Counting Atoms and Moles Using Relative Atomic Masses

Counting Atoms Using Relative Atomic Masses

Introduction

  • The lecture focuses on using relative atomic masses to determine the molecular formula of compounds by figuring out the relative numbers of atoms of each type.
  • The content is based on the concept development study on atomic masses and molecular formula.
  • The objective is to count atoms by assuming known relative atomic masses, caring about relative numbers rather than absolute numbers.
  • The method involves counting particles by measuring mass, a common technique in chemistry.

Counting with Marbles: An Analogy

  • Basic Principle: Illustrates counting particles by measuring mass using glass marbles.
  • Example 1:
    • A single marble weighs 1.5 grams.
    • A sample of 15 grams of marbles implies there are 10 marbles because \frac{15}{1.5} = 10.
  • Example 2:
    • 1.5 kilograms of marbles means there are 1000 marbles since 1.5 \text{ kg} = 1500 \text{ g}, and \frac{1500}{1.5} = 1000.
  • Key Formula: \text{Number of marbles} = \frac{\text{Mass of sample}}{\text{Mass of a single marble}}
  • Example 3:
    • A metric ton (1000 kg) of marbles calculation:
    • \text{Number of marbles} = \frac{1000 \text{ kg} \times 1000 \text{ g/kg}}{1.5 \text{ g/marble}} \approx 667,000 \text{ marbles}

Comparing Numbers of Different Particles

  • Analogy with Marbles:
    • Blue marbles: 1.5 grams each.
    • Red marbles: 2.3 grams each.
    • Bag of blue marbles: 150 grams.
    • Bag of red marbles: 230 grams.
    • The ratio of the masses of the bags (\frac{230}{150}) is the same as the ratio of individual marbles (\frac{2.3}{1.5}), indicating the same number of marbles in each bag.
  • Application to Atoms:
    • Carbon (C) and Sodium (Na) as examples.
    • Relative atomic mass of carbon: 12.
    • Relative atomic mass of sodium: 23.

Mass Ratios and Number of Atoms

  • Ratio of Masses:
    • One carbon atom and one sodium atom have a mass ratio of 12:23.
    • A thousand carbon atoms and a thousand sodium atoms have a mass ratio of 12,000:23,000, which simplifies to 12:23.
    • The mass ratio remains constant as long as the number of atoms is the same.
  • Given Masses:
    • If you have 12 grams of carbon and 23 grams of sodium, the ratio of the masses is 12:23.
    • This indicates that the two samples have the same number of atoms, even without knowing the exact count.

The Mole Concept

  • A mole is defined as the number of atoms in 12 grams of carbon.
  • It's a specific number, similar to a dozen, used for counting atoms.
  • One mole of any element has the same number of atoms as one mole of any other element.
  • For example, a mole is the number of atoms in 23 grams of sodium or 4 grams of helium.
  • The mass of a single mole (molar mass) can be determined by using relative atomic masses and assigning that number in grams.

Molar Mass

  • Definition: Molar mass is the mass of one mole of a substance, expressed in grams.
  • Determination:
    • For an atom (e.g., carbon with 12 amu), the molar mass is 12 grams.
    • For a molecule (e.g., water with 18 amu), the molar mass is 18 grams.
  • Formula:
    • \text{Number of moles} = \frac{\text{Mass of sample}}{\text{Molar mass}}

Importance of Calculating Moles

  • The number of moles reflects the count of particles.
  • Applications:
    • Determining molecular formula (ratio of atom types).
    • Calculating particles produced in chemical reactions.
    • Finding the concentration of a solution.

Example 1: Empirical Formula of Benzene

  • Benzene contains only carbon and hydrogen.
  • Mass composition: 92.3\% carbon and 7.7\% hydrogen.
  • Calculation Steps:
    • Take a 100 gram sample of benzene.
    • Mass of carbon: 92.3 grams.
    • \text{Moles of carbon} = \frac{92.3 \text{ grams}}{12 \text{ grams/mole}} = 7.68 \text{ moles}
    • Mass of hydrogen: 7.7 grams.
    • \text{Moles of hydrogen} = \frac{7.7 \text{ grams}}{1.008 \text{ grams/mole}} \approx 7.7 \text{ moles}
    • Ratio of carbon to hydrogen is approximately 1:1.
    • Empirical formula: C1H1 (Actual molecular formula is C6H6).

Example 2: Molecular Formula of Chloroform

  • Chloroform contains carbon, hydrogen, and chlorine.
  • Mass composition in a 100 gram sample:
    • Carbon: 10.06 grams.
    • Hydrogen: 0.84 grams.
    • Chlorine: 89 grams.
  • Calculation Steps:
    • \text{Moles of carbon} = \frac{10.06 \text{ grams}}{12.01 \text{ grams/mole}} \approx 0.833 \text{ moles}
    • \text{Moles of hydrogen} = \frac{0.84 \text{ grams}}{1.008 \text{ grams/mole}} \approx 0.833 \text{ moles}
    • \text{Moles of chlorine} = \frac{89 \text{ grams}}{35.45 \text{ grams/mole}} \approx 2.51 \text{ moles}
    • Divide each by 0.833 to get the ratio: C:H:Cl = 1:1:3.
    • Molecular formula: CHCl_3.

Relating Reactants to Products

  • Example: Burning Methane (CH_4)
    • Methane reacts to produce carbon dioxide (CO_2).
    • Question: If you start with 1 kilogram of methane, how much carbon dioxide will be produced?
  • Key Concept: One methane molecule produces one carbon dioxide molecule.
  • Calculation Steps:
    • \text{Moles of methane} = \frac{1000 \text{ grams}}{16 \text{ grams/mole}} = 62.46 \text{ moles}
    • Since one mole of methane produces one mole of carbon dioxide:
    • \text{Moles of carbon dioxide} = 62.46 \text{ moles}
    • \text{Mass of carbon dioxide} = 62.46 \text{ moles} \times 44.01 \text{ grams/mole} \approx 2750 \text{ grams} = 2.75 \text{ kg}
    • For every kilogram of methane burned, 2.75 kilograms of carbon dioxide are produced.