Balancing a chemical equation algebraically involves setting up a system of linear equations based on the conservation of mass for each element, assigning variables as coefficients, and solving for these variables.
The algebraic method provides a systematic way to balance chemical equations, which can be particularly useful for more complex reactions. It relies on the principle that the number of atoms of each element must be the same on both sides of the equation.
Here's a breakdown of the process:
Steps for Balancing Using the Algebraic Method
Based on the provided reference information, the steps are as follows:
Step 1: Write an unbalanced chemical equation for the chemical reaction.
Begin with the unbalanced equation representing the reaction. This shows the correct chemical formulas for all reactants and products involved.
Example: The reaction of iron (Fe) with oxygen (O₂) to form iron(III) oxide (Fe₂O₃).
Unbalanced equation: Fe + O₂ → Fe₂O₃
Step 2: Assign the variables as the coefficients to each reactant and product.
Assign an unknown variable (like a, b, c, d, etc.) as the stoichiometric coefficient in front of each chemical formula in the unbalanced equation.
Example: aFe + bO₂ → cFe₂O₃
Step 3: Create algebraic equations.
For each element present in the chemical equation, write a separate algebraic equation. This equation represents the conservation of atoms for that specific element. The total number of atoms of an element on the reactant side must equal the total number of atoms of that same element on the product side.
To do this, multiply the variable coefficient by the subscript of the element in each compound where it appears. Sum these values for the reactant side and set them equal to the sum for the product side.
Example: For aFe + bO₂ → cFe₂O₃
- For Iron (Fe): The total number of Fe atoms on the left is
a(fromaFe). On the right, it's2c(fromcFe₂O₃, where there are 2 Fe atoms per molecule).- Equation for Fe:
a = 2c
- Equation for Fe:
- For Oxygen (O): The total number of O atoms on the left is
2b(frombO₂, where there are 2 O atoms per molecule). On the right, it's3c(fromcFe₂O₃, where there are 3 O atoms per molecule).- Equation for O:
2b = 3c
- Equation for O:
You now have a system of linear equations:
a = 2c2b = 3c
Step 4: Assume the value for variable a is 1 and solve the algebraic equations to get the values of each variable.
To find a unique set of coefficients, assume a value for one of the variables. It's common practice to set the first variable (a) equal to 1.
Example: Set a = 1.
Now substitute this value into the equations and solve for the other variables:
- From
a = 2c: Substitutea=1→1 = 2c. Solving forc, we getc = 1/2. - From
2b = 3c: Substitute the value ofc(1/2) →2b = 3 * (1/2)→2b = 3/2. Solving forb, we getb = (3/2) / 2→b = 3/4.
The values obtained for the variables are:
a = 1b = 3/4c = 1/2
Obtaining Whole Number Coefficients
Chemical equations are conventionally balanced using the smallest possible whole number coefficients. If your solutions are fractions, multiply all variable values by the least common multiple (LCM) of the denominators to convert them into integers.
Example: The denominators are 4 and 2. The LCM of 4 and 2 is 4.
Multiply all values by 4:
a = 1 * 4 = 4b = (3/4) * 4 = 3c = (1/2) * 4 = 2
These are your balanced coefficients.
Final Balanced Equation
Substitute the whole number values back into the equation from Step 2:
aFe + bO₂ → cFe₂O₃ becomes 4Fe + 3O₂ → 2Fe₂O₃
To verify, count the atoms on each side:
| Element | Reactants (4Fe + 3O₂) | Products (2Fe₂O₃) |
|---|---|---|
| Fe | 4 * 1 = 4 | 2 * 2 = 4 |
| O | 3 * 2 = 6 | 2 * 3 = 6 |
The number of atoms for each element is the same on both sides, confirming the equation is balanced.
This algebraic technique is a robust method applicable to any chemical reaction, providing a clear path to balancing even complex equations where balancing by inspection might be challenging.
