Chapter 6: Chemical Composition and the Mole Concept

Chapter 6: Chemical Composition and the Mole Concept

6.1 Introduction: The Importance of Chemical Composition

Chemical composition is fundamental to understanding how much of a substance is present in a sample and how elements combine to form compounds. This knowledge is essential in fields ranging from nutrition and health to environmental science and industrial chemistry.

• Atoms are counted by mass, not individually, due to their small size and large numbers.

• Conversions between moles, grams, and number of atoms/molecules are central to chemical calculations.

• Understanding chemical formulas allows for the determination of the amount of each element in a compound.

Counting Atoms and Molecules: The Mole Concept

6.2 Counting by Mass: The Analogy of Nails

Just as nails are counted by weight in hardware stores, chemists count atoms by mass. Conversion factors are used to relate mass to number of items.

• Conversion factors relate mass to number of items (e.g., 0.150 lb = 1 dozen nails; 1 dozen = 12 nails).

• Example: To find the number of nails in 2.60 lb, use the conversion factors sequentially to get the total count.

6.3 The Mole: Counting Atoms by the Gram

The mole (mol) is a counting unit in chemistry, analogous to a dozen, but much larger. One mole contains Avogadro's number of entities.

• Avogadro's Number: $6.022 \times 10^{23}$ entities per mole.

• 1 mol of any substance contains $6.022 \times 10^{23}$ atoms, molecules, or formula units.

• The mole is defined as the number of atoms in exactly 12 g of carbon-12.

Converting Between Moles and Number of Atoms

• Use the relationships:

  • $1 \text{ mol} = 6.022 \times 10^{23}$ atoms

  • $6.022 \times 10^{23}$ atoms = 1 mol

• Example: Convert 3.5 mol He to atoms:

  • $3.5 \text{ mol He} \times \frac{6.022 \times 10^{23} \text{ He atoms}}{1 \text{ mol He}} = 2.1 \times 10^{24}$ He atoms

Converting Between Grams and Moles of an Element

• Molar mass (g/mol) is numerically equal to atomic mass (amu).

• Example: $0.58 \text{ g C} \times \frac{1 \text{ mol C}}{12.01 \text{ g C}} = 4.8 \times 10^{-2}$ mol C

Converting Between Grams and Number of Atoms

• Step 1: Convert grams to moles using molar mass.

• Step 2: Convert moles to atoms using Avogadro's number.

• Example: $0.58 \text{ g C} \times \frac{1 \text{ mol C}}{12.01 \text{ g C}} \times \frac{6.022 \times 10^{23} \text{ C atoms}}{1 \text{ mol C}} = 2.9 \times 10^{22}$ C atoms

6.4 Counting Molecules by the Gram

The same principles used for atoms apply to molecules and formula units of compounds.

• Molar mass of a compound is the mass of 1 mole of its molecules or formula units.

• Example: For CO2, molar mass = $1 \times 12.01 + 2 \times 16.00 = 44.01$ g/mol.

Converting Between Grams and Moles of a Compound

• Example: $22.5 \text{ g CO}_2 \times \frac{1 \text{ mol CO}_2}{44.01 \text{ g CO}_2} = 0.511$ mol CO2

Converting Between Grams of a Compound and Number of Molecules

• Step 1: Convert grams to moles using molar mass.

• Step 2: Convert moles to molecules using Avogadro's number.

• Example: $22.5 \text{ g CO}_2 \times \frac{1 \text{ mol CO}_2}{44.01 \text{ g CO}_2} \times \frac{6.022 \times 10^{23} \text{ molecules}}{1 \text{ mol CO}_2} = 3.08 \times 10^{23}$ molecules CO2

Chemical Formulas as Conversion Factors

6.5 Using Chemical Formulas for Calculations

Chemical formulas provide the ratio of elements in a compound, which can be used as conversion factors in stoichiometric calculations.

• Example: In CO2, the ratio is 2 O atoms per 1 CO2 molecule, or 2 mol O per 1 mol CO2.

Converting Between Moles of a Compound and Moles of a Constituent Element

• Example: $1.7 \text{ mol CaCO}_3 \times \frac{3 \text{ mol O}}{1 \text{ mol CaCO}_3} = 5.1$ mol O

Converting Between Grams of a Compound and Grams of a Constituent Element

• Step 1: Convert grams of compound to moles of compound.

• Step 2: Use the chemical formula to find moles of the element.

• Step 3: Convert moles of the element to grams using its molar mass.

• Example: $15 \text{ g NaCl} \times \frac{1 \text{ mol NaCl}}{58.44 \text{ g NaCl}} \times \frac{1 \text{ mol Na}}{1 \text{ mol NaCl}} \times \frac{22.99 \text{ g Na}}{1 \text{ mol Na}} = 5.9$ g Na

Mass Percent Composition

6.6 Mass Percent Composition of Compounds

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

• Formula: $\text{mass percent of element X} = \left( \frac{\text{mass of X in sample}}{\text{mass of sample}} \right) \times 100\%$

• Example: For 0.358 g Cr in 0.523 g oxide: $\frac{0.358}{0.523} \times 100 = 68.5\%$ Cr

6.7 Calculating Mass Percent from a Chemical Formula

• Formula: $\text{mass percent of X} = \frac{\text{mass of X in 1 mol compound}}{\text{mass of 1 mol compound}} \times 100\%$

• Example: For CCl2F2:

  • Molar mass = $12.01 + 2 \times 35.45 + 2 \times 19.00 = 120.91$ g/mol

  • Mass percent Cl = $\frac{2 \times 35.45}{120.91} \times 100\% = 58.64\%$

Empirical and Molecular Formulas

6.8 Calculating Empirical Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula is a whole-number multiple of the empirical formula.

• Steps to determine empirical formula:

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

  2. Convert grams to moles for each element.

  3. Divide by the smallest number of moles to get the ratio.

  4. Multiply to get whole numbers if necessary.

• Example: 24.5 g N and 70.0 g O:

  • $24.5 \text{ g N} \times \frac{1 \text{ mol N}}{14.01 \text{ g N}} = 1.75$ mol N

  • $70.0 \text{ g O} \times \frac{1 \text{ mol O}}{16.00 \text{ g O}} = 4.38$ mol O

  • Ratio: N1O2.5 → multiply by 2 → N2O5

6.9 Calculating Molecular Formulas

• The molecular formula is determined from the empirical formula and the molar mass.

• Formula: $n = \frac{\text{molar mass}}{\text{empirical formula molar mass}}$

• Example: For fructose (empirical formula CH2O, molar mass 180.2 g/mol):

  • Empirical formula mass = 30.03 g/mol

  • $n = \frac{180.2}{30.03} = 6$

  • Molecular formula: (CH2O)6 = C6H12O6

Summary Table: Key Conversion Relationships

Conversion	Conversion Factor	Example
Moles ↔ Atoms	$1 \text{ mol} = 6.022 \times 10^{23}$ atoms	$4.8 \text{ mol Cu} \times \frac{6.022 \times 10^{23}}{1} = 2.9 \times 10^{24}$ atoms
Grams ↔ Moles	Molar mass (g/mol)	$6.73 \text{ mol Al} \times 26.98 = 182$ g
Grams ↔ Atoms	Molar mass & Avogadro's number	$48.3 \text{ g Zn} \times \frac{1}{65.39} \times 6.022 \times 10^{23} = 4.45 \times 10^{23}$ atoms
Moles of Compound ↔ Moles of Element	Chemical formula ratio	$7.20 \text{ mol H}_2\text{SO}_4 \times \frac{4}{1} = 28.8$ mol O
Grams of Compound ↔ Grams of Element	Molar mass & formula ratio	$79.2 \text{ g Fe}_2\text{O}_3 \times \frac{1}{159.70} \times 2 \times 55.85 = 55.4$ g Fe
Mass Percent Composition	Percent by mass	$57.2 \text{ g TiO}_2 \times \frac{59.9}{100} = 34.3$ g Ti

Applications and Health Context

• Sodium in Diet: Calculating sodium intake from sodium chloride using mass percent composition.

• Fluoridation: Fluoride in water helps prevent tooth decay, but excessive intake can cause health issues.

• Environmental Chemistry: Determining chlorine content in CFCs is important for ozone depletion studies.

Key Terms and Definitions

• Mole (mol): The SI unit for amount of substance, equal to $6.022 \times 10^{23}$ entities.

• Avogadro's Number: $6.022 \times 10^{23}$, the number of entities in one mole.

• Molar Mass: The mass of one mole of a substance, in grams per mole (g/mol).

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

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

• Mass Percent Composition: The percent by mass of each element in a compound.

Summary of Skills

• Convert between moles, grams, and number of atoms/molecules using molar mass and Avogadro's number.

• Use chemical formulas as conversion factors to relate moles or grams of compounds to their constituent elements.

• Calculate mass percent composition from experimental data or chemical formulas.

• Determine empirical and molecular formulas from mass data and molar mass.