Conservation of Mass
In a chemical reaction, the atoms in the reactants are rearranged to make the products. Bonds break and new bonds form, but not a single atom is lost or made — they are simply swapped around into new combinations.
The law of conservation of mass — no atoms are lost or made during a chemical reaction, so the total mass of the products is equal to the total mass of the reactants.
Because the same atoms are present before and after, a reaction can be written as a balanced symbol equation: one with the same number of atoms of each element on both sides.
🧪 Exam-style questions
In an experiment, 48 g of magnesium reacts completely with 32 g of oxygen. What mass of magnesium oxide is produced?
Show answer
- No atoms are lost or made, so the mass of product equals the total mass of the reactants: 48 + 32 = 80 g. 1 mark
A student burns a piece of magnesium inside a sealed flask and finds that the balance reading does not change. Explain why the mass stays the same, in terms of atoms.
Show answer
- The atoms are only rearranged in the reaction — no atoms are created or destroyed. 1 mark
- The flask is sealed, so the oxygen that combines with the magnesium is still inside the flask (no gas can enter or leave), so the total mass measured is unchanged. 1 mark
Balancing Equations
A balanced symbol equation has the same number of atoms of each element on both sides. Because no atoms are lost or made, we balance an equation by adjusting the big numbers in front of each substance — never by changing the substances themselves.
Reading the numbers in a formula
There are two kinds of number in equations, and they do different jobs:
- A big number in front of a formula (a multiplier, written in normal script) multiplies the whole formula. So 2H₂O means two water molecules — 4 hydrogen atoms and 2 oxygen atoms in total.
- A small subscript number only multiplies the atom (or bracket) immediately before it. So in H₂O the subscript 2 applies only to the hydrogen.
| Formula | Atoms in one formula unit |
|---|---|
| CO₂ | 1 carbon, 2 oxygen |
| Ca(OH)₂ | 1 calcium, 2 oxygen, 2 hydrogen (the 2 multiplies everything in the bracket) |
| Ca(NO₃)₂ | 1 calcium, 2 nitrogen, 6 oxygen (2 × NO₃ = 2 N and 6 O) |
Remember that seven non-metals are written as diatomic molecules whenever they appear on their own: H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂.
Work across the equation one element at a time, and adjust only the big numbers in front — never change a subscript, because that would change the substance itself.
For example, hydrogen burning in oxygen balances to:
2H₂ + O₂ → 2H₂O
There are now 4 hydrogen atoms and 2 oxygen atoms on each side — balanced.
Balance the equation for methane (CH₄) burning completely in oxygen:
CH₄ + O₂ → CO₂ + H₂O
Work across one element at a time, changing only the big numbers in front:
- Carbon — 1 on each side. Already balanced.
- Hydrogen — 4 on the left (CH₄) but only 2 on the right (H₂O). Put a 2 in front of H₂O, giving 4 H on the right.
- Oxygen — the right now has 2 (in CO₂) + 2 (in 2H₂O) = 4 O, but the left has only 2 (O₂). Put a 2 in front of O₂, giving 4 O on the left.
- Re-check — C 1 = 1, H 4 = 4, O 4 = 4. ✓ Balanced.
CH₄ + 2O₂ → CO₂ + 2H₂O
🧱 Interactive — build the balanced equation
Balance the complete combustion of ethane
Change the big number in front of each molecule. That many molecules appear in the particle picture, and the atom tally updates. Keep adjusting until carbon, hydrogen and oxygen each have the same number of atoms on both sides. (Never change a small subscript — that would change the substance.)
| Element | Reactant side | Product side | |
|---|---|---|---|
| C | 2 | 1 | |
| H | 6 | 2 | |
| O | 2 | 3 |
Change the numbers so each element matches on both sides.
🧪 Exam-style questions
Balance the equation for aluminium burning in oxygen.Type a balancing number in each box (leave it as 1 if no number is needed), then press Check. Any correct set of numbers is accepted.
Show answer
4Al + 3O₂ → 2Al₂O₃
- Balancing aluminium and oxygen gives the numbers 4 : 3 : 2. 1 mark
- Check: Al 4 = 4 · O 6 = 6 ✓ (any equal multiple, e.g. 8 : 6 : 4, also balances)
Balance the equation for the complete combustion of ethane, C₂H₆.Type a balancing number in each box, then press Check. Any correct set of numbers is accepted.
Show answer
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
- Balancing carbon, hydrogen and oxygen gives the numbers 2 : 7 : 4 : 6. 1 mark for a correct method, 1 mark for the correct numbers.
- Check: C 4 = 4 · H 12 = 12 · O 14 = 14 ✓ (any equal multiple, e.g. 4 : 14 : 8 : 12, also balances; never change a subscript)
Mass Changes Involving a Gas
Sometimes a reaction looks as though it breaks the law of conservation of mass — the mass on the balance goes up or down. This almost always happens because one of the reactants or products is a gas in an open (non-enclosed) container, so its mass is not being measured.
The two classic cases
Mass appears to increase — when a metal reacts with oxygen, oxygen from the air is added to the solid, so the metal oxide weighs more than the metal did:
2Mg + O₂ → 2MgO
Mass appears to decrease — when a metal carbonate is heated (thermal decomposition), carbon dioxide gas escapes into the air, leaving only the solid metal oxide behind:
CaCO₃ → CaO + CO₂
In a closed system (a sealed container), no gas can enter or leave, so the total mass always stays exactly the same — even when a gas is made.
Mass is never actually lost or created — atoms are not destroyed. You must explain the change in terms of the particle model: the gas particles are still there, they have just left (or joined) the substance on the balance. Saying “the mass disappeared” or “atoms were lost” earns no marks.
A carbonate (marble chips) reacts with dilute acid in an open test tube.
Green copper carbonate is heated in an open crucible until it fully decomposes:
CuCO₃ → CuO + CO₂
The mass of the solid in the crucible decreases. Explain why, in terms of the particle model.
Model answer: One of the products, carbon dioxide, is a gas. In an open crucible the CO₂ molecules escape into the air, so they are no longer sitting on the balance. The mass that is measured therefore falls. No atoms are destroyed — if the gas were trapped in a sealed container, the total mass would stay exactly the same.
🧪 Exam-style questions
A piece of magnesium is heated strongly in an open crucible, where it reacts with oxygen from the air.
The mass of the contents of the crucible increases. Explain this increase in terms of the particle model.
Show answer
- Oxygen molecules from the air combine with (bond to) the magnesium to form magnesium oxide. 1 mark
- The mass of this added oxygen is now part of the solid, so the mass measured on the balance increases. 1 mark
- No atoms are created — the extra mass comes from oxygen that was already present in the air.
Which statement explains why the mass appears to decrease when a metal carbonate is thermally decomposed in an open container?Tick (✓) one box.
Relative Formula Mass (Mr)
Relative atomic mass (Ar) — the average mass of the atoms of an element, compared with carbon-12. It is the larger of the two numbers given for each element on the periodic table.
Relative formula mass (Mr) — the sum of the relative atomic masses of all the atoms shown in the formula of a substance.
Mr = sum of (Ar × number of atoms) for every element in the formula
For example, water (H₂O): (2 × 1) + 16 = 18. Calcium hydroxide, Ca(OH)₂: 40 + (2 × 16) + (2 × 1) = 74.
Because mass is conserved, the sum of the Mr of the reactants equals the sum of the Mr of the products in a balanced equation (remembering to multiply each by its big number).
Percentage by mass of an element
You can use Mr to work out what fraction of a compound's mass comes from one particular element:
percentage mass of an element = Ar × number of atoms of that elementMr of the compound× 100
The relative atomic mass of every element is on the periodic table you are given — it is always the bigger of the two numbers. And don't forget the final × 100 when finding a percentage.
Find the relative formula mass of calcium carbonate, CaCO₃, and then the percentage by mass of calcium it contains. (Ar: C = 12, O = 16, Ca = 40.)
Step 1 — add up the Ar values to get Mr:
Mr(CaCO₃) = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100
Step 2 — put the total mass of calcium over the Mr, then × 100:
percentage of Ca = 40100 × 100 = 40%
🧪 Exam-style questions
Calculate the percentage by mass of magnesium in magnesium carbonate, MgCO₃.Use Ar: C = 12, O = 16, Mg = 24.
Show answer
- Mr(MgCO₃) = 24 + 12 + (3 × 16) = 84, and the mass of magnesium is 24.
- percentage of Mg = 2484 × 100 1 mark = 28.6% (to 3 s.f.) 1 mark
28.57%, 28.6% and 29% are all accepted — any correct rounding earns the mark when the question does not fix the number of figures.
What is the relative formula mass (Mr) of calcium hydroxide, Ca(OH)₂?Use Ar: H = 1, O = 16, Ca = 40. Tick (✓) one box.
Chemical Measurements & Uncertainty
Whenever you take a measurement there is always some uncertainty — no reading is ever perfectly exact. To deal with this, scientists repeat their measurements, discard any anomalous results, and calculate a mean.
The range of the repeated readings (how spread out they are) is used as a simple measure of the uncertainty:
uncertainty = ± range2 where range = largest − smallest reading
A result is then written as the mean ± the uncertainty — for example, 25.40 ± 0.05 cm³.
Always leave out any obviously anomalous reading before you calculate the mean — the mean is taken only from the concordant (close-together) results.
A student measures the volume of acid used in a titration four times and records: 24.00, 24.10, 24.05 and 23.50 cm³. Find the mean volume and the uncertainty.
- Spot the anomaly — 23.50 cm³ lies well outside the tight cluster of the other three, so it is left out before averaging.
- Mean of the concordant results = (24.00 + 24.10 + 24.05) ÷ 3 = 72.15 ÷ 3 = 24.05 cm³.
- Uncertainty — range = 24.10 − 24.00 = 0.10, so uncertainty = ± 0.102 = ± 0.05 cm³.
The result is written as 24.05 ± 0.05 cm³.
A balance with nothing on it should read exactly 0.00 g. If it reads 0.02 g, every measurement comes out 0.02 g too high — a zero error. Because each reading is shifted by the same amount, this is a systematic error, so repeating and averaging will not remove it. Instead, tare (re-zero) the balance before weighing — or subtract the zero reading from every result.
Significant figures
Calculation questions regularly end with “give your answer to 3 significant figures” — a mark you only collect if you can count them. Count from the first non-zero digit:
| Value | Sig. figs | Why |
|---|---|---|
| 0.00456 | 3 | Leading zeros never count — they only set the size of the number. |
| 5.013 | 4 | Zeros trapped between non-zero digits always count. |
| 4.300 | 4 | Trailing zeros after a decimal point count — they signal precision. |
| 4300 | 2 | Trailing zeros in a whole number with no decimal point do not count. |
- Round once, at the very end. Carry every calculator digit through the working — rounding part-way through can drift your final answer outside the mark scheme’s accepted range.
- To round, look only at the next digit — 5 or more rounds up. So 0.018735 to 3 s.f. is 0.0187, and 17.46 to 2 s.f. is 17.
- If the question doesn’t say how to round, match the least precise value you were given: data like 25.0 cm³ and 0.100 mol/dm³ are quoted to 3 s.f., so give your answer to 3 s.f. too.
🧪 Exam-style questions
A student records four titres: 12.30, 12.34, 12.32 and 11.90 cm³. Ignoring any anomalous result, calculate (a) the mean and (b) the uncertainty.
Show answer
- 11.90 cm³ is anomalous, so it is left out.
- (a) mean = (12.30 + 12.34 + 12.32) ÷ 3 = 36.96 ÷ 3 = 12.32 cm³ 1 mark
- (b) range = 12.34 − 12.30 = 0.04, so uncertainty = ± 0.04 ÷ 2 = ± 0.02 cm³ 1 mark
Final result: 12.32 ± 0.02 cm³.
A student repeats a measurement four times and records: 27.5, 27.4, 25.0 and 27.6 cm³. Which result is anomalous and should be left out before calculating the mean?Tick (✓) one box.