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: $C<em>1H</em>1$ (Actual molecular formula is $C<em>6H</em>6$).

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.