chapter 6

6.1: Counting Nails by the Pound

6.1.1 Counting by Weighing and Avogadro's Number

• Molecules are extremely small, making it difficult to count them directly; thus, we use a technique called "counting by weighing."

• Example of Counting Nails:

  • Consider a box of nails weighing:

    • Empty box weight: 213 g

    • Weight with big nails: 1340 g

  • Weight of nails alone = 1340 g - 213 g = 1127 g

  • If each big nail weighs 0.450 g:

    • Number of big nails = 1127 g / 0.450 g = 2504.44 ≈ 2504 big nails

• If instead, the box contains small nails weighing 0.23 g each:

  • Number of small nails = 1127 g / 0.230 g = 4895.65 ≈ 4896 small nails

6.1.2 Conversion Between Mass and Moles

• Use the conversion of how many items in a collective mass (e.g., per dozen) to improve calculations.

• Understanding Molar Mass:

  • Molar mass is the mass of one mole of a substance (g/mol).

  • Avogadro's number (6.022 × 10²³) connects moles to the number of particles, making it critical for chemical calculations.

• If the American system used pounds, Avogadro's number would increase by a factor of 454 (1 pound = 454 grams).

6.1.3 Molar Mass Calculations for Elements and Compounds

• Average atomic mass for an element, e.g., carbon (C) = 12.01 amu = 12.01 g/mol.

• Calculating molar masses of compounds includes adding the masses of constituent elements based on their chemical formulas:

  • Example for Water (H₂O):

    • H: 2 × 1.008 amu + O: 1 × 16.00 amu = 18.016 amu → Molar mass = 18.016 g/mol

  • For compounds like ethyl chloride (C₂H₅Cl), include all constituent atoms in calculations.

6.2: Counting Atoms by the Gram

6.2.1 Understanding Moles and Conversions

• Use Avogadro's number and molar mass to convert between grams and moles.

• Example Calculation:

  • Given carbon atoms: X = 6 × 10²³, determine moles:

  • Moles = X (atoms) / 6.022 × 10²³.

6.2.2 Problem Solving with Molar Mass

• Molar mass for elements (e.g., lithium = 6.94 g, gold = 196.97 g).

• Converting grams to moles using:

  • Moles = Mass (g) / Molar Mass (g/mol).

• Example: Find grams in 0.560 moles of chromium:

  • Given 1 mol Cr = 52.00 g, so calculate:

  • 0.560 mol Cr × 52.00 g/mol = 29.12 g.

6.3: Formula Mass – The Mass of a Molecule or Formula Unit

6.3.1 Calculation of Formula Mass

• Calculate molecular/formula mass by summing atomic masses:

  • Example for Sodium Chloride (NaCl): Na = 22.99 amu, Cl = 35.45 amu → Formula mass = 58.44 amu.

• For ionic compounds with multiple cations or anions, adjust counts during calculation.

6.3.2 Mass of Hydrates

• Hydrates have a specific number of water units per formula unit.

• Example: Copper(II) sulfate pentahydrate (CuSO₄·5H₂O).

6.4: Counting Molecules by the Gram

6.4.1 Molecular and Formula Masses

• Molecular mass: Sum of atom masses in a molecule.

• Formula mass: Sum of atomic masses in empirical formulas of ionic compounds.

6.4.2 Conversions between Grams and Moles

• Use molar mass to convert between substance mass and the number of moles.

6.5: Chemical Formulas as Conversion Factors

6.5.1 Utilizing Ratios in Formulas

• Use formulas to determine ratios of moles between compounds, i.e., H₂O having two H atoms and one O atom.

6.5.2 Mass Relationships in Calculations

• Example: Determine grams of Oxygen in 75.0 g of ethanol (C₂H₆O), using conversion ratios.

6.6: Mass Percent Composition of Compounds

6.6.1 Using Mass to Determine Percent Composition

• Calculate the mass percent of each element in a compound:

  • % Composition = (Mass of element / Mass of compound) × 100%

• Example: If 20.00 g of a zinc-oxygen compound contains 16.07 g of Zn, then O = 20.00 g - 16.07 g = 3.93 g, leading to:

  • %Zn = 80.35%, %O = 19.65%.

6.7: Mass Percent Composition from a Chemical Formula

6.7.1 Determining Percent via Chemical Formula

• Percent composition calculated using:

  • % by mass = (mass of element in one mole / molar mass of compound) × 100%

• Example: For dichlorine heptoxide (Cl₂O₇): calculate Cl and O percentages.

6.8: Calculating Empirical Formulas

6.8.1 Steps for Empirical Formula Calculation

• To derive an empirical formula from mass percentages:

  1. Convert each % to g.

  2. Convert g to moles.

  3. Divide by smallest moles for a ratio.

6.8.2 Example: Iron and Oxygen Compound

• Given percent compositions yield the empirical formula via calculated ratios.

6.9: Molecular Formulas for Compounds

6.9.1 Understanding Empirical vs. Molecular Formulas

• Molecular formulas reflect actual atom counts, while empirical formulas indicate simplest ratios.

• Example: For glucose and sucrose, glucose is C₆H₁₂O₆ and sucrose is C₁₂H₂₂O₁₁.

6.9.2 Steps to Calculate Molecular Formula

1. Calculate empirical formula mass.

2. Divide compound's molar mass by empirical formula mass.

3. Adjust the empirical formula subscripts according to the integer result.