Conservation of Mass and Balanced Equations

Whenever a chemical reaction happens, atoms are not made and they are not destroyed — they are simply rearranged. The same atoms that were there at the start are all still there at the end, just joined together in new ways. This single idea, the law of conservation of mass, is the foundation of the whole of chemistry, and it is what makes a balanced equation possible. This lesson opens Topic C3 (Chemical reactions) of OCR Gateway Science A by showing how to write reactions as word equations and as balanced symbol equations, how to use state symbols, and why a mass reading can sometimes appear to change even though no atoms have 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 a non-enclosed system, and (Higher tier) calculate relative formula mass and write balanced ionic equations.

---

The Law of Conservation of Mass

In a chemical reaction, no atoms are created and no atoms are destroyed. Every atom present in the reactants (the starting substances) must appear somewhere in the products (the substances made). Because the atoms are simply rearranged, the total mass of the reactants equals the total mass of the products. That is the law of conservation of mass.

A simple way to picture it is to imagine the atoms as a fixed set of building bricks. A reaction takes the bricks apart and clicks them back together in a new arrangement, but not one brick is added and not one is thrown away. So if you could weigh everything before and everything after — including any gases — the two masses would be identical.

Exam Tip: The phrase examiners look for is "atoms are not created or destroyed, only rearranged". If you can state that, then say the total mass therefore stays the same, you have the core mark for any conservation-of-mass question.

---

Word Equations

The simplest way to describe a reaction is a word equation. The reactants go on the left, an arrow ($\rightarrow$→) means "reacts to form", and the products go on the right. The arrow is not an equals sign — it shows the direction of change.

For example, when magnesium burns in oxygen:

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

And when methane (natural gas) burns completely:

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

A word equation tells you what reacts and what forms, but it does not tell you the formulae or how many of each particle are involved. For that you need a symbol equation.

---

Symbol Equations and Balancing

A symbol equation replaces the names with chemical formulae. Because atoms are conserved, the number of atoms of each element must be the same on both sides — when that is true, the equation is balanced.

There is one rule that students must never break:

You balance an equation by writing big numbers (multipliers) IN FRONT of formulae. You must NEVER change the small subscript numbers inside a formula.

Changing a subscript changes the substance itself — $\text{H}_2\text{O}$H2​O (water) is not the same as $\text{H}_2\text{O}_2$H2​O2​ (hydrogen peroxide). The subscripts are fixed by the chemistry; only the front multipliers may be adjusted.

Worked Example 1 — Burning magnesium

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

Step 1 — count the atoms as written:

Element	Left	Right
Mg	1	1
O	2	1

Step 2 — oxygen is unbalanced (2 on the left, 1 on the right). Put a 2 in front of $\text{MgO}$MgO to give 2 oxygen atoms on the right:

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

Step 3 — re-count. The right now has 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:

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

Step 2 — put 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 — re-count: right now has 4 H and 2 O. Hydrogen is short on the left, so put 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 (leave oxygen until last, because it appears in two products). Carbon: 1 = 1. Hydrogen: 4 on the left, 2 on the right.

Step 2 — balance 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 put 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 reliable order is to balance metals and non-metals first, leave hydrogen and oxygen until last, and balance the element that appears in only one place on each side before one that appears in several. If you ever feel tempted to change a subscript to make the numbers work, stop — that is always the wrong move.

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 a single unit. 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 chlorine and the nitrate group 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.

---

State Symbols

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

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

For example, the neutralisation of hydrochloric acid by 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)​

Notice that the salt formed (sodium chloride) stays dissolved, so it is (aq), while the water produced is a liquid, (l). State symbols do not change the balancing — they just add information about the conditions.

Exam Tip: "(aq)" specifically means dissolved in water, not just "a liquid". A pure liquid such as water or molten lead is (l); only a substance dissolved in water is (aq).

---

Why Mass Can Appear to Change

If mass is always conserved, why does a mass reading sometimes go up or down during a reaction? The answer is always the same: a gas has entered or escaped from a non-enclosed (open) container, so the balance does not "see" all of the substances.

Mass appears to decrease when a gas is produced and escapes into the air. For example, when a metal carbonate is heated 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 away, so the solid left behind weighs less than the original carbonate — but if you trapped that gas and weighed it too, no mass would have been lost at all.

Mass appears to increase when a substance reacts with a gas taken in from the air. For example, 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 been added to it. Again, nothing has been created — the extra mass is simply the mass of the oxygen that joined in.

In a closed (sealed) system, where no gas can enter or leave, the balance reading does not change, which demonstrates conservation of mass directly.

---

Common Misconceptions

Misconception	The correct idea
"Mass is destroyed when a gas is given off"	The gas still has mass; it has just escaped the open container. Trap it and 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)
"When magnesium burns it loses mass because it turns to ash"	It gains mass — oxygen from the air is added to form magnesium oxide

---

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

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

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

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

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 check conservation of mass with $M_r$Mr​: the total $M_r$Mr​ of the reactants (allowing for the front multipliers) always equals the total $M_r$Mr​ of the products. For example, in $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO (using Mg = 24): left $= 2 \times 24 + 32 = 80$=2×24+32=80; right $= 2 \times (24 + 16) = 80$=2×(24+16)=80. The masses match.

---

Higher Tier: Balanced Ionic Equations

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

For example, when silver nitrate solution reacts with sodium chloride solution, a white precipitate of silver chloride forms. 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−​) stay in solution unchanged — they are spectators. Cancelling them leaves the ionic equation:

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

An ionic equation must be balanced for both atoms and charge. Here the charges are $+1$+1 and $-1$−1 on the left, totalling $0$0, which matches the neutral solid on the right.

---

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 central ideas (a gas is given off and escapes, no atoms destroyed) are present but undeveloped. To climb a band, name the gas as carbon dioxide, give the equation, and state explicitly that the total mass would be unchanged if the gas were collected.

Stronger response: "When copper carbonate is heated it decomposes to form 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 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 state 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 produced is a gas that escapes from the open crucible into the surroundings, so the mass of solid remaining (copper oxide) is less 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, proving that mass 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.

---

Exam Tips

• State the law as "atoms are not created or destroyed, only rearranged, so total mass is conserved".
• Balance using big numbers in front of formulae — never change the small subscripts.
• Balance metals/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 (e.g. oxygen).
• Higher: $M_r$Mr​ is the sum of all the $A_r$Ar​ values; multiply everything inside a bracket by the subscript; ionic equations must balance atoms and charge.

---

Summary

• The law of conservation of mass: atoms are rearranged, never created or destroyed, so the total mass of reactants equals the total mass of products.
• A word equation names 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.

---

This content is aligned with OCR Gateway Science A GCSE Chemistry (J248), Topic C3 Chemical reactions. Refer to the official OCR specification document for the exact wording.