IB Chemistry SL — Stoichiometry & From Models to Materials

Complete Study Guide

Topics Covered

1. The Mole Concept and Molar Mass (R2.1)
2. Avogadro’s Number and Counting Particles (R2.1)
3. Mole Ratios and Balanced Equations (R2.1)
4. Limiting Reagent and Excess Reagent (R2.1)
5. Theoretical Yield, Actual Yield, and Percentage Yield (R2.1)
6. Atom Economy (R2.1)
7. Allotropes of Carbon (S2.4)
8. Polymers — Addition and Condensation (S2.4)
9. Semiconductors and Superconductors (S2.4)

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Prerequisites: The mole concept was introduced in Atomic Structure (Structure 1.4). This guide extends that foundation to reactions — you will apply mole calculations to balanced equations. For the Structure 2.4 section, familiarity with covalent bonding and giant covalent structures from Bonding & Structure (Structure 2) is assumed.

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1. The Mole Concept and Molar Mass (R2.1)

Stoichiometry is the quantitative study of the amounts of substances involved in chemical reactions. The mole is the central unit: it bridges the microscopic world of atoms and molecules with the macroscopic world of masses and volumes that we can measure in the laboratory.

Defining the Mole

One mole of any substance contains exactly $6.022 \times 10^{23}$ elementary entities (atoms, molecules, ions, or formula units). This number is Avogadro’s constant ($N_A$):

$N_A = 6.022 \times 10^{23}\, \text{mol}^{-1}$

The mole is defined so that one mole of carbon-12 atoms has a mass of exactly 12 g. This makes the numerical value of the molar mass equal to the relative atomic (or molecular) mass.

Molar Mass

The molar mass ($M$) is the mass of one mole of a substance, expressed in $\text{g mol}^{-1}$.

• For elements: $M$ numerically equals the relative atomic mass $A_r$ found on the periodic table.
  • $M(\text{Na}) = 22.99\, \text{g mol}^{-1}$
  • $M(\text{O}_2) = 2 \times 16.00 = 32.00\, \text{g mol}^{-1}$
• For compounds: sum the molar masses of all atoms in the formula.
  • $M(\text{H}_2\text{SO}_4) = 2(1.008) + 32.07 + 4(16.00) = 98.09\, \text{g mol}^{-1}$
  • $M(\text{CaCO}_3) = 40.08 + 12.01 + 3(16.00) = 100.09\, \text{g mol}^{-1}$

The Mole–Mass Equation

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

Where:

• $n$ = amount of substance (mol)
• $m$ = mass (g)
• $M$ = molar mass ($\text{g mol}^{-1}$)

Rearrangements: $m = nM$ and $M = \dfrac{m}{n}$

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2. Avogadro’s Number and Counting Particles (R2.1)

The relationship between amount in moles and number of particles is:

$N = n \times N_A$

Where $N$ is the number of particles.

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3. Mole Ratios and Balanced Equations (R2.1)

A balanced chemical equation tells you not just what reacts and what is produced, but also the exact mole ratios in which substances react and are produced. The stoichiometric coefficients ARE the mole ratios.

$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$

This equation means: 1 mol of $\text{N}_2$ reacts with 3 mol of $\text{H}_2$ to produce 2 mol of $\text{NH}_3$.

Strategy for Stoichiometry Problems

1. Write (or check) the balanced equation.
2. Convert the given quantity to moles.
3. Use the mole ratio from the equation to find moles of the target substance.
4. Convert moles of the target substance to the required quantity (mass, volume, number of particles).

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4. Limiting Reagent and Excess Reagent (R2.1)

In most real reactions, the reactants are not mixed in exactly the stoichiometric ratio. The limiting reagent (also called the limiting reactant) is the reactant that is completely used up first — it limits how much product can form. The excess reagent is the reactant that remains after the reaction is complete.

Identifying the Limiting Reagent

Method: Convert all given masses to moles, then divide each by its stoichiometric coefficient. The reactant with the smallest value is the limiting reagent.

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5. Theoretical Yield, Actual Yield, and Percentage Yield (R2.1)

Definitions

• Theoretical yield: The maximum mass of product predicted by stoichiometry, assuming the limiting reagent is completely consumed and no side reactions occur. This is calculated from the balanced equation.
• Actual yield: The mass of product actually obtained in the laboratory. This is always less than or equal to the theoretical yield.
• Percentage yield: The ratio of actual to theoretical yield, expressed as a percentage.

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

Why is the Actual Yield Always Less?

• Some product is lost during transfer, filtration, or purification.
• Side reactions produce unwanted by-products.
• The reaction may not go to completion (equilibrium is established).
• The limiting reagent may not be completely pure.

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6. Atom Economy (R2.1)

Atom economy is a measure of how efficiently atoms in the reactants end up in the desired product. It was introduced by Barry Trost as a principle of green chemistry — maximising atom economy minimises waste.

$\% \text{ atom economy} = \frac{\text{molar mass of desired product(s)}}{\text{total molar mass of all products}} \times 100\%$

Equivalently (since mass is conserved):

$\% \text{ atom economy} = \frac{\text{molar mass of desired product(s)}}{\text{total molar mass of all reactants}} \times 100\%$

Why Atom Economy Matters

• High atom economy → less waste → more sustainable process → lower cost of raw materials.
• A reaction can have a high percentage yield but low atom economy if many by-products are formed.
• Addition reactions always have 100% atom economy (all atoms in the reactants end up in the one product).
• Substitution reactions typically have lower atom economy (a leaving group is produced as waste).

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7. Allotropes of Carbon (S2.4)

An allotrope is one of two or more physically distinct forms of the same element in the same physical state, differing in the way atoms are bonded or arranged. Carbon has four well-known allotropes you must understand for IB SL.

Diamond

• Structure: Each carbon atom is bonded to four other carbon atoms by strong $\sigma$ (single) covalent bonds in a tetrahedral arrangement, forming a giant covalent (macromolecular) network lattice.
• Hybridisation: $sp^3$
• Properties:
  • Hardest natural substance — the rigid 3D network of covalent bonds is extremely difficult to break.
  • Does not conduct electricity — all four valence electrons are used in C-C bonds; there are no free (delocalised) electrons.
  • Very high melting point ($> 3500$°C) — enormous number of strong covalent bonds must be broken.
  • Transparent and colourless.

Graphite

• Structure: Each carbon atom is bonded to three other carbon atoms in flat hexagonal layers (planar sheets). The layers are held together by weak London dispersion (van der Waals) forces.
• Hybridisation: $sp^2$; the remaining $p$ orbital on each carbon is perpendicular to the plane and overlaps with adjacent $p$ orbitals to form a delocalised $\pi$ electron system across each layer.
• Properties:
  • Conducts electricity — the delocalised $\pi$ electrons are free to move along the layers; graphite is unusual among non-metals in being a conductor.
  • Soft and slippery — layers can slide past each other because only weak London dispersion forces hold the layers together; used as a lubricant and in pencil “leads.”
  • High melting point — strong covalent bonds within each layer require a lot of energy to break; but layers can be separated relatively easily.

Graphene

• Structure: A single layer of graphite — one atom thick, arranged in a hexagonal honeycomb lattice. Graphene can be thought of as the building block of graphite.
• Hybridisation: $sp^2$ (same as graphite)
• Properties:
  • Exceptional electrical and thermal conductor — delocalised electrons can move across the entire 2D sheet.
  • Strongest material ever tested (per unit area) — due to the extensive covalent network.
  • Transparent, flexible, and nearly massless per unit area.
  • Applications: flexible electronics, sensors, reinforcement in composites.

Fullerenes (Buckminsterfullerene — $\text{C}_{60}$)

• Structure: Carbon atoms form a closed cage of hexagonal and pentagonal rings. $\text{C}_{60}$ (buckminsterfullerene, “buckyball”) has 60 carbon atoms arranged like a soccer ball: 20 hexagons and 12 pentagons.
• Hybridisation: Approximately $sp^2$; the molecule is nearly spherical.
• Properties:
  • Molecular (not giant covalent) — discrete $\text{C}_{60}$ molecules held together by London dispersion forces in the solid.
  • Relatively low melting point (sublimes around 600°C) compared to diamond or graphite.
  • Semiconducting properties; can become superconducting when doped with alkali metals.
  • Can encapsulate other atoms or molecules inside the cage (“endohedral fullerenes”) — applications in drug delivery.

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8. Polymers — Addition and Condensation (S2.4)

Polymers are giant molecules formed by linking many small repeating units called monomers together. There are two main types of polymerisation reaction at IB SL.

Addition Polymerisation

In addition polymerisation, unsaturated monomers (containing C=C double bonds) join together without losing any atoms. All atoms from the monomers end up in the polymer — atom economy is 100%.

General mechanism: The C=C double bond in each monomer opens up; each monomer unit joins to the next.

Example — poly(ethene) from ethene:

$n\,\text{CH}_2\text{=CH}_2 \rightarrow \text{-(-CH}_2\text{-CH}_2\text{-)}_n\text{-}$

Example — poly(chloroethene) (PVC) from chloroethene:

$n\,\text{CH}_2\text{=CHCl} \rightarrow \text{-(-CH}_2\text{-CHCl-)}_n\text{-}$

Condensation Polymerisation

In condensation polymerisation, monomers join together with the simultaneous loss of a small molecule — usually water ($\text{H}_2\text{O}$) or hydrogen chloride ($\text{HCl}$). This means atom economy is less than 100%.

Two categories:

Polyesters — formed from a diol (two $\text{-OH}$ groups) and a dicarboxylic acid (two $\text{-COOH}$ groups). An ester linkage ($\text{-COO-}$) forms each time, with loss of water.

Example — making Terylene (PET, polyethylene terephthalate) from ethane-1,2-diol and benzene-1,4-dicarboxylic acid:

$\text{HOCH}_2\text{CH}_2\text{OH} + \text{HOOC-C}_6\text{H}_4\text{-COOH} \rightarrow \text{-O-CH}_2\text{CH}_2\text{-OOC-C}_6\text{H}_4\text{-CO-} + n\,\text{H}_2\text{O}$

Polyamides — formed from a diamine (two $\text{-NH}_2$ groups) and a dicarboxylic acid (two $\text{-COOH}$ groups). An amide linkage ($\text{-CO-NH-}$) forms, with loss of water.

Example — Nylon-6,6 from 1,6-diaminohexane and hexanedioic acid:

$\text{H}_2\text{N-(CH}_2)_6\text{-NH}_2 + \text{HOOC-(CH}_2)_4\text{-COOH} \rightarrow \text{-NH-(CH}_2)_6\text{-NH-CO-(CH}_2)_4\text{-CO-} + n\,\text{H}_2\text{O}$

Environmental Considerations

• Many addition polymers (e.g. polyethene, PVC) are non-biodegradable — they persist in the environment for centuries.
• Condensation polymers (e.g. polyesters, polyamides) can sometimes be hydrolysed — broken down by water (and acid or base) — making some more degradable.
• Biodegradable polymers (e.g. polylactic acid, PLA) are increasingly being developed; they contain ester linkages that hydrolyse in the environment.

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9. Semiconductors and Superconductors (S2.4)

Semiconductors

Metals conduct electricity because their valence electrons are delocalised and free to move. Insulators (like diamond) have all electrons in bonding orbitals with no free electrons and a large energy gap to the conduction band.

Semiconductors have an intermediate situation: at low temperatures they behave as insulators, but at higher temperatures (or when doped), electrons gain enough energy to move into the conduction band.

The two most important semiconductors are silicon ($\text{Si}$) and germanium ($\text{Ge}$), both in Group 14. Like diamond, they have giant covalent structures with each atom bonded to four others — but the energy gap to the conduction band is smaller.

Doping — adding tiny amounts of impurity elements — dramatically changes the electrical properties:

Dopant type	Element added	Effect
n-type	Group 15 element (e.g. P, As) — 5 valence electrons	Extra electron (negative charge carrier) per dopant atom
p-type	Group 13 element (e.g. B, Al) — 3 valence electrons	Creates a “hole” (positive charge carrier) per dopant atom

The junction between n-type and p-type semiconductors forms a p-n junction diode, which is the basis of transistors, LEDs, and solar cells.

Superconductors

A superconductor is a material that conducts electricity with zero resistance below a critical temperature ($T_c$, the critical temperature). Below $T_c$, electrical current can flow indefinitely without any energy loss.

Key properties of superconductors:

• Zero electrical resistance below $T_c$.
• The Meissner effect — a superconductor expels magnetic fields from its interior below $T_c$ (perfect diamagnetism). This enables magnetic levitation (maglev trains).
• Classical superconductors: metals like Hg, Pb, Nb at temperatures close to absolute zero (typically below 20 K).
• High-temperature superconductors (HTS): certain metal oxide ceramics (e.g. yttrium barium copper oxide, YBCO) become superconducting at higher temperatures — some above 130 K ($-143$°C), though still very cold by everyday standards.

Applications:

• MRI machines (magnetic resonance imaging) — superconducting magnets create intense, stable magnetic fields.
• Maglev trains — magnetic levitation using superconducting magnets eliminates friction.
• Particle accelerators (e.g. the LHC at CERN) — superconducting electromagnets steer particle beams.
• Power transmission — loss-free electrical cables (under development).

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Practice Questions

Test your understanding with these IB-style questions.

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Question 1

In the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$, 4.20 g of $\text{N}_2$ is reacted with excess $\text{H}_2$. What mass of $\text{NH}_3$ is produced? ($M(\text{N}) = 14.01$, $M(\text{H}) = 1.008$)

A. 5.10 g

B. 8.50 g

C. 2.55 g

D. 17.0 g

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Question 2

Which allotrope of carbon is a discrete molecule (rather than a giant covalent structure)?

A. Diamond

B. Graphite

C. Graphene

D. $\text{C}_{60}$ (buckminsterfullerene)

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Question 3

A reaction produces 8.20 g of product. The theoretical yield, calculated from stoichiometry, is 12.50 g. What is the percentage yield?

A. 34.1%

B. 52.4%

C. 65.6%

D. 84.0%

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Question 4

Which of the following is the best description of a superconductor?

A. A material with very low electrical resistance at all temperatures.

B. A material with zero electrical resistance below a critical temperature.

C. A semiconductor that has been doped with a Group 15 element.

D. A polymer that can conduct electricity along its backbone.

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Question 5

Poly(lactic acid) (PLA) is a biodegradable polymer formed from lactic acid monomers ($\text{CH}_3\text{CH(OH)COOH}$) with loss of water. What type of polymer is PLA?

A. Addition polymer

B. Condensation polymer — polyamide

C. Condensation polymer — polyester

D. Addition polymer formed from an alkene monomer

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Show Answers

1. A — Step 1: $n(\text{N}_2) = 4.20 \div 28.02 = 0.1499\, \text{mol}$. Step 2: Mole ratio $\text{N}_2 : \text{NH}_3 = 1 : 2$, so $n(\text{NH}_3) = 0.1499 \times 2 = 0.2998\, \text{mol}$. Step 3: $M(\text{NH}_3) = 14.01 + 3(1.008) = 17.03\, \text{g mol}^{-1}$; $m = 0.2998 \times 17.03 = 5.11 \approx 5.10\, \text{g}$. B would result from using the ratio 1 : 3 instead of 1 : 2. C would result from using the ratio 2 : 1. D is the molar mass of $\text{NH}_3$, not the mass produced.

2. D — $\text{C}_{60}$ (buckminsterfullerene) consists of discrete closed-cage molecules held together in the solid by London dispersion forces. Diamond (A), graphite (B), and graphene (C) are all giant covalent/macromolecular structures (graphene being a single layer of an infinite 2D covalent network). The molecular nature of $\text{C}_{60}$ is why it has a relatively low melting/sublimation point compared to the other carbon allotropes.

3. C — $\% \text{ yield} = \dfrac{8.20}{12.50} \times 100 = 65.6\%$. A is obtained by dividing incorrectly (12.50 - 8.20 = 4.30; 4.30/12.50 = 34.4%). B results from inverting the fraction ($8.20/12.50 \neq 52.4\%$ — check your arithmetic). D exceeds the calculated value and corresponds to no straightforward error.

4. B — A superconductor is defined by exactly zero electrical resistance below its critical temperature $T_c$. A is incorrect because the zero-resistance state only occurs below $T_c$, not at all temperatures. C describes an n-type semiconductor, a completely different material. D describes a conducting polymer (e.g. polyacetylene), which is unrelated to superconductivity.

5. C — Lactic acid ($\text{CH}_3\text{CH(OH)COOH}$) contains both an alcohol ($\text{-OH}$) and a carboxylic acid ($\text{-COOH}$) functional group. When lactic acid monomers link together with loss of water, an ester linkage ($\text{-COO-}$) forms — this is a polyester, making PLA a condensation polymer — polyester. It is not an addition polymer (no C=C bond is present). It is not a polyamide (no amine group is involved).

Next Topics

• Classification of Matter (Structure 3) — periodic trends and functional group classification complement the material coverage here.
• Rates & Equilibrium (Reactivity 2.2–2.3) — once you can calculate moles and yields, the next step is understanding what controls how fast a reaction proceeds and what determines its equilibrium position.