Quality Core Chemistry Study Guide for First
Semester Final
Chapter 1: An Introduction
1.5 – It is no disaster to make a mistake as long as you learn from it.
Chapter 2: Measurements and Calculations
1.
2.
3.
4.
5.
2.1 – Scientific Notation (number between 1 and 10 x 10n)
2.2 – SI units (mass = kilogram, length = meter, time = seconds,
temperature = Kelvin, quantity = mole)
2.4 – uncertainty (lab)
2.5 - Significant Figures
any non-zero digit is significant (2947 = 4 SF)
any trapped zero is significant (1704902 = 7 SF)
leading zeros after the decimal are not significant (0.00286 = 3 SF)
trailing zeros before the decimal are not significant (120000 = 2 SF)
Can only have as many SF in answer as the fewest SF in problem
2.6 – Dimensional Analysis (conversions)
2.7 – Temperature Conversions
oC + 273 = temp in K
K – 273 = temp in oC
2.8 – Density
Density 
mass
volume
Chapter 3: Matter
3.1 – Matter is anything that has mass and takes up space
Solid, liquid, gas
3.2 – Physical Properties (odor, color, volume, state, melting point,
freezing point, boiling point)
Chemical Properties (a substances ability to form new substances)
Physical Changes – do not affect the composition of the substance
(melting, freezing, boiling, condensation, sublimation, deposition)
Chemical Changes – form new substances and new physical properties
3.3 – Elements are pure substances (cannot be broken down)
Compounds are chemically combined elements that have the same
composition
Mixtures can be separated (filtration, chromatography, distillation)
Homogeneous mixtures – same throughout
Heterogeneous mixtures – unequal distribution
Chapter 4: Chemical Foundations: Elements, Atoms & Ions
4.1
4.2
4.3
4.4
4.5
–
–
–
–
–
elements
element symbols (MEMORIZE THESE!!!)
Dalton’s Atomic Theory (Page 78)
Formulas of Compounds
Structure of the Atom
Electrons – negatively charged, around the nucleus
Protons – positively charged, in the nucleus
Neutrons – no charge, in the nucleus
Most of an atoms mass comes from the protons and neutrons
Rutherford’s Experiment (Page 81-82)
4.6 – Modern Concept of the Atom
4.7 – Isotopes are atoms with the same number of protons but a different
number of neutrons
Number of Neutrons  Atomic Mass - Atomic Number
atomic mass
number of protons
65
30
Zn
Zn
4.8 – Periodic Table
Alkali Metals
Alkali Earth Metals
Transition Metals
Chalcogens
Halogens
Noble Gases
4.9 – Natural States of Elements
Diatomic Molecules – exist in nature a two atoms bonded together
H2, N2, O2, F2, Cl2, Br2, I2
4.10 – ions have either gained or lost electrons
Positive ions are cations
Negative ions are anions
Chapter 5: Nomenclature
Chapter 6: Chemical Reactions: An Introduction
6.1 – indications of chemical reactions
(indicators of chemical change – change in temp or color, formation
of a precipitate, formation of a gas)
6.2 – Chemical equations (reactants, products)
Balancing –
elements that appear once first
Keep polyatomic ions together if possible
Balance hydrogen and oxygen last
Chapter 7: Reactions (Types of Reactions – in notes)
Synthesis
Decomposition
Single Replacement
Double Replacement
Combustion
Chapter 8: Chemical Composition
Percent Composition –
mass of element in 1 mol of compound
 100  % element in compound
molar mass of compound
Example: Calculate the percent composition of the elements in Cu2S.
Step 1: Calculate Molar Mass
63.55 g Cu
2 mol Cu 
 127.1 g Cu
1 mol Cu
1 mol S 
32.07 g S
 32.07 g S
1 mol S
127.1 g Cu
+32.07 g S
159.2 g Cu2S
Step 2: Divide mass of element by molar mass
127.1 g Cu
 100  79.84% Cu
159.2 g Cu 2 S
32.07 g S
 100  20.14% S
159.2 g Cu 2 S
Step 3: Check work (percents should add up to ~100%)
Calculating Empirical Formula Example: Analysis of a compound shows that it contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen. Find the empirical formula.
Step 1: Convert percent composition to mass in grams
Assume 100 gram sample
Na – 32.38 g
S - 22.65 g
O - 44.99 g
Step 2: Convert grams to moles
1 mol Na
 1.408 mol Na
22.99 g Na
1 mol S
22.65 g S 
 0.7063 mol S
32.07 g S
1 mol O
44.99 g O 
 2.812 mol O
16.00 g O
32.38 g Na 
Step 3: Divide all mole amounts by smallest mole amount
1.408 mol Na
 1.993 mol Na, ~ 2.00 mol Na
0.7063
0.7063 mol S
 1.00 mol S
0.7063
2.812 mol O
 3.981 mol O, ~ 4.00 mol O
0.7063
Step 4: (in some cases it may be necessary to multiply mole amounts to get
a whole number – i.e. 2.50 mol x 2 = 5.00 mol or 1.33 mol x 3 = 4.00 mol)
Step 5: Write empirical formula
Na2SO4
Calculating Molecular Formula from Empirical Formula –
molecular formula mass
x
empirical formula mass
Example: The empirical formula of a compound of phosphorus and oxygen was
found to be P2O5. Experimentation shows that the molar mass of this
compound is 283.89 g/mol. What is the compound’s molecular formula?
Step 1: Calculate empirical formula mass
P: 2 x 30.97 = 61.94
O: 5 x 16.00 = 80.00
141.94
Step 2: Divide molecular formula mass by empirical formula mass
283.89
 2.0001  x
141.94
Step 3: multiply empirical formula by x
2.00 x (P2O5) = P4O10
Problems:
1. Determine the percent composition of each of the following compounds:
a. NaCl
b. AgNO3
c. Mg(OH)2
2. Determine the empirical formula of a compound containing 63.50%
silver, 8.25% nitrogen and the rest oxygen.
3. Determine the empirical formula of a compound found to contain 52.11%
carbon, 13.14% hydrogen, and 34.75% oxygen.
4. What is the molecular formula of the molecule that has an empirical
formula of CH2O and a molar mass of 120.12 g/mol?
5. A compound with a formula mass of 42.08 g/mol is found to be 85.64%
carbon and 14.36% hydrogen by mass. Find its molecular formula.
6. Determine the percentage by mass of water in the hydrate CuSO4∙5H2O
Answers:
1. a. 39.34% Na, 60.66% Cl
b. 63.50% Ag, 8.25% N, 28.26% O
c. 41.67% Mg, 54.87% O, 3.46% H
2. AgNO3
3. C2H6O
4. C4H8O4
5. C3H6
6. 36.08%
Chapter 9: Chemical Quantities (Stoichiometry)
amount of given in mol 
coefficent of unknown from eq in moles
 amount of unknown in mol
coefficien t of given from eq in moles
amount of given in grams 
amount of given in mol 
1 mol of given
coefficent of unknown from eq in moles

 amount of unknown in mol
molar mass of given
coefficien t of given from eq in moles
mol of unknown molar mass of unknown

 amount of unknown in grams
mol of given
1 mol of unknown
amount of given in grams 
1 mol of given
mol of unknown molar mass of unknown


 amount of unknown in grams
molar mass of given
mol of given
1 mol of unknown
Limiting Reactant (Limiting Reagent)
Example: The black oxide of iron, Fe3O4, occurs in nature as the mineral
magnetite. This substance can also be made in the laboratory between redhot iron and steam according to the following equation
3Fe(s) + 4H2O(g)--> Fe3O4(s) + 4H2(g)
a)When 36.0 g of H2O is mixed with 167 g of Fe, which is the
limiting reactant?
b)What mass in grams of black iron oxide is produced?
c) What mass in grams of excess reactant remains when the reaction is
completed?
Step 1:convert both given masses to amounts in moles
1 mol H 2 O
36.0 g H 2 O 
 2.0 mol H 2 O
18.02 g H 2 O
167 g Fe 
1 mol Fe
 2.99 mol Fe
55.85 g Fe
Step 2: Convert both given moles to same product
1 mol Fe 3 O 4
2.0 mol H 2 O 
 0.50 mol Fe 3 O 4
4 mol H 2 O
2.99 mol Fe 
1 mol Fe 3 O 4
 1.0 mol Fe 3 O 4
3 mol Fe
Step 3: Identify which reactant produces the least amount of product
H 2 O produces 0.50 mol Fe3 O 4 and is the limiting reactant
Step 4: Convert from moles of product to grams
231.55 g Fe 3 O 4
0.50 mol Fe 3 O 4 
 116 g Fe 3 O 4
1 mol Fe 3 O 4
116 g Fe 3 O 4 is also the THEORETICA L YIELD
Step 5: to determine amount of excess reactant remaining, determine how
much excess reactant will be consumed.
3 mol Fe 55.85 g Fe
2.00 mol H 2 O 

 83.8 g Fe consumed
4 mol H 2 O 1 mol Fe
Step 6: subtract amount of excess reactant consumed from initial amount
167 g Fe initially present - 83.8 g Fe consumed  83.2 g Fe remaining
Problems:
1. Zinc and sulfur react to form zinc sulfide according to the following
equation:
8Zn(s) + S8(s)--> 8ZnS(s)
a) if 2.00 mol of Zn are heated with 1.00 mol of S8, identify the
limiting reactant
b) how many moles of excess reactant remain?
c) How many moles of product are formed
2. Carbon reacts with steam, H2O, at high temperatures to produce
hydrogen and carbon monoxide.
a) if 2.40 mol of carbon are exposed to 3.10 mol of steam, identify the
limiting reactant.
b) How many moles of each product are formed?
c) What mass of each product is formed?
Answers:
1.a) Zn
b) 0.75 mol S8 remain
c) 2.00 mol ZnS
2.a) carbon
b) 2.40 mol H2 and 2.40 mol CO c) 4.85 g H2 and 67.2 g CO
Percent Yield
actual yield
 100
theoretica l yield
was collected what is the percent yield?
percent yield 
Example: if 87.0 grams of Fe3O4
87.0
 100  75%
116
Problems:
1. Methanol can be produced through the reaction of CO and H2 in the
presence of a catalyst.
CO(g)  2H 2 (g) catalyst
 CH 3 OH(l )
If 75.0 g of CO reacts to produce 68.4 g CH3OH, what is the percent
yield of CH3OH?
2. Aluminum reacts with excess copper(II) sulfate according to the
reaction given below. If 1.85 g of Al react and the percent yield of
Cu is 56.6%, what mass of Cu is produced?
Al(s)  CuSO 4 (aq) 
 Al 2 (SO 4 ) 3 (aq)  Cu(s)
(unbalance d)
Answers:
1. 79.8%
2. 3.70 g
Chapter 13: Gases
13.1 – Pressure is force per unit area
Measured with a barometer or manometer
Units can be; mm Hg, torr, atmospheres, pascal
13.2 – Boyles Law
P1V1  P2 V2
Pressure can be any unit
Volume can be any unit
13.3 – Charles’s Law
V1 V2

T1 T2
Volume can be any unit
Temperature must be in Kelvin
NOT IN BOOK – Guy-Lussac’s Law
P1 P2

T1 T2
Pressure can be any unit
Temperature must be in Kelvin
13.4 – Avogadro’s Law
V1 V2

n1 n 2
Volume can be any unit
n must be in moles
13.5 Ideal Gas Law
PV  nRT
Pressure must be in atmospheres
Volume must be in liters
n must be in moles
L  atm
R = gas constant 0.0821
mol  K
Temperature must be in K
NOT IN BOOK – Combined Gas Law
P1V1 P2 V2

T1
T2
Pressure can be any unit
Volume can be any unit
Temperature must be in Kelvin
13.6 – Dalton’s Law of Partial Pressures
Partial Pressure is the pressure a gas would exert if it were
alone in the container
Ptot  P1  P2  P3 ...
For a mixture of gases, the total pressure is equal to the sum
of the partial pressures
Collecting Gas Over Water (extension of Dalton’s Law of PP)
Ptot  PH 2O  PGAS
Sometimes gas is collected by water displacement as show below
In the space above the water there are two gasses; water vapor
and the gas generated by the chemical reaction.
Using Dalton’s Law of Partial Pressures, we can find the
pressure exerted by the collected gas by subtracting the
pressure of the water vapor.
Since water vapor pressure is directly proportional to the
temperature of the water, using a table of water vapor
pressures we can determine the pressure due to the water vapor.
Problems:
1. A container holds three gases: oxygen, carbon dioxide, and helium.
The partial pressures of the three gases are 2.00 atm, 3.00 atm, and
4.00 atm, respectively. What is the total pressure inside the
container?
2. A container with two gases, helium and argon, is 30.0% by volume
helium. Calculate the partial pressure of helium and argon if the
total pressure inside the container is 4.00 atm.
3. If 60.0 L of nitrogen is collected over water at 40.0 °C when the
atmospheric pressure is 760.0 mm Hg, what is the partial pressure of
the nitrogen?
4. 80.0 liters of oxygen is collected over water at 50.0 °C. The
atmospheric pressure in the room is 96.00 kPa. What is the partial
pressure of the oxygen?
5. A tank contains 480.0 grams of oxygen and 80.00 grams of helium at a
total pressure of 7.00 atmospheres. Calculate the following.
a)
b)
c)
d)
e)
How many moles of O2 are in the tank?
How many moles of He are in the tank?
Total moles of gas in tank.
Mole fraction of O2.
Mole fraction of He.
f) Partial pressure of O2.
g) Partial pressure of He.
6. A mixture of 14.0 grams of hydrogen, 84.0 grams of nitrogen, and 2.0
moles of oxygen are placed in a flask. When the partial pressure of
the oxygen is 78.00 mm of mercury, what is the total pressure in the
flask?
7. Some hydrogen gas is collected over water at 20.0oC. The levels of
water inside and outside the gas collection bottle are the same. The
partial pressure of hydrogen is 742.5 torr. What is the barometric
pressure at the time the gas is collected? (water vapor pressure at
20.0oC = 17.5 mm Hg)
8. Helium gas is collected over water at 25oC. What is the partial
pressure of the helium, given that the barometric pressure is 750.0
mm Hg? (water vapor pressure at 25.0oC = 23.8 mm Hg)
Answers:
1. 9.00 atm.
2. PHe = 0.300 x 4.00 atm = 1.20 atm. PAr = 4.00
3. 760.0 mmHg minus 55.3 mmHg
4. 96.00 kPa minus 12.33 kPa
5.
a)
b)
c)
d)
e)
f)
g)
480.0 g O2 / 32.0 g/mol
80.00 g He / 4.00 g/mol
35.0 moles
15.0 mol O2 / 35.0 mol
20.0 mol He / 35.0 mol
7.00 atm x 0.4286
7.00 atm
6.
(14.0 g / 2.00 g/mol) + (84.0 g /28.0 g/mol) + (2.0 moles) = 12.0 moles
total
2.0/ 12.0 = 0.167 of the total pressure. 78.00 is to 0.167 as the total
pressure is to one, so 468 mmHg is the answer.
7. 760 torr
8. 726.2 mm Hg