How much? The amount of chemical change

An equation with the **same number of each kind of atom** on both sides — atoms are conserved.

The study of the **whole-number ratios** in which substances react and are formed, read from a balanced equation.

The ratio of the **coefficients** in a balanced equation — how many moles of one substance react with or form another.

Only the **coefficients** (the big numbers in front) — **never** a subscript inside a formula.

Changing a subscript changes the **substance** itself (e.g. H_{2}O → H_{2}O_{2}), so it no longer describes the same reaction.

**(s)** solid, **(l)** pure liquid, **(g)** gas, **(aq)** aqueous (dissolved in water).

**Aqueous** — the substance is **dissolved in water** (different from a pure liquid, (l)).

The ratio N_{2} : H_{2} : NH_{3} is **1 : 3 : 2** — 1 mol N_{2} reacts with 3 mol H_{2} to make 2 mol NH_{3}.

Balance **C first, then H, then O last** (oxygen appears in more than one product), then reduce to smallest whole numbers.

**CH_{4} + 2 O_{2} → CO_{2} + 2 H_{2}O** — smallest whole-number coefficients.

The C : CO_{2} ratio is 1 : 1, so **0.5 mol** of CO_{2}.

Changing a **formula** (subscript) instead of a coefficient, or forgetting the **state symbols** when asked.

The reactant that **runs out first** — it controls (limits) the amount of product that can form.

The reactant **left over** once the limiting reactant has been used up.

The **maximum** amount (or mass) of product, calculated from the **limiting** reactant.

Convert each reactant mass to **moles**, divide each by its **coefficient**, and the **smallest** result is limiting.

It compares the reactants fairly against the **mole ratio** in the balanced equation, so you can see which runs out first.

Always the **limiting** reactant — never the one in excess.

$n = \dfrac{m}{M}$ — convert every mass to moles before using the mole ratio.

Balanced equation → mass to **moles** (n = m/M) → scale by the **mole ratio** → moles back to **mass** (m = nM).

From the **coefficients** of the balanced equation (e.g. N_{2} + 3H_{2} → 2NH_{3} is 1 : 3 : 2).

Working out the product from the reactant in **excess**, or forgetting the **mole ratio** when coefficients are not 1 : 1.

**A** (0.3 mol runs out first); B is in excess by 0.2 mol.

$\%\text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$ — how much product you actually obtained versus the maximum predicted by the equation.

The amount of product predicted from the balanced equation if the **limiting reactant** reacted completely.

The amount of product you really obtain — always **less** than theoretical, due to side reactions, reversible reactions and losses.

Side reactions, reversible reactions not going to completion, and losses during separation/purification.

$\%\text{ AE} = \dfrac{M(\text{desired product})}{M(\text{all reactants})} \times 100$ — the fraction of reactant atoms ending up in the wanted product.

Yield = **how much product you made**; atom economy = **how little reactant mass you wasted** as by-products. They are independent.

**Addition** reactions — all reactants combine into a single product, so there are no by-products.

Sum the molar masses of **all** reactants, each multiplied by its **coefficient** in the balanced equation.

Fewer atoms wasted as by-products → less raw material used and less waste to treat → more **sustainable and economical**.

No — actual yield cannot beat the theoretical maximum. A value over 100% signals an error (e.g. impure/wet product).

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

Putting only **one** reactant (or forgetting coefficients) on the bottom — you must sum **every** reactant's molar mass.

At the **same temperature and pressure**, equal **volumes** of gases contain equal numbers of **moles** — so the volume ratio equals the coefficient ratio.

Because at fixed T and P volume is **proportional to amount**, so the balanced **coefficients** give the **volume ratio** — no moles needed.

**22.7 dm³ mol⁻¹** at STP (273 K, 100 kPa) — given in the data booklet.

$n = \dfrac{V}{V_{m}}$ with $V_{m} = 22.7$ dm³ mol⁻¹ (volume in dm³).

Multiply the amount by the molar volume: $V = n\,V_{m} = n \times 22.7$ dm³.

**1000 cm³** — divide a cm³ value by 1000 before using the molar volume 22.7 dm³ mol⁻¹.

**2 volumes** of NH_{3} (the volume ratio matches the 1 : 3 : 2 coefficients).

Subtract the volume that **reacted** (from the coefficient ratio) from the volume **supplied**.

**No** — only **gases** contribute; liquids and solids (like condensed water) add zero volume.

Forgetting to **subtract the gas that reacted** when asked for the volume remaining, or counting **liquid** products as gas.

**273 K and 100 kPa** (standard temperature and pressure).

A precise technique to find an **unknown concentration** by reacting it with a **standard solution** to the **end point** (an indicator colour change).

A solution of **precisely known concentration**, made up in a **volumetric flask**.

Delivers a **fixed, exact** volume of the solution being analysed (e.g. 25.0 cm³).

Delivers the **variable** volume of titrant (the **titre**), read to ±0.05 cm³.

$n = CV$ — amount (mol) = concentration (mol dm⁻³) × volume (**dm³**). Given in the data booklet.

**(1)** n = CV on the known reagent → mol. **(2)** Cross by the **mole ratio**. **(3)** C = n/V (or M = m/n) on the unknown.

The volume must be in **dm³** — divide a cm³ titre by **1000** first.

Titres that **agree** (typically within 0.10 cm³). Only the concordant titres are **averaged** — a rough trial is ignored.

Add a **known excess** of a reagent, let it react, then titrate the **leftover** excess. Amount reacted = **added − leftover**.

When the reaction is **slow** or the sample is an **insoluble solid** (e.g. a carbonate), making a direct titration impractical.

**2 : 1** — sulfuric acid is diprotic, so it needs **two** moles of NaOH per mole of acid.

Forgetting the **mole ratio** from the balanced equation, or leaving a volume in **cm³** instead of dm³.