Chapter 7 – Chemical Reactions and Quantities: Study Notes

Chapter 7 – Chemical Reactions and Quantities

Section 7.1: Equations for Chemical Reactions

Chemical reactions involve the transformation of substances into new substances with different formulas and properties. For example, when iron (Fe) reacts with oxygen (O2), the product is rust (Fe2O3).

• Chemical change: A process in which a substance is converted into one or more new substances with different formulas and properties.

• Possible observations:

  • Formation of bubbles

  • Color change

  • Production of a solid (precipitate)

  • Heat produced or absorbed

Symbols Used in Chemical Equations

• An arrow (→) separates reactants from products.

• Reactants are written on the left side of the arrow; products on the right.

• Multiple reactants or products are separated by a + sign.

• The delta symbol (Δ) above the arrow indicates heat is used to start the reaction.

• Physical states are denoted in parentheses:

  • (s) = solid

  • (l) = liquid

  • (g) = gas

  • (aq) = aqueous (dissolved in water)

Example: $\mathrm{C}(s) + \mathrm{O}_2(g) \xrightarrow{\Delta} \mathrm{CO}_2(g)$

Identifying a Balanced Equation

• Law of Conservation of Matter: No atoms are lost or created in a chemical reaction.

• The number of atoms of each element on the reactant side must equal the number on the product side.

Example: $\mathrm{C}(s) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)$ (1 C atom and 2 O atoms on both sides)

Guide to Balancing a Chemical Equation

1. Write the equation using correct formulas for reactants and products.

2. Count the atoms of each element in reactants and products.

3. Use coefficients to balance each element (never change subscripts).

4. Check the final equation to confirm it is balanced (lowest whole number ratio).

Example: Formation of aluminum sulfide: $2\mathrm{Al}(s) + 3\mathrm{S}(s) \rightarrow \mathrm{Al}_2\mathrm{S}_3(s)$

Practice Problems

• Balance: $\mathrm{P}_4(s) + 6\mathrm{Br}_2(l) \rightarrow 4\mathrm{PBr}_3(g)$

• Balance: $\mathrm{Fe}_2\mathrm{O}_3(s) + 3\mathrm{H}_2(g) \rightarrow 2\mathrm{Fe}(s) + 3\mathrm{H}_2\mathrm{O}(l)$

Section 7.2: Types of Reactions

Chemical reactions can be classified into several types based on the patterns of reactants and products.

• Combination (Synthesis) Reaction: Two or more elements/compounds combine to form one product. Example: $2\mathrm{Mg}(s) + \mathrm{O}_2(g) \rightarrow 2\mathrm{MgO}(s)$

• Decomposition Reaction: One substance splits into two or more simpler substances. Example: $2\mathrm{KClO}_3(s) \rightarrow 2\mathrm{KCl}(s) + 3\mathrm{O}_2(g)$

• Single Replacement Reaction: One element replaces another in a compound. Example: $\mathrm{Fe}(s) + \mathrm{CuSO}_4(aq) \rightarrow \mathrm{FeSO}_4(aq) + \mathrm{Cu}(s)$

• Double Replacement Reaction: Ions in two compounds exchange places. Example: $\mathrm{AgNO}_3(aq) + \mathrm{NaCl}(aq) \rightarrow \mathrm{AgCl}(s) + \mathrm{NaNO}_3(aq)$

• Combustion Reaction: A hydrocarbon reacts with oxygen to produce carbon dioxide and water, releasing energy. Example: $\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(g) + \text{energy}$

Section 7.3: Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between substances.

• Oxidation: Loss of electrons (or gain of oxygen).

• Reduction: Gain of electrons (or loss of oxygen).

Example: Formation of copper(II) oxide: $2\mathrm{Cu}(s) + \mathrm{O}_2(g) \rightarrow 2\mathrm{CuO}(s)$

Section 7.4: The Mole

The mole is a counting unit in chemistry, similar to terms like dozen or gross. It allows chemists to count atoms, molecules, or ions by relating them to a measurable quantity.

• Avogadro's Number: $6.02 \times 10^{23}$ particles per mole.

• 1 mole of any substance contains $6.02 \times 10^{23}$ entities (atoms, molecules, ions, etc.).

Conversion Factors:

• $1\ \text{mole} = 6.02 \times 10^{23}\ \text{particles}$

• $\frac{1\ \text{mole}}{6.02 \times 10^{23}\ \text{particles}}$ or $\frac{6.02 \times 10^{23}\ \text{particles}}{1\ \text{mole}}$

Guide to Calculating Atoms or Molecules

1. State the given and needed quantities.

2. Use Avogadro's number to write conversion factors.

3. Set up the problem to cancel units and calculate the number of particles.

Example: How many CO2 molecules are in 0.500 mole of CO2? $0.500\ \text{mol} \times \frac{6.02 \times 10^{23}\ \text{molecules}}{1\ \text{mol}} = 3.01 \times 10^{23}\ \text{molecules}$

Moles of Elements in a Formula

Subscripts in a chemical formula indicate the number of atoms of each element in one molecule and the number of moles of each element in one mole of the compound.

Example: Aspirin, C9H8O4

• 1 molecule: 9 C, 8 H, 4 O atoms

• 1 mole: 9 moles C, 8 moles H, 4 moles O

Conversion Factors:

• $\frac{9\ \text{moles C}}{1\ \text{mole C}_9\text{H}_8\text{O}_4}$, $\frac{8\ \text{moles H}}{1\ \text{mole C}_9\text{H}_8\text{O}_4}$, $\frac{4\ \text{moles O}}{1\ \text{mole C}_9\text{H}_8\text{O}_4}$

Guide to Calculating Moles of an Element in a Compound

1. State the given and needed quantities.

2. Use Avogadro's number and subscripts to write conversion factors.

3. Set up the problem to cancel units and calculate the number of particles.

Example: How many atoms of O are there in 0.150 moles of C9H8O4? $0.150\ \text{mol} \times \frac{4\ \text{mol O}}{1\ \text{mol C}_9\text{H}_8\text{O}_4} \times \frac{6.02 \times 10^{23}\ \text{atoms}}{1\ \text{mol}} = 3.61 \times 10^{23}\ \text{atoms O}$

Section 7.5 & 7.6: Molar Mass and Calculations

The molar mass of an element or compound is the mass of one mole of that substance, expressed in grams per mole (g/mol). It is numerically equal to the atomic or formula mass from the periodic table.

• Examples:

  • 1 mole C = 12.01 g C

  • 1 mole Li = 6.941 g Li

Guide to Calculating Molar Mass of a Compound:

1. Obtain the molar mass of each element from the periodic table.

2. Multiply each molar mass by the number of atoms (subscript) in the formula.

3. Add the masses of all elements to get the total molar mass.

Example: Lithium carbonate, Li2CO3: $2 \times 6.941\ \text{g Li} + 12.01\ \text{g C} + 3 \times 16.00\ \text{g O} = 73.89\ \text{g/mol}$

Calculations Using Molar Mass for Compounds

• Molar mass is used as a conversion factor between grams and moles.

• Example: 1 mole Li2CO3 = 73.89 g Li2CO3

• Conversion factors: $\frac{73.89\ \text{g}}{1\ \text{mol}}$ or $\frac{1\ \text{mol}}{73.89\ \text{g}}$

Guide to Calculating Moles from Mass or Mass from Moles

1. State the given and needed quantities.

2. Write conversion factors, including molar mass.

3. Set up the problem to cancel units and calculate the needed quantity.

Example: How many moles of NaCl are in 737 g of NaCl? $737\ \text{g} \times \frac{1\ \text{mol}}{58.44\ \text{g}} = 12.6\ \text{mol}$

Map: Mass–Moles–Particles

Mass	Moles	Particles
Grams of element	Moles of element	Atoms (or ions)
Grams of compound	Moles of compound	Molecules (or formula units)

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

Section 7.7: Mole-to-Mole Relationships in Chemical Equations

Balanced chemical equations provide the mole ratios needed to relate quantities of reactants and products. These ratios are used as conversion factors in stoichiometric calculations.

• Law of Conservation of Mass: The total mass of reactants equals the total mass of products.

• Example: $2\mathrm{Ag}(s) + \mathrm{S}(s) \rightarrow \mathrm{Ag}_2\mathrm{S}(s)$

Information from a Balanced Equation:

Reactants	Products
2 Ag atoms + 1 S atom	1 Ag2S formula unit
2 moles Ag + 1 mole S	1 mole Ag2S
2 × 107.9 g Ag + 32.07 g S	247.9 g Ag2S

Mole-Mole Factors from a Chemical Equation (Stoichiometry)

Mole-to-mole relationships between reactants and products are written as conversion factors using the coefficients from the balanced equation.

Example: $2\mathrm{Fe}(s) + 3\mathrm{S}(s) \rightarrow 1\mathrm{Fe}_2\mathrm{S}_3(s)$ Conversion factors: $\frac{2\ \text{mol Fe}}{3\ \text{mol S}}$, $\frac{1\ \text{mol Fe}_2\mathrm{S}_3}{2\ \text{mol Fe}}$, etc.

Guide to Calculating Quantities of Reactants and Products

1. State the given and needed quantities.

2. Write conversion factors (use coefficients for mole-to-mole ratios).

3. Set up the problem to cancel units and calculate the needed quantity.

Example: How many moles of Fe are needed for the reaction of 12.0 moles of O2? $4\mathrm{Fe}(s) + 3\mathrm{O}_2(g) \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3(s)$ $12.0\ \text{mol O}_2 \times \frac{4\ \text{mol Fe}}{3\ \text{mol O}_2} = 16.0\ \text{mol Fe}$

Practice Problems

• The reaction between H2 and O2 produces 13.1 moles of water. How many moles of O2 reacted? $2\mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 2\mathrm{H}_2\mathrm{O}(g)$ $13.1\ \text{mol H}_2\mathrm{O} \times \frac{1\ \text{mol O}_2}{2\ \text{mol H}_2\mathrm{O}} = 6.55\ \text{mol O}_2$

Additional info:

• These notes cover the foundational concepts of chemical equations, types of reactions, the mole, molar mass, and stoichiometry, which are essential for GOB Chemistry students.