# Stoichiometric relationships

> IB Chemistry HL · IB Chemistry HL 2025 Syllabus
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u5-stoichiometric-relationships/

Stoichiometric relationships use the mole concept to quantify reactants and products in chemical reactions. These core calculations are the foundation for all quantitative topics in IB Chemistry HL.

**Prerequisites:** [Basic atomic structure and chemical formulae](https://www.owlsprep.com/study/ib-chemistry-hl-u2-atomic-structure/)

## Learning objectives

- Calculate amount of substance, mass and molar mass using the mole relationship
- Determine empirical and molecular formulas from experimental composition data
- Use balanced chemical equations to calculate reacting masses and product yields
- Calculate percentage yield and identify sources of error in yield calculations

## The Mole and Molar Mass

**Mole** — The SI base unit for amount of substance. 1 mole of any substance contains exactly $6.022 \times 10^{23}$ elementary entities (atoms, molecules, ions), equal to Avogadro's constant ($N_A$).

*Notation:* mol

*Example:* 1 mol of carbon atoms contains $6.022 \times 10^{23}$ C atoms.

The core relationship between amount of substance ($n$, in mol), mass ($m$, in g) and molar mass ($M$, in g mol⁻¹) is:

$$n = \frac{m}{M}$$

**Worked example:** Calculate the amount of sodium chloride (NaCl) in 12.5 g of pure solid NaCl. Given $M(Na) = 23.0$ g mol⁻¹, $M(Cl) = 35.5$ g mol⁻¹.

1. First calculate the molar mass of NaCl:

   $$M(NaCl) = M(Na) + M(Cl) = 23.0 + 35.5 = 58.5 \text{ g mol}^{-1}$$
2. Substitute values into the mole formula to solve for $n$:

   $$n = \frac{m}{M} = \frac{12.5 \text{ g}}{58.5 \text{ g mol}^{-1}} = 0.214 \text{ mol}$$

> **Exam tip:** Always include units in your final answer; IB exam markers deduct 1 mark per question for missing units.

## Empirical and Molecular Formulas

**Empirical Formula** — The simplest whole number ratio of atoms of each element present in a compound.

To find the empirical formula from mass or percentage composition data, follow these four steps:

1. Divide the mass (or percentage mass) of each element by its molar mass to get moles
2. Divide each mole value by the smallest mole value from the first step
3. Multiply all values by a whole number if needed to get whole number ratios
4. Write the ratio as subscripts to get the empirical formula

**Worked example:** A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula.

1. Assume 100 g of compound, so masses are 40.0 g C, 6.7 g H, 53.3 g O. Calculate moles of each element:

   $$n(C) = \frac{40.0}{12.0} = 3.33 \text{ mol}, \quad n(H) = \frac{6.7}{1.0} = 6.7 \text{ mol}, \quad n(O) = \frac{53.3}{16.0} = 3.33 \text{ mol}$$
2. Divide all values by the smallest mole value (3.33):

   $$C: \frac{3.33}{3.33} =1, \quad H: \frac{6.7}{3.33} ≈ 2, \quad O: \frac{3.33}{3.33} = 1$$
3. The ratio 1:2:1 is already whole numbers, so the empirical formula is:

To get the molecular formula, you need the actual molar mass of the compound. Calculate the empirical formula mass, divide the actual molar mass by the empirical mass to get a multiplier, then multiply all subscripts by this multiplier.

## Balanced Equations and Reaction Stoichiometry

A balanced chemical equation has equal numbers of each atom on both reactant and product sides, following the law of conservation of mass. The coefficients in a balanced equation give the mole ratio of all reactants and products.

**Worked example:** 2.5 g of calcium carbonate ($CaCO_3$) reacts completely with hydrochloric acid (HCl). Calculate the mass of $CO_2$ produced.

1. Write and balance the full chemical equation:

   $$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$$
2. Calculate moles of $CaCO_3$, $M(CaCO_3) = 100.1$ g mol⁻¹:

   $$n(CaCO_3) = \frac{2.5}{100.1} ≈ 0.025 \text{ mol}$$
3. Use the 1:1 mole ratio from the balanced equation: 1 mol $CaCO_3$ produces 1 mol $CO_2$, so $n(CO_2) = 0.025$ mol
4. Calculate mass of $CO_2$, $M(CO_2) = 44.0$ g mol⁻¹:

   $$m(CO_2) = n \times M = 0.025 \times 44.0 = 1.1 \text{ g}$$

> **tip**
>
> For limiting reactant problems, always calculate moles of each reactant first, then compare to the mole ratio to find which reactant is limiting. All product calculations use moles of the limiting reactant.

## Percentage Yield

**Percentage Yield** — A measure of how much product is actually obtained compared to the maximum theoretical amount predicted by stoichiometry, calculated as: $$\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$$

*Notation:* % yield

*Example:* An 82% yield means 82% of the expected product was collected from the reaction.

**Worked example:** In the reaction of calcium carbonate above, 0.9 g of $CO_2$ was collected. Calculate the percentage yield.

1. We already calculated the theoretical yield of $CO_2$ as 1.1 g. The actual yield collected is 0.9 g.
2. Substitute into the percentage yield formula:

   $$\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{0.9}{1.1} \times 100 = 82\%$$

Percentage yield is always less than 100% due to practical losses during filtration, transfer or purification, or incomplete reactions.

## Common pitfalls

- **Wrong:** Forgetting to count all atoms when calculating molar mass, e.g. $M(H_2O) = 1 + 16 = 17$ g mol⁻¹
  - Why it fails: Misses the two hydrogen atoms in the water molecule, leading to an incorrect molar mass and wrong final answer
  - Correct: Count all atoms in the formula: $M(H_2O) = 2(1.0) + 16.0 = 18.0$ g mol⁻¹
- **Wrong:** Using excess reactant moles for product yield calculations instead of limiting reactant moles
  - Why it fails: The reaction stops when the limiting reactant is completely consumed; excess reactant does not contribute to forming more product
  - Correct: Always identify the limiting reactant first before calculating any product yields
- **Wrong:** Rounding intermediate calculation values too early, leading to inaccurate final answers
  - Why it fails: Early rounding introduces cumulative error that can change the final answer enough to lose marks in exams
  - Correct: Keep at least one extra significant figure in intermediate steps, only round the final answer to the correct number of sig figs
- **Wrong:** Confusing empirical and molecular formulas when answering exam questions
  - Why it fails: Many compounds share the same empirical formula but have different molecular formulas and molar masses
  - Correct: Always confirm what the question asks for, and use the given actual molar mass to find the molecular formula if required
- **Wrong:** Using the mass of an impure reactant directly in mole calculations
  - Why it fails: Impure samples contain non-reacting contaminants that add mass but do not participate in the reaction
  - Correct: Multiply the total mass of the impure sample by the percentage purity to get the mass of pure reactant before calculating moles

## Cheatsheet

| Concept | Core Relationship | Units |
| --- | --- | --- |
| Amount of substance | $n = \frac{m}{M} = \frac{N}{N_A}$ | $n$: mol, $m$: g, $M$: g mol⁻¹ |
| Empirical formula steps | Mass → divide by M → divide by smallest → whole number ratio | None |
| Percentage yield | $\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$ | % |
| Mole ratio | Equal to coefficient ratio in balanced equation | None |
| Molecular formula | $\text{Multiplier} = \frac{M_{\text{actual}}}{M_{\text{empirical}}}$ | None |

## What's next

Stoichiometry is the foundation of all quantitative chemistry, so the skills you learned here will be applied in every subsequent topic in IB Chemistry HL, from thermochemistry enthalpy calculations to organic reaction yield calculations. Mastering these core calculations now will prevent common errors in more complex topics later. Next, you will extend these ideas to gas stoichiometry and solution stoichiometry, where you apply the mole concept to gas volumes and solution concentrations, both commonly tested in Paper 1 and Paper 2 exams. You will also use these skills in your practical internal assessment when calculating yields from your own experiments.

- [Limiting and excess reactants](https://www.owlsprep.com/study/ib-chemistry-hl-u5-limiting-and-excess-reactants/)
- [Reaction rate and rate expressions](https://www.owlsprep.com/study/ib-chemistry-hl-u5-reaction-rate-and-rate-expressions/)
- [Collision theory](https://www.owlsprep.com/study/ib-chemistry-hl-u5-collision-theory/)

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