Chap 6: Chemical Reactions: Mole and Mass Relationships

Chapter Six: Chemical Reactions – Mole and Mass Relationships

Introduction and Definitions of the Mole

The concept of the mole is fundamental in chemistry, serving as the bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. The mole allows chemists to count entities by weighing them, making it possible to relate chemical equations to laboratory measurements.

• The Mole (mol): The SI unit for the amount of a substance. One mole contains exactly $6.022 \times 10^{23}$ entities (Avogadro's number).

• Avogadro’s Number ($N_A$): $6.022 \times 10^{23}$, representing the number of particles (atoms, molecules, or formula units) in one mole of a substance.

• Definition: A mole is the amount of a substance whose mass in grams is numerically equal to its molecular or formula weight.

• Example: 1 mole of water ($H_2O$) = 18.02 g = $6.022 \times 10^{23}$ molecules of water.

Additional info: Avogadro’s number is extremely large; for perspective, it is much greater than the population of Earth or the number of seconds in the age of the Earth.

Matter: Macroscopic versus Microscopic

The mole concept links the microscopic scale (atoms, molecules) to the macroscopic scale (grams, liters) used in the laboratory. The following table illustrates this relationship for elements, molecular compounds, and ionic compounds:

	Elements	Molecular Compounds	Ionic Compounds
Microscopic	1 atom Na	1 molecule of water	1 formula unit of NaCl
Macroscopic	1 mole Na is $6.022 \times 10^{23}$ Na atoms	1 mole of water is $6.022 \times 10^{23}$ molecules	1 mole of NaCl is $6.022 \times 10^{23}$ formula units
Mass (microscopic)	22.99 amu	18.02 amu	58.44 amu
Mass (macroscopic)	22.99 grams	18.02 grams	58.44 grams
Term	Atomic mass	Molecular mass	Formula mass

Key Terms and Definitions

• Molecular weight: The sum of atomic weights of all atoms in a molecule.

• Formula weight: The sum of atomic weights of atoms in one formula unit of any compound.

• Mole: The amount of substance whose mass in grams equals its molecular or formula weight.

• Avogadro’s number ($N_A$): The number of molecules or formula units in a mole ($6.022 \times 10^{23}$).

Mole and Mass Relationships and Calculations

Gram–Mole Conversions

Molar mass is the mass of one mole of a substance, numerically equal to its molecular or formula weight in grams. It serves as a conversion factor between the number of moles and the mass of a substance.

• Molar mass: $\text{Molar mass} = \text{mass of 1 mole of a substance}$

• Conversion factors:

  • From moles to grams: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

  • From grams to moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• Example: The molar mass of water is 18.0 g/mol. To convert 0.50 mol of water to grams: $0.50 \text{ mol} \times 18.0 \text{ g/mol} = 9.0 \text{ g}$

Mole Relationships in Chemical Equations

Balanced chemical equations show the mole-to-mole relationships between reactants and products. The coefficients in a balanced equation represent the number of moles of each substance involved.

• Mole ratios: Derived from the coefficients in a balanced equation and used as conversion factors.

• Example: For the reaction $N_2 + 3H_2 \rightarrow 2NH_3$:

  • Mole ratio of $H_2$ to $N_2$: $3:1$

  • Mole ratio of $NH_3$ to $N_2$: $2:1$

  • Conversion factor: $\frac{3 \text{ mol } H_2}{1 \text{ mol } N_2}$

• Important: Equations must be interpreted in terms of moles, not grams.

Types of Stoichiometric Calculations

There are three main types of stoichiometric conversions:

1. Mole to mole: Use mole ratios from the balanced equation.

2. Mole to mass (or mass to mole): Use molar mass as a conversion factor.

3. Mass to mass: Convert mass of A to moles of A, use mole ratio to find moles of B, then convert to mass of B.

Example (Mole to Mole): For $4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)$, how many moles of $Fe_2O_3$ are formed from 6.2 mol of Fe?

• Set up: $6.2 \text{ mol Fe} \times \frac{2 \text{ mol } Fe_2O_3}{4 \text{ mol Fe}} = 3.1 \text{ mol } Fe_2O_3$

Example (Mole to Mass): If a bottle contains 0.082 mol of ibuprofen (molecular weight 206.3 g/mol), the mass is:

• $0.082 \text{ mol} \times 206.3 \text{ g/mol} = 16.9 \text{ g}$

Mass-to-Mass Calculations

To determine the mass of a product or reactant, follow these steps:

1. Write the balanced chemical equation.

2. Convert the given mass to moles using molar mass.

3. Use the mole ratio from the balanced equation to find moles of the desired substance.

4. Convert moles of the desired substance to grams using its molar mass.

Example: If 0.022 g of CaC2O4 is produced, how much CaCl2 was used? (Molar mass CaC2O4 = 128.1 g/mol, CaCl2 = 111.0 g/mol)

• Convert grams CaC2O4 to moles, use mole ratio, then convert to grams CaCl2.

Percent Yield, Actual Yield, Theoretical Yield, and Limiting Reagents

Yield in Chemical Reactions

The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants, calculated using stoichiometry. The actual yield is the amount of product actually obtained from a reaction. The percent yield expresses the efficiency of a reaction.

• Percent yield formula:

  • $\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$

• Example: If the theoretical yield of CO2 is 88.0 g and the actual yield is 72.4 g:

  • $\text{Percent yield} = \frac{72.4}{88.0} \times 100\% = 82.3\%$

Limiting Reagent

In a chemical reaction, the limiting reagent is the reactant that is completely consumed first, thus limiting the amount of product formed. The other reactants are in excess.

• To identify the limiting reagent, compare the mole ratios of reactants used to those required by the balanced equation.

• Example: For $A_2 + 2B_2 \rightarrow 2AB_2$, if you have 8 mol $A_2$ and 8 mol $B_2$, $B_2$ is the limiting reagent because the reaction requires twice as much $B_2$ as $A_2$.

Summary of Key Concepts

• Chemical equations must be balanced; the number and kinds of atoms must be the same on both sides.

• A mole refers to Avogadro’s number of formula units of a substance. One mole has a mass equal to its formula weight in grams.

• Molar masses act as conversion factors between numbers of molecules and masses in grams.

• The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products.

• Mole ratios relate amounts of reactants and/or products. Using molar masses and mole ratios in calculations relates unknown masses to known masses or molar amounts.

• The yield is the amount of product obtained. The percent yield is the amount of product obtained divided by the amount theoretically possible, multiplied by 100%.