Module 1  Matter and its Properties
Matter  has mass and occupies space. 
3 States of Matter
State 
Definition 
Examples 
Solid 
rigid; has a fixed shape and volume 
ice cube, diamond, iron bar 
Liquid 
has a definite volume but takes the shape of its container 
gasoline, water, blood 
Gas 
has no fixed volume or shape; takes the shape of its container 
air, helium, oxygen 
Elements and Compounds
Elements 
cannot be broken down into other substances by chemical means 
iron, aluminum, oxygen, and hydrogen 
Compound 
substances that have the same composition no matter where we find them; can be broken down into elements 
Water (H20), Salt (NaCl), Ammonia (NH3) 
Physical and Chemical Properties and Changes
Physical Properties 
odor, color, volume, state (gas, liquid, or solid), density, melting point, boiling point 
Chemical Properties 
burning, digestion, fermentation, rusting, electrolysis 
Other Properties
Extensive 
changes when the amount of material changes 
mass, length, volume, shape 
Intrinsive 
does not depend on the size of the material 
temperature, odor, color, hardness, density 
Mixture and Pure Substances
Mixture 
has variable composition 

Homogenous 
also called a solution; does not vary in composition from one region to another 

Heterogenous 
contains regions that have different properties from those of other regions 
Pure Substance 
always have the same composition; either elements or compounds 
Types of bonds
Ionic 
when one atom shifts or transfers an electron to another atom; metals + nonmetals 
Na^{+} (1A) and Cl^{} (7A) creates a stable bond (octet rule) 
Covalent 
atoms share electrons; nonmetals 
O^{2}(6A) and 2 atoms of H^{+}(1A) = H₂O 
Metallic 
a metal shares an electron with another metal; positively charged ions in electrons 
Module 2  Isotopes, Compounds, Empirical Formula
Atoms have a constant or fixed number of protons
Atomic Number  gives the protons in the nucleus of an atom; represented as Z
Neutral Atom  number of protons is equal to the number of electrons
Z = nuclear charge = number of protons = number of electrons in neutral form
Mass Number  sum of the number of protons and neutrons; represented by A
An atom can be represented by the nuclear symbol ^{A}zE
Nucleons  protons + neutrons 
John Dalton's Atomic Theory
All atoms of an element have the same mass, although isotopes are atoms of the same element but has different numbers of protons
Ex: All carbons atoms (Z=6) have 6 protons and electrons, but only 98.89% of naturally occuring carbon atoms have 6 neutrons (A=12) 
Chemical Compounds
Radicals/Polyatomic Ions  stable groups which form chemical bonds as an intact unit.
The valence numbe is taken as one.
If a molecule contains more than one radical (At least two unpaired electrons), the formula uses parentheses. Calcium Phosphate  Ca₃(PO₄)₂ 
Some Polyatomic Ions
Monovalent (1^{} ) 

Bivalent (2^{} ) 

Trivalent (3^{} ) 
Ammonium 
NH₄^{+} 
Carbonate 
CO₃ 
Phosphate 
PO₄ 
Acetate 
C₂H₃O₂ 
Chromate 
CrO₄ 
Borate 
BO₃ 
Chlorate 
ClO₃ 
Oxalate 
C₂O₄ 
Chlorite 
ClO₂ 
Sulfate 
SO₃ 
Bicarbonate 
HCO₃ 
Sulfite 
SO₂ 
Biculfate 
HSO₄ 
Peroxide 
O₂ 
Hydroxide 
OH 
Nitrate 
NO₃ 
Nitrite 
NO₂ 
Diatomic Molecules
H₂ 
hydrogen 
N₂ 
nitrogen 
F₂ 
fluorine 
O₂ 
oxygen 
I₂ 
iodine 
Cl₂ 
chlorine 
Br₂ 
bromine 
CrissCross Method
Determine the charge or valence number of the elements
Exchange their valence numbers
Reducing by their gcf is possible
Calculating Empirical Formula
Percentage Composition  amounts of the elements for a given amount of compound
Empirical Formula  simpest formula of any compound (smallest ratio of moles); derived from mass analysis
 Determine the given number of moles in each element
 Divide each by the smallest number of moles given
 Multiply each by the smallest number that will turn them into whole numbers. 
Calculating Empirical Formula with Molar mass
Given 
Mass number 
Quotient 
Quotient (x/0.75) 
Smallest Ratio (y*2) 
28.03% Mg 
24.035 
1.15 
1.53≈1.5 
3 
21.6% Si 
28.086 
0.75 
1 
2 
1.16% H 
1.16 
1.15 
1.53≈1.5 
3 
49.21% O 
49.21 
3.08 
4.1≈4 
8 
Answer = Mg₃Si₂H₃O₈ 
Divide the given percent composition to the mass number of each element
Divide each quotient by the smallest number among them
Multiply the quotients by the smallest number that will make them whole
Calculating Molecular Formula By Empirical Formula
Empirical Composition 
Mass number 
Product (rounded off) 
Product (emp* (mass/x)) 
Mg₃ 
24.305 
73 
6 
Si₂ 
28.086 
56 
4 
H₃ 
1.008 
3 
6 
O₈ 
15.999 
128 
16 


Σ = 260 
Suppose the molar mass is 520.8; divide it by the summation (520.8/260 ≈ 2). Multiply 2 by the empirical compostion of each element. Answer = Mg₆Si₄H₆O₁₆ 
Get the summation summation of the product of each empirical composition to their mass number
Divide the summation from the molar mass
Multiply the quotient to the empirical composition of each element
Module 3  Molar Mass, Chem Reactions, Eq
Mole(mol)  SI unit for determining molar mass; amount of substance that contains the same number of atoms in 12g of Carbon12
Avogadro's number  6.02214076 × 10^{23}
Elements  mass in amu of 1 atom of an element is the same as the mass in grams of 1 mole of atoms of the element
Mass of S (32.07 amu) is equal to the mass of 1 mol (6.02214076 × 10^{23}) of S (32.07 amu) 
Calculating Molecular Mass/Weight
Composition 
Number of Atoms 
Mass Number (amu) 
Product (amu) 
H₂ 
2 
1.008 
2.02 
O 
1 
16.00 
16.00 



Σ = 18.02 
Determine the number of atoms of each element then multiply to their corresponding mass number
Get the summation of the products
Writing and Balancing Chem Eq
Law of Conservation of Mass  mass is neither created nor destroyed in a chemical reaction
Antoine Lavoisier  French chemist; proponent
Reactants  starting material in a chemical reaction
Product  substance formed in a chemical reaction
Reactants → Products
"to yield" or "to form" (→)
"to react with" or "to combine with" (+) 
Examples
CH₄ + O₂ → CO₂ + H₂O ⟹ CH₄ + 2O₂ → CO₂ + 2H₂O 
Al + BaO → Al₂O₃ + Ba ⟹ 2Al + 3BaO → Al₂O₃ + 3Ba 
Cl₂ + KBr → KCl + Br₂ ⟹ Cl₂ + 2KBr → 2KCl + Br₂ 
Types of Chemical Reactions
Type 
Definition 
Example 
Combination/Synthesis 
two or more reactants combine to form a single product 
2Mg + O₂ → 2MgO 
Decomposition 
one reactant breaks down into two or more products 
CaCO₃ → CaO + CO₂ 
Single Displacement 
one element is substituted for another element in a compound 
K + NaCl → KCl + Na 
Double Displacement/salt metathesis 
two substances react by exchanging ions to produce two new molecules 
AgNo₃ + NaCl → AgCl + NaNo₃ 


Module 4  Mass Relationships in Chem Reactions
Stoichiometry  quantitative relationship between reactants and products in a chemical reaction
Stoichiometric coefficient  added before an element, ion, or molecule to balance chemical reactions
Mole method using molemole factor:
2Na(s) + 2HCl(aq) → 2NaCl(aq) + H₂(g)
2 moles Na ≅ 2 moles NaCl; hence, 
Calculating Amount of Product and Reactant
3Hg₂(g) + N₂(g) → 2NH₃(g)
How many moles of H₂ are needed to produce 26.5 moles of NH₃?
26.5 moles NH₃ x (3 moles H₂)/(2 moles NH₃) = 39.8 moles of H₂
How many moles of NH₃ will be produced if 33.7 moles of N₂ reacts completely with H₂
33.7 moles of N₂ x (2 moles of NH₃)/(1 mole of N₂) = 67.4 moles of NH₃ 
In using molemole factor, the arrangement of fractions is done in a way that there is cancellation of similar units
Calculating " " with Molar Mass
2LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)
How many grams of CO₂ can be absorbed by 236.1 g of LiOH?
236.1g LiOH x (1 mole of LiOH)/(23.95g LiOH) x (1 mole CO₂) /(2 moles LiOH) x (44.01g of CO₂)/ (1 mole CO₂) = 221.1g CO₂ 
Determine the mass of each element and add each to the given compounds
(Li=6.941, O=15.999, H=1.008, C=12.011)
Since the given number has 4 significant figures, the numbers also have 4 significant figures
Limiting and Excess Reagent
In chemical reactions, the amount of reactants isn't always stoichiometrically exact, so scientists use cheaper reactants (excess)
Limiting Reagent  Reagent that is completely reacted or used up
Excess Reagent  Reactant present with higher quantity than what is required to react in a limiting reagent 
Example
3H₂ + 2N₂ → 2NH₃
Suppose 6 moles of H₂ was mixed with 4 moles of N₂ . To determine which is the limiting reagent, the amount of NH₃ must be computed given the moles of H₂ and N₂ and the molemole factor of the equation 
Solution
Simplify the number of moles by multiplying each final no. of moles of reagent by the proportion of the initial number of moles and given reagent
The reagent with lesser number of moles of NH₃ is the limiting reagent and vise versa; in this case, H₂ is the limiting reagent and N₂ is the excess reagent
Calculate the excess
To determine how much of 4 moles of N₂ is in excess, use molemole factor of N₂ and H₂
6 moles of H₂ x (1 mole N₂)/(3 moles H₂) = 2 moles N₂ in excess
The number of moles of N₂ required to react with 6 moles of H₂ is only 2, thus, 6 moles of N₂ has an excess of 4 moles
6 moles of N₂  2 moles of N₂ in 6 moles of H₂ = 4 moles excess N₂ 
Limiting and Excess Reagent with Molar Mass
Mg₂Si + 4H₂O → 2Mg(OH)₂ + SiH₄
If we start with 50.0 g of each reactant, how much in grams SiH₄ can be formed?
50g Mg₂Si x (1 mol Mg₂Si)/(76.7 g Mg₂Si) x (1 mol SiH₄)/(1 mol Mg₂Si) x (32.1 g SiH₄)/(1 mol SiH₄) = 20.9 SiH₄
50 g H₂O x (1 mole H₂O)/(18.0 g H₂O) x (1 mol SiH₄)/(4 mol H₂O) x (32.1 g SiH₄)/(1 mol SiH₄) = 22.3 g SiH₄
50g Mg₂Si = 20.9 SiH₄ (Limiting reactant)
50 g H₂O = 22.3 g SiH₄ (Excess reactant) 
Divide the number of initial molar mass of compound by the final/given molar mass and multiply with the molar mass of required compound to convert
Determine how much of 50 g of H₂O is in excess by 50 g H₂O x (1 mol Mg₂Si)/(4 mol H₂O) = 12.5 g Mg₂Si
50 g Mg₂Si  12.5 g Mg₂Si excess in 50 g H₂O = 37.5 g excess Mg₂Si
Theoretical and Percent Yield
Percent Yield  ratio of actual yield to the theoretical yield expressed as a percentage
Percent Yield = (Actual Yield)/(Theoretical Yield) x 100%
Theoretical Yield  maximum/expected amount of product produced from the given amount of reactant
Actual Yield  actual amount of product produced from the given amount of reactant (determined experimentally) 
Calculating Theoretical and Percent Yield
In calculating theoretical yield, always use the limiting reactant.from the previous example
Mg₂Si + 4H₂O 2Mg(OH)₂ + SiH₄
If 19.87 g SiH₄ is formed, what is the percent yield of the reaction?
50g Mg₂Si = 20.9 SiH₄ (Limiting reactant) = Theoretical yield
Percent Yield = (Actual Yield)/(Theoretical Yield) x 100%
= 19.87 g SiH₄/20.9 SiH₄ x 100% = 94.89%
Percent error = 100%  98.89% = 5.11% 
Module 5  Gases I
Pressure  amount of force exerted per unit area
Standard atmosphere (atm)  widely used unit for pressure; 1 atm = 760mmHg
Torr (or mmHg)  milliliter of mercury equal to 1 atmosphere; named after Italian scientist Evangelista Torricelli (invented barometer)
Pounds per square inch (psi)  amount of pressure in pounds that gas exerts in a container per square inch of unit area
kilopascal (kPa)  equal to 1000 Pa, modern unit for pressure/default
Conversion Factor:
1 atm = 760mmHg = 760 Torr = 101.3 kPa = 14.7 psi = 1k Pa = 1000 Pa 
Gas Laws
Boyle's Law 
Volume is inversely proportional to its pressure at a constant temparature; ↑V ⟺ ↓P 
P₁V₁ = P₂V₂ 
Charles' Law 
Volume is directly proportional to its absolute temperature and constant pressure; ↑V ⟺ ↑T 
V₁/T₁ = V₂/T₂ 
Avogadro's Law 
Volume is directly proportional to the number of moles contained in the volume at constant temperature and pressure; ↑V ⟺ ↑n 
V₁/n₁ = V₂/n₂ 
GayLussac's Law/Ideal Gas Law 
Sums up and combines Boyle's, Charles', and Avogadro's Laws 
PV = nRT (where R or universal gas constant = 0.0821 atm.L/mol.K) 
Gas Mixtures
Different gases can be present in a container and can be represented as n₁ (gas 1), n₂ (gas 2), or n₃ (gas 3), etc.; the total number of moles as nₜₒₜₐₗ
The pressure exerted by the mixture can be interpreted as Pₘᵢₓₜᵤᵣₑ = (nₜₒₜₐₗRT)/V
It can be simplified as P₁= (n₁RT)/V; P₂ = (n₂RT)/V; P₃ = (n₃RT)/V
Pressures P₁, P₂, and P₃ are partial pressure of each gas
Dalton's Law of Partial Pressure  pressure exerted by the mixture is the sum of the pressures exerted by each component
Get the partial pressure of gas 1 by P₁ = Pₘᵢₓₜᵤᵣₑ X₁ (wherein X₁ is the mole fraction of gas 1) 
Module 6  Gases II
Stoichiometric ratio  dictates the ratio of components to start the reaction
Standart Temperature and Pressure (STP) = 0°C (273 K) and pressure of 1 atm
Amount of gaseous products are determined using n = (PVₛₜₚ)/(RT) wherein Vₛₜₚ is the volume of gases involved measured in STP in liters (L)
n = (PVₛₜₚ)/(0.0821)(273 K) = Vₛₜₚ/22.4
n = Vₛₜₚ/22.4
Gases are also measured in Standard Ambient Temperature and Pressure (SATP) which is more accurate than STP which is at 25°C (298 K) and 1 atm.
n = (PVₛₐₜₚ)/(0.0821)(298 K) = Vₛₐₜₚ/24.5
n = Vₛₐₜₚ/24.5 
Temperature Conversion
Celsius 
→ 
Kelvin 
C + 273.15 
Celsius 
→ 
Farenheight 
C (9/5) + 32 
Farenheight 
→ 
Kelvin 
(F  32)(5/9) + 273.15 
Farenheight 
→ 
Celsius 
(F  32)(5/9) 
Kelvin 
→ 
Celsius 
K  273.15 
Kelvin 
→ 
Farenheight 
(K  273.15)(9/5) + 32 
Kinetic Molecular Theory
1. Gases are very small molecules separated by expansive space between them
2. Force of attraction between particles is negligible
3. The molcules are in constant motion and move randomly in all directions
4. Sometimes particles collide with each other or with the walls of container
5. The collisions are perfectly elastic, hence, there is no change in momentum
6. The average kinetic energy is determined only by the absolute temparature of the gas
To determine the kinetic energy of gas particles, the rootmeansquare velocity is used:
vᵣₘₛ = √(3RT/M)
where R = ideal gas constant, T = absolute temperature in K, M = molar mass in g/mol
To compare the velocities of gases with different molar masses at the same absolute temperature:
vᵣₘₛ₁/vᵣₘₛ₂ = √(M₂)/√(M₁)
where M₁ or M₂ = molar mass of gas 1 or 2
This expression is also known as Graham's Law of Diffusion which states that the diffusion rate (rate at which the gas moves), is inversely proportional to the square root of its molar mass 
YEY! you finished q1, I am so proud of you :)
Calculating Empirical Formula with Molar mass
Given 
Mass number 
Quotient 
Quotient (x/0.75) 
Smallest Ratio (y*2) 
Divide the given percent composition to the mass number of each element
Divide each quotient by the smallest number among them
Multiply the quotients by the smallest number that will make them whole
Calculating Empirical Formula with Molar mass
Given 
Mass number 
Quotient 
Quotient (x/0.75) 
Smallest Ratio (y*2) 
28.03% Mg 
24.305 
1.15 
1.53≈1.5 
3 
21.6% Si 
21.16 
1.15 
1.53≈1.5 
3 
1.16% H 
1.16 
0.75 
1 
2 
49.21% O 
49.21 
3.08 
4.1≈4 
8 
Divide the given percent composition to the mass number of each element
Divide each quotient by the smallest number among them
Multiply the quotients by the smallest number that will make them whole
Calculating Empirical Formula with Molar mass
Given 
Mass number 
Quotient 
Quotient (x/0.75) 
Smallest Ratio (y*2) 
28.03% 
24.035 
1.15 
1.5 
3 
1.6% 
28.086 
0.75 
1 
2 
1.16% 
1.008 
1.15 
1.5 
3 
49.21 
15.999 
3.08 
4 
8 
Answer = Mg₃Si₃H₂O₈ 
Divide the given percent composition to the mass number of each element
Divide each quotient by the smallest number among them
Multiply the quotients by the smallest number that will make them whole
