Relative Formula Mass, Mass Changes When a Reactant or Product is a Gas, and Chemical Measurements

SPEC 4.3.1.2–4 • QUANTITATIVE CHEMISTRY • PAPER 1

M_r, Mass Changes & Chemical Measurements

Three connected ideas: how to calculate relative formula mass (with the bracket trap), how to explain apparent mass changes when a gas escapes or is absorbed, and how to handle uncertainty in repeated measurements.

Relative Formula Mass (M_r)

M_r is the gateway to every mole calculation. The definition is short — but the calculation has two traps that examiners exploit every year.

Relative Formula Mass

The relative formula mass (M_r) of a compound is the sum of the relative atomic masses (A_r) of the atoms in the numbers shown in the formula.

The Two Rules

Subscripts are multipliers. The small number after an element symbol tells you how many of that atom there are. H₂ means 2 hydrogen atoms.

A bracket subscript multiplies the entire group inside. Mg(NO₃)₂ means the whole NO₃ group is doubled.

Exam Tip

You'll be given a periodic table in the exam — it contains every A_r you need. There is no separate data sheet at GCSE Chemistry. Read A_r values directly from the periodic table.

Worked Examples — Simple M_r

Three increasing-difficulty examples. Practice each in your head before checking the answer.

CompoundAtoms in formulaM_r calculationM_r

H₂O2 H, 1 O2(1) + 1618

CO₂1 C, 2 O12 + 2(16)44

CaCO₃1 Ca, 1 C, 3 O40 + 12 + 3(16)100

NaCl1 Na, 1 Cl23 + 35.558.5

⚠️ COMMON SLIP: writing M_r(H₂O) as "2 + 16 = 18" is numerically correct but the working is wrong — the "2" is the subscript count, not the A_r. Always show 2(1) so the multiplication is explicit. Examiners want to see the method.

The Bracket Trap — M_r

When a formula has brackets, the subscript after the bracket multiplies everything inside. This is the most common A_r/M_r error in past papers.

M_r of Mg(NO₃)₂ = ?

1

Read the formula carefully. Mg(NO₃)₂ contains 1 Mg, plus the NO₃ group twice. So that's 2 N atoms and 6 O atoms in total (3 oxygens × 2).

2

Add up the A_r values. A_r(Mg) = 24, A_r(N) = 14, A_r(O) = 16.

M_r = 24 + 2(14) + 6(16) = 24 + 28 + 96 = 148

3

Sanity check. An alternative working: 24 + 2(14 + 3×16) = 24 + 2(14 + 48) = 24 + 2(62) = 24 + 124 = 148. ✓ Same answer.

⚠️ EXAMINER ALERT: the recurring error is forgetting that the 2 outside the bracket multiplies BOTH the N and the O inside it. Students often double the N but forget to double the O. Always count the inner atoms first, then apply the bracket.

M_r in Balanced Equations

This is the bridge from Lesson 3.01 (conservation of mass) to mole calculations: in a balanced equation, total M_r on each side is equal.

Spec Statement

In a balanced chemical equation, the sum of the M_r of the reactants in the quantities shown equals the sum of the M_r of the products in the quantities shown.

CaCO₃ → CaO + CO₂

Reactants

1 × CaCO₃ = 1 × 100 = 100

Products

1 × CaO = 1 × (40 + 16) = 56

1 × CO₂ = 1 × 44 = 44

Total = 56 + 44 = 100 ✓

Why It Matters

This is why you can predict reaction masses from M_r alone. If you start with 100 g of CaCO₃ and the reaction goes to completion, you'll have 56 g of CaO + 44 g of CO₂ — totalling exactly 100 g. The M_r ratios are the mass ratios.

Percentage by Mass of an Element

Examiners use this to test M_r indirectly — you have to compute M_r first, then build the percentage.

% mass of element = (A_r × number of atoms) ÷ M_r of compound × 100

Worked Example — % Ca in CaCO₃

A_r(Ca) = 40, number of Ca atoms = 1, so total mass of Ca = 40

M_r(CaCO₃) = 100 (from previous slide)

% Ca = (40 ÷ 100) × 100 = 40%

Worked Example — % N in NH₄NO₃ (ammonium nitrate)

M_r(NH₄NO₃) = 14 + 4(1) + 14 + 3(16) = 14 + 4 + 14 + 48 = 80

Total N atoms = 2 (one in NH₄, one in NO₃); total mass of N = 2 × 14 = 28

% N = (28 ÷ 80) × 100 = 35%

⚠️ COMMON SLIP: students count only ONE nitrogen in NH₄NO₃ — forgetting that nitrogen appears in both ions. Always list every atom of the target element, even if they're separated.

Mass Changes — Open vs Closed System

Conservation of mass always holds. But on a balance, the mass can appear to change if a gas escapes or enters — that's the key idea of spec 4.3.1.3.

The Key Distinction

Open system → gas can leave or enter, so the balance shows an apparent mass change. Closed (sealed) system → no gas exchange, so the balance reads the same before and after.

Mass Decreases — Gas Escapes

The classic spec example: thermal decomposition of a metal carbonate. Carbon dioxide gas is produced and escapes into the atmosphere, leaving the metal oxide as the only solid product.

CaCO₃(s) → CaO(s) + CO₂(g)↑

SubstanceStateM_rOn the balance?

CaCO₃ (start)solid100YES — full 100

CaO (product)solid56YES — only 56 left

CO₂ (product)gas44NO — escapes into the air

In Words

The total mass is still 100 g (56 + 44). But you only see the 56 g of CaO solid on the balance. The 44 g of CO₂ gas has left the system. Conservation still holds — you just have to account for the gas.

Particle-Model Explanation

The CO₂ molecules formed in the reaction are a gas — their particles move freely with lots of empty space between them. In an open container, those gas particles can leave the surface of the solid and escape into the air. The atoms aren't destroyed — they're now in CO₂ gas in the atmosphere instead of being part of the solid being weighed.

Mass Increases — Gas Absorbed

The other spec example: a metal reacting with oxygen. Oxygen from the air joins the metal, so the solid product weighs more than the metal you started with.

2Mg(s) + O₂(g) → 2MgO(s)

Worked Example

Start: 2 × Mg = 2 × 24 = 48 g of magnesium on the balance

Plus: 1 × O₂ = 32 g of oxygen comes in from the air (NOT on the balance to start)

End: 2 × MgO = 2 × 40 = 80 g of magnesium oxide on the balance

Mass increase: 80 − 48 = 32 g — exactly the mass of oxygen absorbed

Particle-Model Explanation

Particles of oxygen gas from the air collide with the metal and chemically bond to it, forming the metal oxide. The total number of atoms in the system is unchanged — but more of them are now in the solid on the balance, so the reading goes up.

PEE Chain — Explaining a Mass Change

The structured answer for any "explain why the mass increased / decreased" question. This is a guaranteed 2-mark question on Paper 1 — get the chain right and the marks are automatic.

PPoint: No atoms are lost or made — total mass is conserved.

EEvidence: A gas was either produced and escaped (mass decrease) OR absorbed from the air (mass increase). Name the gas: carbon dioxide / water vapour / oxygen / etc.

EExplain: Because the system is open (not sealed), the gas can leave or enter — so the balance shows an apparent change even though total mass is unchanged.

⚠️ EXAMINER ALERT (P2 Jun22 Q02.2): "the contents evaporated" is insufficient — you must name the substance (water vapour / steam / CO₂) AND say it escaped. Generic answers do not score.

Chemical Measurements — The Mean

Whenever a measurement is repeated, you calculate a mean. The spec is explicit about uncertainty — and the mark scheme is strict about how the mean is calculated.

Mean

The sum of all the values divided by the number of values.

mean = (x₁ + x₂ + ... + xₙ) ÷ n

⚠️ EXAMINER ALERT (P1 Jun24 Q04.5): students averaged only the highest and lowest values. This gives the median (or midrange), not the mean. Always sum every value and divide by the total count. There is no shortcut.

Why repeats?

Single measurements contain random error. Taking several readings and averaging cancels out random scatter — the mean is closer to the true value than any individual reading.

Range and Uncertainty

The mean tells you the central value of your repeats. The uncertainty tells you the spread on either side — how confident you can be in that mean.

range = max − min uncertainty = range ÷ 2

What "range about the mean" means

The spec wording — "use the range of a set of measurements about the mean as a measure of uncertainty" — describes the spread on each side of the mean. Half the total range gives that spread, so:

uncertainty = ½ × (max − min)

Reporting a Result

Quote your answer as: mean ± uncertainty with appropriate units.

Example: mean = 24.5 °C, range = 0.3 °C, uncertainty = 0.15 °C → report as 24.5 ± 0.15 °C.

What it tells you

A small uncertainty means precise measurements (data clustered tightly around the mean). A large uncertainty means a lot of scatter — repeats vary widely.

Worked Example — Mean & Uncertainty

Authentic Paper 1 style. Four temperature trials from a displacement reaction (modelled on AQA P1 Jun24 Q04.5).

The Data

Trial 1Trial 2Trial 3Trial 437.6 °C37.2 °C37.8 °C37.4 °CCalculate the mean highest temperature, including the uncertainty.

1

Mean. Sum all four: 37.6 + 37.2 + 37.8 + 37.4 = 150.0. Divide by n = 4: mean = 37.5 °C

2

Range. max − min = 37.8 − 37.2 = 0.6 °C

3

Uncertainty. range ÷ 2 = 0.6 ÷ 2 = ± 0.3 °C

37.5 ± 0.3 °C

Random vs Systematic Errors

Two error types, two completely different solutions. The exam tests whether you can tell them apart from a graph or a description.

⚠️ EXAMINER ALERT (P1 Jun23 Q02.6): Random error = scatter both above and below the best-fit line. Systematic error = every point shifted in the same direction. Repeats reduce random error; they do not reduce systematic error — you'd need to recalibrate.

Significant Figures

Final answers should be quoted to a sensible number of significant figures (sig figs). The convention examiners follow is simple.

The Convention

Match the precision of the least precise input in the calculation.

Data given to 3 sig figs → answer to 3 sig figs

Data given to 2 sig figs → answer to 2 sig figs

If inputs are mixed, use the lowest

Worked Example

Mean = 37.5 °C (3 sig figs). Uncertainty calculated as 0.3 °C — this is to 1 sig fig (which matches the precision of the difference).

Reported answer: 37.5 ± 0.3 °C. Both values are quoted to 1 decimal place — matching the precision of the original data.

Exam Tip

Don't write 37.50000 ± 0.300000. Don't write 38 ± 0.3. Match the original data's decimal places — usually 1 d.p. for temperature, 2 d.p. for mass in grams. The examiner wants appropriate precision, not the calculator's full output.

Practice Question 1

Calculate the percentage by mass of nitrogen in ammonium sulfate, (NH₄)₂SO₄.

A_r values: N = 14, H = 1, S = 32, O = 16. 3 marks

P

Plan: Need M_r of (NH₄)₂SO₄ first, then total mass of N, then percentage.

E

Calculate M_r. The bracket subscript "2" multiplies the entire NH₄ group.

M_r = 2(14 + 4×1) + 32 + 4(16) = 2(18) + 32 + 64 = 36 + 32 + 64 = 132

E

Total mass of N. 2 N atoms × A_r(14) = 28

E

Percentage. (28 ÷ 132) × 100 = 21.2% (to 3 s.f.)

Practice Question 2

A student measured the volume of gas produced in a reaction five times:

24.3, 24.6, 24.5, 24.5, 24.6 cm³

Calculate the mean and uncertainty in the volume measurement. 3 marks

P

Plan: Sum all five values for the mean. Find max − min for the range. Halve the range for the uncertainty.

E

Mean. 24.3 + 24.6 + 24.5 + 24.5 + 24.6 = 122.5; ÷ 5 = 24.5 cm³1 mark

E

Range. 24.6 − 24.3 = 0.3 cm³1 mark

E

Uncertainty. 0.3 ÷ 2 = ± 0.15 cm³1 mark

24.5 ± 0.15 cm³

Common Errors — Mark-Scheme Verbatim

Sit these in your memory. The right-hand column is what the examiner is looking for.

Students often writeCorrect versionWhy it matters

M_r(Mg(NO₃)₂) = 24 + 14 + 3(16) = 86M_r = 24 + 2(14) + 6(16) = 148Bracket multiplies the WHOLE group — don't forget the inner atoms (Examiner Intelligence)

"Mass decreased — the contents evaporated"" Water vapour / steam escaped " (or name the specific gas)Generic answers do not score (P2 Jun22 Q02.2)

"Mass gained" — no reason given"Oxygen from the air was absorbed / reacted with the metal"Must explain WHY (Question Pattern Library)

Mean = (highest + lowest) ÷ 2Mean = sum of all values ÷ nAveraging only extremes gives the median (P1 Jun24 Q04.5)

Calling a graph "random error" when all points are shifted the same wayAll points shifted same way = SYSTEMATIC . Random = scatter both sides of best-fit lineMislabelling the error type (P1 Jun23 Q02.6)

Answer given to wrong number of sig figsMatch the precision of the least-precise inputConvention examiners enforce (Examiner Intelligence)

Summary

Key Definitions

Relative formula mass (M_r): sum of A_r values of all atoms in the formula

Uncertainty: the range about the mean within which the true value is likely to lie (= half the range)

Random error: scatter both sides of the line of best fit; reduced by repeats

Systematic error: all points shifted the same way; reduced by recalibration

M_r Calculation

Subscripts multiply the atom they follow. Brackets multiply the entire group inside. Mg(NO₃)₂ = 148, NOT 86.

% By Mass

(A_r × number of atoms) ÷ M_r × 100. Count every atom of the target element across the whole formula.

Mass Changes

Open system — gas leaves (mass ↓) or enters (mass ↑). Closed system — no change. Always name the gas.

Mean & Uncertainty

Mean = sum ÷ n. Uncertainty = range ÷ 2. Quote as mean ± uncertainty with appropriate sig figs.

Exam Tip

The three most-tested errors here all share one theme: show the working method, not just the answer. Examiners give 1 mark for the method (= what you've calculated and why) and 1 mark for the value. Always show every step.

Ready for 3.03

Next up (Higher Tier only): Moles — Avogadro's constant, mass ↔ moles conversion, and the foundation for every quantitative calculation in Module 3.