Standard form
Chemistry runs on numbers that are either enormous or tiny: one mole is 602 000 000 000 000 000 000 000 particles, and a bond length is around 0.000 000 000 15 m. Writing those out is hopeless — so chemists write every number as A × 10n, where A is between 1 and 10. Big numbers get a positive power; numbers smaller than 1 get a negative power.
Worked example — write 0.00082 in standard form
- Move the decimal point until one non-zero digit sits in front of it: 0.00082 → 8.2 (four places to the right).
- Count the moves for the power: four places, and the number is smaller than 1, so the power is negative: 10−4.
0.00082 = 8.2 × 10−4
Which is 47 500 in standard form? Tick (✓) one box.
(2.0 × 104) × (3.0 × 10−2) = ? Give your answer as an ordinary number.
Worked steps
- Multiply the numbers: 2.0 × 3.0 = 6.0
- Add the powers: 104 × 10−2 = 102
- 6.0 × 102 = 600
6.02 × 1023 ÷ 2.0 × 1022 — roughly how big is the answer? Tick (✓) one box.
Write 0.0000715 in standard form: give the POWER only.
Worked steps
- Move the decimal 5 places right to get 7.15
- 0.0000715 = 7.15 × 10−5
Significant figures
Significant figures start at the first non-zero digit: the leading zeros in 0.00408 are placeholders, not precision. But trailing zeros after a decimal point do count — writing 2.50 g instead of 2.5 g is a claim that you measured to the nearest 0.01 g.
Then there's the rule that actually earns marks. At A-Level, calculation answers are quoted to the same number of significant figures as the least precise data value in the question. If the question gives you 25.00 (4 s.f.) and 0.010 (2 s.f.), your answer gets 2 s.f. — that's what mark schemes will expect.
Keep every digit on the calculator between steps and round once, at the final answer. Rounding mid-calculation is one of the quietest ways to drift off the mark scheme's accepted range.
Worked example — how many s.f. should the answer get?
6125 × 384 = 2 352 000
- Check the data: 6125 is 4 s.f.; 384 is 3 s.f. — the least precise is 3 s.f.
- Quote to 3 s.f.: 2.35 × 106
How many significant figures in 0.004080? Tick (✓) one box.
Round 6.0961 to 3 s.f. Tick (✓) one box.
25.00 × 0.010 = ? Quote to the appropriate number of s.f.
Worked steps
- 25.00 × 0.010 = 0.25
- 0.010 has 2 s.f. → quote 0.25
13.6 ÷ 4.0 = ? Appropriate s.f.
Worked steps
- = 3.4 exactly
- 4.0 is 2 s.f. → 3.4
Units & prefixes
SI prefixes are the same in every science, and chemistry leans on a small set of them. Two conversions dominate everything you'll do in the first term: cm3 → dm3 is ÷ 1000, and dm3 → m3 is ÷ 1000 again — plus the mass ladder mg → g → kg. A useful sanity check: converting to a smaller unit needs a bigger number, and vice versa.
| Prefix | Symbol | Meaning |
|---|---|---|
| giga | G | × 109 |
| mega | M | × 106 |
| kilo | k | × 103 |
| centi | c | × 10−2 |
| milli | m | × 10−3 |
| micro | µ | × 10−6 |
| nano | n | × 10−9 |
cm3 → ÷1000 → dm3 → ÷1000 → m3
Worked example — convert 45 600 J to kJ
- k means ×1000, so going J → kJ divides by 1000.
45 600 ÷ 1000 = 45.6 kJ
Convert 3.5 kg to g.
Convert 1.5 dm3 to cm3.
A pipette measures 25.0 cm3. Express this in dm3.
Enthalpy changes are quoted in kJ mol−1. 5 200 J mol−1 = ? Tick (✓) one box.
Which is the same as 1 nm? Tick (✓) one box.
Rearranging equations
Formula triangles are not recommended at A Level. They work when an equation has exactly three quantities — but in the first weeks of Year 12 you'll meet PV = nRT, which has five. No triangle can help you there.
You will need to rearrange equations as taught in GCSE Maths: do the same thing to both sides. If the quantity you want is multiplied by something, divide both sides by it; if it's divided, multiply.
It's the same skill, on the same equations, you used all through GCSE quantitative chemistry — n = m ÷ Mr, c = n ÷ V and the rest — and you'll be rearranging them in every A-Level calculation from here.
Worked example — make Mr the subject of n = m ÷ Mr
- Multiply both sides by Mr (it's currently dividing):
n × Mr = m
- Divide both sides by n (it's multiplying the Mr we want):
Mr = mn
Make n the subject of PV = nRT. Tick (✓) one box.
q = mcΔT. Make ΔT the subject. Tick (✓) one box.
c = n ÷ V. Find V when n = 0.050 mol and c = 0.20 mol/dm3.
Worked steps
- V = n ÷ c
- 0.050 ÷ 0.20
- = 0.25 dm3 (= 250 cm3)
BODMAS on a calculator
Your calculator follows the order of operations with no mercy: multiplication and division before addition and subtraction, brackets first. Type an Mr calculation carelessly and the calculator cheerfully answers a different question. The fix is calculator brackets — especially around anything you're dividing by.
Worked example — Mr of Ca(OH)2 on a calculator
- Bracket first: (16.0 + 1.0) = 17.0 — one OH unit.
- Then the sum:
40.1 + 2 × (16.0 + 1.0)
The ×2 hits the whole bracket, not just the O. - = 40.1 + 34.0 = 74.1
Without a calculator: 4 + 6 × 5 = ?
Mr of Ca(OH)2. Ar: Ca 40.1, O 16.0, H 1.0.
Worked steps
- (16.0 + 1.0) = 17.0
- 40.1 + 2 × 17.0
- = 74.1
Which keystrokes give 6.0 ÷ (2.0 × 1.5) correctly? Tick (✓) one box.
How did you do?
Answer the questions above and your score appears here.
Next up: writing chemical formulae from memory — the skill your first A-Level lessons will quietly assume you have.
That's fixable before September — I take on a small number of A-Level students each year →