Reacting masses and volumes

IB Chemistry SL· Reactivity 1: Stoichiometry and Periodicity· 20 min read

1. Molar Mass and Mole Ratios★★☆☆☆⏱ 5 min

📘 Definition

Mole ratio

The whole number ratio of the amount (in moles) of each reactant and product in a balanced chemical equation, derived from stoichiometric coefficients

Example:

For $2 H_{2} + O_{2} \to 2 H_{2} O$ , the mole ratio of $H_{2} : O_{2} : H_{2} O$ is 2:1:2

To calculate reacting masses, first find the molar mass of all relevant substances by summing relative atomic masses from the periodic table, then convert the given mass of a substance to moles using $n = \frac{m}{M}$ .

📐 Worked Example

What is the molar mass of calcium hydroxide, $C a (O H)_{2}$ ? Given $A_{r}$ : Ca=40.1, O=16.0, H=1.0

1
Count the number of each atom in the formula:
$1 \times Ca, 2 \times O, 2 \times H$
2
Sum the relative atomic masses to get molar mass:
$M_{Ca(OH)_2} = (1 \times 40.1) + (2 \times 16.0) + (2 \times 1.0) = 74.1 \ g \ mol^{-1}$

Exam tip:

Always check the number of atoms in polyatomic ions; multiply atomic masses by the subscript outside parenthesis.

2. Reacting Mass Calculations★★☆☆☆⏱ 6 min

To find the mass of a product formed or reactant consumed, follow this 4-step process: 1) write the balanced chemical equation, 2) calculate moles of the known substance, 3) use the mole ratio to find moles of the unknown, 4) convert moles back to mass.

📐 Worked Example

What mass of carbon dioxide is produced when 10.0 g of methane ( $C H_{4}$ ) undergoes complete combustion? $A_{r}$ : C=12.0, H=1.0, O=16.0

1
Write the balanced complete combustion equation:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
2
Calculate moles of known $C H_{4}$ :
$M_{CH_4} = 12.0 + (4 \times 1.0) = 16.0 \ g \ mol^{-1}, n_{CH_4} = \frac{10.0}{16.0} = 0.625 \ mol$
3
Use mole ratio to find moles of $C O_{2}$ :
$n_{CH_4}:n_{CO_2} = 1:1, \text{ so } n_{CO_2} = 0.625 \ mol$
4
Convert moles of $C O_{2}$ to mass:
$M_{CO_2} = 12.0 + (2 \times 16.0) = 44.0 \ g \ mol^{-1}, m_{CO_2} = 0.625 \times 44.0 = 27.5 \ g$

3. Volumes of Gases in Reactions★★★☆☆⏱ 5 min

📘 Definition

Molar volume of a gas

V_m

At standard temperature and pressure (STP: 273 K, 100 kPa), one mole of any ideal gas occupies 22.7 dm³, per IB SL syllabus

Example:

0.5 mol of gas at STP has a volume of 11.35 dm³

For gas reactions at constant temperature and pressure, the mole ratio of gases equals the volume ratio. Relate moles of gas to volume using $n = \frac{V}{V _{m}}$ .

📐 Worked Example

What volume of $C O_{2}$ is produced at STP when 10.0 g of $C a C O_{3}$ decomposes? $M (C a C O_{3}) = 100.1 g m o l^{- 1}$ , $V_{m} = 22.7 d m^{3} m o l^{- 1}$

1
Write the balanced decomposition equation:
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
2
Calculate moles of $C a C O_{3}$ :
$n_{CaCO_3} = \frac{10.0}{100.1} \approx 0.100 \ mol$
3
Mole ratio $C a C O_{3} : C O_{2} = 1 : 1$ , so $n (C O_{2}) = 0.100$ mol
4
Calculate volume at STP:
$V = n \times V_m = 0.100 \times 22.7 = 2.27 \ dm^3$

Exam tip:

IB uses 22.7 dm³ mol⁻¹ for molar volume at STP; old sources use 22.4 dm³ mol⁻¹, always use 22.7 in IB exams.

4. Limiting and Excess Reactants★★★☆☆⏱ 6 min

When reactants are not mixed in the exact mole ratio from the balanced equation, one reactant will be completely consumed first. This is the limiting reactant, and it determines the maximum amount of product that can form. The remaining reactant is in excess.

📐 Worked Example

5.00 g of iron (Fe) reacts with 10.0 g of sulfur (S) to form iron(II) sulfide (FeS). What mass of FeS is formed? $A_{r}$ : Fe=55.8, S=32.1

1
Write the balanced equation:
$Fe(s) + S(s) \rightarrow FeS(s)$
2
Calculate moles of each reactant:
$n_{Fe} = \frac{5.00}{55.8} \approx 0.0896 \ mol, n_S = \frac{10.0}{32.1} \approx 0.312 \ mol$
3
Compare to 1:1 mole ratio: Fe is the limiting reactant, so we use its moles to calculate product yield
4
Mole ratio $F e : F e S = 1 : 1$ , so $n (F e S) = 0.0896$ mol
5
Calculate mass of FeS:
$M_{FeS} = 55.8 + 32.1 = 87.9 \ g \ mol^{-1}, m_{FeS} = 0.0896 \times 87.9 \approx 7.88 \ g$

5. Common Pitfalls

Wrong move:

Forgetting to multiply atomic masses by subscripts when calculating molar mass

Why:

Students often miss that polyatomic groups with subscripts outside parenthesis require multiplying all inner atoms by the subscript, leading to incorrect molar mass

Correct move:

Always count the total number of each atom in the full chemical formula before summing atomic masses

Wrong move:

Using mole ratios from an unbalanced chemical equation

Why:

Unbalanced equations have incorrect stoichiometric coefficients, leading to wrong mole ratios and incorrect final values

Correct move:

Always balance the chemical equation first before starting any calculation

Wrong move:

Using 22.4 dm³ mol⁻¹ for molar volume at STP in IB exams

Why:

IB uses the modern STP definition of 100 kPa, which gives 22.7 dm³ mol⁻¹, so the old value will lose marks

Correct move:

Memorize that $V_{m} = 22.7 d m^{3} m o l^{- 1}$ at STP for IB Chemistry SL

Wrong move:

Using the excess reactant to calculate the mass of product formed

Why:

The excess reactant is not fully consumed, so this will overestimate the amount of product produced

Correct move:

Always use the moles of the limiting reactant to calculate the maximum yield of product

6. Quick Reference Cheatsheet

Concept	Formula	Key IB Value
Moles from mass	n = \frac{m}{M}
Moles from gas volume (STP)	n = \frac{V}{V_m}	V_m = 22.7 \ dm^3 \ mol^{-1}
Mole Ratio	From balanced equation coefficients	Whole numbers only
Limiting Reactant	Compare moles / coefficient for each reactant	Lowest value = limiting
Mass of product	m = n_{limiting} \times ratio \times M_{product}

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Calculate mass of product formed
2024 · 2
Gas volume limiting reactant problem

Going deeper

syllabusOfficial IB SL Chemistry SyllabusReactivity 1 Stoichiometry reference

What's Next

Reacting masses and volumes form the foundation of all stoichiometric calculations in IB Chemistry, which you will apply to more complex topics including solution stoichiometry, titration calculations, and theoretical percentage yield. Mastery of this sub-topic is essential for almost every calculation-based question in both Paper 1 and Paper 2 of your SL exam, so it is important to practice these core step-by-step processes until they become automatic. The problem-solving skills you learn here will also help you when you tackle limiting reactant problems in more advanced reaction topics like chemical equilibrium and enthalpy change calculations.