Module 2 Lecture 5: Atomic and Intro to Molecular Mass

Core Fundamentals of Atomic Structure

• Subatomic Composition: Atoms are composed of three primary subatomic particles: protons, neutrons, and electrons.

• The Nucleus: This central dense region of the atom contains the protons and the neutrons.

• Electrons: These particles are located outside of the nucleus, characterized as rapidly moving. They typically occupy regions known as electron "shells" or orbitals.

Understanding Average Atomic Mass

• Definition of Atomic Weight: The atomic weight reported on periodic tables is not the mass of a single atom, but rather the average atomic mass (also referred to as the standard atomic weight).

• Basis of Calculation: This value is determined based on the relative natural abundance of the isotopes of a specific element as found on Earth.

• Rationale for Weighted Averages:

  • A measurable sample of an element contains an immense number of atoms (often exceeding quadrillions).

  • These atoms are divided among various naturally occurring isotopes.

  • Since it is impossible to count every individual isotope in a sample to determine mass, scientists use a weighted average of naturally occurring isotopes.

• Natural Percent Abundance: This refers to the percentage chance that any given atom of an element found on Earth will be a specific isotope.

• The Cow Analogy: To illustrate the concept of averages across a population, the lecture notes the average mass of cows (e.g., comparing groups weighing $1000\,lbs$ and $1200\,lbs$) to show how individual variation contributes to a collective average.

• Mathematical Formula for Weighted Average:     $\text{Weighted Average} = \frac{\% \text{ isotope}_1 \times \text{mass isotope}_1 + \% \text{ isotope}_2 \times \text{mass isotope}_2 \dots}{100\%}$

Isotopic Examples and Calculations

• Chlorine (Cl):

  • Isotope 1: Cl-35 represents approximately $76\%$ of naturally occurring Chlorine atoms.

  • Isotope 2: Cl-37 represents approximately $24\%$ of naturally occurring Chlorine atoms.

  • The resulting weighted average (atomic weight) for Chlorine is $35.45\,amu$, rather than a simple whole number like 35 or 37.

• Silver (Ag) Calculation Example:

  • Given Data:

    • Ag-107: $51.84\%$ abundance with a mass of $106.905\,amu$.

    • Ag-109: $48.16\%$ abundance with a mass of $108.904\,amu$.

  • Calculation Step 1 (Numerators):         $(51.84\% \times 106.905\,amu) + (48.16\% \times 108.904\,amu) = 1078677.2\, \%\,amu$

  • Calculation Step 2 (Final Average):         $\frac{1078677.2\% \,amu}{100\%} = 107.87\,amu$

  • Conclusion: This calculated value matches the average atomic mass reported on the standard periodic table.

Chemical Definitions: Molecules and Compounds

• Molecule: A structure consisting of two or more atoms held together by chemical bonds.

• Compound: A chemical substance consisting of two or more different elements chemically combined in a fixed, consistent ratio.

• Molecular Element: A molecule that consists of only one type of element (atoms of the same element bonded together).

• Formula Unit: This term describes the smallest unit of a non-molecular chemical, specifically used for ionic compounds.

• Chemical Formula: A representation that provides the lowest whole number ratios of the various atoms making up a chemical substance.

Chemical Formulas and Bonding

• Covalent Bonding: Molecules consist of atoms that are covalently bonded, meaning electrons are shared between the atoms to form the bond.

• Subscript Notation: Compounds and molecules are represented by elemental symbols followed by subscripts.

  • The subscript indicates the number of atoms of that element present in the compound.

  • If the number of atoms is 1, no subscript is used.

• General Formula Template:     $X_n Y_m Z_o$

Determing Molecular and Formula Mass

• Definition: The molar mass of a molecule is the sum of the atomic masses of every atom that constitutes the molecule.

• Calculation Requirement: One must account for the specific quantity of each atom as dictated by the simplest chemical formula.

• General Mathematical Mass Formula:     $\text{Mass of } X_n Y_m Z_o = (n \times \text{Mass}_X) + (m \times \text{Mass}_Y) + (o \times \text{Mass}_Z)$

Group Question: Practical Application (The Car Example)

• Scenario Components and Mass:

  • 1 Engine = $1500\,lbs$

  • 1 Chassis = $400\,lbs$

  • 4 Tires = $25\,lbs$ each (Total for tires = $4 \times 25 = 100\,lbs$)

  • Total Mass for 1 Car = $1500 + 400 + 100 = 2000\,lbs$

• Problem 1: Mass of 10 Cars:

  • Calculation: $10\, \text{cars} \times 2000\,lbs/\text{car} = 20,000\,lbs$

• Problem 2: Unit Conversion and Quantity Determination:

  • Given: A total mass of $3.909 \times 10^4 \,kg$ of cars.

  • Conversion Factor: $1\,kg = 2.2\,lbs$

  • Step A (Convert total kg to lbs):         $3.909 \times 10^4 \,kg \times 2.2\,lbs/kg = 86,000\, (\text{approx.})\, lbs$

  • Step B (Calculate number of cars): Divide total weight in lbs by the weight of one car ($2000\,lbs$).