Lecture 5

To-Dos

• Finish reading Chapters R & 1
• Homework:
  • Chapter R - Due 2/5 @ 11:55pm
  • Chapter 1 - Due 2/5 @ 11:55pm
• Friday, Jan. 31st
  • Have Calculators & Clickers ready
  • Session ID
• Extra Credit Due TONIGHT by 11:55pm

Problem from Last Class

• Problem: A 22.65g sample of solid is placed in a flask. An unknown liquid, in which the solid is insoluble, is added to the flask so that the total volume of solid and liquid together is 30.00mL. The solid and liquid together weigh 38.58 g. The density of the solid is 2.48 g/mL. What is the density of the liquid?
• Solution:
  • Volume of solid: $V_{\text{solid}} = \frac{22.65 \text{ g}}{2.48 \text{ g/mL}} = 9.13 \text{ mL}$
  • Volume of liquid: $V_{\text{liquid}} = 30.00 \text{ mL} - 9.13 \text{ mL} = 20.87 \text{ mL}$
  • Mass of liquid: $m_{\text{liquid}} = 38.58 \text{ g} - 22.65 \text{ g} = 15.93 \text{ g}$
  • Density of liquid: $d_{\text{liquid}} = \frac{15.93 \text{ g}}{20.87 \text{ mL}} = 0.7633 \text{ g/mL}$

Three Important Laws

• Law of Conservation of Mass:
  • Mass is neither created nor destroyed.
• Law of Multiple Proportions:
  • When two elements form a series of compounds, the ratios of the masses of the second element that combine with 1 gram of the first element can always be reduced to small whole numbers.
• Law of Definite Proportion:
  • A given compound always contains exactly the same proportion of elements by mass.

Law of Definite Proportion (Formerly Proust’s Law)

• Example: Sodium Sulfate (Na2SO4)
  • 1.4 Parts by mass Sodium (Na)
  • 1 Part by mass Sulfur (S)
  • 2 Parts by mass Oxygen (O)
  • Note: This is by mass and not by the subscript numbers in the molecule.
• Example: Nitric Acid (HNO3)
  • 1 Part by mass Hydrogen (H)
  • 14 Parts by mass Nitrogen (N)
  • 48 Parts by mass Oxygen (O)

Law of Multiple Proportions

• Example: If Oxygen (O2) and Hydrogen (H2) react, it will generate 2 compounds: Water (H2O) and Hydrogen Peroxide (H2O2).
  • Hydrogen + Oxygen → Water
    • 2g + 16g → 18g
  • Hydrogen + Oxygen → Hydrogen Peroxide
    • 2g + 32g → 34g
• Water Ratio :
  • $\frac{16 \text{g Oxygen}}{2 \text{g Hydrogen}} = 8$
• Hydrogen Peroxide Ratio:
  • $\frac{32 \text{g Oxygen}}{2 \text{g Hydrogen}} = 16$
• Multiple Proportions Ratio:
  • $\frac{8 \text{ Water}}{16 \text{ Hydrogen Peroxide}} = \frac{1}{2}$
  • Simple whole numbers

Law of Multiple Proportions Example

• Two elements, R and Q, combine to form two binary compounds. In the first compound, 14.0 g of R combines with 3.00 g of Q. In the second compound, 7.00 g of R combines with 4.50 g of Q. Show that this data is in accord with the law of multiple proportions
• Compound 1: $\frac{14 \text{g R}}{3 \text{g Q}} = 4.67$
• Compound 2: $\frac{7 \text{g R}}{4.5 \text{g Q}} = 1.56$
• Law of Multiple Proportions: $\frac{4.67 \text{ Compound 1}}{1.56 \text{ Compound 2}} = \frac{3}{1}$
• If the formula for the second compound is RQ, what is the formula for the first compound?
  • Compound 1: $\frac{14 \text{g R}}{3 \text{g Q}} = \frac{4.67 \text{ R}}{ \text{Q}}$
  • Compound 2: $\frac{7 \text{g R}}{4.5 \text{g Q}} = \frac{1.56 \text{ R}}{ \text{Q}}$
  • $\frac{4.67}{1.56} = \frac{3}{1}$
  • Compound 1 must have 3x as much R as Compound 2.
  • Compound 1 is R3Q

Dalton’s Atomic Theory

• Each element is made up of tiny particles called atoms.
• The atoms of a given element are identical; the atoms of different elements are different in some fundamental way or ways.
• Chemical compounds are formed when atoms of different elements combine with each other. A given compound always has the same relative numbers and types of atoms.
• Chemical reactions involve reorganization of the atoms – changes in the way they are bound together. The atoms themselves are not changed in a chemical reaction.

Determining the Composition of Compounds

• Joseph Gay-Lussac & Amadeo Avogadro studied the combination of different volumes of gases and the resulting products.

Periodic Table of the Elements

• The image shows a typical periodic table layout including:
  • Atomic Number
  • Symbol
  • Name
  • Atomic Mass

The Electron

• J.J. Thomson conducted experiments using cathode-ray tubes
• High voltage applied to partially evacuated tubes
• Assumed that the cathode ray comprised of electrons
• Determined the charge-to-mass ratio of an electron
  • $\frac{e}{m}$, where:
    • e is the charge of the electron in coulombs
    • m is the mass in grams

Millikan's Oil Drop Experiment

• Robert Millikan performed experiments involving charged oil drops
• Determined the magnitude of the charge on a single electron
• Calculated the mass of the electron
  • $9.11 × 10^{-31} \text{ kg}$

Plum Pudding Model

• Hypothesized by J. J. Thomson
• Atoms are made of a positively charged cloud
• Negative electrons are embedded at random

Testing the Plum Pudding Model

• Ernest Rutherford – 1911 – Metal Foil Experiment
• Radioactivity:
  • Accidentally discovered by Henri Becquerel in 1896
  • A uranium-containing mineral produced its image on a photographic plate in the dark
• Types of radioactive emission
  • Gamma rays (γ) - High-energy “light”
  • Beta particles (β) - High-speed electrons
  • Alpha particles (α) - Possess a charge twice that of the electron (-1), with the opposite sign (+2)

Nuclear Atom

• Protons (+)
• Neutrons
• Electrons (-)
• $1 \text{ Atomic Mass Unit (1 amu)} = 1.66 x 10^{-24} \text{ g}$

Isotopes

• Isotope Notation Example:
  • $^{12}_{6}C$
    • Element symbol: C
    • Mass number (# protons + neutrons): 12
    • Atomic number (# protons): 6
• ISOTOPE: Carbon-14
  • 6 protons
  • 8 neutrons

Phosphorous Isotope Symbol

• What is the symbol for Phosphorous (P) with 17 neutrons?
• Answer: $^{32}_{15}P$

Atomic Mass

• Atomic Mass: weighted average of isotopic masses of element's naturally occurring isotopes.
• The image shows a typical periodic table entry including:
  • Atomic number
  • Symbol
  • Atomic mass

Weighted Average Example - Carbon

• Carbon-12 = 98.892 %
• Carbon-13 = 1.108 %
• Carbon-14 = 1x10-10 %
• Carbon-12 = 12.00000 amu
• Carbon-13 = 13.00335 amu
• Carbon-14 = 14.00324 amu
• Weighted Average = (0.98892)(12.0000)+(0.01108)(13.00335)+(1x10^{-12})(14.00324)
• Weighted Average = 12.0107 amu = atomic weight of Carbon
• Individual isotope masses & Mass Number are NOT on periodic table!

Clicker Question - Chlorine

• Calculate mass (in amu) of 23 Chlorine atoms.
• Solution:
  • $23 \text{ Cl atoms} \times \frac{35.45 \text{ amu}}{1 \text{ Cl atom}} = 815.4 \text{ amu}$

Silver Isotopes Problem

• Silver has two isotopes with an average atomic mass of 107.868 amu. The isotope Ag-109 weighs 108.905 amu and is 48.16% in natural abundance. What is the atomic weight of the second isotope of silver (in amu)?
• I am going to cover this on Monday