Experiment 6: Boyle’s Law
 It states that at a constant temperature, the volume of gas is inversely proportional
to the pressure.
V α 1/P at constant temperature
K = VP
o If the pressure is decreased, the volume will increase.
P1V1 = P2V2
 Pressure is usually expressed in millimeters of mercury or mmHg.
760 mmHg = 1 atm
760 torr = 1 atm
1.013 x 105 pascals = 1 atm
Sample Problem: What will a 500 mL of gas initially at 25°C at 750 mmHg occupy when
conditions change to 25°C at 650 mmHg?
V1 = 500 mL
V2 = ?
V1 P1 = V 2 P2
V2 = V1P1
P2
V2 = (500 mL) (750 mmHg)
650 mmHg
V2 = 577 mL
P1 = 750 mmHg
P2 = 650 mmHg
Experiment 7: Charles Law
 It states that if the pressure remains constant, the volume of gas is directly
proportional to its temperature in Kelvin.
V α T at constant pressure
k = V/T
o If the temperature is increased, the volume will also increase.
V1 = V2
T1 = T2
 Kelvin is Celsius temperature + 273
Sample Problem: A sample of gas occupies 368 mL at 27°C and 600 mmHg. What will be
the volume of gas at 127°C at 600 mmHg?
V1 = 368 mL
V2 = ?
V1 = V2
T1 = T2
V2 = V1T2
T1
V2 = (368 mL) (400 K)
300 K
V2 = 491 mL
T1 = 27°C (27 + 273 = 300 K)
T2 = 127°C (27 + 273 = 400 K)
Experiment 8: Laws of Chemical Combination
 Law of Conservation of Mass – Matter can neither be created nor destroyed but it
can be converted from one form to another.
o In ordinary reactions, no detectable changes in mass can be noted before and
after the reaction.
 Law of Definite Proportion – A pure compound is always composed of the same
element combined in a fixed proportion by weight in whole numbers.
o Percentage Composition
% Composition = __ _mass of the element __ _
x 100
molecular mass of the compound


Mass of the element = (atomic weight of the element) (subscript)
Molecular mass = sum of all the atomic weights in a compound.
Sample Problem: Compute the percentage composition of each element in Na2CO3.
% Composition = __ _mass of the element __ _
molecular mass of the compound
x 100
Molecular mass = (23) (2) + 12 + (16) (3)
= 106
Na = (23) (2) _
106
= 43%
C =
12 _
106
= 11%
O = (16) (3) _
106
= 45%
x 100
x 100
x 100
o Empirical Formula – the simplest whole number ratio of the combination of
the different atoms in a compound.
1) Convert grams to moles
EF = Weight of the element x ____number of moles___
Atomic weight of the element
2) Find the empirical formula by dividing the answers by the smallest
number of moles.
Sample Problem: A compound contains 74.0 g of carbon, 8.65 g of hydrogen, and 17.3 g of
nitrogen. What is its empirical weight?
EF = Weight of the element x ____number of moles___
Atomic weight of the element
C = 74.0 g x __1 mol_
12 g
H = 8.65 g x __1 mol_
1g
N = 17.3 g x __1 mol_
14 g
= 6.17 mol_
1.24 mol
= 8.65 mol_
1.24 mol
= 1.24 mol _
1.24 mol
= 5
= 7
= 1
EF = C5H7N
o Molecular Formula – the total number of atoms in a single molecule.
MF = ____given molar mass___
Empirical formula mass
1) Solve for the empirical formula
2) Solve for molecular formula
 Empirical formula mass is the sum of the atomic weight of the
empirical formula multiplied by the subscript.
Sample Problem: A compound contains 74.0 g of carbon, 8.65 g of hydrogen, and 17.3 g of
nitrogen. What is its empirical weight? What is its molecular weight if it has a molar mass
of 81?
1. Solve for EF
EF = Weight of the element x ____number of moles___
Atomic weight of the element
C = 74.0 g x __1 mol_
12 g
H = 8.65 g x __1 mol_
1g
N = 17.3 g x __1 mol_
14 g
= 6.17 mol_
1.24 mol
= 8.65 mol_
1.24 mol
= 1.24 mol _
1.24 mol
= 5
= 7
= 1
EF = C5H7N
2. Solve for MF
MF = ____given molar mass___
Empirical formula mass
MF =
81
___
(12) (5) + (1) (7) + 14 (1)
= 1
MF = C5H7N
A. Law of Conservation of Mass
K2CrO4 + Pb(NO3)2  KNO3 + PbCrO4
Potassium chromate + lead nitrate = colorless liquid + yellow precipitate

There was no detectable change in mass before and after the reaction.
B. Law of Definite Proportion



Magnesium oxide (MgO), which is a white ash, was formed when magnesium was
heated.
Oxygen combined with magnesium when it was ignited in the crucible.
It was necessary to cover the crucible to prevent the excessive loss of product.
Experiment 9: Periodic Table
 Rows are called periods.
o Rows are indicated by Arabic numbers.
 Columns are called group or family.
o Groups are labeled with Roman numerals.
 Elements are arranged in order of increasing atomic numbers in rows.
 Elements with similar chemical properties are arranged in the same column.
o They have the same number of valence electron.
 Elements in the same horizontal row have the same energy levels but cannot be
expected to behave in similar ways.
o Elements in Group I-A are called active metals.
 They have one electron in the outermost shell.
o Elements in Group II-A are called alkaline earth metals.
 They are the least active elements.
o Elements in Group VII-A are called halogens.
 They exist as diatomic molecules.
o Elements in Group VIII-A are called noble gases.
 Inert gases
 Unreactive.
 Elements at the left of the zigzag line are metals.
A. Reaction of Active Metals with Water

Active metals form ionic hydroxides and hydrogen
2Na + H2O  2NaOH + H2
2K + H2O  2KOH + H2
2Mg + H2O  2MgOH + H2
Magnesium + water + heat + phenolphthalein = pink/ violet solution = BASIC
B. Properties of Carbonates of Group I-A and Group II-A

Most carbonates are only slightly soluble in water.
NaCO3 = soluble
KCO3 = soluble
CaCO3 = insoluble
BaCO3 = insoluble
C. Properties of Metal Oxides (Group II-A)

Metallic oxides + water = base (metal hydroxides)
Mg + O2  MgO
MgO + H2O  Mg(OH)2 = dark pink
Mg ribbon
= solid, silver strip
MgO
= white ash
Magnesium oxide + water + phenolphthalein = pink/ violet solution = BASE
CaO + H2O  Ca(OH)2
Calcium oxide + water + phenolphthalein = pink/ violet solution = BASE
D. Properties of Non-Metal Oxides (Group V-A)

Non-metal oxides + water = acid (oxyacid)
S + O2  SO2
SO2 + H2O  H2SO3
Sulfur + heat + H2O = yellow precipitate + gas
= blue litmus paper changed to red = ACID
Cu + 4HNO3  Cu(NO3)2 + 2H2O + 2NO2
HNO3 + Cu wire = brown gas + green solution
= blue litmus paper changed to red = ACID
E. Metalloid Property

Amphoterism – the capacity of a substance to act both as an acid and a base.
AlCl3 + blue litmus paper
= Red = ACID
AlCl3 + red litmus paper
= Red = ACID
AlCl3 + HCl + blue litmus paper
= Red = ACID
AlCl3 + NaOH + red litmus paper = Blue = BASE
F. Ionization Energy

The ionization energy increases in rows and decreases in groups.