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Chemical equations and conservation of mass
What a balanced equation tells you, balancing harder equations including combustion, polyatomic ions and net ionic equations, the law of conservation of mass in open and closed systems, and solving mass-change problems from practical data.
Put a fizzing antacid tablet in water on a kitchen scale and the reading slowly drops. Did some mass disappear? No: carbon dioxide drifted out of the glass. Trap the gas and the reading holds steady to the last decimal place. That simple observation, that atoms are never created or destroyed in a reaction, is what makes every calculation in this module possible.
What a balanced equation tells you
A balanced equation is a recipe written in particles. Take the combustion of propane, the fuel in barbecue gas bottles:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
- Qualitatively it names the reactants and products and their states: (s), (l), (g) and (aq).
- Quantitatively the coefficients give the ratio in which particles react: 1 molecule of propane reacts with 5 molecules of oxygen to make 3 molecules of CO₂ and 4 molecules of water. Because you can scale any ratio up, the same numbers apply to moles: 1 mol of propane reacts with 5 mol of O₂. This mole ratio is the bridge for every stoichiometry problem in 2D, 2G and 2I.
Stoichiometry is the study of these quantitative relationships between the amounts of reactants and products.
Balancing harder equations
You balanced simple equations in Year 10. At Stage 6 the equations are longer, so work systematically:
- Write correct formulae for every species. Never change a subscript to balance an equation: that changes the substance.
- Balance elements that appear in only one reactant and one product first. Leave free elements (O₂, H₂, Fe) until last, because you can change them without upsetting anything else.
- If a polyatomic ion such as SO₄²⁻, NO₃⁻ or PO₄³⁻ passes through unchanged, balance it as one unit.
- If you end up with an odd number of O atoms needed from O₂, double every coefficient (or use a fraction temporarily and then clear it).
- Check every element, and check the total charge for ionic equations.
Balance C₄H₁₀(g) + O₂(g) → CO₂(g) + H₂O(l).
Carbon: 4 C on the left, so 4CO₂. Hydrogen: 10 H, so 5H₂O.
Oxygen on the right: 4 × 2 + 5 × 1 = 13 O atoms, which needs $\tfrac{13}{2}$ O₂. Double everything to clear the fraction:
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)
Check: C 8 = 8; H 20 = 20; O 26 = 16 + 10 ✓
Balance Al₂(SO₄)₃(aq) + NaOH(aq) → Al(OH)₃(s) + Na₂SO₄(aq).
Treat SO₄ and OH as units. Al: 2Al(OH)₃. That needs 6 OH, so 6NaOH. Na: 6 Na, so 3Na₂SO₄. SO₄: 3 on each side ✓
Al₂(SO₄)₃(aq) + 6NaOH(aq) → 2Al(OH)₃(s) + 3Na₂SO₄(aq)
Net ionic equations
In solution, ionic compounds exist as separate ions. Ions that are present but do not change are spectator ions, and leaving them out gives a net ionic equation. For the reaction above, Na⁺ and SO₄²⁻ are spectators:
Al³⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s)
A net ionic equation must balance both atoms and charge: here (+3) + 3(−1) = 0 on the left and 0 on the right.
The law of conservation of mass
In the 1770s and 1780s Antoine Lavoisier showed by careful weighing in sealed vessels that the total mass of substances is the same before and after a reaction. In particle terms the reason is simple: a reaction only rearranges atoms, so the same atoms, with the same masses, are there at the end.
total mass of reactants = total mass of products
Whether you see conservation on a balance depends on the system:
- In an open system matter can cross the boundary. If a gas is produced and escapes, the balance reading falls by exactly the mass of gas lost. If a gas from the air is taken in (magnesium burning in oxygen, iron rusting), the reading rises.
- In a closed system no matter enters or leaves, so the reading stays constant.
That apparent "loss" is useful: the change in mass of an open flask measures the mass of gas produced, which you can turn into moles (2B) and compare with the equation.
Solving mass-change problems
Marble chips (CaCO₃) are added to excess hydrochloric acid in an open flask on a balance. The reading falls from 175.82 g to 175.16 g. What mass of calcium carbonate reacted?
Mass of CO₂ lost = 175.82 − 175.16 = 0.66 g.
$n(\text{CO}_2) = \dfrac{m}{M} = \dfrac{0.66}{44.01} = 0.015\ \text{mol}$
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g): the ratio CaCO₃ : CO₂ is 1 : 1, so n(CaCO₃) = 0.015 mol.
$m(\text{CaCO}_3) = nM = 0.015 \times 100.09 = 1.5\ \text{g}$ (2 significant figures, matching the 0.66 g mass loss).
2.50 g of copper(II) carbonate is heated until no further change. CuCO₃(s) → CuO(s) + CO₂(g). Predict the mass of black residue and the mass lost.
M(CuCO₃) = 63.55 + 12.01 + 3 × 16.00 = 123.56 g mol⁻¹, so n = 2.50 ÷ 123.56 = 0.02023 mol.
Ratio 1 : 1, so n(CuO) = 0.02023 mol and m(CuO) = 0.02023 × 79.55 = 1.61 g.
Mass lost = mass of CO₂ = 2.50 − 1.61 = 0.89 g. Check: 0.02023 × 44.01 = 0.890 g ✓. Mass is conserved: 1.61 + 0.89 = 2.50 g.
These examples use moles, which are explained fully in 2B. Come back to them once you've read it.
- With the Open flask and 1.00 g of chips, press Add the chips. The final change should be −0.44 g, which is 0.0100 mol of CO₂. Does that match 1.00 g ÷ 100.09 g mol⁻¹?
- Repeat with 2.00 g. Does the mass lost double (to 0.88 g)? What does the shape of the graph tell you about the rate?
- Try Balloon on top with 1.00 g. The reading falls by about 0.29 g even though no atoms escape. Read the explanation: why does the balloon make the flask seem lighter?
- Try the Rubber stopper with 1.00 g (no change in reading; the pressure rises by about 55 kPa) and then with 2.00 g. What happens, and why is this a safety issue?
Test yourself
What is the coefficient of O₂ when C₅H₁₂ + O₂ → CO₂ + H₂O is balanced with the smallest whole numbers?
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O: oxygen on the right is 10 + 6 = 16 atoms, so 8O₂.
Magnesium ribbon is burned in an open crucible. What happens to the balance reading, and why?
The product contains oxygen atoms that came from outside the crucible, so the mass of solid rises.
Which is the correct net ionic equation for the reaction between silver nitrate and sodium chloride solutions?
Na⁺ and NO₃⁻ are spectators; the product is a solid precipitate.
An open flask containing excess acid and 3.00 g of zinc loses 0.09 g as the zinc dissolves: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). What amount of hydrogen was produced?
n(H₂) = 0.09 ÷ 2.016 = 0.045 mol. (Check: 3.00 g Zn is 0.0459 mol, which would give 0.0459 mol H₂, so the measurement agrees to within the ±0.01 g precision of the balance.)
In the balloon version of the marble-chip experiment, the balance reading drops a little even though no gas escapes. The best explanation is that
The balloon pushes aside about 1.2 g of air for every litre it inflates; the upthrust reduces the reading. CO₂ is actually denser than air.
An open container on a balance: what happens to the reading?
Label the open and closed systems.
When balancing equations, only the may be changed. Ions that do not take part are called ions. Mass is conserved because atoms are , never destroyed. In an system a gas can escape, so the mass lost equals the mass of produced.
A student places a flask containing 50 mL of dilute hydrochloric acid and a weighing boat holding 1.50 g of sodium hydrogen carbonate (NaHCO₃) on a balance. The total reading is 168.24 g. The solid is tipped in and the reading falls to 167.46 g.
(a) Write a balanced equation for the reaction. (b) Calculate the amount of gas produced. (c) Explain why the result supports, rather than contradicts, the law of conservation of mass.
(a) NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g)
(b) m(CO₂) = 168.24 − 167.46 = 0.78 g; n(CO₂) = 0.78 ÷ 44.01 = 0.018 mol.
(c) The flask is an open system. The carbon dioxide produced left the flask and went into the air, so the balance only weighed the substances that stayed behind. The fall in mass equals the mass of CO₂ that escaped. If the CO₂ is included, the total mass of products equals the total mass of reactants. This is confirmed by the mole ratio: 1.50 g of NaHCO₃ is 1.50 ÷ 84.01 = 0.0179 mol, which should produce 0.0179 mol of CO₂, matching the 0.018 mol measured.