## Module 8: Calculations Involving Formulas and Equations This module covers essential calculations used in chemistry, focusing on formula weight, moles, percent composition, and stoichiometry. Let's dive in! ### 8.1 Formula Weight and Molecular Weight The terms "formula weight" (FW) and "molecular weight" (MW) are often used interchangeably, but there's a subtle difference: * **Formula Weight (FW):** The sum of the atomic weights (in atomic mass units, amu) of all atoms in a compound's formula. This applies to *both* ionic and molecular compounds. Let's calculate the formula weight of NaCl (sodium chloride): * Atomic weight of Na: 22.99 amu * Atomic weight of Cl: 35.45 amu * FW(NaCl) = 22.99 amu + 35.45 amu = 58.44 amu * **Molecular Weight (MW):** The sum of the atomic weights of all atoms in a *molecular* compound, expressed in amu. This is specifically for covalent compounds. While the calculation is the same, the term "molecular weight" emphasizes the discrete molecules present in covalent compounds. Remember that 1 amu is approximately equal to 1 gram per mole (g/mol). This is crucial for converting between amu and grams. ### 8.2 The Mole The mole (mol) is a fundamental unit in chemistry. It's defined as the amount of a substance that contains the same number of entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. This number is Avogadro's number, approximately $6.022 \times 10^{23}$. Think of a mole as a convenient counting unit for atoms, molecules, or ions, just like a dozen is a counting unit for eggs. The molar mass of a substance is its mass in grams per mole (g/mol), numerically equal to its formula weight or molecular weight in amu. The molar mass of NaCl is 58.44 g/mol. This means that one mole of NaCl weighs 58.44 grams. ### 8.3 Percent Composition Percent composition tells us the relative amounts of each element in a compound. It's calculated as: $$ \text{Percent Composition of Element X} = \frac{\text{Mass of Element X}}{\text{Total Mass of Compound}} \times 100\% $$ Let's find the percent composition of each element in NaCl: * Mass of Na in one mole of NaCl: 22.99 g * Mass of Cl in one mole of NaCl: 35.45 g * Total mass of one mole of NaCl: 58.44 g Percent Na = $\frac{22.99}{58.44} \times 100\% \approx 39.34\%$ Percent Cl = $\frac{35.45}{58.44} \times 100\% \approx 60.66\%$ ### 8.4 Stoichiometry Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. It relies heavily on balanced chemical equations. A common mistake is forgetting to balance the chemical equation before performing stoichiometric calculations. Always ensure your equation is balanced! Let's look at two examples: **Example 1:** $2KClO_3 \rightarrow 2KCl + 3O_2$ How many moles of KCl will be formed from 1.65 moles of KClO3? From the balanced equation, we see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, 1.65 moles of KClO3 will produce 1.65 moles of KCl. **Example 2:** $4Fe + 3O_2 \rightarrow 2Fe_2O_3$ How many grams of Fe2O3 are produced when 32.9 grams of O2 are reacted? 1. **Convert grams of O2 to moles:** The molar mass of O2 is 32.00 g/mol. So, $\frac{32.9 \text{ g}}{32.00 \text{ g/mol}} \approx 1.03 \text{ mol } O_2$. 2. **Use mole ratio from balanced equation:** 3 moles of O2 produce 2 moles of Fe2O3. Therefore, 1.03 moles of O2 will produce $\frac{2}{3} \times 1.03 \text{ mol} \approx 0.69 \text{ mol } Fe_2O_3$. 3. **Convert moles of Fe2O3 to grams:** The molar mass of Fe2O3 is 159.70 g/mol. So, $0.69 \text{ mol} \times 159.70 \text{ g/mol} \approx 110 \text{ g } Fe_2O_3$. ### 8.5 Key Takeaways * Formula weight and molecular weight are crucial for understanding the mass composition of compounds. * The mole is a fundamental unit for counting atoms, molecules, and ions. * Percent composition describes the relative amounts of each element in a compound. * Stoichiometry uses balanced chemical equations to determine the quantitative relationships between reactants and products. Mastering mole ratios is key to success in stoichiometry problems.