Chemical Quantities and the Mole: Chapter 7 Study Notes

Chapter 7: Chemical Quantities

The Mole Concept

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. It allows chemists to count particles (atoms, molecules, ions, etc.) by weighing, rather than by direct counting, which is impractical for such small entities.

• Definition: A mole (mol) is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 grams of carbon-12.

• Avogadro’s Number: The number of particles in one mole is 6.022 × 1023. This is known as Avogadro’s number.

• Analogy: Just as a dozen means 12 items, a mole means 6.022 × 1023 items (atoms, molecules, formula units, ions, etc.).

• Example: 1 mole of H2O contains 6.022 × 1023 molecules of water.

Counting Particles Using the Mole

Avogadro’s number allows conversion between the number of moles and the number of particles in a sample.

• Conversion:

  • Number of particles = moles × Avogadro’s number

  • Moles = number of particles / Avogadro’s number

• Example Calculation: To find the number of atoms in 4.00 moles of iron:

  • $4.00 \text{ mol Fe} \times 6.022 \times 10^{23} \text{ atoms/mol} = 2.41 \times 10^{24} \text{ atoms Fe}$

Table: Number of Particles in Sample Quantities

This table shows how one mole of various substances contains Avogadro’s number of particles, regardless of the substance.

Sample	Number and Type of Particles
Aluminum (Al)	6.022 × 1023 atoms of Al
Iron (Fe)	6.022 × 1023 atoms of Fe
Water (H2O)	6.022 × 1023 molecules of H2O
Vitamin C (C6H8O6)	6.022 × 1023 molecules of vitamin C
Sodium chloride (NaCl)	6.022 × 1023 formula units of NaCl
Potassium phosphate (K3PO4)	6.022 × 1023 formula units of K3PO4

Molar Mass

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

• Definition: Molar mass is the mass of 1 mole of a substance, measured in g/mol.

• Calculation: For elements, the molar mass is the atomic mass from the periodic table (in g/mol). For compounds, it is the sum of the atomic masses of all atoms in the formula.

• Example: The atomic mass of carbon is 12.01 amu, so the molar mass is 12.01 g/mol.

• Formula:

  • $\text{Molar mass} = \text{Sum of atomic masses of all atoms in the formula}$

Conversions Using Molar Mass

Molar mass is used to convert between mass and moles of a substance.

• Conversion:

  • Moles = mass (g) / molar mass (g/mol)

  • Mass (g) = moles × molar mass (g/mol)

• Example Calculation: To find the mass of 0.713 moles of cobalt (Co):

  • $0.713 \text{ mol Co} \times 58.93 \text{ g/mol} = 42.04 \text{ g Co}$

  • Additional info: 58.93 g/mol is the molar mass of cobalt.

Mole Ratios in Chemical Formulas

Chemical formulas indicate the ratio of atoms of each element in a compound. These ratios can be expressed in terms of moles.

• Example: The formula for aspirin is C9H8O4.

  • 1 molecule of aspirin contains 9 atoms of C, 8 atoms of H, and 4 atoms of O.

  • In terms of moles: 9 moles of C, 8 moles of H, and 4 moles of O per mole of C9H8O4.

Percent Composition by Mass

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

• Definition: Percent composition by mass is the percent of the total mass of a compound contributed by each element.

• Formula:

  • $\text{Percent by mass of element} = \frac{\text{mass of element in 1 mole of compound}}{\text{molar mass of compound}} \times 100\%$

• Example: For H2O:

  • Mass of H in 1 mole = 2 × 1.008 g = 2.016 g

  • Mass of O in 1 mole = 16.00 g

  • Molar mass of H2O = 18.016 g

  • Percent H = (2.016 g / 18.016 g) × 100% = 11.19%

  • Percent O = (16.00 g / 18.016 g) × 100% = 88.81%

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in a molecule.

• Empirical Formula: Simplest ratio of elements in a compound.

• Molecular Formula: Actual number of atoms in a molecule; may be a multiple of the empirical formula.

• Relationship:

  • $\text{Molecular formula} = n \times (\text{Empirical formula})$

  • $n = \frac{\text{Molar mass of molecular formula}}{\text{Molar mass of empirical formula}}$

• Example: If the empirical formula is CH2 and the molar mass is 56 g/mol, then:

  • Molar mass of CH2 = 14 g/mol

  • $n = 56 / 14 = 4$

  • Molecular formula = C4H8

Steps to Determine Empirical Formula from Experimental Data

To find the empirical formula from percent composition or mass data:

1. Assume a 100 g sample (so percent equals grams).

2. Convert grams of each element to moles using their molar masses.

3. Divide all mole values by the smallest number of moles to get the simplest ratio.

4. If necessary, multiply all ratios by a whole number to obtain whole numbers.

5. Write the empirical formula using these whole-number ratios.

• Example: A compound contains 30.4 g nitrogen and 69.6 g oxygen. Find the empirical formula.

  • Convert grams to moles: N: 30.4 g / 14.01 g/mol = 2.17 mol; O: 69.6 g / 16.00 g/mol = 4.35 mol

  • Divide by smallest: N: 2.17/2.17 = 1; O: 4.35/2.17 = 2

  • Empirical formula: NO2

Summary Table: Key Conversions in Chemical Quantities

Conversion	Equation
Moles to particles	$\text{Number of particles} = \text{moles} \times 6.022 \times 10^{23}$
Particles to moles	$\text{Moles} = \frac{\text{Number of particles}}{6.022 \times 10^{23}}$
Mass to moles	$\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
Moles to mass	$\text{Mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Additional info: These notes cover the essential concepts of chemical quantities, the mole, molar mass, percent composition, and empirical/molecular formulas, as relevant to an introductory college chemistry course.