Lesson 3 · 25 min

Mole ratios and mass calculations

Coefficient ratios are mole ratios. Follow the grams → moles → moles → grams road to find how much carbon dioxide one kilogram of petrol makes.

In this lesson

Read the mole ratio of any two species off a balanced equation
Carry out the three-step grams → moles → moles → grams calculation
Write the calculation as one factor-label chain and report it to the right significant figures
Explain why mass ratios are not coefficient ratios

Burn 1 kg of petrol in a car: how much carbon dioxide comes out? Intuition says “less than a kilogram — surely not more than the fuel”. In fact it is over 3 kg. Where does the extra mass come from? The balanced equation holds the answer: every carbon atom grabs two oxygen atoms from the air. Turning that into a reliable calculation takes one tool — the mole ratio.

Predict first

Petrol can be represented by octane, $\ce{C8H18}$ . Roughly how much carbon dioxide does 1 kg of octane make when it burns completely?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Less than 1 kg About 1 kg About 3 kg About 10 kg

Coefficient ratio = mole ratio

In a balanced equation the coefficients are ratios of particles. Multiply every count by $N_\mathrm{A}$ and the ratios stay the same, so coefficient ratios are mole ratios. For $\ce{N2} + \ce{3H2} \mathrel{\ce{->}} \ce{2NH3}$ , the $\ce{H2}$ consumed is always three times the $\ce{N2}$ , and the $\ce{NH3}$ formed twice the $\ce{N2}$ .

But mass ratios are not coefficient ratios, because different substances have different molar masses. So whenever a mass is given and another mass is asked for, convert to moles first, cross the bridge with the coefficient ratio in the “world of moles”, and convert back to mass.

m_\mathrm{B} = \frac{m_\mathrm{A}}{M_\mathrm{A}} \times \frac{\nu_\mathrm{B}}{\nu_\mathrm{A}} \times M_\mathrm{B}

grams → moles → moles → grams.

$m_\mathrm{A},\ m_\mathrm{B}$: mass of the given species A and of the wanted species B
$M_\mathrm{A},\ M_\mathrm{B}$: their molar masses
$\nu_\mathrm{A},\ \nu_\mathrm{B}$: their coefficients in the balanced equation (exact integers)

grams → moles → moles → grams

The mole road map

Reaction

$\ce{2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(l)}$

Given Mass / g Find

m( $\ce{Fe2O3}$ ) 100.0 g
÷ molar mass ÷ 159.69 g/mol
n( $\ce{Fe2O3}$ ) 0.62623 mol
× mole ratio × 2/1
n( $\ce{Al}$ ) 1.2525 mol
× molar mass × 26.98 g/mol
m( $\ce{Al}$ ) 33.79 g

× N_A → molecules (or formula units): 7.542 × 10²³

Mole ratio = coefficient ratio: 1 : 2

Fe₂O₃ 0.626 mol

Al 1.25 mol

As one factor-label chain

100.0\ \text{g}\ \ce{Fe2O3} \times \frac{1\ \text{mol}\ \ce{Fe2O3}}{159.69\ \text{g}\ \ce{Fe2O3}} \times \frac{2\ \text{mol}\ \ce{Al}}{1\ \text{mol}\ \ce{Fe2O3}} \times \frac{26.98\ \text{g}\ \ce{Al}}{1\ \text{mol}\ \ce{Al}} = 33.79\ \text{g}\ \ce{Al}

Answer (4 significant figures): 33.79 g

Try other reactions: roasting limestone, the sodium azide airbag, the blast furnace. Notice that the middle step’s ratio comes only from the coefficients, never from masses.

Worked example

How much aluminium for a thermite reaction?

Thermite welding reduces iron oxide with aluminium powder: $\ce{2Al} + \ce{Fe2O3} \mathrel{\ce{->}} \ce{Al2O3} + \ce{2Fe}$ . What mass of aluminium is needed to reduce 100.0 g of $\ce{Fe2O3}$ ?

Concept check

Ammonia synthesis, $\ce{N2} + \ce{3H2} \mathrel{\ce{->}} \ce{2NH3}$ : 6.0 mol of $\ce{H2}$ reacts completely. How much $\ce{NH3}$ forms?

2.0 mol 4.0 mol 6.0 mol 12 mol

Try the calculation

10.0 g of methane burns completely ( $\ce{CH4} + \ce{2O2} \mathrel{\ce{->}} \ce{CO2} + \ce{2H2O}$ ). What mass of oxygen is needed?

Key takeaways

Coefficient ratios in a balanced equation are ratios of particles and of moles.
The road: $m_\mathrm{A} \xrightarrow{\div M_\mathrm{A}} n_\mathrm{A} \xrightarrow{\times \nu_\mathrm{B}/\nu_\mathrm{A}} n_\mathrm{B} \xrightarrow{\times M_\mathrm{B}} m_\mathrm{B}$ .
Mass ratios are not coefficient ratios; total reactant mass equals total product mass, a useful check.
Coefficients are exact; the significant figures of the answer come from the measured data.

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Data and references

IUPAC abridged standard atomic weights (2021), via the VisualChem element table.
P. Atkins, L. Jones, L. Laverman, Chemical Principles, 7th ed., Fundamentals L (reaction stoichiometry).
人民教育出版社《普通高中教科书化学必修第一册》（2019），第二章第三节物质的量。