Balancing equations computations

Balancing is therefore an exercise in arithmetic and ratios. A coefficient before a formula tells you how many particles or formula units are involved. For example,

$3\mathrm{H_2O}$

represents three water molecules. Each contains two H atoms and one O atom, so the totals are

• H: $3\times2=6$ atoms
• O: $3\times1=3$ atoms.

The coefficient multiplies the whole formula. By contrast, the subscript 2 in $\mathrm{H_2O}$ is part of water's formula. It must not be changed: $\mathrm{H_2O_2}$ is hydrogen peroxide, a different substance.

A reliable balancing method

Begin with correct formulae for all reactants and products. Then:

1. List the elements present.
2. Count the atoms of each element on both sides, multiplying each subscript by its coefficient. An unwritten coefficient or subscript is 1.
3. Choose an element that is not balanced. It is often helpful to begin with an element in the most complicated formula and leave free $\mathrm{O_2}$ or $\mathrm{H_2}$ until later.
4. Change coefficients to make the totals equal. Never alter a subscript.
5. Recount, because changing one coefficient may affect several elements.
6. Reduce the coefficients to the smallest possible whole-number ratio.
7. Check every element independently.

Using multiplication to balance an equation

Consider magnesium burning in oxygen:

$\mathrm{Mg + O_2 \rightarrow MgO}$

There are two O atoms on the left but only one on the right. A coefficient of 2 before $\mathrm{MgO}$ supplies $2\times1=2$ O atoms on the right:

$\mathrm{Mg + O_2 \rightarrow 2MgO}$

That coefficient also gives two Mg atoms on the right, so the magnesium coefficient must also become 2:

$\mathrm{2Mg + O_2 \rightarrow 2MgO}$

The coefficient ratio is $2:1:2$. A final count gives two Mg and two O atoms on each side.

Using ratios in combustion equations

Methane burns completely to form carbon dioxide and water:

$\mathrm{CH_4 + O_2 \rightarrow CO_2 + H_2O}$

Carbon is already balanced. Four H atoms on the left require two water molecules because $2\times2=4$:

$\mathrm{CH_4 + O_2 \rightarrow CO_2 + 2H_2O}$

The products now contain four O atoms altogether: two in $\mathrm{CO_2}$ and $2\times1=2$ in water. Since every oxygen molecule supplies two O atoms, two $\mathrm{O_2}$ molecules are required:

$\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$

The ratio $1:2:1:2$ means one methane molecule reacts with two oxygen molecules to form one carbon dioxide molecule and two water molecules.

Odd and even atom totals

Sometimes one formula contains an odd number of an atom while another supplies that atom in pairs. For

$\mathrm{Fe + O_2 \rightarrow Fe_2O_3}$

oxygen appears in groups of 2 on the left and 3 on the right. The lowest common multiple of 2 and 3 is 6. Use $3\mathrm{O_2}$ and $2\mathrm{Fe_2O_3}$ to give six O atoms on each side. Two units of $\mathrm{Fe_2O_3}$ contain $2\times2=4$ Fe atoms, so four Fe atoms are needed on the left:

$\mathrm{4Fe + 3O_2 \rightarrow 2Fe_2O_3}$

The final ratio is $4:3:2$. This lowest-common-multiple approach is often quicker than repeatedly guessing coefficients.

Equations with an element in several products

For the combustion of butanol,

$\mathrm{C_4H_{10}O + O_2 \rightarrow CO_2 + H_2O}$

balance carbon first: four C atoms require $4\mathrm{CO_2}$. Next, ten H atoms require $5\mathrm{H_2O}$. The products then contain $4\times2+5\times1=13$ O atoms. One O atom is already present in the butanol molecule, leaving 12 to be supplied by oxygen gas. Since each $\mathrm{O_2}$ supplies two O atoms, $12\div2=6$ oxygen molecules are needed:

$\mathrm{C_4H_{10}O + 6O_2 \rightarrow 4CO_2 + 5H_2O}$

This shows why the total for an element must include every compound containing it. Oxygen is often balanced last because it commonly appears in several formulae.

When fractions appear temporarily

A fractional coefficient can be useful during working, but the final equation is normally written using whole numbers. If an intermediate result contains $\tfrac{3}{2}\mathrm{O_2}$, multiply every coefficient in the entire equation by 2. This preserves the reaction ratio while removing the fraction. After doing so, check that the coefficients have no common factor greater than 1.

Checking the completed equation

For each element, calculate

$\text{total atoms}=\text{coefficient}\times\text{atoms in one formula}.$

Construct two columns—reactants and products—and compare the totals element by element. Equal overall numbers of particles are not required; only equal numbers of each type of atom are required. A balanced equation may therefore have different total numbers of particles on its two sides.