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 relationships between reactants and products.

• Chemical equations use formulas to represent substances involved in a reaction.

• Arrows (→) separate reactants (starting materials, left side) from products (ending materials, right side).

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

Example: The reaction of hydrogen and oxygen to form water:

• $2H_2 + O_2 \rightarrow 2H_2O$

Balancing Equations

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

• Balance by changing coefficients (numbers in front of formulas), not subscripts (numbers within formulas).

• Start with elements that appear in only one reactant and one product.

• Use trial and error, adjusting coefficients as needed, and check all elements at the end.

Example: Combustion of methane:

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

Why Use Coefficients Instead of Changing Subscripts?

• Changing subscripts alters the identity of the substance (e.g., $H_2O$ vs. $H_2O_2$ are different compounds).

• Coefficients adjust the quantity, not the identity, of substances.

Example:

• $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$ (water)

• $H_2(g) + O_2(g) \rightarrow H_2O_2(l)$ (hydrogen peroxide)

Required Symbols in Chemical Equations

• States of matter are indicated in parentheses:

  • (g) = gas

  • (l) = liquid

  • (s) = solid

  • (aq) = aqueous (dissolved in water)

Example: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$

3.2 Simple Patterns of Chemical Reactivity

Chemical reactions can be classified into several types. Three broad classes are:

• Combination reactions

• Decomposition reactions

• Combustion reactions

Combination Reactions

Two or more substances react to form one product.

• General form: $A + B \rightarrow C$

Combination Reactions	Description
$C(s) + O_2(g) \rightarrow CO_2(g)$	Carbon reacts with oxygen to form carbon dioxide.
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$	Nitrogen reacts with hydrogen to form ammonia.
$CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq)$	Calcium oxide reacts with water to form calcium hydroxide.

Combination Reaction Prediction: Metal and Nonmetal

• Products can often be predicted using common charges for groups in the periodic table.

• Example: Magnesium reacts with oxygen to form magnesium oxide: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

Decomposition Reactions

One substance breaks down into two or more substances.

• General form: $C \rightarrow A + B$

Decomposition Reactions	Description
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$	Potassium chlorate decomposes to potassium chloride and oxygen.
$PbCO_3(s) \rightarrow PbO(s) + CO_2(g)$	Lead(II) carbonate decomposes to lead(II) oxide and carbon dioxide.
$Cu(OH)_2(s) \rightarrow CuO(s) + H_2O(g)$	Copper(II) hydroxide decomposes to copper(II) oxide and water.

Example: In automobile airbags, sodium azide ($NaN_3$) decomposes rapidly to release nitrogen gas ($N_2$).

Predicting Decomposition Reactions: Heating a Metal Carbonate

• Metal carbonates decompose when heated to give off 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 and usually involve oxygen as a reactant.

• When burning hydrocarbons (compounds with C and H), the products are always $CO_2$ and $H_2O$.

• 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.

• For elements, the formula weight is the atomic weight (e.g., Na: 23.0 amu).

• For ionic compounds, use the empirical formula.

• Example: Sulfuric acid ($H_2SO_4$):

• $FW(H_2SO_4) = 2(1.0) + 32.1 + 4(16.0) = 98.1$ amu

Molecular Weight (MW)

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

• Example: Glucose ($C_6H_{12}O_6$):

• $MW(C_6H_{12}O_6) = 6(12.0) + 12(1.0) + 6(16.0) = 180.0$ amu

Percent Composition

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

• Equation:

• $\%\ \text{Element} = \left( \frac{\text{number of atoms of element} \times \text{atomic weight of element}}{\text{FW of the substance}} \right) \times 100$

Example: Percent C in glucose ($C_6H_{12}O_6$):

• $\%C = \frac{6 \times 12.0}{180.0} \times 100 = 40.0\%$

3.4 Avogadro’s Number

Avogadro’s number defines the number of particles in one mole of a substance.

• One mole (mol) contains $6.02 \times 10^{23}$ particles (atoms, molecules, or ions).

• This is the number of particles in exactly 12 g of carbon-12.

Molar Mass

• The molar mass is the mass of 1 mole of a substance (g/mol).

• For elements, the 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 the same number as the molar mass in g/mol.

Mole Relationships

Name of Substance	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 ions	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 units

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.

• The number of atoms of an element in a mole of a compound is the number of atoms in the formula times Avogadro’s number.

Converting Amounts Using Moles

• Moles provide a bridge from the molecular scale to the real-world scale.

• Conversions can be made between mass, moles, and number of particles using molar mass and Avogadro’s number.

Example: How many atoms in 3 g of copper (Cu)?

• $3\ \text{g Cu} \times \frac{1\ \text{mol Cu}}{63.5\ \text{g Cu}} \times \frac{6.02 \times 10^{23}\ \text{atoms}}{1\ \text{mol Cu}} = 3 \times 10^{22}$ 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.

• Steps to determine empirical formula (assuming 100 g sample):

  1. Convert mass percent to grams.

  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) analysis:

• C: 61.31 g, H: 5.14 g, N: 10.21 g, O: 23.33 g

• Convert to moles:

  • C: $61.31 \div 12.01 = 5.105$ mol

  • H: $5.14 \div 1.01 = 5.09$ mol

  • N: $10.21 \div 14.01 = 0.7288$ mol

  • O: $23.33 \div 16.00 = 1.456$ mol

• Divide by smallest (0.7288):

  • C: $5.105 \div 0.7288 = 7$

  • H: $5.09 \div 0.7288 = 7$

  • N: $0.7288 \div 0.7288 = 1$

  • O: $1.456 \div 0.7288 = 2$

• Empirical formula: $C_7H_7NO_2$

Molecular Formulas from Empirical Formulas

• The molecular formula is a whole-number multiple of the empirical formula.

• Equation:

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

Example: Empirical formula CH, molar mass 78 g/mol:

• FW(CH) = 13

• Whole-number multiple = $78 \div 13 = 6$

• Molecular formula: $C_6H_6$

3.6 Quantitative Information from Balanced Equations

Balanced equations provide quantitative relationships between reactants and products.

• Coefficients indicate relative numbers of molecules and moles, which can be converted to mass.

Example: $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$

• 2 molecules (or moles) of $H_2$ react with 1 molecule (or mole) of $O_2$ to produce 2 molecules (or moles) of $H_2O$.

Stoichiometric Calculations

• Stoichiometry uses mole ratios from balanced equations to relate quantities of reactants and products.

• General steps:

  1. Convert grams of substance A to moles of A (using molar mass).

  2. Use coefficients to convert moles of A to moles of B.

  3. Convert moles of B to grams of B (using molar mass).

Example: How many grams of water from 1.00 g glucose?

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

• Step 1: $1.00\ \text{g} \times \frac{1\ \text{mol}}{180.0\ \text{g}} = 0.00556$ mol $C_6H_{12}O_6$

• Step 2: $0.00556\ \text{mol} \times \frac{6\ \text{mol } H_2O}{1\ \text{mol } C_6H_{12}O_6} = 0.0333$ mol $H_2O$

• Step 3: $0.0333\ \text{mol} \times 18.0\ \text{g/mol} = 0.600\ \text{g } H_2O$

Heat and Stoichiometry

• If heat is involved, it does not appear in the balanced equation but may be indicated by the Greek symbol delta ($\Delta$) over the arrow.

• Amounts of heat are related to stoichiometry (see Chapter 5).

3.7 Limiting Reactants

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

• Other reactants are called excess reactants.

• To identify the limiting reactant, compare the mole ratios of reactants used in the balanced equation to the amounts available.

Example: If 10 molecules of $H_2$ and 7 molecules of $O_2$ are mixed, $H_2$ is the limiting reactant if it is used up first.

Theoretical Yield and Percent Yield

• Theoretical yield: The maximum amount of product that can be formed from the given amounts of reactants, calculated using stoichiometry.

• Actual yield: The amount of product actually obtained from a reaction.

• Percent yield: The ratio of actual yield to theoretical yield, expressed as a percentage.

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