1. Chemistry Unit 3 Review
I. History Atomic Theory: Briefly explain the contributions made by each…
A) Democritus.- Atoms of matter are different sizes and shape.
B) Dalton.- Atoms of different elements are different.
C) Thomson.- Cathod Ray Tube exp. Positive body with negative charge
throughout.
D) Rutherford.- Gold foil experiment, atom is mostly empty space with
dense core.
E) Bohr.- Planetary model.
F) Schrodinger, et al.- Electron cloud model.
II. Subatomic Particles: Complete the following chart.
Atomic Mass Number of Number of Number of
Isotope Name Number Number Protons Neutrons Electrons
Gold – 197 79 197 79 118 79
Arsenic – 74 33 74 33 41 33
Neon - 20 10 20 10 10 10
Iodine - 53 127 53 74 53
Magnesium - 12 28 12 16 12
Tungsten - 184 74 184 74 110 74
Carbon –14 6 14 6 8 6
III. Bohr Models: Draw Bohr models for Neon, Magnesium and Carbon.
Mg
Ne C
10p+ 12p+ 6p+
10n 12n 6n
2. IV. Short Answer:
Name and describe the current model of the atom including the three
subatomic particles that make up an atom, their charges, relative masses
and locations within the atom. Electron cloud model: Electrons are found in a
cloud that surrounds the nucleus.
V. Vocabulary: Define the following terms.
Isotope, Ion, Atomic number, Average atomic mass, Percent Abundance, Mass
number, Electron Configuration, Valence Electrons, Quantum Number, Shells,
Energy Levels, Periods, Groups.
VI. Average Atomic Mass: Calculate the following.
1. The element magnesium consists of three naturally occurring isotopes
with masses 23.98504, 24.98584 and 25.98259 amu. The relative
abundances of these three isotopes are 78.70, 10.13 and 11.17 percent,
respectively. From this data, calculate the average atomic mass of
magnesium.
(23.98504*.7870) + (24.98584*.1013) + (25.98259*.1117) = 24.31
2. Lead has four naturally occurring isotopes.
Pb-204 has a mass of 203.973amu and is 1.48% abundant.
Pb-206 has a mass of 205.9745amu and is 23.6% abundant.
Pb-207 has a mass of 206.9759amu and is 22.6% abundant.
Pb-208 has a mass of 207.9766amu and is 52.3% abundant. Calculate the
average atomic mass of lead.
207.18
VII. Electron Configurations: Write the electron configuration for each of
the following elements.
1) Lithium: 1s² 2s¹
2) Sulfur: 1s² 2s²2p6 3s² 3p4
3) Zinc: (Ar) 4s² 3d10
4) Scandium: (Ar) 4s² 3d¹
5) Potassium: 1s² 2s² 2p6 3s² 3p6 4s¹
3. VIII. Radioactivity- Complete the following problems.
1. After 16.0 days, the quantity of a certain radioisotope is reduced from 55.6
grams to 13.9 grams. Calculate the half-life of the isotope.
a. Half-life= Total Mass/ number of cycles = 16/2 = 8 days
2. The half-life of a certain isotope is 7.90 seconds. Calculate the quantity of
the isotope remaining after 39.5 seconds if the sample originally contained
36.0 grams.
a. Total mass / 2n = remaining mass 36/25
3. The quantity of 14C in a sample of once-living material is found to be 5.70g.
The quantity that was present in the sample when the organism died was
91.2g. How old is the sample? (The half-life of carbon-14 is 5730 years.)
a. 5.7=91.7 / 2 n = 4
n
b. 573011460171902292028650 years old
4. Understand the products and how they are derived of alpha, beta, and
gamma decay. Alpha = Helium Nucleus Beta= electron Gamma = energy
IX. Short Answer
1) Two isotopes of iodine (atomic number 53) are I-127 and I-131. Compare
these two isotopes in terms of:
a. number of protons 53
b. mass number 127,131
c. number of neutrons 74,78
2) What is special about the number of electrons, relative to the number of
protons? Why must this be so?
a. The number of protons and neutrons are equal in a neutral atom.
3) What is the term used to describe the normal energy level an electron
occupies? What does it mean if an electron becomes excited? What
happens to the energy that is given off when an excited electron falls back
to its normal level?
a. Quantum Energy level, the excited state is reached when the
electrons are promoted to a higher orbital through the introduction
of energy.
4. XII. Calculate the percent abundance for each element of the following
compounds.
a. H2O H=11% O=89%
b. NH3 H= 18% N= 82%
c. C6H12O6 C= 37% H=7% O= 56%
d. Ca(OH)2 Ca= 54% O=43% H=3%