The Law of Conservation of Matter

Translating Reality into Equations: Balancing Chemical Reactions

Because matter is conserved, we cannot write a chemical equation that is "unbalanced." If an equation shows four hydrogen atoms on the left side (the reactants) but only two on the right side (the products), that equation is physically impossible because it implies that two hydrogen atoms vanished into thin air.

To reflect physical reality, we use coefficients—numbers placed in front of chemical formulas—to ensure that the number of atoms for each element is identical on both sides of the equation.

The Anatomy of a Balanced Equation

Consider the synthesis of water. If we simply wrote: $H_{2} + O_{2} \rightarrow H_{2}O$

This equation is unbalanced. On the reactant side, we have two oxygen atoms, but on the product side, we only have one. To fix this, we must adjust the coefficients so that the "atom count" matches.

1. Step 1: Balance the Oxygen. We need two oxygens on the right, so we place a coefficient of 2 in front of $H_{2}O$: $H_{2} + O_{2} \rightarrow 2H_{2}O$
2. Step 2: Balance the Hydrogen. Now the right side has four hydrogens ($2 \times 2 = 4$). We must go back to the left side and place a 2 in front of $H_{2}$: $2H_{2} + O_{2} \rightarrow 2H_{2}O$

Now, both sides have exactly 4 hydrogens and 2 oxygens. The equation is balanced and obeys the Law of Conservation of Matter.

Worked Example: Balancing a Combustion Reaction

Combustion reactions (burning something in oxygen) are classic examples where conservation is vital for stoichiometry calculations. Let's balance the combustion of propane ($C_{3}H_{8}$).

Unbalanced Equation: $C_{3}H_{8} + O_{2} \rightarrow CO_{2} + H_{2}O$

Step-by-Step Solution:

1. Count the atoms on each side:

   • Reactants: $C=3, H=8, O=2$
   • Products: $C=1, H=2, O=3$ (2 from $CO\_{2}$and 1 from$H\_{2}O$)

2. Balance Carbon (C): Place a 3 in front of $CO_{2}$ to match the 3 carbons in propane. $C_{3}H_{8} + O_{2} \rightarrow 3CO_{2} + H_{2}O$

3. Balance Hydrogen (H): Place a 4 in front of $H_{2}O$ to match the 8 hydrogens in propane ($4 \times 2 = 8$). $C_{3}H_{8} + O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

4. Balance Oxygen (O): Now, count the oxygens on the product side: $(3 \times 2) + (4 \times 1) = 10$ oxygen atoms. To get 10 oxygens on the reactant side, place a 5 in front of $O_{2}$. $C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

Final Verification:

• Left side: $C=3, H=8, O=10$
• Right side: $C=3, H=8, O=10$

The equation is balanced, meaning the mass of the propane plus the mass of the oxygen used will exactly equal the mass of the carbon dioxide and water vapor produced.

Summary Table: Reactants vs. Products

By mastering the Law of Conservation of Matter, you gain the ability to predict exactly how much of a chemical you need to create a specific amount of product—a skill that forms the backbone of all chemical engineering and laboratory science.