Real Reactions: Balancing Equations

Outline

• Applying the Conservation of Mass to Real Reactions
  • Writing Equations and Balancing Them
  • When is a Law not a Law?
• Practice with Concepts
• Discussion Questions
• Optional Website Reading

Applying the Conservation of Mass to Real Reactions

The chemistry chant of the day is: in a chemical reaction, no atoms are created or destroyed. Keep this firmly in mind, and you can't go far astray. In a chemical reaction, all the atoms on one side of the reaction arrow (equal sign) must show up on the other side. You are not allowed to lose them or change them into something else (that's physics; we do that next year); you must balance the equation, just like in algebra.

Writing Equations and Balancing Them

Here's what happens when you burn propane in air:

C₃H₈ (g) + O₂(g)--> CO₂(g) + H₂O(l)

Notice that we indicate the physical state -- gas, liquid, solid, in aqueous solution -- of each reactant (original compounds) and product. This equation only shows us what molecules and elements are involved, but not how many atoms of each kind.

For a more accurate picture, we need to balance each element, so that there are the same number of product element atoms as reactant element atoms. WE DON'T CHANGE SUBSCRIPTS! These indicate the ratio of elements in the molecule, and define the kind of compound; if we change them, we change the products or reactants. We have to change the number of molecules instead.

Since the oxygen exists on the reactant side as in a oxygen-only molecule (non-compound), we will leave it until last, so we can adjust the oxygen without affecting anything else. Balance C first (this choice is arbitrary, we could have done H first). There are 3 Cs in the propane, and only one currently on the product side, so we have to multiply the product molecule containing C by 3:

C₃H₈ (g) + O₂(g)--> 3CO₂(g) + H₂O(l)

There are 8 hydrogens on the reactant side, and only 2 on the product side, so we need to multiply the water molecule by 4 to get 8 on the product side:

C₃H₈ (g) + O₂(g)--> 3CO₂(g) + 4H₂O(l)

Now we need to balance the oxygens. There are 3*2 + 4*1 oxygens on the product side, for a total of 10. So we need to make sure we have 10 on the reactant side. Multiply the O₂ molecule by 5:

C₃H₈ (g) + 5O₂(g)--> 3CO₂(g) + 4H₂O(l)

So 1 molecule of propane reacts with 5 molecules of oxygen gas to form 3 molecules of carbon dioxide and 4 molecules of water.

When is a law not a law?

Answer: when you have a nonstoichiometric solid. Such solids violate the law of constant composition. If certain crystals are irradiated, (exposed to high-energy ultraviolet light), exposed to high heat, or otherwise put in a high-energy situation, some of their component ions will escape the crystal, leaving only an electron behind to hold their place. The ratio of whole atoms of one element ot whole atoms of another element is upset. Not only do such molecules bend our perception of the basic laws of chemistry, but they exhibit very odd behaviors as well. Many nonstoichiometric metal oxides are superconductors, offering no measurable resistance to low levels of electrical current.

Practice with the Concepts

Discussion Questions

• Conservation of matter drives how we balance equations, but we must also conserver energy. Why is an equation that considers only atom rearrangement incomplete?
• What is the point of having natural laws if not all situations conform to the law as stated?

Optional Readings

• There is a nice short interactive at Exploration Series: Chemical Interactions, if you want to practice balancing chemistry reactions, just to see how this works....but we'll have more extensive interactives later.

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