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Balancing Chemical Equations: Rules & Examples | Chemca
Foundation Chemistry
Balancing Simple Chemical Equations
By Chemca Editorial Team•March 2026•8 min read
According to the Law of Conservation of Mass, matter can neither be created nor destroyed in a chemical reaction. Therefore, the total number of atoms of each element must be the same on both the reactant and product sides of a chemical equation.
1. The Hit and Trial Method
The most basic method for balancing simple chemical equations is the Hit and Trial method. Here, we use the smallest whole number coefficients to equalize the number of atoms.
Core Rules:
Never change subscripts: Changing $H_2O$ to $H_2O_2$ changes the substance itself. Only change the coefficients (the numbers in front).
Balance polyatomic ions as a whole: If $SO_4$ appears on both sides, count it as one unit to save time.
Save Oxygen and Hydrogen for last: Usually, balancing metals and non-metals first makes the process easier.
2. Step-by-Step Example
Let's balance the combustion of Propane ($C_3H_8$):
$$ C_3H_8 + O_2 \rightarrow CO_2 + H_2O $$
Balance Carbon: There are 3 carbons on the left, so put a '3' in front of $CO_2$.
$$ C_3H_8 + O_2 \rightarrow \mathbf{3}CO_2 + H_2O $$
Balance Hydrogen: There are 8 hydrogens on the left, so put a '4' in front of $H_2O$ ($4 \times 2 = 8$).
$$ C_3H_8 + O_2 \rightarrow 3CO_2 + \mathbf{4}H_2O $$
Balance Oxygen: Count oxygens on the right: $(3 \times 2) + (4 \times 1) = 10$. To get 10 on the left, put a '5' in front of $O_2$.
$$ C_3H_8 + \mathbf{5}O_2 \rightarrow 3CO_2 + 4H_2O $$
3. Common Balancing Mistakes
Subscript Error: Writing $H_2 + O_2 \rightarrow H_2O_2$ to balance the oxygen. While mathematically "balanced," this is chemically wrong because $H_2O$ is water and $H_2O_2$ is hydrogen peroxide.
Dieckmann Condensation: The Intramolecular Claisen
By Chemca Editorial Team•March 2026•9 min read
The Dieckmann Condensation is the base-catalyzed intramolecular chemical reaction of diesters to yield cyclic $\beta$-keto esters. It is essentially the intramolecular version of the Claisen condensation.
1. General Reaction
A diester (typically a 1,6 or 1,7-diester) reacts with a base (such as Sodium Ethoxide, $NaOEt$) in an alcoholic solvent to form a 5 or 6-membered cyclic product.
The formed enolate acts as a nucleophile and attacks the carbonyl carbon of the second ester group within the same molecule, forming a cyclic tetrahedral intermediate.
Step 3: Elimination of Alkoxide
The tetrahedral intermediate collapses, expelling an ethoxide ion ($EtO^-$) and reforming the $C=O$ double bond to create the cyclic $\beta$-keto ester.
Step 4: Deprotonation (Driving Force)
The acidic $\alpha$-hydrogen between the two carbonyl groups is removed by the base. This step is irreversible and drives the equilibrium toward the product.
Crucial: If the product does not have an acidic $\alpha$-hydrogen to be removed in Step 4, the Dieckmann condensation will not occur effectively.
Strategy: Convert carboxylic acid to amide. Then use the Hoffmann Bromamide Degradation to remove the carbonyl carbon and obtain an amine with one less carbon.
NCERT Ex 12.15 Solved: Aldehydes & Ketones Conversions | ChemCa
ChemCa
NCERT Exercise 12.15 Solved
Detailed "2-step" conversion strategies for Aldehydes, Ketones, and Carboxylic Acids. Master reagents like Grignard, Rosenmund Catalyst, and Aldol Condensation.
Strategy: Cross-Aldol condensation with acetaldehyde gives Cinnamaldehyde. Catalytic hydrogenation reduces both the double bond and the aldehyde to alcohol.
(viii) Benzaldehyde to $\alpha$-Hydroxyphenylacetic acid
Strategy: Nitration first (-COOH is meta directing). Then reduce the carboxylic acid to alcohol (Diborane is preferred as it doesn't reduce the nitro group easily).
Concept: Add HBr (Markovnikov) to move the leaving group to the secondary carbon. Elimination yields the more substituted, stable alkene (But-2-ene) as the major product.
