Chapter 4 part 1 CHEM1311

Main Ideas

1. Why Chemical Equations Must Be Balanced

Balancing follows the law of conservation of matter: matter is not created or destroyed in a chemical reaction; atoms are rearranged. Therefore, each element must have the same number of atoms among the reactants and products.

The reactants are written to the left of the reaction arrow, and the products are written to the right.

2. Subscripts and Coefficients

A subscript is part of a compound’s formula and indicates how many atoms are in each molecule or formula unit. Subscripts must not be changed when balancing, because changing one changes the substance.

A coefficient is the number placed in front of a formula. It indicates how many molecules or formula units are present and multiplies every atom in that formula. If no coefficient is written, it is understood to be 1.

3. Balancing Equations

A practical approach is to:

1. Count each type of atom on both sides.
2. Adjust coefficients only until the counts match.
3. Recount all atoms after each adjustment, since a coefficient affects every element in that formula.
4. When possible, balance an unchanged polyatomic ion as a unit. In some examples, leave oxygen until later.
5. If a fractional coefficient is needed, use it temporarily, then multiply all coefficients by the same number to obtain whole numbers.

Examples:

• Methane combustion: [ \mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O} ]

• Ethane combustion: [ \mathrm{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O} ] The lesson first obtains a fractional coefficient for oxygen, then multiplies every coefficient by 2.

• Sodium bromide and phosphoric acid reaction: [ \mathrm{3NaBr + H_3PO_4 \rightarrow Na_3PO_4 + 3HBr} ]

• Calcium hydroxide and aluminum sulfate reaction: [ \mathrm{3Ca(OH)_2 + Al_2(SO_4)_3 \rightarrow 3CaSO_4 + 2Al(OH)_3} ]

4. Using a Solubility Table to Assign States

The table is provided and does not need to be memorized. Students need to know how to use it to determine whether compounds are aqueous (aq) or solid (s).

An aqueous compound is dissolved in water. The dissolved substance is the solute, and water is the solvent. Liquid and gas are other states of matter.

The instructor’s method is to find the compound’s anion in the table and apply its default behavior:

• Anions in the table’s upper section are generally treated as soluble and aqueous, unless paired with a listed exception.
• Anions in the lower section are generally treated as forming solids, unless paired with a listed exception. An exception makes the compound aqueous.

The instructor emphasizes that the table’s alkali-metal-cation exceptions include the elements in Group 1, such as lithium, sodium, and potassium, even if not every one is explicitly listed.

The lesson applies the table to NaBr, H₃PO₄, Na₃PO₄, HBr, calcium hydroxide, aluminum sulfate, calcium sulfate, and aluminum hydroxide.

5. Writing a Net Ionic Equation

The lesson gives a four-step procedure:

1. Balance the molecular equation.
2. Determine the states of matter. These first two steps may be done in either order.
3. Write the complete ionic equation.
   • Split aqueous compounds into their ions.
   • Keep solids and liquids intact.
   • Include the correct coefficients, charges, and state labels.
   • When a dissolved compound separates into ions, the formula’s subscripts indicate how many ions of each type are present; they do not mean those ions remain bonded together.
4. Cancel spectator ions—ions that appear unchanged on both sides—and rewrite what remains as the net ionic equation.

Example from the lesson:

• Balanced molecular equation: [ \mathrm{CaCl_2(aq) + 2AgNO_3(aq) \rightarrow Ca(NO_3)_2(aq) + 2AgCl(s)} ]

• Calcium and nitrate ions appear unchanged on both sides, so they are spectators. Silver and chloride form the solid precipitate.

• The lesson writes the net ionic equation as: [ \mathrm{2Cl^-(aq) + 2Ag^+(aq) \rightarrow 2AgCl(s)} ] Its equivalent simplest whole-number form is: [ \mathrm{Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)} ]

The net ionic equation highlights the particles actually involved in the reaction, rather than including ions that remain unchanged in solution.

Speaker and Source

• Aleksandra Chuprin — instructor and narrator.