Chemistry Speed Tricks

0. Choosing a method

1. Conservation: skip the intermediate steps

• Element conservation: the number of atoms of a given element does not change.
• Electron conservation: electrons lost = electrons gained.
• Charge conservation: total positive charge = total negative charge in solution.
• Mass conservation: mass in = mass out.

Example 1: mass conservation as a check

8 g of CH₄ burns completely to give 22 g CO₂ and 18 g H₂O. How much O₂ was consumed? By mass conservation: 22 + 18 − 8 = 32 g — with no equation balancing at all.

Example 2: charge conservation

A solution contains Na⁺ 0.1 mol, Mg²⁺ 0.2 mol and Cl⁻ 0.3 mol; the rest is SO₄²⁻. Positive charge = 0.1 + 0.4 = 0.5; chloride supplies 0.3, so sulfate must supply 0.2 → n(SO₄²⁻) = 0.1 mol.

Example 3: electron conservation

2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻: each Fe²⁺ loses one electron, each Cl₂ gains two, so 0.1 mol Fe²⁺ consumes 0.05 mol Cl₂.

2. Difference method: treat the change as a new substance

When the question only gives a mass difference before and after the reaction, read that difference as a substance in its own right — its "molar mass" is the difference between the two sides of the equation.

Example 1: iron displacing copper

Fe + CuSO₄ → FeSO₄ + Cu. Every 56 g of Fe becomes 64 g of Cu, so the solid gains 8 g per mole. Put 5.6 g of Fe into excess CuSO₄: 5.6 g = 0.1 mol → the solid gains 0.1 × 8 = 0.8 g (final mass 6.4 g).

Example 2: decomposing calcium carbonate

CaCO₃ → CaO + CO₂↑. Per 100 g of CaCO₃ the solid loses 44 g, which is exactly the mass of CO₂ released — so any solid loss can be read directly as CO₂.

3. Average-value method: the mixture lies between its components

Any average over a mixture (average molar mass, average atomic mass, average mass fraction) must fall between the values of the components. That single fact eliminates options faster than any calculation.

Example 1: proving both components are present

A gas mixture has an average molar mass of 36 g/mol and contains only CO (28) and CO₂ (44). Since 36 lies between 28 and 44, both gases must be present.

Example 2: cross-multiplication by hand

The average 36 is 8 away from 28 and 8 away from 44 → the ratio is 8 : 8 = 1 : 1. When the two gaps are equal, the two components are present in equal amounts — a rule worth memorising.

4. Unified mole method: one currency for everything

Convert grams, litres and concentrations into moles once, then work entirely in moles and convert back at the very end. Never mix units inside the same calculation.

Example 1: metal plus acid

Fe + 2HCl → FeCl₂ + H₂↑. 5.6 g of Fe = 0.1 mol → 0.1 mol H₂ → 0.1 × 22.4 = 2.24 L at STP.

Example 2: preparing and diluting solutions

250 mL of 0.2 mol/L NaOH: n = 0.25 × 0.2 = 0.05 mol → 0.05 × 40 = 2 g. Diluting 100 mL of 1 mol/L to 500 mL: 100 × 1 = 500 × c → c = 0.2 mol/L.

5. Practice

1. 8 g CH₄ burnt completely: how much O₂ is consumed?
2. Na⁺ 0.1 mol, Mg²⁺ 0.2 mol, Cl⁻ 0.3 mol: how much SO₄²⁻ is present?
3. 5.6 g Fe dropped into excess CuSO₄: how much does the solid gain?
4. Average molar mass 36 g/mol from CO and CO₂: what is the ratio?
5. 5.6 g Fe plus excess HCl: what volume of H₂ at STP? And how many grams of NaOH in 250 mL of 0.2 mol/L solution?

6. Speed habits that work in every subject

• Units first: carry units through the algebra; if the units are wrong the numbers do not matter.
• Estimate the magnitude: predicting the order of magnitude filters out about 90% of place-value errors.
• Always check: verify with the DS digital root — two seconds per question.
• Ten minutes a day: skills decay. Speed, like fitness, is a frequency game.

All four subject areas are now complete. Pick a mode in the practice centre, or take a timed test to see where you stand.

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