Module 7 Lesson 3: Writing Chemical Equations Study Notes for Chemical Equations

Overview of Module 7: Lesson 3

This lesson focuses on the process of writing chemical equations, specifically transitioning from word equations to formula equations.
The lesson is structured into two parts:
- Part 1: Initial instruction and practice.
- Part 2: Continued practice and the final submission of notes on the second day.

Fundamental Steps for Writing Chemical Equations

To convert a word chemical equation into a formula chemical equation, follow these four primary steps:

Recall Formula Naming Rules: Apply naming conventions for both ionic and covalent compounds.
Account for Diatomic Elements: Remember that there are seven specific elements that only exist in diatomic form in nature.
Include Appropriate Symbols: Add state symbols and reaction indicators (e.g., physical states, heat, catalysts).
Balance the Equation: The final step is always to ensure the number of atoms for each element is equal on both sides of the equation.

Standard Symbols in Chemical Equations

Various symbols are utilized within chemical equations to provide specific context regarding the reaction conditions and states of matter:

Plus Sign ( $+$ ): Separates two reactants or two products.
Yields Arrow ( $\rightarrow$ ): Separates the reactant side from the product side; means "yields" or "produces."
Double-Ended Arrow ( $\rightleftharpoons$ ): Indicates a reversible reaction that can proceed in both the forward and backward directions.
Parenthetical States of Matter:
- $(s)$ : Solid state.
- $(l)$ : Liquid state.
- $(g)$ : Gas state.
- $(aq)$ : Aqueous solution; indicates a compound or molecule is dissolved in water.
$\Delta$ (Triangle) or the word "Heat": Placed above the yields arrow to signify that heat is required for the reaction to occur or was supplied to the reaction.
Element/Substance Name above the Arrow: Indicates a catalyst. A catalyst is an element or compound added to speed up a reaction without being consumed or rearranged into the products themselves. (Note: Enzymes are biological catalysts to be discussed in later lessons).

Review of Oxidation Numbers

Oxidation numbers are critical for writing ionic formulas. An oxidation number indicates how many electrons an atom will gain or lose when participating in a chemical bond.

Periodic Table Assignments:
- Group 1: $+1$ (Loses one electron).
- Group 2: $+2$ (Loses two electrons).
- Group 13: $+3$ (Loses three electrons).
- Group 14: $+4$ or $-4$ .
- Group 15: $-3$ (Gains three electrons).
- Group 16: $-2$ (Gains two electrons).
- Group 17: $-1$ (Gains one electron).
- Group 18 (Noble Gases): $0$ (Stable; does not gain or lose electrons).
Transition Metals: Often have multiple oxidation numbers requiring Roman numerals, with specific exceptions like Silver ( $Ag$ ), which is always $+1$ .

Example 1: Beryllium Chloride and Silver Nitrate reacting in Water

The Scenario: Dissolved beryllium chloride reacts with dissolved silver nitrate in water to produce aqueous beryllium nitrate and silver chloride powder.

Step 1: Identify Ionic Compounds:
- Beryllium Chloride: Beryllium ( $Be^{2+}$ ) and Chloride ( $Cl^{1-}$ ). Using the "criss-cross" method, the formula is $BeCl_2$ . Because it is dissolved, the state is $(aq)$ .
- Silver Nitrate: Silver ( $Ag^{1+}$ ) and Nitrate ( $NO_3^{1-}$ from the polyatomic ion chart on page 7). The formula is $AgNO_3$ . State is $(aq)$ .
- Beryllium Nitrate: Beryllium ( $Be^{2+}$ ) and Nitrate ( $NO_3^{1-}$ ). The formula is $Be(NO_3)_2$ . State is $(aq)$ .
- Silver Chloride: Silver ( $Ag^{1+}$ ) and Chloride ( $Cl^{1-}$ ). The formula is $AgCl$ . Since it is described as a "powder," the state is $(s)$ .
Step 2: Balancing the Equation:
- Unbalanced: $BeCl_2(aq) + AgNO_3(aq) \rightarrow Be(NO_3)_2(aq) + AgCl(s)$
- Inventory Tracking: There are 2 chlorides on the left, so add a 2 in front of $AgCl$ . This creates 2 silvers, so add a 2 in front of $AgNO_3$ .
- Check: 1 Beryllium, 2 Chlorines, 2 Silvers, and 2 Nitrates on both sides.
Final Balanced Equation: $BeCl_2(aq) + 2AgNO_3(aq) \rightarrow Be(NO_3)_2(aq) + 2AgCl(s)$

Example 2: Combustion of Isopropanol

The Scenario: Isopropanol ( $C_3H_8O$ ) burns in oxygen; carbon dioxide, water, and heat are produced.

Key Identification: This is a combustion reaction. Anything burning in the presence of oxygen producing carbon dioxide, water, and heat fits this classification.
Reactants:
- Isopropanol: Given as $C_3H_8O$ . This is a covalent compound (non-metals), so oxidation numbers are not used.
- Oxygen: Oxygen is a diatomic element. It must be written as $O_2$ , never just $O$ .
Products:
- Carbon Dioxide: $CO_2$ .
- Water: $H_2O$ .
- Heat: Represented as "Heat" at the end to show the reaction is exothermic.
Balancing Process:
- Unbalanced: $C_3H_8O + O_2 \rightarrow CO_2 + H_2O + \text{Heat}$
- Step-by-Step:
  1. 3 Carbons on left -> $3CO_2$ .
  2. 8 Hydrogens on left -> $4H_2O$ .
  3. Oxygen Count: Products have 6 (from $CO_2$ ) + 4 (from $H_2O$ ) = 10 oxygens. Reactants have 1 (from isopropanol) and 2 (from $O_2$ ). This requires 9 more oxygens from the $O_2$ source.
  4. Using the fraction method: $9/2$ for $O_2$ , then multiplying the entire equation by 2 to remove the fraction.
Final Balanced Equation: $2C_3H_8O + 9O_2 \rightarrow 6CO_2 + 8H_2O + \text{Heat}$

Example 3: Sodium Metal and Iron (II) Chloride

The Scenario: Sodium metal reacts with iron (II) chloride to form iron metal and sodium chloride.

State Rules: All metals are solid at room temperature except for Mercury ( $Hg$ ).
Reactants:
- Sodium metal: Elements by themselves have an oxidation number of 0. Formula: $Na(s)$ .
- Iron (II) Chloride: Iron ( $Fe^{2+}$ —indicated by the Roman numeral II) and Chloride ( $Cl^{1-}$ ). Formula: $FeCl_2(s)$ .
Products:
- Iron metal: $Fe(s)$ .
- Sodium Chloride: Sodium ( $Na^{1+}$ ) and Chloride ( $Cl^{1-}$ ). Formula: $NaCl(s)$ .
Balancing Process:
- Unbalanced: $Na(s) + FeCl_2(s) \rightarrow Fe(s) + NaCl(s)$
- Step-by-Step: 2 Chlorines on the left require a coefficient of 2 for $NaCl$ . This resulting 2 sodiums on the right require a coefficient of 2 for the reactant $Na$ .
Final Balanced Equation: $2Na(s) + FeCl_2(s) \rightarrow Fe(s) + 2NaCl(s)$

Important Diatomic Element Reminder

Diatomic elements must always be written with a subscript of 2 when they are alone in a reaction. They can be remembered by starting at Hydrogen and then forming a "7" shape on the periodic table:

Hydrogen ( $H_2$ )
Nitrogen ( $N_2$ )
Oxygen ( $O_2$ )
Fluorine ( $F_2$ )
Chlorine ( $Cl_2$ )
Bromine ( $Br_2$ )
Iodine ( $I_2$ )