IB Chemistry SL Topic 7 — Mathematics Paper 1 & 2 Core skill ~12 min read

Applying General Mathematics

Chemistry has four different percentages, and they all look identical until you notice what each one divides by. Getting the denominator wrong is the most common way to lose a mark you had already earned.

📚 What you need to know

The maths chemistry actually uses

The list is short, and every item shows up in a specific place. Knowing where means you recognise what a question is asking before you read the numbers.

Type of mathsWhere it turns up
Decimalsalmost everything — concentrations, masses, volumes
Fractionsuncertainty work; convert to a decimal before writing the final answer
Percentagesyield, atom economy, change, difference, error, uncertainty
Ratiosmole calculations where the equation is not 1 : 1
Reciprocalsgas laws (1 / V) and rate against 1 / T
LogarithmspH, and the Arrhenius equation
Section 1 of the data booklet gives you many of the equations, but not all of them. Percentage yield is a good example — it is examined and it is not printed, so it has to be learned. And a fraction left as a fraction in a final answer normally loses the mark, so find the decimal conversion button on your calculator now rather than in the exam.

Mean and range

The mean is straightforward arithmetic. The judgement is in deciding what goes into it. In a titration, the rough titre is deliberately an overshoot, and any titre that is not concordant with the others is treated as anomalous. Neither belongs in the average.

WHAT GOES INTO THE AVERAGE, AND WHAT DOES NOTfour titres from the same titrationmean = 23.90range = 0.10rough, discarded23.8024.0024.2024.40titre / cm³the rough result is excluded before the mean and the range are takenan anomaly left in distorts both, and the range most of all
Include the rough titre and the mean shifts to 24.01 — but the range jumps from 0.10 to 0.50, a fivefold exaggeration of how spread out the data really is.

The two mean calculations that come up most often in chemistry are the average titre, and the relative atomic mass of an element from its isotopic abundances. The second is a weighted mean: each isotope counts in proportion to how common it is.

Scientific notation and orders of magnitude

WHICH WAY DID THE DECIMAL POINT MOVE?BIGGER THAN 10SMALLER THAN 130 000 0000.000 023 × 10⁷2 × 10⁻⁵point moves left, n is positivepoint moves right, n is negativea must be at least 1 and below 10n simply counts how many places the point travelledrounding to significant figures changes a, never the powerso 4.37 × 10⁶ to 2 significant figures is 4.4 × 10⁶
The order of magnitude is just the power of ten on its own. Two quantities differ by one order of magnitude if one is roughly ten times the other.

Four percentages, four denominators

This is the section worth reading twice. All four formulas multiply by 100 and all four look like a difference over something. The something is different every time.

THE DIFFERENCE IS ALWAYS THE DENOMINATORfour different denominators, four different answersPERCENTAGE CHANGEhow much a value moved(final − initial) ÷ initial × 100PERCENTAGE DIFFERENCEhow far apart two values are(value 1 − value 2) ÷ mean × 100PERCENTAGE ERRORhow far from the accepted value(accepted − experimental) ÷ accepted × 100PERCENTAGE UNCERTAINTYhow precise the measurement isuncertainty ÷ measured value × 100the numbers can be identical; only what you divide by changesso read which comparison the question is actually asking for
The bottom two are the pair most often confused. Percentage error asks “how right was I?”; percentage uncertainty asks “how precisely did I measure?” — and they are calculated from completely different information.
Percentage difference has no fixed direction, so a negative sign carries no meaning — quote the size and drop the sign. Percentage change does have a direction: a negative answer means the quantity fell, and saying so is part of the answer.

When an effect can be ignored

Approximations are not sloppiness; they are a deliberate decision that a term is too small to matter. The standard one at SL is the weak acid calculation.

A weak acid dissociates only slightly, so almost all of it is still undissociated at equilibrium. That lets you make two simplifications: the equilibrium concentration of the acid is taken as its initial concentration, and because each molecule that does dissociate gives one H+ and one A, those two concentrations are equal.

The simplification Ka = [H+][A] ÷ [HA]
with [H+] = [A], this becomes
Ka = [H+]2 ÷ [HA]
Two honest caveats to keep in your back pocket. The assumption gets worse as Ka gets larger, because a stronger weak acid dissociates more and the “almost all undissociated” claim weakens. And it also ignores the small number of H+ ions coming from the water itself. If a question asks for the full expression, give the full one.
WORKED EXAMPLE

Copper has two isotopes: copper-63, of isotopic mass 62.93 and abundance 69.15%, and copper-65, of isotopic mass 64.93 and abundance 30.85%. Calculate the relative atomic mass of copper.

What kind of mean is this? A weighted one. Each isotope contributes in proportion to how common it is, so you cannot simply average 62.93 and 64.93. Multiply each mass by its abundance (62.93 × 69.15) + (64.93 × 30.85) = 4351.6 + 2003.1 Divide by the total abundance 6354.7 ÷ 100 = 63.547 Ar = 63.55 Dividing by 100 works because the abundances are percentages and therefore add to 100. If they are given as a ratio instead, divide by the total number of parts.
WORKED EXAMPLE

The concentration of a reactant falls from 0.250 mol dm–3 to 0.180 mol dm–3. Calculate the percentage change, and the percentage difference between the two values.

Percentage change — divide by the initial value (0.180 − 0.250) ÷ 0.250 × 100 −28.0% The minus sign matters here: the concentration decreased by 28.0%. Percentage difference — divide by the mean mean = (0.250 + 0.180) ÷ 2 = 0.215 0.070 ÷ 0.215 × 100 32.6% Same two numbers, and the answers differ by nearly five percentage points. That is the whole reason the distinction is examined.
WORKED EXAMPLE

A student measures the enthalpy of combustion of methanol as –520 kJ mol–1. The accepted value is –726 kJ mol–1. Calculate the percentage error and comment on the result.

Use the accepted value as the denominator (726 − 520) ÷ 726 × 100 206 ÷ 726 × 100 = 28.37 28.4% Comment The experimental value is much less exothermic than it should be, which points to heat being lost to the surroundings rather than to the water. Work with the magnitudes and say which way the result is out. An error this large is far too big to be explained by the precision of a thermometer.

💡 Exam tip

⚠️ Common mix-up

Up next: Units, Symbols and Numerical Values — the arithmetic above is only half of it. A number in chemistry is meaningless until it carries the right unit, and most calculation errors are really unit errors in disguise.

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