Moles and Chemical Composition: The Mole Concept, Molar Mass, and Chemical Calculations

Moles and Chemical Composition

Introduction

The concept of the mole is fundamental in chemistry, allowing chemists to count atoms, molecules, and ions by weighing them. This section introduces the mole, Avogadro's number, and the calculations involving chemical composition, including molar mass and conversions between mass, moles, and number of particles.

Learning Objectives

• Use Avogadro’s number to determine the number of particles in a given mass of a substance.

• Calculate formula mass, molecular mass, and molar mass of a substance.

• Convert between mass and moles of a substance.

• Identify and balance chemical equations.

• Calculate the mass or number of moles of a reactant or product from another reactant or product.

• Calculate the concentration (molarity, percent by volume, or percent by mass) of a solute in solution.

• Calculate the amount of solute or solution given the concentration and the other amount.

What and How Much?

Quantifying Substances in Chemistry

To understand chemical reactions and properties, it is essential to know both the amount of material present and its composition. This is crucial in various applications:

• Determining the amount of sodium in sodium chloride for dietary needs.

• Calculating the amount of iron in iron ore for steel production.

• Measuring the amount of hydrogen in water for hydrogen fuel.

• Estimating the amount of chlorine in freon to assess ozone depletion.

Counting by Mass: The Analogy of Nails

Counting Large Numbers of Objects

When dealing with large numbers of small objects (like nails, atoms, or molecules), it is often more practical to count them by mass rather than individually. For example, hardware stores sell nails by the kilogram, not by individual count.

• Example: If a dozen nails has a mass of 0.07 kg, and a customer buys 1.18 kg of nails, the number of nails can be calculated as follows:

Solution Map:

• Given: 1.18 kg nails

• Conversion: 1 dozen nails = 0.07 kg; 1 dozen = 12 nails

• Calculation:

  • Number of dozens = $\frac{1.18\ \text{kg}}{0.07\ \text{kg/dozen}} = 16.86$ dozen

  • Number of nails = $16.86 \times 12 = 202$ nails

This analogy helps understand how chemists count atoms and molecules by weighing them and using conversion factors.

The Mole Concept

Definition and Importance

The mole (mol) is the SI unit for the amount of substance. It is defined as the amount of a substance that contains exactly $6.022 \times 10^{23}$ particles (Avogadro's number). This number is analogous to a 'dozen,' but much larger, and is used for counting atoms, molecules, ions, or other specified entities.

• Avogadro's Number: $6.022 \times 10^{23}$ particles/mol

• Works for all objects: atoms, molecules, ions, electrons, etc.

• 1 mole of any element contains $6.022 \times 10^{23}$ atoms of that element.

Example: 1 mole of carbon-12 atoms has a mass of exactly 12 g and contains $6.022 \times 10^{23}$ atoms.

Molar Mass and Mass Relationships

Definition of Molar Mass

The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). For elements, the molar mass in grams is numerically equal to the atomic mass in atomic mass units (amu).

• Formula: $\text{Molar mass (g/mol)} = \text{Atomic or molecular mass (amu)}$

Table: Moles and Masses of Elements

Substance	Pieces in 1 mole	Weight of 1 mole (g)
Hydrogen	$6.022 \times 10^{23}$ atoms	1.008
Carbon	$6.022 \times 10^{23}$ atoms	12.01
Oxygen	$6.022 \times 10^{23}$ atoms	16.00
Sulfur	$6.022 \times 10^{23}$ atoms	32.06
Calcium	$6.022 \times 10^{23}$ atoms	40.08
Chlorine	$6.022 \times 10^{23}$ atoms	35.45
Copper	$6.022 \times 10^{23}$ atoms	63.55

Empirical and Molecular Formulas

Definitions and Differences

• Empirical Formula: The simplest whole-number ratio of elements in a compound.

• Molecular Formula: The actual number of atoms of each element in a molecule.

Examples:

• Hydrogen Peroxide: Molecular formula = H2O2, Empirical formula = HO

• Glucose: Molecular formula = C6H12O6, Empirical formula = CH2O

• Benzene: Molecular formula = C6H6, Empirical formula = CH

Calculations Involving Moles

Conversions Between Moles, Mass, and Number of Particles

• To convert from moles to number of particles:

  • $\text{Number of particles} = \text{moles} \times 6.022 \times 10^{23}$

• To convert from mass to moles:

  • $\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• To convert from moles to mass:

  • $\text{Mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Example Calculations

• Example 1: How many moles are in $1.1 \times 10^{22}$ silver atoms?

  • $\text{Moles} = \frac{1.1 \times 10^{22}}{6.022 \times 10^{23}} = 1.8 \times 10^{-2}$ moles

• Example 2: How many moles of sulfur are in 57.8 g of S?

  • Molar mass of S = 32.06 g/mol

  • $\text{Moles} = \frac{57.8}{32.06} = 1.80$ moles

• Example 3: How many aluminum atoms are in a 16.2 g aluminum can?

  • Molar mass of Al = 26.98 g/mol

  • Moles of Al = $\frac{16.2}{26.98} = 0.601$ moles

  • Number of atoms = $0.601 \times 6.022 \times 10^{23} = 3.62 \times 10^{23}$ atoms

Chemical Formulas as Conversion Factors

Using Formulas to Relate Moles of Elements and Compounds

• 1 H2O molecule = 2 H atoms + 1 O atom

• 1 mol H2O = 2 mol H + 1 mol O

• 1 mol CaCO3 = 1 mol Ca + 1 mol C + 3 mol O

Percent Composition

Definition and Calculation

Percent composition is the percentage by mass of each element in a compound. It can be determined from the chemical formula or by experimental analysis.

• Formula: $\text{Percent} = \frac{\text{part}}{\text{whole}} \times 100\%$

• For NaCl, if it is 39% Na by mass, then 100 g NaCl contains 39 g Na.

Percent composition can be used as a conversion factor in stoichiometric calculations.

Summary Table: Key Conversions in Chemical Calculations

From	To	Conversion Factor
Mass (g)	Moles	$\frac{1\ \text{mol}}{\text{molar mass (g)}}$
Moles	Mass (g)	$\text{molar mass (g)}$
Moles	Particles	$6.022 \times 10^{23}$ particles/mol
Particles	Moles	$\frac{1\ \text{mol}}{6.022 \times 10^{23}\ \text{particles}}$