Chapter 3: Chemical Reactions and Reaction Stoichiometry – Study Notes

Chapter 3: Chemical Reactions and Reaction Stoichiometry

3.1 Chemical Equations

Chemical equations are the symbolic representation of chemical reactions, showing the transformation of reactants into products. They are essential for communicating chemical changes and for quantitative calculations in chemistry.

• Chemical equations use arrows to separate reactants (starting materials, left side) from products (ending materials, right side).

• The plus sign (+) separates multiple reactants or products.

• Example: $2H_2 + O_2 \rightarrow 2H_2O$

Balancing Equations

Balancing chemical equations ensures the Law of Conservation of Mass is obeyed, meaning the number of atoms of each element is the same on both sides of the equation.

• Start with an element that appears in only one reactant and one product.

• Balance by changing coefficients, not subscripts.

• Proceed by trial and error, adjusting coefficients for other elements as needed.

• Check all elements at the end to confirm balance.

• Example: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

Why Use Coefficients Instead of Changing Subscripts?

Changing subscripts alters the identity of the substance, while coefficients adjust the quantity. For example, water ($H_2O$) and hydrogen peroxide ($H_2O_2$) have different properties and formulas.

• Example: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$ vs. $H_2(g) + O_2(g) \rightarrow H_2O_2(l)$

Required Symbols in Chemical Equations

States of matter are indicated in parentheses:

• (g) = gas

• (l) = liquid

• (s) = solid

• (aq) = aqueous (dissolved in water)

• Example: $CH_4(g) + 2O_2(g) \xrightarrow{\Delta} CO_2(g) + 2H_2O(g)$

3.2 Simple Patterns of Chemical Reactivity

Chemical reactions can be classified into broad types, which help predict products and understand reaction mechanisms.

• Combination reactions

• Decomposition reactions

• Combustion reactions

Combination Reactions

Two or more substances react to form one product.

• General form: $A + B \rightarrow C$

• Example: $C(s) + O_2(g) \rightarrow CO_2(g)$

Combination Reactions	Description
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$	Formation of ammonia from nitrogen and hydrogen
$CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq)$	Formation of calcium hydroxide from calcium oxide and water

Combination Reaction Prediction: A Metal and a Nonmetal

Products of combination reactions between metals and nonmetals can be predicted using common charges for groups in the periodic table.

• Example: $2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$

Decomposition Reactions

In decomposition reactions, a single compound breaks down into two or more simpler substances.

• General form: $C \rightarrow A + B$

• Example: $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$

Decomposition Reactions	Description
$PbCO_3(s) \rightarrow PbO(s) + CO_2(g)$	Decomposition of lead(II) carbonate
$Cu(OH)_2(s) \rightarrow CuO(s) + H_2O(g)$	Decomposition of copper(II) hydroxide

Predicting Decomposition Reactions: Heating a Metal Carbonate

Metal carbonates decompose upon heating to yield carbon dioxide and a metal oxide.

• Example: $CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$

Combustion Reactions

Combustion reactions are rapid reactions that produce a flame, typically involving oxygen as a reactant.

• Example: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$

3.3 Formula Weight (FW)

The formula weight is the sum of the atomic weights of all atoms in a chemical formula, expressed in atomic mass units (amu).

• For elements, FW is the atomic weight.

• For ionic compounds, use the empirical formula.

• Example for $H_2SO_4$: $2(1.0) + 32.1 + 4(16.0) = 98.1$ amu

Molecular Weight (MW)

If the substance is a molecule, the formula weight is also called the molecular weight.

• MW is the sum of atomic weights of all atoms in a molecule.

• Example for glucose ($C_6H_{12}O_6$): $6(12.0) + 12(1.0) + 6(16.0) = 180.0$ amu

Percent Composition

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

• Formula: $\%\,\text{Element} = \left( \frac{\text{number of atoms of element} \times \text{atomic weight of element}}{\text{formula weight of substance}} \right) \times 100$

• Example for carbon in glucose: $\%C = \left( \frac{6 \times 12.0}{180.0} \right) \times 100 = 40.0\%$

3.4 Avogadro’s Number

Avogadro’s number ($6.02 \times 10^{23}$) is the number of particles (atoms, molecules, ions) in one mole of a substance.

• One mole is the amount of substance containing as many entities as there are atoms in exactly 12 g of carbon-12.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

• For elements, molar mass is the atomic weight from the periodic table.

• For diatomic elements, it is twice the atomic weight.

• The formula weight in amu is numerically equal to the molar mass in g/mol.

Mole Relationships

Mole relationships allow conversion between mass, moles, and number of particles using Avogadro’s number and molar mass.

Name	Formula	Formula Weight (amu)	Molar Mass (g/mol)	Number and Kind of Particles in One Mole
Atomic nitrogen	N	14.0	14.0	$6.02 \times 10^{23}$ N atoms
Molecular nitrogen	N2	28.0	28.0	$6.02 \times 10^{23}$ N2 molecules
Silver	Ag	107.9	107.9	$6.02 \times 10^{23}$ Ag atoms
Silver ion	Ag+	107.9	107.9	$6.02 \times 10^{23}$ Ag+ ions
Barium chloride	BaCl2	208.2	208.2	$6.02 \times 10^{23}$ BaCl2 formula units

Converting Amounts Using Moles

Moles serve as a bridge between the molecular scale and the real-world scale, allowing conversion between mass and number of particles.

• Example: How many atoms in 3 g of copper (Cu)? $3\,\text{g}\,Cu \times \left( \frac{1\,\text{mol}\,Cu}{63.5\,\text{g}\,Cu} \right) \times \left( \frac{6.02 \times 10^{23}\,\text{atoms}}{1\,\text{mol}\,Cu} \right) = 3 \times 10^{22}\,\text{atoms}$

3.5 Empirical Formulas from Analysis

The empirical formula gives the simplest whole-number ratio of atoms in a compound. It can be determined from percent composition data.

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

2. Convert grams to moles for each element.

3. Calculate the mole ratio by dividing by the smallest number of moles.

Example: Para-aminobenzoic acid (PABA)

• C: 61.31 g $\rightarrow$ $\frac{61.31}{12.01} = 5.105$ mol

• H: 5.14 g $\rightarrow$ $\frac{5.14}{1.01} = 5.09$ mol

• N: 10.21 g $\rightarrow$ $\frac{10.21}{14.01} = 0.7288$ mol

• O: 23.33 g $\rightarrow$ $\frac{23.33}{16.00} = 1.456$ mol

• Divide each by 0.7288: C: $\frac{5.105}{0.7288} \approx 7$ H: $\frac{5.09}{0.7288} \approx 7$ N: $\frac{0.7288}{0.7288} = 1$ O: $\frac{1.456}{0.7288} \approx 2$ Empirical formula: $C_7H_7NO_2$

Molecular Formulas From Empirical Formulas

The molecular formula is a whole-number multiple of the empirical formula. If the molar mass is known, the molecular formula can be determined.

• Whole number multiple = $\frac{\text{Molecular weight (MW)}}{\text{Empirical formula weight (FW)}}$

• Example: Empirical formula CH, MW = 78 g/mol, FW = 13 g/mol $\frac{78}{13} = 6$ Molecular formula: $C_6H_6$

Combustion Analysis

Combustion analysis is used to determine the empirical formula of compounds containing C, H, and O by measuring the masses of CO2 and H2O produced.

1. Mass of C is determined from the mass of CO2 produced.

2. Mass of H is determined from the mass of H2O produced.

3. Mass of O is determined by subtracting the masses of C and H from the total mass of the sample.

3.6 Quantitative Information from Balanced Equations

Balanced equations provide quantitative relationships between reactants and products, allowing calculation of masses, moles, and molecules involved.

• Coefficients indicate relative numbers of molecules and moles.

• Example: $2H_2 + O_2 \rightarrow 2H_2O$ means 2 molecules (or moles) of H2 react with 1 molecule (or mole) of O2 to produce 2 molecules (or moles) of H2O.

Stoichiometric Calculations

Stoichiometry involves using mole ratios from balanced equations to relate quantities of reactants and products.

• Convert grams to moles using molar mass.

• Use mole ratios from the balanced equation to convert between substances.

• Convert moles back to grams if needed.

Example: Grams of Water from Glucose

• Reaction: $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$

• Given: 1.00 g glucose

• Step 1: $1.00\,g\,C_6H_{12}O_6 \times \left( \frac{1\,mol}{180.0\,g} \right) = 0.00556\,mol$

• Step 2: $0.00556\,mol\,C_6H_{12}O_6 \times \left( \frac{6\,mol\,H_2O}{1\,mol\,C_6H_{12}O_6} \right) = 0.0333\,mol\,H_2O$

• Step 3: $0.0333\,mol\,H_2O \times 18.0\,g/mol = 0.600\,g\,H_2O$

Limiting Reactants

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

• Identify the limiting reactant by comparing the mole ratios of reactants used to those required by the balanced equation.

• The other reactant(s) are in excess.

Theoretical Yield and Percent Yield

Theoretical yield is the maximum amount of product that can be formed from given reactants, calculated using stoichiometry. Percent yield compares the actual yield to the theoretical yield.

• Theoretical yield: Maximum possible product from stoichiometric calculations.

• Percent yield: $\text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100$

Additional info: These notes cover the essential concepts and calculations of reaction stoichiometry, including empirical and molecular formulas, combustion analysis, and quantitative relationships in chemical equations, as required for a General Chemistry college course.