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Edexcel GCSE Chemistry (1CH0) · Topic 5 — Separate chemistry 1
Mini-Lesson

Separate chemistry 1

This mini-lesson walks you through the whole of Edexcel Topic 5 — Separate chemistry 1: transition metals, corrosion & alloys, titrations & concentration, the molar gas volume, percentage yield and atom economy, empirical formulae, dynamic equilibrium (the Haber process) and chemical & fuel cells.

transition metals fuel cells 11 spec areas Separate / Triple only

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Screens marked HT are Higher-tier only. Press Start when you're ready.

5.1 — Transition metals

The block in the middle

Most metals are transition metals — they sit in the central block of the periodic table, between Group 2 and Group 3. Their typical properties are:

  • High melting points and high densities (iron melts at ~1538 °C).
  • They form coloured compounds (e.g. blue Cu²⁺, green Fe²⁺, orange Fe³⁺).
  • They (and their compounds) are useful as catalysts (e.g. iron in the Haber process).
  • They have variable oxidation states, forming more than one ion (e.g. Fe²⁺ and Fe³⁺).
G1 G2 transition metals Sc Ti V Cr Mn Fe Co Ni Cu Zn … G3 G0 between Group 2 and Group 3 soft, reactive hard, dense, less reactive non-metals
Transition metals fill the central block. Iron is the spec's named example of a catalyst.
5.1 — vs Group 1

Compared with Group 1

A favourite exam comparison: transition metals against the Group 1 alkali metals.

Transition metals • high melting points / dense • coloured compounds • variable oxidation states • good catalysts • react slowly / not at all w/ water Group 1 metals • low melting points / soft • white compounds • always form +1 ions • not used as catalysts • react vigorously with water
Transition metals are harder, denser, higher-melting and far less reactive than Group 1.

Watch out: a common error is to call transition metals "more reactive" because they're famous. They are actually less reactive than Group 1 — sodium and potassium fizz and burn on water; iron and copper do not.

Quick check

Spot the transition-metal property

?Which property is typical of a transition metal but not of a Group 1 metal?
5.2–5.3 — Corrosion

Rusting needs air AND water

Corrosion is the oxidation of a metal at its surface. For iron this is rusting, which forms hydrated iron(III) oxide. Rusting needs BOTH oxygen and water:

iron + oxygen + water → hydrated iron(III) oxidethe classic experiment: nail in water + air rusts; nails sealed from air OR from water stay shiny
water + air RUSTS water, no air no rust dry air, no water no rust Only tube 1 rusts → both oxygen and water are needed
Remove the air (oil layer on boiled water) OR the water (drying agent) and the iron will not rust.
5.3 — Preventing rust

Barriers and sacrifice

Edexcel lists two ideas: keep oxygen/water out, or use a more reactive metal to corrode instead.

  • Barrier methods — paint, oil/grease, or plastic coating exclude oxygen and water.
  • Galvanising — coating iron with zinc. Zinc is a barrier and, because it is more reactive, it gives sacrificial protection.
  • Sacrificial protection — attach blocks of a more reactive metal (zinc or magnesium). It oxidises in preference to the iron, so the iron is protected even if scratched.
sea water iron / steel hull Zn / Mg more-reactive block corrodes first the zinc/magnesium is "sacrificed" so the iron survives
The sacrificial metal must be more reactive than iron — that's the whole point.

Watch out: the sacrificial metal is the more reactive one, not the less reactive one. A less reactive coating (like tin on a tin can) protects only as a barrier — scratch it and the iron rusts faster.

Quick check

Protecting a steel pipe

?A buried steel pipe is protected by bolting on blocks of metal that corrode instead of it. For this sacrificial protection to work, the blocks must be made of a metal that is…
5.5–5.7 — Alloys

Why alloys are harder

An alloy is a mixture of a metal with other elements. Steels are alloys of iron with carbon (and sometimes other metals). Alloys are usually harder than the pure metal.

pure metal alloy neat rows slide easily → softer different-size atoms distort rows → harder
Different-sized atoms disrupt the regular layers, so they can no longer slide over each other — the alloy is harder.

Iron is alloyed to make alloy steels (e.g. adding chromium and nickel makes stainless steel, which resists corrosion). Other metals are matched to uses by their properties: aluminium (low density), copper (conducts), gold (unreactive); alloys include brass (copper + zinc) and magnalium (aluminium + magnesium).

5.8 — Concentration · HT only

Concentration of a solution

Concentration says how much solute is dissolved per dm³ (1 dm³ = 1000 cm³ = 1 litre). You meet two units:

conc (mol/dm³) = moles ÷ volume (dm³)and conc (g/dm³) = conc (mol/dm³) × molar mass (g/mol)

To convert g/dm³ → mol/dm³, divide by the molar mass (Mr in g/mol). To go back the other way, multiply.

Worked example

0.50 mol of NaOH is dissolved to make 0.25 dm³ of solution.

conc = 0.50 ÷ 0.25 = 2.0 mol/dm³

In g/dm³: 2.0 × 40 = 80 g/dm³  (Mr of NaOH = 23+16+1 = 40)

Calculate · HT only

Your turn — concentration

10.40 mol of sodium chloride is dissolved to make 2.0 dm³ of solution. Calculate the concentration in mol/dm³.
mol/dm³
Hint: conc = moles ÷ volume = 0.40 ÷ 2.0.
5.9 — Core practical

The titration

A titration finds an unknown concentration accurately, e.g. a strong acid against a strong alkali. You measure a fixed volume of one solution with a pipette into a conical flask, add an indicator, then run in the other from a burette until the colour just changes (the end-point).

burette acid (known conc) tap conical flask alkali + indicator (pipette) repeat until concordant titres within 0.10 cm³
Repeat until two or more concordant titres (within 0.10 cm³) are obtained, then average only those.

Watch out: the mean titre uses only the concordant results — discard the rough trial and any anomalies. Never just average every reading.

5.10 — Titration calc · HT only

Titration calculations

Use the balanced equation. For a strong acid–strong alkali like HCl + NaOH → NaCl + H₂O the ratio is 1 : 1.

moles = conc (mol/dm³) × volume (dm³)find moles of the known solution → use the ratio → divide by the other volume
Worked example

25.0 cm³ of 0.100 mol/dm³ NaOH is exactly neutralised by 20.0 cm³ of HCl. Find the concentration of the HCl.

moles NaOH = 0.100 × (25.0/1000) = 0.00250 mol

ratio 1:1, so moles HCl = 0.00250 mol

conc HCl = 0.00250 ÷ (20.0/1000) = 0.125 mol/dm³

Calculate · HT only

Your turn — titration

2In HCl + NaOH → NaCl + H₂O, 25.0 cm³ of 0.200 mol/dm³ NaOH is neutralised by 20.0 cm³ of HCl. Calculate the concentration of the HCl in mol/dm³.
mol/dm³
Hint: moles NaOH = 0.200 × 25.0/1000 = 0.00500; ratio 1:1; ÷ (20.0/1000).
5.11–5.12 — Percentage yield

Percentage yield

The percentage yield compares what you actually made with the theoretical maximum from the equation:

% yield = (actual yield ÷ theoretical yield) × 100actual = mass really obtained · theoretical = mass the equation predicts

The actual yield is usually less than the theoretical because of:

  • Incomplete reactions (some reactant is left over).
  • Practical losses when transferring or filtering.
  • Competing (side) reactions making unwanted products.
Worked example

A reaction should make 8.0 g of product but only 6.0 g is obtained.

% yield = (6.0 ÷ 8.0) × 100 = 75%

Calculate

Your turn — percentage yield

3A reaction has a theoretical yield of 50 g but the actual yield is 36 g. Calculate the percentage yield.
%
Hint: (36 ÷ 50) × 100.
5.13–5.14 — Atom economy

Atom economy

The atom economy measures how much of the reactant mass ends up as the useful product (the rest is wasted by-products):

atom economy = (Mr of desired product ÷ total Mr of all products) × 100a high atom economy means less waste and a more sustainable, efficient process
Worked example

CaCO₃ → CaO + CO₂, making CaO.  (Mr: CaO = 56, CO₂ = 44)

atom economy = (56 ÷ (56 + 44)) × 100 = (56 ÷ 100) × 100 = 56%

% yield vs atom economy: they are different. Percentage yield is about how much you actually got in the lab; atom economy is about how much of the atoms in your equation are useful — even a 100%-yield reaction can have a poor atom economy.

Calculate

Your turn — atom economy

4Iron is produced in the blast furnace by the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂. The desired product is Fe. Total Mr of products: 2Fe = 112, 3CO₂ = 132. Calculate the atom economy for making iron.
%
Hint: (112 ÷ (112 + 132)) × 100 = (112 ÷ 244) × 100.
5.16–5.18 — Molar gas volume · HT only

The molar volume of a gas

At room temperature and pressure (rtp), one mole of any gas occupies the same volume — the molar volume:

1 mol of gas = 24 dm³ (24 000 cm³) at rtpvolume of gas (dm³) = moles × 24  ·  moles = volume ÷ 24

By Avogadro's law, equal volumes of gases (at the same temperature and pressure) contain equal numbers of molecules, so the mole ratio in the equation is also the volume ratio.

Worked example

What volume does 0.50 mol of carbon dioxide occupy at rtp?

volume = 0.50 × 24 = 12 dm³

Calculate · HT only

Your turn — gas volume

5Calculate the volume occupied by 0.25 mol of hydrogen gas at rtp. (Molar volume = 24 dm³/mol.)
dm³
Hint: volume = moles × 24 = 0.25 × 24.
Empirical & molecular formula

Empirical formula from masses

The empirical formula is the simplest whole-number ratio of atoms. From masses (or % composition):

  • Divide each element's mass ÷ its Ar to get moles.
  • Divide each by the smallest to get the simplest ratio.

The molecular formula is a whole-number multiple of the empirical formula (find it by comparing the empirical Mr with the real Mr).

Worked example

4.6 g sodium combines with 1.6 g oxygen.  (Ar: Na = 23, O = 16)

Na: 4.6 ÷ 23 = 0.20 mol  ·  O: 1.6 ÷ 16 = 0.10 mol

divide each by the smallest (0.10): Na = 2, O = 1 → ratio 2 : 1

empirical formula = Na₂O

Calculate

Your turn — empirical formula

?2.4 g of magnesium combines with 1.6 g of oxygen. Using Ar(Mg) = 24 and Ar(O) = 16, what is the empirical formula?
5.19 — Dynamic equilibrium

The Haber process

In a sealed container a reversible reaction reaches dynamic equilibrium: forward and backward reactions still occur, but at equal rates, so concentrations stay constant. The Haber process makes ammonia for fertilisers:

N₂ + 3H₂ ⇌ 2NH₃nitrogen (from air) + hydrogen (from natural gas) ⇌ ammonia · reversible
N₂ + 3H₂ reactants 2NH₃ ammonia forward rate backward rate at equilibrium the two rates are equal

Typical industrial conditions: about 450 °C, 200 atmospheres, with an iron catalyst.

5.20–5.21 — Le Chatelier · HT only

Choosing the conditions

Changing the conditions shifts the equilibrium position (Le Chatelier's idea). For the Haber process the chosen conditions are a compromise between yield, rate and cost:

  • Pressure ~200 atm — high pressure favours the side with fewer gas molecules (the NH₃ side, 4 → 2), giving more ammonia; but very high pressure is expensive and unsafe.
  • Temperature ~450 °C — the forward reaction is exothermic, so a lower temperature gives a higher yield, but too low is far too slow. 450 °C is a compromise giving an acceptable yield at an acceptable rate.
  • Iron catalyst — speeds up equilibrium being reached; it does not change the position/yield.

Watch out: a catalyst speeds up how fast equilibrium is reached but does not change the equilibrium yield. The 450 °C is a compromise, not the temperature that gives the most ammonia.

Quick check

Reading the equilibrium

?In the Haber process, why is an iron catalyst used?
5.25–5.27 — Cells & fuel cells

Chemical cells and fuel cells

A simple chemical cell is two different metal electrodes in an electrolyte. The more reactive metal loses electrons more readily, so a voltage is produced — until one reactant is used up (then it goes flat).

A hydrogen–oxygen fuel cell is fed hydrogen and oxygen continuously and produces a voltage with water as the only product:

2H₂ + O₂ → 2H₂Ohydrogen + oxygen → water · the cell keeps working while fuel is supplied
H₂ in O₂ in anode (−) cathode (+) electrolyte (H⁺ ions move) V e⁻ → H₂O out (only product)
Hydrogen at the (−) electrode, oxygen at the (+); electrons flow round the external circuit doing the work.
5.27 — Evaluate fuel cells

Fuel cells vs rechargeable cells

Strengths • only product is water • no recharging — keeps going while fuel is supplied • no toxic metals to dispose of • lighter than many batteries Weaknesses • hydrogen is hard to store (flammable, high pressure) • H₂ often made from fossil fuels / using electricity • few refuelling stations
Unlike a rechargeable cell, a fuel cell does not store charge — it runs as long as hydrogen and oxygen are supplied.

Compared with rechargeable cells: fuel cells don't run down or need recharging and have no toxic metals, but hydrogen is hard to store/transport and is often produced from fossil fuels.

Quick check

The fuel-cell product

?What is the only product of a hydrogen–oxygen fuel cell?
Recap

Topic 5 at a glance

Transition metals: high mp/density, coloured compounds, variable oxidation states, catalysts (vs soft, +1, reactive Group 1)

Corrosion: rusting needs oxygen AND water; prevent by barriers or sacrificial protection (more reactive metal) / galvanising

Alloys: different-sized atoms disrupt layers → harder; e.g. steels

Concentration: mol/dm³ = moles ÷ volume; × Mr for g/dm³ HT

Titration: burette/pipette/indicator; mean of concordant titres

Gas volume: moles × 24 dm³ at rtp HT

% yield: actual ÷ theoretical × 100  ·  Atom economy: Mr product ÷ total Mr × 100

Empirical formula: mass ÷ Ar, then ÷ smallest

Equilibrium: Haber N₂ + 3H₂ ⇌ 2NH₃; conditions are a compromise HT

Cells: chemical cell gives a voltage until a reactant is used up; H₂–O₂ fuel cell → water only

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