Exam 3 Study Guide: Chemical Composition, Reactions, and Stoichiometry

Chapter 6: Chemical Composition

Calculating Molar Mass of Compounds

The molar mass of a compound is the mass of one mole of its molecules or formula units, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in the chemical formula.

• Step 1: Identify the number of each type of atom in the formula.

• Step 2: Multiply the number of atoms by their respective atomic masses (from the periodic table).

• Step 3: Add the total masses for all elements.

Example: For H2O: (2 × 1.01 g/mol) + (1 × 16.00 g/mol) = 18.02 g/mol

Conversions Between Moles, Atoms, Molecules, and Grams

Conversions between these quantities use Avogadro's number and molar mass as conversion factors.

• Moles to Atoms/Molecules: $\text{Number of particles} = \text{moles} \times 6.022 \times 10^{23}$

• Grams to Moles: $\text{Moles} = \frac{\text{grams}}{\text{molar mass}}$

• Moles to Grams: $\text{Grams} = \text{moles} \times \text{molar mass}$

Example: How many molecules are in 2.00 mol of CO2? $2.00 \text{ mol} \times 6.022 \times 10^{23} = 1.20 \times 10^{24}$ molecules$

Conversions Between Compound Mass, Moles, and Atoms

To convert between mass, moles, and number of atoms in a compound, use the following sequence:

• Mass → Moles → Molecules → Atoms

Example: How many hydrogen atoms are in 5.00 g of H2O?

1. Find moles: $\frac{5.00 \text{ g}}{18.02 \text{ g/mol}} = 0.277 \text{ mol}$

2. Find molecules: $0.277 \text{ mol} \times 6.022 \times 10^{23} = 1.67 \times 10^{23}$ molecules

3. Find H atoms: $1.67 \times 10^{23} \times 2 = 3.34 \times 10^{23}$ H atoms

Percent Composition of Compounds

Percent composition is the percentage by mass of each element in a compound.

• Formula: $\%\,\text{Element} = \frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100\%$

Example: Percent O in H2O: $\frac{16.00}{18.02} \times 100\% = 88.8\%$

Determination of Molecular and Empirical Formula

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms.

• Step 1: Convert mass % to grams (assume 100 g sample).

• Step 2: Convert grams to moles for each element.

• Step 3: Divide by the smallest number of moles to get the ratio.

• Step 4: Multiply to get whole numbers if necessary.

• Molecular formula: $\text{Molecular formula} = n \times \text{Empirical formula}$, where $n = \frac{\text{molar mass}}{\text{empirical formula mass}}$

Example: A compound with 40% C, 6.7% H, and 53.3% O has an empirical formula of CH2O.

Chapter 7: Chemical Reactions

Writing and Balancing Chemical Equations

Chemical equations represent chemical reactions. They must be balanced to obey the law of conservation of mass.

• Step 1: Write correct formulas for reactants and products.

• Step 2: Balance atoms by adjusting coefficients.

Example: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$ becomes $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Reaction Classification

Chemical reactions can be classified into several types:

• Combination (Synthesis): $A + B \rightarrow AB$

• Decomposition: $AB \rightarrow A + B$

• Single Replacement: $A + BC \rightarrow AC + B$

• Double Replacement: $AB + CD \rightarrow AD + CB$

• Combustion: Hydrocarbon + O2 → CO2 + H2O

Writing Molecular, Total, and Net Ionic Equations

Reactions in aqueous solution can be represented in three ways:

• Molecular equation: Shows all compounds as neutral formulas.

• Total ionic equation: Shows all strong electrolytes as ions.

• Net ionic equation: Shows only the species that change during the reaction.

Example: For NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq): Net ionic: $\text{Ag}^+ (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s)$

Writing and Balancing Acid/Base Equations

Acid-base reactions involve the transfer of H+ ions. They are balanced like other reactions, ensuring both mass and charge are conserved.

Example: $\text{HCl} (aq) + \text{NaOH} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2\text{O} (l)$

Writing and Balancing Gas Evolution Equations

Some reactions produce a gas as a product, such as CO2, H2, or SO2. These equations are balanced by the same principles.

Example: $\text{Na}_2\text{CO}_3 (aq) + 2\text{HCl} (aq) \rightarrow 2\text{NaCl} (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g)$

Chapter 8: Quantities in Chemical Reactions (Stoichiometry)

Using Balanced Chemical Equations for Calculations

Balanced equations provide the mole ratios needed for stoichiometric calculations.

Mole-to-Mole Conversion

• Use coefficients from the balanced equation to convert between moles of reactants and products.

Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ 2 mol H2 produces 2 mol H2O

Mass-to-Mass Conversion

• Convert mass of reactant to moles, use mole ratio, then convert to mass of product.

Example: How many grams of H2O are produced from 4.00 g H2?

1. Find moles H2: $\frac{4.00}{2.02} = 1.98$ mol

2. Mole ratio: 1.98 mol H2 × $\frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2}$ = 1.98 mol H2O

3. Mass H2O: $1.98 \times 18.02 = 35.7$ g

Determining Limiting Reactant

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Calculate the amount of product possible from each reactant; the smallest amount indicates the limiting reactant.

Determining Theoretical and Percent Yield

• Theoretical yield: Maximum amount of product possible, calculated from limiting reactant.

• Percent yield: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Example: If theoretical yield is 10.0 g and actual yield is 8.5 g: $\frac{8.5}{10.0} \times 100\% = 85\%$

Enthalpy of Reaction

Enthalpy change ($\Delta H$) is the heat absorbed or released during a chemical reaction at constant pressure.

• Exothermic: $\Delta H$ is negative (heat released)

• Endothermic: $\Delta H$ is positive (heat absorbed)

Example: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$, $\Delta H = -890$ kJ

Additional info: Academic context and examples have been added to ensure completeness and clarity for exam preparation.