Laws of Chemical Combinations: Definition, Common Mistakes to Avoid & Real-Life Applications

Proposed by Joseph Proust in 1799, this law states that a given chemical compound always contains its constituent elements in a fixed ratio by mass, regardless of the source or preparation method. For instance, in carbon dioxide ( ${\mathrm{C}\mathrm{O}}_2$ ), the mass ratio of carbon to oxygen is always: $\mathrm{C}:\mathrm{O}=12:32=3:8$

Whether ${\mathrm{C}\mathrm{O}}_2$ is formed from burning coal or limestone decomposition, this ratio remains constant. JEE questions often involve calculating mass ratios in compounds to confirm this law.

Proposed by Antoine Lavoisier in 1789, this law states that the total mass of reactants equals the total mass of products in a chemical reaction. Matter is neither created nor destroyed during a reaction. For example, in the combustion of carbon: $12\mathrm{g}{\mathrm{C}}^{+}32\mathrm{g}{\mathrm{O}}_2\to 44\mathrm{g}{\mathrm{C}\mathrm{O}}_2$

The mass of reactants ( $12\mathrm{g}+32\mathrm{g}=44\mathrm{g}$ ) equals the mass of the product ( 44 g ). This law is foundational for balancing chemical equations, ensuring the same number of atoms on both sides. For JEE, verify mass balance in stoichiometric problems.

Proposed by John Dalton in 1803, this law applies when two elements form more than one compound. The masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers. For example, hydrogen and oxygen form water $\left({\mathrm{H}}_2\mathrm{O}\right)$ and hydrogen peroxide $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ :

1. In ${\mathrm{H}}_2\mathrm{O}:2\mathrm{g}\mathrm{H}$ combines with 16 g O .

2. In ${\mathrm{H}}_2{\mathrm{O}}_2:2\mathrm{g}\mathrm{H}$ combines with 32 g O .

The masses of oxygen ( 16 g and 32 g ) for a fixed 2 g of hydrogen are in the ratio $16:32=1:2$ . This law supports Dalton's atomic theory, as it implies atoms combine in simple ratios. JEE problems may ask for mass ratios in compounds like CO and ${\mathrm{C}\mathrm{O}}_2$ or ${\mathrm{N}}_2\mathrm{O}$ and ${\mathrm{N}\mathrm{O}}_2$ .

Proposed by Joseph Louis Gay-Lussac in 1808, this law states that when gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at the same temperature and pressure. For example, in the formation of water vapor: $100\mathrm{m}\mathrm{L}{\mathrm{H}}_2+50\mathrm{m}\mathrm{L}{\mathrm{O}}_2\to 100\mathrm{m}\mathrm{L}{\mathrm{H}}_2\mathrm{O}\text{(vapor)}$

The volumes of hydrogen and oxygen ( 100 mL and 50 mL ) combine in a 2:1 ratio. This law is effectively the law of definite proportions applied to gas volumes. JEE questions often involve calculating volume ratios in gaseous reactions, ensuring constant temperature and pressure.

Proposed by Amedeo Avogadro in 1811, this law states that equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules. For the reaction of hydrogen and oxygen to form water: $2\text{volumes}{\mathrm{H}}_2+1\text{volume}{\mathrm{O}}_2\to 2\text{volumes}{\mathrm{H}}_2\mathrm{O}\text{(vapor)}$

Each volume contains the same number of molecules, explaining why volume ratios are simple whole numbers. Avogadro distinguished between atoms and molecules, suggesting hydrogen and oxygen are diatomic ( ${\mathrm{H}}_2,{\mathrm{O}}_2$ ), which clarified Gay-Lussac's observations. For JEE, apply Avogadro's law to relate gas volumes to moles, using the molar volume at STP ( $22.7\mathrm{L}/\mathrm{m}\mathrm{o}\mathrm{l}$ ).

The laws of chemical combinations are critical for solving JEE Main problems, particularly in:

1. Stoichiometry: Use the law of conservation of mass to balance equations and calculate reactant/product masses.

2. Compound Composition: Apply the law of definite proportions to find mass percentages in compounds.

3. Multiple Compounds: Use the law of multiple proportions to determine mass ratios in compounds formed by the same elements.

4. Gas Reactions: Apply Gay-Lussac's and Avogadro's laws to calculate volume ratios or moles in gaseous reactions.

Example problem in JEE Main Exam: Calculate the mass of ${\mathrm{C}\mathrm{O}}_2$ produced from 12 g of carbon burning in excess oxygen.

$\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{C}\mathrm{O}}_2$

Moles of $\mathrm{C}=\frac{12}{12}=1\mathrm{m}\mathrm{o}\mathrm{l}$ . From the equation, 1 mol C produces $1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{C}\mathrm{O}}_2\left(44\mathrm{g}\right)$ . Thus, 44 g ${\mathrm{C}\mathrm{O}}_2$ is produced, satisfying conservation of mass.