4.2 Balancing chemical reaction equations
A chemical equation is balanced when it shows equal numbers of atoms of each element on both sides. Remember that this follows from atomic theory since atoms cannot be created nor destroyed. We use coefficients in front of formulae in chemical reactions to denote the relative number of atoms or molecules or moles or formula units involved in the reaction. We do not, however, change the subscripts (chemical formulae themselves) to balance a chemical equation, since that would change the substances. The absence of a coefficient is understood to imply a “1”.
Consider the reaction between methane (CH4) and bromine (Br2):
CH4(g) + Br2(l) -> CBr4(g) + HBr(l) (unbalanced)
| Count: | 1 C atom | 1 C atom |
| 4 H atoms | 1 H atom | |
| 2 Br atoms | 5 Br atoms |
As written, this equation is not balanced because it violates the Law of Conservation of Mass. Although one carbon atom is indicated on both the left side and right side of the equation, two bromine atoms (one Br2 molecule) appear on the left and five bromine atoms appear on the right. Likewise, four hydrogen atoms are shown on the left and one hydrogen atom on the right. The order in which you balance the atoms does not matter, so count the atoms on each side of the chemical equation and use coefficients to make those values equal. We start by balancing the number of H atoms on each side of the equation. If we put the coefficient 4 in front of hydrogen bromide (HBr) we will have equal numbers of hydrogen atoms on each side of the arrow:
CH4(g) + Br2(l) -> CBr4(g) + 4HBr(l) (unbalanced)
| Count: | 1 C atom | 1 C atom |
| 4 H atoms | 4 H atom | |
| 2 Br atoms | 8 Br atoms |
At this point, we have balanced the number of carbon atoms and hydrogen atoms in the equation. But the bromine atoms are not balanced. We can balance the equation by placing a 4 in front of the bromine molecule. Thus,
CH4(g) + 4Br2(l) -> CBr4(g) + 4HBr(l)
| Count: | 1 C atom | 1 C atom |
| 4 H atoms | 4 H atom | |
| 8 Br atoms | 8 Br atoms |
In a more complexexample, let’s balance the following equation:
MoF6 + Mo(CO)6 -> MoF3 + CO
When we inspect this equation we see that none of the 4 types of atoms are balanced.
MoF6 + Mo(CO)6 -> MoF3 + CO (unbalanced)
| Count: | 6 F atoms | 3 F atoms |
| 6 C atoms | 1 C atoms | |
| 2 Mo atoms | 1 Mo atom | |
| 6 O atoms | 1 O atoms |
So we start with an atom that appears in a species only once on each side of the equation, such as fluorine. We can get 6 F atoms on the right by placing a 2 in front of MoF3
MoF6 + Mo(CO)6 -> 2MoF3 + CO (unbalanced)
This change balances the Mo on both sides.
We now balance C. We have 6 atoms on the left and 1 C on the right. We can balance the C atoms by placing a 6 in front of CO:
MoF6 + Mo(CO)6 ® 2MoF3 + 6CO (balanced)
Finally, let’s balance the reaction when potassium chlorate (KClO3) is heated, and oxygen gas is evolved and solid potassium chloride (KCl) is formed. First write the chemical reaction equation in terms of the species described:
KClO3(s) -> KCl(s) + O2(g) (unbalanced)
Next, count the numbers of each type of elements on the reactants and products sides:
| Count: | 1 K atom | 1 K atom |
| 1 Cl atom | 1 Cl atom | |
| 3 O atoms | 2 O atoms |
1. Notice that the number of K atoms and Cl atoms are balanced.
KClO3(s) -> KCl(s) + O2(g) (unbalanced)
2. Since the reactants have 3 O atoms and the oxygen on the product side is a diatomic molecule, we must use a fractional coefficient 3/2 to balance the O atoms (see step 3c above).
KClO3(s) -> KCl(s) + 3/2O2(g) (balanced)
3. Atoms do not occur as fractions and so the final balanced equation must include only whole number coefficients. Multiply all coefficients by the denominator in the fraction (in this case 2) to get whole number coefficients.
2KClO3(s) -> 2KCl(s) + 3O2(g) (balanced)
In summary, there is no right or wrong way to balance a chemical equation. The following simple steps can be helpful:
1. Write the chemical equation with correct formulas. Do not change these formulas in the course of balancing the equation.
2. Count the number of each type of atom on both sides of the chemical equation.
3. Try different coefficients (whole numbers) in front of each substance to give equal numbers of each atom on both sides of the arrow:
a) for equations that include a substance that has a more complex formula than the other substances, try to balance the number of atoms in that substance by changing coefficients for the substance(s) that also contain those atom types because changing coefficients for simpler substances affects fewer elements; then finish by changing coefficients for atoms that occur as free elements; or
b) start changing the coefficient for a substance that appears the fewest times on each side for the chemical equation. Often, this will be a “heavy” atom (non-hydrogen atom).
c) if a diatomic molecule is a reactant or product, leave that substance to balance last and use a fractional coefficient (such as 3/2, 5/2, etc.) to give the correct number of that type of atom.
4. If you end up with fractional coefficients, multiply the entire equation through by the denominator (usually 2) to get whole number coefficients.
Combustion reactions are particularly nice examples of balancing chemical equations. Combustion is the process of burning an organic (carbon containing) compound in the presence of oxygen to liberate carbon dioxide and water. This means that all combustion reactions have O2 as a reactant and CO2 and H2O as products. Moreover, the combustion of fuels is everywhere in our lives: the energy derived from petroleum products to power our cars involves combustion of hydrocarbons in the cylinders; our bodies derive the energy they need to function through the combustion of glucose from carbohydrate in our diets; the combustion of fat to produce water for hydration sustains bears during hibernation and (since their humps are largely made of fat) camels on long trips across the desert.
Consider the combustion of propane, the fuel used by camping stoves and barbeque tanks. Propane is a hydrocarbon (made of carbon and hydrogen only) with the molecular formula C3H8. The first step is to write the unbalanced chemical equation with correct formulas:
C3H8(g) + O2(g) -> CO2(g) + H2O(l) (unbalanced)
| Count: | 3 C atoms | 1 C atom |
| 8 H atoms | 2 H atom | |
| 2 O atoms | 3 O atoms |
To balance the equation for a complicated reaction, a good rule of thumb is to keep the coefficient of the most complex substance unchanged (see step 3a above). In the present example, the most complex substance is propane; so the steps are:
1. C3H8 is the most complex substance so we will balance the C and H atoms by placing coefficients in front of the other substances.
2. Balance the carbon atoms by placing 3 in front of CO2
C3H8(g) + O2(g) -> 3CO2(g) + H2O(l) (unbalanced
3. Balance the hydrogen atoms by placing a 4 in front of H2O
C3H8(g) + O2(g) -> 3CO2(g) + 4H2O(l) (unbalanced)
4. Lastly, balance the oxygen atoms (10 on the right hand side) using the coefficient 5 for O2 to give the equation
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l) (balanced)