Lavoisier carefully weighed the reactants and products of his experiments and found something that now seems obvious: the total mass does not change. No atom appears or disappears in a chemical reaction; atoms are simply rearranged into different substances. That law, the conservation of mass, is the ultimate reason why we balance equations: a wrongly placed coefficient would be claiming that atoms have been created or destroyed out of nothing.
The law of conservation of mass and what you must not change
If you burn $10\ \text{g}$ of steel wool inside a sealed flask on a balance, the needle does not move during the combustion; however, if you do it on an open dish, the final residue weighs $14\ \text{g}$: has matter appeared from nothing, or how does chemistry explain this apparent change?
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Count the atoms of each element on both sides of equation $\text{CH}4 + \text{O}2 \rightarrow \text{CO}2 + \text{H}2\text{O}$, as it is written. The figure shows it…
Count the atoms of each element on both sides of equation $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$, as it is written. The figure shows it using the particle model:
Reason whether it is balanced or not. If it is not, state which elements fail.
The thin layer that protects the aluminium parts of an aeroplane from the air is aluminium oxide. The figure shows the particles involved: Check, by counting the atoms…
The thin layer that protects the aluminium parts of an aeroplane from the air is aluminium oxide. The figure shows the particles involved:
Check, by counting the atoms of each element on both sides, whether this equation is balanced: $4\ \text{Al} + 3\ \text{O}_2 \rightarrow 2\ \text{Al}_2\text{O}_3$.
The trial-and-error method: an algorithm, not a lucky guess
Balancing "by eye" works for very simple equations, but as soon as there are three or four substances it is best to follow an orderly procedure so that you do not have to go back again and again.
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Hydrochloric acid is obtained industrially by reacting gaseous hydrogen and chlorine. Balance the equation: $\text{H}2 + \text{Cl}2 \rightarrow \text{HCl}$.
Hydrochloric acid is obtained industrially by reacting gaseous hydrogen and chlorine. Balance the equation: $\text{H}_2 + \text{Cl}_2 \rightarrow \text{HCl}$.
Metallic calcium reacts with oxygen in the air to form calcium oxide, quicklime. Balance the equation: $\text{Ca} + \text{O}2 \rightarrow \text{CaO}$.
Metallic calcium reacts with oxygen in the air to form calcium oxide, quicklime. Balance the equation: $\text{Ca} + \text{O}_2 \rightarrow \text{CaO}$.
In the laboratory, oxygen can be obtained by heating potassium chlorate, which decomposes into potassium chloride and oxygen. Balance the equation:…
In the laboratory, oxygen can be obtained by heating potassium chlorate, which decomposes into potassium chloride and oxygen. Balance the equation: $\text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2$.
Equations with several substances: the right order avoids backtracking
The more substances are involved, the easier it is to unbalance something you had already balanced when you adjust another part of the equation. Following strictly the order from the previous section (metals, remaining non-metals, H, O) minimises that backtracking.
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When a piece of aluminium is dropped into a test tube with hydrochloric acid, dissolved aluminium chloride and hydrogen bubbles are formed. Balance the equation:…
When a piece of aluminium is dropped into a test tube with hydrochloric acid, dissolved aluminium chloride and hydrogen bubbles are formed. Balance the equation: $\text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2$.
When baking a cake, the sodium hydrogen carbonate in the baking powder decomposes with heat into sodium carbonate, water and the carbon dioxide that makes the dough…
When baking a cake, the sodium hydrogen carbonate in the baking powder decomposes with heat into sodium carbonate, water and the carbon dioxide that makes the dough rise. Balance the equation: $\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$.
To weld railway tracks, the thermite reaction is used: aluminium powder mixed with iron(III) oxide reacts, giving off so much heat that the iron formed remains molten.…
To weld railway tracks, the thermite reaction is used: aluminium powder mixed with iron(III) oxide reacts, giving off so much heat that the iron formed remains molten.
a) Balance the equation: $\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}$.
b) Explain why it is convenient to balance iron and aluminium first and leave oxygen until the end.
Fractional coefficients and the smallest-whole-number criterion
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Ethane, $\text{C}2\text{H}6$, accompanies methane in natural gas. a) Balance its combustion, $\text{C}2\text{H}6 + \text{O}2 \rightarrow \text{CO}2 + \text{H}2\text{O}$,…
Ethane, $\text{C}_2\text{H}_6$, accompanies methane in natural gas.
a) Balance its combustion, $\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$, keeping the coefficient of ethane equal to $1$ for now: you will find that the coefficient for oxygen turns out to be a fraction.
b) Then multiply all the coefficients by the appropriate number to write the equation with the lowest possible integer coefficients.
Oxy-acetylene torches in workshops burn acetylene, $\text{C}2\text{H}2$, with oxygen and reach about $3\,000\ \celsius$. a) Balance the combustion equation:…
Oxy-acetylene torches in workshops burn acetylene, $\text{C}_2\text{H}_2$, with oxygen and reach about $3\,000\ \celsius$.
a) Balance the combustion equation: $\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$.
b) If a fraction appeared while balancing it, explain how to convert the balanced equation into lowest integer coefficients and write the final equation.
When balancing the combustion of propane, a student wrote: $2\ \text{C}3\text{H}8 + 10\ \text{O}2 \rightarrow 6\ \text{CO}2 + 8\ \text{H}2\text{O}$. a) Check, by…
When balancing the combustion of propane, a student wrote: $2\ \text{C}_3\text{H}_8 + 10\ \text{O}_2 \rightarrow 6\ \text{CO}_2 + 8\ \text{H}_2\text{O}$.
a) Check, by counting atoms, that their equation respects the conservation of atoms.
b) Even so, the teacher marks it as "could be improved". Explain what it means that the coefficients must be the lowest possible integers and rewrite the equation correctly.
The mole ratio and the general formula for a hydrocarbon
This reading in moles is not just a curiosity: if we generalise the balancing of a combustion to any hydrocarbon $\text{C}_x\text{H}_y$, we obtain a formula that works for all the particular cases you have balanced one by one.
This same idea, read backwards, lets you deduce an unknown formula: if you know the products of a balanced combustion, you can count back how many C and H atoms the starting fuel had.
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The industrial synthesis of ammonia is written as $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$, as shown in the figure using the particle model: Interpret it in…
The industrial synthesis of ammonia is written as $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$, as shown in the figure using the particle model:
Interpret it in two ways:
a) Using molecules: how many molecules of nitrogen and hydrogen react and how many molecules of ammonia are formed.
b) Using moles: how many moles of each gas are consumed when $2\ \text{mol}$ of ammonia are formed.
In the complete combustion of an unknown hydrocarbon $\text{X}$, the balanced equation is: $2\ \text{X} + 9\ \text{O}2 \rightarrow 6\ \text{CO}2 + 6\ \text{H}2\text{O}$.…
In the complete combustion of an unknown hydrocarbon $\text{X}$, the balanced equation is: $2\ \text{X} + 9\ \text{O}_2 \rightarrow 6\ \text{CO}_2 + 6\ \text{H}_2\text{O}$. The figure represents it using the particle model (one molecule of each type):
a) Deduce, by counting the carbon and hydrogen atoms that must appear in the products, the molecular formula of $\text{X}$.
b) Write the complete equation using the formula found.
The complete combustion of any hydrocarbon $\text{C}x\text{H}y$ follows the same pattern. a) Write the balanced general equation in terms of $x$ and $y$:…
The complete combustion of any hydrocarbon $\text{C}_x\text{H}_y$ follows the same pattern.
a) Write the balanced general equation in terms of $x$ and $y$: $\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\ \text{O}_2 \rightarrow x\ \text{CO}_2 + \frac{y}{2}\ \text{H}_2\text{O}$, and justify it by counting atoms.
b) Apply it to predict how many $\text{O}_2$ molecules are needed to burn one molecule of heptane, $\text{C}_7\text{H}_{16}$, a component of petrol, and verify the result by balancing its combustion equation.
An effervescent tablet in almost freezing water takes several minutes to be used up, fizzing sluggishly, but in hot water it disappears in a few seconds amid violent foaming: why is the same mass of reactant transformed at such different rates depending on the temperature?
Raising the temperature usually increases the rate because a larger fraction of collisions overcomes the barrier. Increasing the concentration of reactants in solution or gases usually increases the number of encounters. Crushing a solid increases the surface area in contact. A catalyst provides a pathway with a lower activation energy: it takes part in intermediate steps and is regenerated; it does not provide net energy or increase the equilibrium yield by itself.
Design a comparison using a whole effervescent tablet and a crushed one: same mass, water, temperature and container. Repeat and measure the time until the same end point; change only the surface area. Do it under supervision, wearing goggles, without swallowing any products or sealing the container. Record hypotheses, variables, spread and limitations: also varying the temperature would make it impossible to attribute the change to a single factor.
The diagram links the main ideas:
Let's practice! 8 medium exercises in this section