Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products. This is important because a chemical equation must obey the law of conservation of mass and the law of constant proportions, i.e. the same number of atoms of each element must exist on the reactant side and the product side of the equation.
Two quick and easy methods of balancing a chemical equation are discussed in this article. The first method is the traditional balancing method and the second one is the algebraic balancing method.
Table of Content
Related Terminology |
The Traditional Balancing Method (Method 1) |
The Algebraic Balancing Method (Method 2) |
Solved Examples |
Related Terminology
Chemical Equation
- A chemical equation is a symbolic representation of a chemical reaction in which the reactants and products are denoted by their respective chemical formulae.
- An example of a chemical equation is 2H_{2} + O_{2} â†’ 2H_{2}O which describes the reaction between hydrogen and oxygen to form water
- The reactant side is the part of the chemical equation to the left of the â€˜â†’â€™ symbol whereas the product side is the part to the right of the arrow symbol.
Stoichiometric Coefficient
- A stoichiometric coefficient describes the total number of molecules of a chemical species that participate in a chemical reaction.
- It provides a ratio between the reacting species and the products formed in the reaction.
- In the reaction described by the equation CH_{4} + 2O_{2} â†’ CO_{2} + 2H_{2}O, the stoichiometric coefficient of O_{2} and H_{2}O is 2 whereas that of CH_{4} and CO_{2} is 1.
- The total number of atoms of an element present in a species (in a balanced chemical equation) is equal to the product of the stoichiometric coefficient and the number of atoms of the element in one molecule of the species.
- For example, the total number of oxygen atoms in the reacting species â€˜2O_{2}â€™ is 4.
- While balancing chemical equations, stoichiometric coefficients are assigned in a manner that balances the total number of atoms of an element on the reactant and product side.
The Traditional Balancing Method
The first step that must be followed while balancing chemical equations is to obtain the complete unbalanced equation. In order to illustrate this method, the combustion reaction between propane and oxygen is taken as an example.
Step 1
- The unbalanced equation must be obtained from the chemical formulae of the reactants and the products (if it is not already provided).
- The chemical formula of propane is C_{3}H_{8}. It burns with oxygen (O_{2}) to form carbon dioxide (CO_{2}) and water (H_{2}O)
- The unbalanced chemical equation can be written as C_{3}H_{8} + O_{2} â†’ CO_{2} + H_{2}O
Step 2
The total number of atoms of each element on the reactant side and the product side must be compared. For this example, the number of atoms on each side can be tabulated as follows.
Chemical Equation: C_{3}H_{8} + O_{2} â†’ CO_{2} + H_{2}O | |
Reactant Side | Product Side |
3 Carbon atoms from C_{3}H_{8} | 1 Carbon atom from CO_{2} |
8 Hydrogen atoms from C_{3}H_{8} | 2 Hydrogen atoms from H_{2}O |
2 Oxygen atoms from O_{2} | 3 Oxygen atoms, 2 from CO_{2} and 1 from H_{2}O |
Step 3
- Now, stoichiometric coefficients are added to molecules containing an element which has a different number of atoms in the reactant side and the product side.
- The coefficient must balance the number of atoms on each side.
- Generally, the stoichiometric coefficients are assigned to hydrogen and oxygen atoms last.
- Now, the number of atoms of the elements on the reactant and product side must be updated.
- It is important to note that the number of atoms of an element in one species must be obtained by multiplying the stoichiometric coefficient with the total number of atoms of that element present in 1 molecule of the species.
- For example, when the coefficient 3 is assigned to the CO_{2} molecule, the total number of oxygen atoms in CO_{2} becomes 6. In this example, the coefficient is first assigned to carbon, as tabulated below.
Chemical Equation: C_{3}H_{8} + O_{2} â†’ 3CO_{2} + H_{2}O | |
Reactant Side | Product Side |
3 Carbon atoms from C_{3}H_{8} | 3 Carbon atoms from CO_{2} |
8 Hydrogen atoms from C_{3}H_{8} | 2 Hydrogen atoms from H_{2}O |
2 Oxygen atoms from O_{2} | 7 Oxygen atoms, 6 from CO_{2} and 1 from H_{2}O |
Step 4
Step 3 is repeated until all the number of atoms of the reacting elements are equal on the reactant and product side. In this example, hydrogen is balanced next. The chemical equation is transformed as follows.
Chemical Equation: C_{3}H_{8} + O_{2} â†’ 3CO_{2} + 4H_{2}O | |
Reactant Side | Product Side |
3 Carbon atoms from C_{3}H_{8} | 3 Carbon atoms from CO_{2} |
8 Hydrogen atoms from C_{3}H_{8} | 8 Hydrogen atoms from H_{2}O |
2 Oxygen atoms from O_{2} | 10 Oxygen atoms, 6 from CO_{2} and 4 from H_{2}O |
Now that the hydrogen atoms are balanced, the next element to be balanced is oxygen. There are 10 oxygen atoms on the product side, implying that the reactant side must also contain 10 oxygen atoms.
Each O_{2} molecule contains 2 oxygen atoms. Therefore, the stoichiometric coefficient that must be assigned to the O_{2} molecule is 5. The updated chemical equation is tabulated below.
Chemical Equation: C_{3}H_{8} + 5O_{2} â†’ 3CO_{2} + 4H_{2}O | |
Reactant Side | Product Side |
3 Carbon atoms from C_{3}H_{8} | 3 Carbon atoms from CO_{2} |
8 Hydrogen atoms from C_{3}H_{8} | 8 Hydrogen atoms from H_{2}O |
10 Oxygen atoms from O_{2} | 10 Oxygen atoms, 6 from CO_{2} and 4 from H_{2}O |
Step 5
- Once all the individual elements are balanced, the total number of atoms of each element on the reactant and product side are compared once again.
- If there are no inequalities, the chemical equation is said to be balanced.
- In this example, every element now has an equal number of atoms in the reactant and product side.
- Therefore, the balanced chemical equation is C_{3}H_{8} + 5O_{2} â†’ 3CO_{2} + 4H_{2}O.
The Algebraic Balancing Method
This method of balancing chemical equations involves assigning algebraic variables as stoichiometric coefficients to each species in the unbalanced chemical equation. These variables are used in mathematical equations and are solved to obtain the values of each stoichiometric coefficient. In order to better explain this method, the reaction between glucose and oxygen that yields carbon dioxide and water has been considered as an example.
Step 1
- The unbalanced chemical equation must be obtained by writing the chemical formulae of the reactants and the products.
- In this example, the reactants are glucose (C_{6}H_{12}O_{6}) and oxygen (O_{2}) and the products are carbon dioxide (CO_{2}) and water (H_{2}O)
- The unbalanced chemical equation is C_{6}H_{12}O_{6} + O_{2} â†’ CO_{2} + H_{2}O
Step 2
Now, algebraic variables are assigned to each species (as stoichiometric coefficients) in the unbalanced chemical equation. In this example, the equation can be written as follows.
aC_{6}H_{12}O_{6} + bO_{2} â†’ cCO_{2} + dH_{2}O
Now, a set of equations must be formulated (between the reactant and product side) in order to balance each element in the reaction. In this example, the following equations can be formed.
The equation for Carbon
- On the reactant side, â€˜aâ€™ molecules of C_{6}H_{12}O_{6} will contain â€˜6aâ€™ carbon atoms.
- On the product side, â€˜câ€™ molecules of CO_{2} will contain â€˜câ€™ carbon atoms.
- In this equation, the only species containing carbon are C_{6}H_{12}O_{6} and CO_{2}.
Therefore, the following equation can be formulated for carbon: 6a = c
Â The equation for Hydrogen
- The species that contain hydrogen in this equation are C_{6}H_{12}O_{6} and H_{2}
- â€˜aâ€™ molecules of C_{6}H_{12}O_{6} contains â€˜12aâ€™ hydrogen atoms whereas â€˜dâ€™ H_{2}O molecules will contain â€˜2dâ€™ hydrogen atoms.
- Therefore, the equation for hydrogen becomes 12a = 2d.
Simplifying this equation (by dividing both sides by 2), the equation becomes:
Â 6a = d
The equation for Oxygen
Every species in this chemical equation contains oxygen. Therefore, the following relations can be made to obtain the equation for oxygen:
- For â€˜aâ€™ molecules of C_{6}H_{12}O_{6}, there exist â€˜6aâ€™ oxygen atoms.
- â€˜bâ€™ molecules of O_{2} contain a total of â€˜2bâ€™ oxygens.
- â€˜câ€™ molecules of CO_{2} contain â€˜2câ€™ number of oxygen atoms.
- â€˜dâ€™ molecules of H_{2}O hold â€˜dâ€™ oxygen atoms.
Therefore, the equation for oxygen can be written as:
6a + 2b = 2c+ d
Step 3
The equations for each element are listed together to form a system of equations. In this example, the system of equations is as follows:
6a = c (for carbon); 6a = d (for hydrogen); 6a + 2b = 2c + d (for oxygen)
This system of equations can have multiple solutions, but the solution with minimal values of the variables is required. To obtain this solution, a value is assigned to one of the coefficients. In this case, the value of a is assumed to be 1. Therefore, the system of equations is transformed as follows:
a = 1
c = 6a = 6*1 = 6
d = 6a = 6
Substituting the values of a,c, and d in the equation 6a + 2b = 2c + d, the value of â€˜bâ€™ can be obtained as follows:
6*1 + 2b = 2*6 + 6
2b = 12; b = 6
It is important to note that these equations must be solved in a manner that each variable is a positive integer. If fractional values are obtained, the lowest common denominator between all the variables must be multiplied with each variable. This is necessary because the variables hold the values of the stoichiometric coefficients, which must be a positive integer.
Step 4
- Now that the smallest value of each variable is obtained, their values can be substituted into the chemical equation obtained in step 2.
- Therefore, aC_{6}H_{12}O_{6} + bO_{2} â†’ cCO_{2} + dH_{2}O becomes: C_{6}H_{12}O_{6} + 6O_{2} â†’ 6CO_{2} + 6H_{2}O
- Thus, the balanced chemical equation is obtained.
The algebraic method of balancing chemical equations is considered to be more efficient than the traditional method. However, it can yield fractional values for the stoichiometric coefficients, which must then be converted into integers.
Solved Examples
Some examples describing the balancing of chemical equations are provided in this subsection. These equations have been balanced using both the methods described above.
Example 1
Unbalanced chemical equation: Al + O_{2} â†’ Al_{2}O_{3}
Traditional Method
Following the traditional method, the reaction can be balanced as follows:
Equation: Al + O_{2} â†’ Al_{2}O_{3} | |
Reactant Side | Product Side |
1 Aluminum atom | 2 Aluminum Atoms |
2 Oxygen atoms | 3 Oxygen atoms |
First, the aluminum atoms are balanced. The equation becomes 2Al + O_{2} â†’ Al_{2}O_{3}
Now, the oxygen atoms must be balanced, there are two oxygen atoms on the reactant side and 3 on the product side. Therefore, there must be 3 O_{2} molecules that yield 2 Al_{2}O_{3} atoms. The chemical equation is transformed into 2Al + 3O_{2} â†’ 2Al_{2}O_{3}
Since the number of aluminum atoms on the product side has doubled, so must the number on the reactant side.
Equation: 4Al + 3O_{2} â†’ 2Al_{2}O_{3} | |
Reactant Side | Product Side |
4 Aluminum Atoms | 4 Aluminum Atoms |
6 Oxygen Atoms | 6 Oxygen Atoms |
Since each element is balanced, the balanced chemical equation is found to be 4Al + 3O_{2} â†’ 2Al_{2}O_{3}
Algebraic Method
Using the algebraic method of balancing chemical equations, the following variables can be assigned to the unbalanced equation.
aAl + bO_{2} â†’ cAl_{2}O_{3}
The equation for Aluminum: a = 2c
The equation for Oxygen: 2b = 3c
Assuming a = 1, we get:
c = a/2 ; c = 1/2
2b = 3*(Â½) = 3/2 ; b = Â¾
Since fractional values of b and c are obtained, the lowest common denominator between the variables a, b, and c must be found and multiplied with each variable. Since the lowest common denominator is 4, each of the variables must be multiplied by 4.
Therefore, a = 4*1 = 4 ;Â b = (Â¾)*4 = 3 ; c = (Â½)*4 = 2
Substituting the values of a, b, and c in the unbalanced equation, the following balanced chemical equation is obtained.
4Al + 3O_{2} â†’ 2Al_{2}O_{3}
Example 2
Unbalanced chemical equation: N_{2} + H_{2} â†’ NH_{3}
Traditional Method
In this reaction, the nitrogen atoms are balanced first. The reactant side has two nitrogen atoms, implying that 2 molecules of NH_{3} must be formed for each N_{2} molecule.
Chemical Equation: N_{2} + H_{2} â†’ 2NH_{3} | |
Reactant Side | Product Side |
2 nitrogen atoms | 2 nitrogen atoms |
2 hydrogen atoms | 6 hydrogen atoms |
Each H_{2} molecule contains 2 hydrogen atoms. In order to balance the number of hydrogen atoms in the equation, the total number of hydrogen atoms must be equal to 6. Therefore, the stoichiometric coefficient that must be assigned to hydrogen is 3.
Chemical Equation: N_{2} + 3H_{2} â†’ 2NH_{3} | |
Reactant Side | Product Side |
2 nitrogen atoms | 2 nitrogen atoms |
6 hydrogen atoms | 6 hydrogen atoms |
Thus, the balanced chemical equation is N_{2} + 3H_{2} â†’ 2NH_{3}
Algebraic Method
The variables a, b, and c must be assigned to N_{2}, H_{2}, and NH_{3} respectively. The chemical equation can be written as:
aN_{2} + bH_{2} â†’ cNH_{3}
The equation for nitrogen: 2a = c
The equation for hydrogen: 2b = 3c
Assuming a = 1, the values of b and c can be obtained as follows.
c = 2a = 2
2b = 3c = 3*2 = 6; b = 6/2 = 3
Since a, b, and c have no common multiples, they can be substituted into the equation as follows.
N_{2} + 3H_{2} â†’ 2NH_{3}
This is the balanced form of the given chemical equation.
Exercises
In order to practice different methods of balancing chemical equations, the following unbalanced equations can be worked on.
- FeCl_{3} + NaOH â†’ NaCl + Fe(OH)_{3}
- Zn + HCl â†’ ZnCl_{2} + H_{2}
- P_{2}O_{5} + H_{2}O â†’ H_{3}PO_{4}
- FeSO_{4} + NaOH â†’ Na_{2}SO_{4} + Fe(OH)_{2}
- Mg + HCl â†’Â MgCl_{2} + H_{2}
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