking

SACE expects you to explain absorption in terms of bond vibrations and dipole change, distinguish greenhouse from non-greenhouse gases, and link combustion to rising $\text{CO}_2$ .

Lead worked calculation

Comparing the CO2 from two fuels per mole

Compare the $\text{CO}_2$ released per mole burned for methane and octane. Methane: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ . Octane: $2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}$ .

Read the carbon-to-CO2 ratio

Methane gives $1\ \text{mol}$ of $\text{CO}_2$ per mole burned. Octane gives $\tfrac{16}{2} = 8\ \text{mol}$ of $\text{CO}_2$ per mole burned.

Per gram comparison (more meaningful)

For methane, $\dfrac{1 \times 44.0}{16.0} = 2.75\ \text{g CO}_2$ per gram of fuel. For octane, $\dfrac{8 \times 44.0}{114.0} = 3.09\ \text{g CO}_2$ per gram.

Interpret

Methane releases less $\text{CO}_2$ per gram because it has the highest hydrogen-to-carbon ratio of common fuels, so a larger share of its energy comes from forming water rather than $\text{CO}_2$ . This is one reason natural gas is a lower-carbon fuel than petrol.

Why some gases absorb infrared and others do not

The Earth absorbs visible sunlight and re-emits energy as infrared. Whether a gas captures this IR depends on its molecular structure.

$\text{CO}_2$ : although linear and symmetric overall, its bending vibration and asymmetric stretch create a temporary changing dipole, so it absorbs IR strongly.
$\text{H}_2\text{O}$ : bent and polar; its stretches and bend all change the dipole, making it the most significant natural greenhouse gas.
$\text{N}_2$ and $\text{O}_2$ : homonuclear diatomics with non-polar bonds; stretching keeps the dipole at zero, so they are IR-transparent despite being abundant.

The enhanced greenhouse effect

The natural greenhouse effect keeps Earth habitable, around $33\ ^\circ\text{C}$ warmer than it would otherwise be. The enhanced greenhouse effect is the extra warming from human activity raising greenhouse-gas concentrations, chiefly through combustion of fossil fuels: $\text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ . More $\text{CO}_2$ means more IR is absorbed and re-radiated downward, raising average temperatures.

Monitoring greenhouse gases

Atmospheric $\text{CO}_2$ is measured continuously (famously at Mauna Loa) using infrared gas analysers, which exploit the very IR absorption that makes $\text{CO}_2$ a greenhouse gas: the more $\text{CO}_2$ in the sample, the more IR it absorbs. This neatly ties the monitoring technique back to the molecular property being studied.

Why it matters for monitoring

Linking molecular structure to IR absorption explains both why certain trace gases drive climate change and how instruments detect them. It connects combustion chemistry (the source of extra $\text{CO}_2$ ) to the physical mechanism of warming, the foundation of climate monitoring and emissions policy.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20214 marksExplain, in terms of molecular vibrations and bond polarity, why carbon dioxide and water vapour are greenhouse gases but nitrogen (

\text{N}_2

) and oxygen (

\text{O}_2

) are not, even though

\text{N}_2

and

\text{O}_2

make up most of the atmosphere.

Show worked answer →

A molecule absorbs infrared radiation only if a vibration (a stretching or bending of its bonds) causes a change in its dipole moment. (1 mark)

$\text{CO}_2$ and $\text{H}_2\text{O}$ contain polar bonds; when these bonds vibrate, the distribution of charge changes, creating a changing dipole that can absorb IR photons of matching energy. (1 mark)

$\text{N}_2$ and $\text{O}_2$ are diatomic molecules made of two identical atoms, so their bonds are non-polar; stretching them does not change the dipole (which stays zero). With no changing dipole, they cannot absorb IR. (1 mark)

Therefore, despite being the most abundant gases, $\text{N}_2$ and $\text{O}_2$ are transparent to the IR re-emitted by Earth, while trace $\text{CO}_2$ and $\text{H}_2\text{O}$ trap it. (1 mark)

SACE 20194 marksThe complete combustion of octane is

2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}

. Calculate the mass of

\text{CO}_2

produced when

1.00\ \text{kg}

of octane is burned completely. (

M(\text{C}_8\text{H}_{18}) = 114.0

M(\text{CO}_2) = 44.0\ \text{g mol}^{-1}

Show worked answer →

Step 1: $n(\text{C}_8\text{H}_{18}) = \dfrac{1000}{114.0} = 8.77\ \text{mol}$ . (1 mark)

Step 2: from the $2:16$ (i.e. $1:8$ ) ratio, $n(\text{CO}_2) = 8 \times 8.77 = 70.2\ \text{mol}$ . (1 mark)

Step 3: $m(\text{CO}_2) = nM = 70.2 \times 44.0 = 3.09 \times 10^{3}\ \text{g} = 3.09\ \text{kg}$ . (2 marks)