CHEM 108 COMPREHENSIVE STUDY GUIDE
A systematic exploration of the fundamental laws, theories and mathematical
concepts of inorganic, organic and biological chemistry designed to contain
comprehensive information needed for health professions study.
Credits: 4
Prerequisites: High school chemistry (recommended, but not required)
Course Topics
Module 1: Matter, metric measurements, atomic theory, periodic table, naming
and writing of formulas.
Module 2: Balancing/writing molecular and ionic reactions, redox balancing,
molarity, stoichiometric calculations, percent composition, empirical formula.
Module 3: Kinetic-molecular theory, gas laws, quantum theory of atoms, electron
configuration, periodic table, periodic properties.
Module 4: Ionic and molecular bonding, octet rule, Lewis structures, molecular
geometry.
Module 5: Rate laws, reaction mechanisms, activation energy, catalysis, chemical
equilibria.
Module 6: Acids and bases, pH calculations, titration calculations, acid-base
equilibria, buffers.
Module 7: Organic chemistry, spectroscopy.
Module 8: Biochemistry, nuclear reactions, radiation, half-life.
Lab Topics
Lab 1: Equipment / Safety / Mass - Volume measurement
Lab 2: Density determination
Lab 3: Quantitative and Qualitative analysis
Lab 4: Reaction chemistry
Lab 5: Bonding Type by Solubility, Distillation and Spectroscopy
Lab 6: Measuring Kc and Shifting Chemical Equilibria
Lab 7: Acid-Base titrations
Lab 8: Organic Synthesis
Lab 9: Inorganic Synthesis
Lab 10: Urinalysis
Module 1: Matter, metric measurements, atomic theory, periodic table, naming and
writing of formulas.
#1
Convert 0.0000726 to exponential form and explain your answer.
Convert 0.0000726 = smaller than 1 = negative exponent, move decimal 5 places
= 7.26 x 10-5
Convert 5.82 x 103 to ordinary form and explain your answer.
Convert 5.82 x 103 = positive exponent = larger than 1, move decimal 3 places =
5820
#2
,CHEM 108 COMPREHENSIVE STUDY GUIDE
Using the following information, complete the two conversions shown below,
showing all work:
1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts
1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints
kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10)
2.73 liters = ? ml
2.73 liters x 1000 ml / 1 liter = 2730 ml
8.6 pts = ? qts
8.6 pts x 1 qt / 2 pts = 4.3 qts
#3
Do the three conversions shown below, showing all work:
28oC = ? K
28oC + 273 = 301 K oC → K (make larger) +273
158oF = ? oC
158oF - 32 ÷ 1.8 = 70 oC oF → oC (make smaller) -32 ÷1.8
343K = ? oF
343K - 273 = 70 oC x 1.8 + 32 = 158 oF K → oC → oF
#4
Complete the two problems below. Be sure to show the correct number of
significant figures in each calculation.
Show the calculation of the density of bromine if 40.3 g occupies 12.9 ml.
D = M / V = 40..9 = 3.12 g/ml
Show the calculation of the volume of 28.6 grams of kerosene with density of
0.817 g/ml.
V = M / D = 28..817 = 35.0 ml
#5
Complete the three problems below:
13.010 contains ? significant figures.
13.010 contains 5 significant figures.
0.041 contains ? significant figures.
0.041 contains 2 significant figures.
13.010 - 0.041 = ? (give answer to correct number of significant figures)
13.010 - 0.041 = 12.969 = 12.969 (to the thousandth place for 13.010)
#6
Classify each of the following 3 as an element, compound, solution or
heterogeneous mixture and explain your answer.
Wine
Wine - is not on periodic table (not element) - no element names (not compound)
appears to be one substance = Solution
Tomato soup
Tomato soup - is not on periodic table (not element) - no element names (not
compound) appears to be one substance = Solution
Tin
Tin - is on periodic table = Element
#7
,CHEM 108 COMPREHENSIVE STUDY GUIDE
Classify each of the following 3 as a chemical change or a physical change.
Grape juice ferments to wine
Grape juice ferments to wine - sugar forms alcohol = chemical change
Paper burns
Paper burns - burning always = chemical change
Peeling the shell from a hard-boiled egg
Peeling the shell from a hard-boiled egg - shell is still (broken) shell = physical
change
#8
Show the full Nuclear symbol including any + or - charge (n), the atomic number
(y), the mass number (x) and the correct element symbol (Z) for each element for
which the protons, neutrons and electrons are shown - symbol should appear as
follows: xZy+/- n
12 protons, 12 neutrons, 10 electrons
12 protons = Mg12, 12 neutrons = 24Mg12,
10 electrons = (+12 - 10 = +2) = 24Mg 12+2
#9
Name each of the following 3 chemical compounds. Be sure to name all acids as
acids (NOT for instance as binary compounds).
N2O4
N2O4 - binary molecular = dinitrogen tetroxide
SnO2
SnO2 - binary ionic = tin (IV) oxide
HClO4
HClO4 - nonbinary acid = perchloric acid
#10
Write the formula for each of the following 3 chemical compounds explaining the
answer with appropriate charges and/or prefixes and/or suffixes.
Chromium (III) sulfide
Chromium (III) sulfide - ide - binary of Cr+3 and S-2 = Cr2S3
Hydroselenic acid
Hydroselenic acid - hydro = binary acid of H+ and Se-2 = H2Se
Barium iodide
Barium iodide - ide = binary of Ba+2 and I- = BaI2
Module 2: Balancing/writing molecular and ionic reactions, redox balancing,
molarity, stoichiometric calculations, percent composition, empirical formula.
#1
Show the calculation of the molecular weight for the following two compounds,
reporting your answer to 2 places after the decimal.
Ca3(PO4)2
3Ca + 2P + 8O = 310.18
C9H8NO4Cl
9C + 8H + N + 4O + Cl = 229.61
, CHEM 108 COMPREHENSIVE STUDY GUIDE
#2
Show the calculation of the number of moles in the given amount of the following
2 substances. Report your answer to 3 significant figures.
12.0 grams of Ca3(PO4)2
Moles = grams / molecular weight = 12..18 = 0.0387 mole
15.0 grams of C9H8NO4Cl
Moles = grams / molecular weight = 15..61 = 0.0653 mole
#3
Show the calculation of the number of grams in the given amount of the following
2 substances.
Report your answer to 1 place after the decimal.
1.05 moles of Ca3(PO4)2
Grams = Moles x molecular weight = 1.05 x 310.18 = 325.7 grams
1.18 moles of C9H8NO4Cl
Grams = Moles x molecular weight = 1.18 x 229.61 = 270.9 grams
#4
Show the calculation of the percent of each element present in the following 2
compounds.
Report your answer to 2 places after the decimal.
(NH4)2CrO4
%N = 2 x 14.01/152.08 x 100 = 18.43%
%H = 8 x 1.008/152.08 x 100 = 5.30%
%Cr = 1 x 52.00/152.08 x 100 = 34.20%
%O = 4 x 16.00/152.08 x 100 = 42.08%
C8H8NOI
%C = 8 x 12.01/261.05 x 100 = 36.80%
%H = 8 x 1.008/261.05 x 100 = 3.09%
%N = 1 x 14.01/261.05 x 100 = 5.37%
%O = 1 x 16.00/261.05 x 100 = 6.13%
%I = 1 x 126.9/261.05 x 100 = 48.61%
#6
Show the calculation of the empirical formula for each compound whose elemental
composition is shown below.
57.66% C, 11.61% H, 30.73% O
57.66% C / 12.01 = 4..921 = 2.5 x 2 = 5
11.61% H / 1.008 = 11..921 = 6 x 2 = 12
30.73% O / 16.00 = 1..921 = 1 x 2 = 2
C5H12O2
#7
Classify each of the following 3 reactions as either:
Combination
Decomposition
Combustion
Double
A systematic exploration of the fundamental laws, theories and mathematical
concepts of inorganic, organic and biological chemistry designed to contain
comprehensive information needed for health professions study.
Credits: 4
Prerequisites: High school chemistry (recommended, but not required)
Course Topics
Module 1: Matter, metric measurements, atomic theory, periodic table, naming
and writing of formulas.
Module 2: Balancing/writing molecular and ionic reactions, redox balancing,
molarity, stoichiometric calculations, percent composition, empirical formula.
Module 3: Kinetic-molecular theory, gas laws, quantum theory of atoms, electron
configuration, periodic table, periodic properties.
Module 4: Ionic and molecular bonding, octet rule, Lewis structures, molecular
geometry.
Module 5: Rate laws, reaction mechanisms, activation energy, catalysis, chemical
equilibria.
Module 6: Acids and bases, pH calculations, titration calculations, acid-base
equilibria, buffers.
Module 7: Organic chemistry, spectroscopy.
Module 8: Biochemistry, nuclear reactions, radiation, half-life.
Lab Topics
Lab 1: Equipment / Safety / Mass - Volume measurement
Lab 2: Density determination
Lab 3: Quantitative and Qualitative analysis
Lab 4: Reaction chemistry
Lab 5: Bonding Type by Solubility, Distillation and Spectroscopy
Lab 6: Measuring Kc and Shifting Chemical Equilibria
Lab 7: Acid-Base titrations
Lab 8: Organic Synthesis
Lab 9: Inorganic Synthesis
Lab 10: Urinalysis
Module 1: Matter, metric measurements, atomic theory, periodic table, naming and
writing of formulas.
#1
Convert 0.0000726 to exponential form and explain your answer.
Convert 0.0000726 = smaller than 1 = negative exponent, move decimal 5 places
= 7.26 x 10-5
Convert 5.82 x 103 to ordinary form and explain your answer.
Convert 5.82 x 103 = positive exponent = larger than 1, move decimal 3 places =
5820
#2
,CHEM 108 COMPREHENSIVE STUDY GUIDE
Using the following information, complete the two conversions shown below,
showing all work:
1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts
1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints
kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10)
2.73 liters = ? ml
2.73 liters x 1000 ml / 1 liter = 2730 ml
8.6 pts = ? qts
8.6 pts x 1 qt / 2 pts = 4.3 qts
#3
Do the three conversions shown below, showing all work:
28oC = ? K
28oC + 273 = 301 K oC → K (make larger) +273
158oF = ? oC
158oF - 32 ÷ 1.8 = 70 oC oF → oC (make smaller) -32 ÷1.8
343K = ? oF
343K - 273 = 70 oC x 1.8 + 32 = 158 oF K → oC → oF
#4
Complete the two problems below. Be sure to show the correct number of
significant figures in each calculation.
Show the calculation of the density of bromine if 40.3 g occupies 12.9 ml.
D = M / V = 40..9 = 3.12 g/ml
Show the calculation of the volume of 28.6 grams of kerosene with density of
0.817 g/ml.
V = M / D = 28..817 = 35.0 ml
#5
Complete the three problems below:
13.010 contains ? significant figures.
13.010 contains 5 significant figures.
0.041 contains ? significant figures.
0.041 contains 2 significant figures.
13.010 - 0.041 = ? (give answer to correct number of significant figures)
13.010 - 0.041 = 12.969 = 12.969 (to the thousandth place for 13.010)
#6
Classify each of the following 3 as an element, compound, solution or
heterogeneous mixture and explain your answer.
Wine
Wine - is not on periodic table (not element) - no element names (not compound)
appears to be one substance = Solution
Tomato soup
Tomato soup - is not on periodic table (not element) - no element names (not
compound) appears to be one substance = Solution
Tin
Tin - is on periodic table = Element
#7
,CHEM 108 COMPREHENSIVE STUDY GUIDE
Classify each of the following 3 as a chemical change or a physical change.
Grape juice ferments to wine
Grape juice ferments to wine - sugar forms alcohol = chemical change
Paper burns
Paper burns - burning always = chemical change
Peeling the shell from a hard-boiled egg
Peeling the shell from a hard-boiled egg - shell is still (broken) shell = physical
change
#8
Show the full Nuclear symbol including any + or - charge (n), the atomic number
(y), the mass number (x) and the correct element symbol (Z) for each element for
which the protons, neutrons and electrons are shown - symbol should appear as
follows: xZy+/- n
12 protons, 12 neutrons, 10 electrons
12 protons = Mg12, 12 neutrons = 24Mg12,
10 electrons = (+12 - 10 = +2) = 24Mg 12+2
#9
Name each of the following 3 chemical compounds. Be sure to name all acids as
acids (NOT for instance as binary compounds).
N2O4
N2O4 - binary molecular = dinitrogen tetroxide
SnO2
SnO2 - binary ionic = tin (IV) oxide
HClO4
HClO4 - nonbinary acid = perchloric acid
#10
Write the formula for each of the following 3 chemical compounds explaining the
answer with appropriate charges and/or prefixes and/or suffixes.
Chromium (III) sulfide
Chromium (III) sulfide - ide - binary of Cr+3 and S-2 = Cr2S3
Hydroselenic acid
Hydroselenic acid - hydro = binary acid of H+ and Se-2 = H2Se
Barium iodide
Barium iodide - ide = binary of Ba+2 and I- = BaI2
Module 2: Balancing/writing molecular and ionic reactions, redox balancing,
molarity, stoichiometric calculations, percent composition, empirical formula.
#1
Show the calculation of the molecular weight for the following two compounds,
reporting your answer to 2 places after the decimal.
Ca3(PO4)2
3Ca + 2P + 8O = 310.18
C9H8NO4Cl
9C + 8H + N + 4O + Cl = 229.61
, CHEM 108 COMPREHENSIVE STUDY GUIDE
#2
Show the calculation of the number of moles in the given amount of the following
2 substances. Report your answer to 3 significant figures.
12.0 grams of Ca3(PO4)2
Moles = grams / molecular weight = 12..18 = 0.0387 mole
15.0 grams of C9H8NO4Cl
Moles = grams / molecular weight = 15..61 = 0.0653 mole
#3
Show the calculation of the number of grams in the given amount of the following
2 substances.
Report your answer to 1 place after the decimal.
1.05 moles of Ca3(PO4)2
Grams = Moles x molecular weight = 1.05 x 310.18 = 325.7 grams
1.18 moles of C9H8NO4Cl
Grams = Moles x molecular weight = 1.18 x 229.61 = 270.9 grams
#4
Show the calculation of the percent of each element present in the following 2
compounds.
Report your answer to 2 places after the decimal.
(NH4)2CrO4
%N = 2 x 14.01/152.08 x 100 = 18.43%
%H = 8 x 1.008/152.08 x 100 = 5.30%
%Cr = 1 x 52.00/152.08 x 100 = 34.20%
%O = 4 x 16.00/152.08 x 100 = 42.08%
C8H8NOI
%C = 8 x 12.01/261.05 x 100 = 36.80%
%H = 8 x 1.008/261.05 x 100 = 3.09%
%N = 1 x 14.01/261.05 x 100 = 5.37%
%O = 1 x 16.00/261.05 x 100 = 6.13%
%I = 1 x 126.9/261.05 x 100 = 48.61%
#6
Show the calculation of the empirical formula for each compound whose elemental
composition is shown below.
57.66% C, 11.61% H, 30.73% O
57.66% C / 12.01 = 4..921 = 2.5 x 2 = 5
11.61% H / 1.008 = 11..921 = 6 x 2 = 12
30.73% O / 16.00 = 1..921 = 1 x 2 = 2
C5H12O2
#7
Classify each of the following 3 reactions as either:
Combination
Decomposition
Combustion
Double
