Conservation of Mass and Equations

Strike a match, rust a nail, bake a cake — in every case a chemical reaction is taking place, and in every case the atoms you started with are still there at the end. They have not vanished and none have appeared from nowhere; they have simply been taken apart and joined back together in a fresh arrangement. That single, powerful idea is the law of conservation of mass, and it is what allows chemists to write reactions as balanced equations. This lesson opens Topic C3 (Chemical reactions) of your OCR Gateway Combined Science course. You will learn to write reactions as word equations and as balanced symbol equations, to add state symbols, and to explain the puzzle of why a balance reading can seem to rise or fall even though not one atom has been lost.

By the end of this lesson you should be able to state the law of conservation of mass, write word equations and balanced symbol equations, add state symbols, explain apparent changes in mass in an open container, and (Higher tier) work out a relative formula mass and write a balanced ionic equation.

This lesson builds AO1 recall of the conservation law and equation notation, AO2 application when you balance symbol equations and compute a relative formula mass, and AO3 interpretation when you explain why a mass reading appears to change in an open container.

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The Law of Conservation of Mass

In any chemical reaction, atoms are never made and never destroyed. Every atom in the reactants (the substances you start with) has to turn up somewhere among the products (the substances that form). Because the atoms are only rearranged, the total mass of the products is exactly the same as the total mass of the reactants. This is the law of conservation of mass.

Think of the atoms as a fixed collection of building bricks. A reaction pulls some of the bricks apart and clips them together in a new pattern — but it never adds a brick and never bins one. So if you could put everything from before the reaction on one balance and everything from after on another (gases included), the two readings would match exactly.

Exam Tip: The exact phrase examiners want is "atoms are not created or destroyed, only rearranged". Follow it with "so the total mass stays the same" and you have secured the central mark on any conservation-of-mass question.

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Word Equations

The quickest way to record a reaction is a word equation. The reactants sit on the left, the products on the right, and an arrow ($\rightarrow$→) between them means "reacts to form". The arrow points the way the change goes — it is not an equals sign.

When magnesium burns in oxygen, for instance:

$\text{magnesium} + \text{oxygen} \rightarrow \text{magnesium oxide}$magnesium+oxygen→magnesium oxide

And when methane (the main gas in natural gas) burns completely in plenty of air:

$\text{methane} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water}$methane+oxygen→carbon dioxide+water

A word equation tells you which substances react and which form, but it says nothing about their formulae or how many particles of each are involved. For that, you need a symbol equation.

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Symbol Equations and Balancing

A symbol equation swaps the names for chemical formulae. Since atoms are conserved, there must be the same number of atoms of each element on both sides. When that is true, the equation is balanced.

One rule must never be broken:

You balance an equation only by placing big numbers (multipliers) in FRONT of formulae. You must NEVER alter the small subscript numbers inside a formula.

Changing a subscript changes the substance itself: $\text{H}_2\text{O}$H2​O (water) is a completely different chemical from $\text{H}_2\text{O}_2$H2​O2​ (hydrogen peroxide). The subscripts are set by the chemistry — only the front numbers are yours to adjust.

Worked Example 1 — Burning magnesium

Balance: $\text{Mg} + \text{O}_2 \rightarrow \text{MgO}$Mg+O2​→MgO

Step 1 — count the atoms as they stand:

Element	Left	Right
Mg	1	1
O	2	1

Step 2 — oxygen does not match (2 on the left, 1 on the right). Put a 2 in front of $\text{MgO}$MgO so the right-hand side has 2 oxygen atoms:

$\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$Mg+O2​→2MgO

Step 3 — recount. The right now shows 2 Mg, so put a 2 in front of $\text{Mg}$Mg on the left:

$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO

Step 4 — final check: 2 Mg and 2 O on each side. Balanced. With state symbols added:

$2\text{Mg}_{(s)} + \text{O}_{2(g)} \rightarrow 2\text{MgO}_{(s)}$2Mg(s)​+O2(g)​→2MgO(s)​

Worked Example 2 — Hydrogen and oxygen

Balance: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$H2​+O2​→H2​O

Step 1 — count: the left has 2 H and 2 O; the right has 2 H and 1 O. Oxygen is short on the right.

Step 2 — place a 2 in front of $\text{H}_2\text{O}$H2​O:

$\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$H2​+O2​→2H2​O

Step 3 — recount: the right now has 4 H and 2 O. Hydrogen is now short on the left, so place a 2 in front of $\text{H}_2$H2​:

$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$2H2​+O2​→2H2​O

Step 4 — final check: 4 H and 2 O on each side. Balanced.

Worked Example 3 — Complete combustion of methane

Balance: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$CH4​+O2​→CO2​+H2​O

Step 1 — count carbon and hydrogen first, and save oxygen for last because it turns up in two different products. Carbon: 1 = 1. Hydrogen: 4 on the left, 2 on the right.

Step 2 — fix hydrogen with a 2 in front of $\text{H}_2\text{O}$H2​O:

$\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$CH4​+O2​→CO2​+2H2​O

Step 3 — now count oxygen on the right: 2 (in $\text{CO}_2$CO2​) + 2 (in $2\text{H}_2\text{O}$2H2​O) = 4. The left has only 2, so place a 2 in front of $\text{O}_2$O2​:

$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$CH4​+2O2​→CO2​+2H2​O

Step 4 — final check: C 1 = 1, H 4 = 4, O 4 = 4. Balanced.

Exam Tip: A dependable routine is to balance metals and non-metals first and leave hydrogen and oxygen until last, dealing with any element that appears in a single place before one spread across several. If you are ever tempted to change a subscript to force the numbers, stop — that move is always wrong.

Worked Example 4 — A precipitation reaction

Balance: $\text{AgNO}_3 + \text{MgCl}_2 \rightarrow \text{AgCl} + \text{Mg(NO}_3)_2$AgNO3​+MgCl2​→AgCl+Mg(NO3​)2​

Step 1 — treat the nitrate group $\text{NO}_3$NO3​ as one unit that stays together. Count: Ag 1 left / 1 right; Cl 2 left / 1 right; Mg 1 left / 1 right; $\text{NO}_3$NO3​ 1 left / 2 right.

Step 2 — balance the chlorine and the nitrate group together by putting 2 in front of $\text{AgCl}$AgCl and 2 in front of $\text{AgNO}_3$AgNO3​:

$2\text{AgNO}_3 + \text{MgCl}_2 \rightarrow 2\text{AgCl} + \text{Mg(NO}_3)_2$2AgNO3​+MgCl2​→2AgCl+Mg(NO3​)2​

Step 3 — final check: Ag 2 = 2, $\text{NO}_3$NO3​ 2 = 2, Mg 1 = 1, Cl 2 = 2. Balanced.

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State Symbols

A state symbol is written in brackets straight after a formula to show its physical state during the reaction:

Symbol	Meaning
(s)	solid
(l)	liquid
(g)	gas
(aq)	aqueous — dissolved in water

For example, neutralising hydrochloric acid with sodium hydroxide solution:

$\text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)}$HCl(aq)​+NaOH(aq)​→NaCl(aq)​+H2​O(l)​

The salt that forms (sodium chloride) stays dissolved, so it is labelled (aq), while the water made is a pure liquid, (l). State symbols never change how you balance an equation — they just add detail about the conditions.

Exam Tip: "(aq)" means specifically dissolved in water, not simply "a liquid". A pure liquid such as water or molten lead is (l); only something dissolved in water earns (aq). This is a common misconception worth pinning down now.

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Why Mass Can Appear to Change

If mass is always conserved, why does a balance reading sometimes climb or drop during a reaction? The reason is always the same: a gas has either escaped from, or been drawn into, an open (non-enclosed) container, so the balance is not "seeing" every substance.

Mass seems to fall when a gas is produced and drifts away into the air. Heating a metal carbonate is a classic case — it gives off carbon dioxide:

$\text{CaCO}_{3(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)}$CaCO3(s)​→CaO(s)​+CO2(g)​

The carbon dioxide floats off, so the solid left behind weighs less than the carbonate you started with. Trap that gas and weigh it too, though, and you would find no mass has been lost at all.

Mass seems to rise when a substance reacts with a gas taken in from the air. When magnesium burns, it combines with oxygen:

$2\text{Mg}_{(s)} + \text{O}_{2(g)} \rightarrow 2\text{MgO}_{(s)}$2Mg(s)​+O2(g)​→2MgO(s)​

The magnesium oxide weighs more than the original magnesium because oxygen atoms from the air have joined it. Nothing has been created — the extra mass is simply the mass of the oxygen that reacted.

In a closed (sealed) system, where no gas can get in or out, the balance reading does not change at all, which is a direct demonstration of conservation of mass.

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Common Misconceptions

Misconception	The correct idea
"Mass is destroyed when a gas is given off"	The gas still has mass; it has only escaped the open container. Trap it and the total mass is unchanged
"You balance by changing the small subscript numbers"	Never change subscripts — that changes the substance. Only put big numbers in front
"The arrow means equals"	The arrow means "reacts to form" and shows the direction of change
"(aq) just means a liquid"	(aq) means dissolved in water; a pure liquid is (l)
"Burning magnesium loses mass because it turns to ash"	It gains mass — oxygen from the air is added to form magnesium oxide

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Higher Tier: Relative Formula Mass ($M_r$Mr​)

Higher tier: The relative formula mass ($M_r$Mr​) of a compound is the sum of the relative atomic masses ($A_r$Ar​) of every atom in its formula. You simply add up the $A_r$Ar​ of each atom shown.

Worked Example 5 — $M_r$Mr​ of water and carbon dioxide

Using $A_r$Ar​: H = 1, C = 12, O = 16.

Water, $\text{H}_2\text{O}$H2​O: $(2 \times 1) + (1 \times 16) = 2 + 16 = 18$(2×1)+(1×16)=2+16=18.

Carbon dioxide, $\text{CO}_2$CO2​: $(1 \times 12) + (2 \times 16) = 12 + 32 = 44$(1×12)+(2×16)=12+32=44.

Worked Example 6 — $M_r$Mr​ with a bracket

Calcium hydroxide, $\text{Ca(OH)}_2$Ca(OH)2​. The subscript 2 multiplies everything inside the bracket. Using $A_r$Ar​: Ca = 40, O = 16, H = 1.

$M_r = 40 + 2 \times (16 + 1) = 40 + 2 \times 17 = 40 + 34 = 74$Mr​=40+2×(16+1)=40+2×17=40+34=74

You can also check conservation of mass with $M_r$Mr​: the total $M_r$Mr​ of the reactants (including the front multipliers) always equals the total $M_r$Mr​ of the products. In $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO (with Mg = 24): the left comes to $2 \times 24 + 32 = 80$2×24+32=80 and the right to $2 \times (24 + 16) = 80$2×(24+16)=80. The two sides match.

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Higher Tier: Balanced Ionic Equations

Higher tier: Many reactions in solution actually happen between ions. An ionic equation shows only the ions that genuinely take part and ignores the spectator ions — the ones that are present but unchanged on both sides.

When silver nitrate solution is mixed with sodium chloride solution, a white precipitate of silver chloride appears. The full equation is:

$\text{AgNO}_{3(aq)} + \text{NaCl}_{(aq)} \rightarrow \text{AgCl}_{(s)} + \text{NaNO}_{3(aq)}$AgNO3(aq)​+NaCl(aq)​→AgCl(s)​+NaNO3(aq)​

The sodium ions ($\text{Na}^+$Na+) and nitrate ions ($\text{NO}_3^-$NO3−​) drift about unchanged in solution — they are spectators. Cancel them and you are left with the ionic equation:

$\text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)} \rightarrow \text{AgCl}_{(s)}$Ag(aq)+​+Cl(aq)−​→AgCl(s)​

An ionic equation has to balance for both atoms and charge. Here the charges are $+1$+1 and $-1$−1 on the left, adding to $0$0, which matches the neutral solid on the right.

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Model Answer — Explaining an Apparent Mass Change

Question (6 marks): A student heats a sample of copper carbonate in an open crucible. The mass of the solid decreases. Explain why the mass decreases, and explain why this does not break the law of conservation of mass.

Mid-band response: "The copper carbonate breaks down and gives off a gas. The gas escapes so the solid weighs less. No atoms are destroyed."

Examiner-style commentary: The core ideas — a gas is given off and escapes, and no atoms are destroyed — are all present but left undeveloped. To move up a band, name the gas as carbon dioxide, give the equation, and say clearly that the total mass would be unchanged if the gas were collected.

Stronger response: "When copper carbonate is heated it decomposes into copper oxide and carbon dioxide gas. The carbon dioxide escapes from the open crucible into the air, so the solid left behind has a smaller mass. The law of conservation of mass is not broken because the atoms are only rearranged, not destroyed — the missing mass is simply the mass of the carbon dioxide that has left."

Examiner-style commentary: A strong, accurate answer that names the products and explains the escaping gas. To reach the top band, include the balanced equation with state symbols and add that in a sealed container the mass would not change.

Top-band response: "Heating copper carbonate causes thermal decomposition: $\text{CuCO}_{3(s)} \rightarrow \text{CuO}_{(s)} + \text{CO}_{2(g)}$CuCO3(s)​→CuO(s)​+CO2(g)​. The carbon dioxide made is a gas, so it escapes from the open crucible into the surroundings, leaving a smaller mass of solid (copper oxide) than the mass of copper carbonate at the start. This does not break the law of conservation of mass, because no atoms are created or destroyed — they are only rearranged. The apparent loss is simply the mass of the carbon dioxide that has left the crucible. If the reaction were carried out in a sealed container so the gas could not escape, the total mass would be unchanged, which shows that mass really is conserved."

Examiner-style commentary: Full marks. It names the process, gives the balanced equation with state symbols, identifies the escaping gas as the cause, and clinches the point with the sealed-container argument — exactly the reasoning examiners reward.

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Exam Tips

• State the law as "atoms are not created or destroyed, only rearranged, so the total mass is conserved".
• Balance using big numbers in front of formulae — never touch the small subscripts.
• Balance metals and non-metals first, hydrogen and oxygen last, and check every element on both sides.
• Add state symbols (s)(l)(g)(aq); remember (aq) means dissolved in water.
• A mass decrease in an open container means a gas escaped; a mass increase means a gas was taken in (such as oxygen).
• Higher: $M_r$Mr​ is the sum of all the $A_r$Ar​ values; multiply everything inside a bracket by its subscript; ionic equations must balance atoms and charge.

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Summary

• The law of conservation of mass: atoms are rearranged, never created or destroyed, so the total mass of the reactants equals the total mass of the products.
• A word equation names the reactants and products; the arrow means "reacts to form".
• A balanced symbol equation has equal numbers of each kind of atom on both sides; balance only with front multipliers, never by changing subscripts.
• State symbols are (s), (l), (g) and (aq) (dissolved in water).
• Mass appears to fall when a gas escapes and to rise when a gas is taken in; in a sealed system the mass stays constant.
• Higher: relative formula mass $M_r$Mr​ is the sum of the $A_r$Ar​ values; ionic equations show only the reacting ions and balance both atoms and charge.

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This content is aligned with OCR Gateway Combined Science A (J250), Topic C3 Chemical reactions. Refer to the official OCR specification for exact wording.