Balancing chemical equations
Picture a campfire at a Yellowstone campground: the log turns to ash and smoke, and it looks like matter vanished. Lavoisier carefully weighed the reactants and products of his experiments and found that 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. By the end of this page you will know where the campfire's "missing" mass went.
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 reading does not change 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?
Particle diagrams show the same thing. Here is methane burning, $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$: count the atoms on each side.
The steel wool puzzle is solved: on the open dish, oxygen from the air joins the iron to form iron oxide, so the solid gains mass. In the sealed flask nothing enters or leaves, so the total mass stays the same. The campfire works the same way: the "missing" mass left as $\text{CO}_2$ and water vapor in the air.
Why $\text{O}_2$ but just $\text{Hg}$?
Some elements exist as diatomic molecules: $\text{H}_2$, $\text{N}_2$, $\text{O}_2$, $\text{F}_2$, $\text{Cl}_2$, $\text{Br}_2$, $\text{I}_2$. Their atoms come in bonded pairs, so the subscript $2$ is part of the formula. Metals such as mercury, iron or aluminum are written as single atoms ($\text{Hg}$, $\text{Fe}$, $\text{Al}$) because they form a continuous lattice, not separate pairs.
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 aluminum parts of an airplane from the air is aluminum oxide. The figure shows the particles involved: Check, by counting the atoms of…
The thin layer that protects the aluminum parts of an airplane from the air is aluminum oxide. The figure shows the particles involved:
Check, by counting the atoms of each element on both sides, whether this equation is balanced: $4\ \text{Al} + 3\ \text{O}_2 \rightarrow 2\ \text{Al}_2\text{O}_3$.
Lavoisier and Priestley studied oxygen by heating mercury oxide, a red powder that decomposes into liquid mercury and oxygen gas. The figure shows how the particles of…
Lavoisier and Priestley studied oxygen by heating mercury oxide, a red powder that decomposes into liquid mercury and oxygen gas. The figure shows how the particles of each substance are formed:
- a) Balance the equation: $\text{HgO} \rightarrow \text{Hg} + \text{O}_2$.
- b) Mercury is written as $\text{Hg}$ and not $\text{Hg}_2$, whereas oxygen does have a subscript. Explain, using the particle model, the reason for this difference.
A car's catalytic converter finishes burning carbon monoxide, a very toxic gas, transforming it into carbon dioxide. Balance the equation:…
A car's catalytic converter finishes burning carbon monoxide, a very toxic gas, transforming it into carbon dioxide. Balance the equation: $\text{CO} + \text{O}_2 \rightarrow \text{CO}_2$.
In the electrolysis of water, electric current decomposes it into hydrogen and oxygen gases. Balance the equation: $\text{H}2\text{O} \rightarrow \text{H}2 + \text{O}2$.
In the electrolysis of water, electric current decomposes it into hydrogen and oxygen gases. Balance the equation: $\text{H}_2\text{O} \rightarrow \text{H}_2 + \text{O}_2$.
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.
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: $\text{Ca} + \text{O}_2 \rightarrow \text{CaO}$.
Sodium is a soft metal that reacts violently with chlorine to form common salt. Balance the equation: $\text{Na} + \text{Cl}2 \rightarrow \text{NaCl}$.
Sodium is a soft metal that reacts violently with chlorine to form common salt. Balance the equation: $\text{Na} + \text{Cl}_2 \rightarrow \text{NaCl}$.
In the lab, oxygen can be obtained by heating potassium chlorate, which decomposes into potassium chloride and oxygen. Balance the equation:…
In the lab, oxygen can be obtained by heating potassium chlorate, which decomposes into potassium chloride and oxygen. Balance the equation: $\text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2$.
The phosphorus on the match head burns with oxygen in the air to form phosphorus(V) oxide. Balance the equation: $\text{P} + \text{O}2 \rightarrow \text{P}2\text{O}5$.
The phosphorus on the match head burns with oxygen in the air to form phosphorus(V) oxide. Balance the equation: $\text{P} + \text{O}_2 \rightarrow \text{P}_2\text{O}_5$.
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) minimizes that backtracking.
Camping gas canisters contain propane, $\text{C}3\text{H}8$, which burns with oxygen in the air to form carbon dioxide and water vapor. The figure shows the molecules…
Camping gas canisters contain propane, $\text{C}_3\text{H}_8$, which burns with oxygen in the air to form carbon dioxide and water vapor. The figure shows the molecules involved:
Balance the equation by inspection: $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$. Remember to leave oxygen until the end.
When a piece of aluminum is dropped into a test tube with hydrochloric acid, dissolved aluminum chloride and hydrogen bubbles are formed. Balance the equation:…
When a piece of aluminum is dropped into a test tube with hydrochloric acid, dissolved aluminum 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 batter…
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 batter rise. Balance the equation: $\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$.
To weld railroad tracks, the thermite reaction is used: aluminum powder mixed with iron(III) oxide reacts, giving off so much heat that the iron formed remains molten.…
To weld railroad tracks, the thermite reaction is used: aluminum 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 aluminum first and leave oxygen until the end.
Fractional coefficients and the smallest-whole-number criterion
When the fuel already contains oxygen, count it too.
The bioethanol mixed with gas, $\text{C}2\text{H}6\text{O}$, burns to form carbon dioxide and water. Balance the equation:…
The bioethanol mixed with gas, $\text{C}_2\text{H}_6\text{O}$, burns to form carbon dioxide and water. Balance the equation: $\text{C}_2\text{H}_6\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$. Notice that the fuel itself already contains oxygen in its formula.
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 whole-number coefficients.
Your cells "burn" glucose from the blood in cellular respiration, obtaining energy and releasing carbon dioxide and water. Balance the equation:…
Your cells "burn" glucose from the blood in cellular respiration, obtaining energy and releasing carbon dioxide and water. Balance the equation: $\text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$. Check the result by carefully counting the oxygen atoms on both sides.
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 generalize 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.
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 gasoline, and verify the result by balancing its combustion equation.
Reaction rate and collision theory
An antacid tablet in ice 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 makes particles move faster, so there are more collisions per second, and a larger fraction of them have enough energy to overcome the barrier. Increasing the concentration of reactants in solution or gases packs more particles into the same volume, so the number of collisions per second increases, and with it the number of collisions with enough energy. Rate depends on both the number of collisions and the energy of each one. 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 antacid 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:
Sulfur dioxide coming out of some smokestacks is oxidized in the atmosphere to sulfur trioxide, a step prior to the formation of acid rain. Balance the equation:…
Sulfur dioxide coming out of some smokestacks is oxidized in the atmosphere to sulfur trioxide, a step prior to the formation of acid rain. Balance the equation: $\text{SO}_2 + \text{O}_2 \rightarrow \text{SO}_3$.
Car engines produce nitrogen monoxide, $\text{NO}$, which oxidizes when released into the air to form the brownish nitrogen dioxide seen in pollution episodes. Balance…
Car engines produce nitrogen monoxide, $\text{NO}$, which oxidizes when released into the air to form the brownish nitrogen dioxide seen in pollution episodes. Balance the equation: $\text{NO} + \text{O}_2 \rightarrow \text{NO}_2$.