Balancing Chemical Equations

Balancing Chemical Equations

• Balancing chemical equations is crucial for reflecting the laws of conservation of mass and charge.
• The mass of reactants consumed must equal the mass of products generated.
• The number of atoms of each element should be the same on both sides of the equation.
• Stoichiometric coefficients indicate the relative number of moles of each species.

Stoichiometric Coefficients

• Coefficients are placed in front of each compound.
• They indicate the relative number of moles of a given species.
• Generally, stoichiometric coefficients are whole numbers.
• Example: Combustion of nonane: $C<em>9H</em>{20}(g) + 14O<em>2(g) \rightarrow 9CO</em>2(g) + 10H_2O(l)$
  • 1 mole of $C<em>9H</em>{20}$ reacts with 14 moles of $O<em>2$ to produce 9 moles of $CO</em>2$ and 10 moles of $H_2O$.

Steps for Balancing Chemical Equations

• Ensures accurate calculations regarding the reaction.

Example: Balancing $C<em>4H</em>{10}(l) + O<em>2(g) \rightarrow CO</em>2(g) + H_2O(l)$

Method 1:

1. Balance Carbons:
   • Start with carbon as it appears only once on each side.
     $C<em>4H</em>{10}(l) + O<em>2(g) \rightarrow 4CO</em>2(g) + H_2O(l)$
2. Balance Hydrogens:
   • Hydrogen also appears only once on each side.
     $C<em>4H</em>{10}(l) + O<em>2(g) \rightarrow 4CO</em>2(g) + 5H_2O(l)$
3. Balance Oxygens:
   • Oxygen appears in multiple reactants and products, so balance it last.
   • Now there are 13 oxygen atoms on the product side.
     $C<em>4H</em>{10}(l) + \frac{13}{2}O<em>2(g) \rightarrow 4CO</em>2(g) + 5H_2O(l)$
4. Produce Whole Number Ratio:
   • Multiply each coefficient by 2 to eliminate the fraction.
     $2C<em>4H</em>{10}(l) + 13O<em>2(g) \rightarrow 8CO</em>2(g) + 10H_2O(l)$
5. Check:
   • Verify that all elements and total charges are balanced.
   • If there is a charge difference, balance the charge as well.

Method 2: If in Doubt, Take a Guess

1. Assume a coefficient:
   • Assume 4 moles of $C<em>4H</em>{10}$.
     $4C<em>4H</em>{10}(l) + O<em>2(g) \rightarrow 16CO</em>2(g) + H_2O(l)$
2. Balance Hydrogens:
   • 40 hydrogen atoms on the reactant side.
     $4C<em>4H</em>{10}(l) + O<em>2(g) \rightarrow 16CO</em>2(g) + 20H_2O(l)$
3. Balance Oxygens:
   • 52 oxygen atoms on the product side.
     $4C<em>4H</em>{10}(l) + 26O<em>2(g) \rightarrow 16CO</em>2(g) + 20H_2O(l)$
4. Simplify to Simplest Whole Number Ratio:
   • Divide all coefficients by the greatest common factor (2 in this case).
     $2C<em>4H</em>{10}(l) + 13O<em>2(g) \rightarrow 8CO</em>2(g) + 10H_2O(l)$
5. Check:
   • Verify all elements and charges are balanced.

Final Notes

• Both methods yield correct ratios.
• Simpler numbers facilitate easier stoichiometric calculations.
• Balancing charge in oxidation-reduction reactions: see chapter 11 of NCAT general chemistry review.