Chemistry

-study of matter
-anything that has mass or occupies space
- liquid, solid, and gas, and plasma à 4 states of matter
- plasma à melting form of gas

Physical properties: characteristics of something

Physical properties of matter:
1. Density: how much “stuff” is in something
-mass per unit volume (mass/volume)
2. melting/boiling point
3. solubility: the ability of a substance to dissolve in a solvent.
4. conductivity: the ability of a substance to conduct electricity or heat
5. hardness: measure of a substance to resist deformation

Chemical Properties of matter:

1. Reactivity: tendency of a substance to undergo chemical reaction
2. Flammability- the ability of a substance to burn in the presence of oxygen
3. Acidity
Atom: the smallest particle of an element that can under go a chemical reaction
-proton, neutron, electron
9/6/24
Periodic table: chart that shows all chemical elements
H: Hydrogen
He: Helium
Li: Lithium
Be: Beryllium

Horizontal: periods
Vertical: groups/families
7 periods on periodic table
Hydrogen and helium are in the second period
18 groups on periodic table

Gold: 79/AU
Metals: left
Non-metals: right
Metalloids: middle

Parts of Periodic Table

1- Atomic Number
-above
-Number of protons (inside nucleus)
-number of electrons (=protons)
2-Atomic Mass
-below
-the quantity

The electron configuration tells us how the electrons are arranged around the nucleus of an
atom

1) Alkali metals (group 1) very reactive

2) Alkaline earth metals (group 2) reactive
3) Halogens (group 17) reactive
4) Noble Gases (group 18) non-reactive/stable
Mass number: total number of protons and neutrons in an atom
=protons + neutrons
Number of protons = Atomic number. Number of electrons = Atomic number. Number of
neutrons = mass number - atomic number.

Ionization energy
Electron negativity: increase across a period (horizontal)
Atom radius: decrease across a period

Periodic table 1-20 test

1)hydrogen
2)helium
3)lithium
4)
5)boron
6)
7)
8)oxygen
9)
10)
11) neon
12)
13)
14)
15)phosphorus
16)
17)
18)
19)potassium
20)calcium

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Energy levels
K = 1 2(1)^2 = 2
L= 2 2(2)^2 = 8
M = 3 2(3)^2 = 18
N = 4 2(4)^2 = 32
2n^2 = shows how many electrons something can occupy
n = number of energy level

Orbitals: shell of an electron

s-Orbital: can hold 2 electrons/nucleus
p-Orbital: can hold 6 electrons/dumbbell shaped
d- Orbital: can hold 10 electrons/double dumbbell
f-Orbital: can hold 14 electrons
According to this principle electrons are filled in the following: 1s 2s 2- 3s 3p 4s 3d 4p 5s 4d 6s 4f
6p 7s 5s 6d 7p….

Calcium = 20
1s2 2s2 2p6 3s2 3p6 4s2
Zinc = 30
1s2 2s2 2p6 3s2 3p6 4s2 3d10

9/10 CK-12 lesson

Macroscopic: substances and objects that can be seen
Microscopic: substances and objects that cannot be seen
Areas of Chemistry
 Physical chemistry
-Study of macroscopic properties, atomic properties, and phenomena in chemical systems
-chemical reactions, energy transfers that occur in reactions
 Organic chemistry
-study of chemicals containing carbon
-carbon is one of the most abundant elements on Earth
-most of all chemicals found in all living organisms are based on carbon
 Inorganic chemistry
-study of chemicals that do not contain carbon
-rocks/minerals
-design and properties of materials involved in energy and information tech

 Analytical chemistry
-The study of the composition of matter
-Separating, identifying, and quantifying chemicals in samples of matter
 Biochemistry
-study of chemical processes that occur in living things
-basic cellular processes up to understanding disease states so better treatments can be
developed

Pure and Applied Chemistry

 Pure chemistry does research primarily to advance mankinds understanding of
chemistry
-concerned with a greater understanding of the theories behind how matter is changing
chemical reactions
-less applications of the research they are doing
 Applied chemistry is directed towards specific practical goal or application
-uses applied research (issue, data, analysis)
 In-between pure and applied chemistry
-biochemistry in some cases
-sometimes theres no clear line between pure and applied

9/23
Shell Subshells and Orbitals
 Bohr Model: simplified atom model
Circular lines: energy levels
N= principal quantum numbers
KLMN=1234
1- 2 electrons
2- 8 electrons
3- 18 electrons
Formula: 2n^2 = maximum number of electrons
N=shell or energy level
Subshells: SPDF

Quantum Numbers:

Ex.
n= 1
l=0
ml =0
n= energy level
l=orbital type
ml= specific orbital
10/7/2024
How to name compounds
Oxygen: oxo
Sulphur: sulphate
Carbon: carbonate
Phosphorus: phosphate
Nitrogen: nitrate
1: mono
2: di
3: tri
4: tetra
5: penta
6: hexa
7: hepta
8: Octa
9: nona
10: deca
-H2CO2: di oxo carbonate (ii) acid
Homework:
H3PO4 and HNO3
H3OPO4 = tetra oxo phosphate (v) acid
HNO3= tri oxo nitrate (v) acid
1. Hcl
2. HClO3
3. KNO3 this is a salt
Chemical bonding is the process by which atoms combine form to combine atoms.
Types of chemical bonds
-electrovalent bonding or ionic bonding
-in ionic bonding, one atom transfers one or more electrons to another atom. This usually
occurs between a metal and non-metal (Na + Cl)

The following characteristics are found in ionic bonded molecules to the existence of a strong
force of attraction between cations and anions
-an ionic bond is the most powerful of all bonds
-the ionic bond is the most reactive of all existing bonds in an appropriate medium since it
possesses charge separation
-the melting and boiling points of ionic bonds compounds are pretty high
-ionic-bonded molecules are strong
conductors of electricity in their
aqueous solutions or molten form. This
is because of ions which function as large
carriers
Li(3) + F(9) = Lithium+ and Flourine-
Covalent bonding: two or more
atoms share electrons to achieve
stability. This type of bonding usually occurs between non metals. The shared electrons allow
both atoms to have a full outer shell

Metallic bonding: metal atoms share their electrons freely. These shared electrons are not tied
to any specific atom and can move throughout the entire structure. This is why metals are good
conductors of electricity and heat
Ex.: Aluminum (Al)
-in aluminum metal, the aluminum atoms are held together by a metallic bond where the
valence electrons are free to move throughout the metal.
-this gives aluminum its properties like ductility (ability to be stretched into wires) and
malleability (ability to be hammered into thin sheets)

Hydrogen bonding: a type of attractive interaction between a hydrogen atom covalently bonded
to a highly electronegative atom (nitrogen, oxygen, or fluorine) and another electronegative
atom.

HW: Na + Br = Ionic bond because sodium is a metal and bromine is a nonmetal so sodium gives
one electron to bromine
Na+ Br- = NaBr

Chemical Equations and Balancing of Chemical Equations:

Law
of Conservation of Mass:
-mass is neither created not destroyed in a chemical reaction
-the mass of reactants must equal the mass of products
Steps to Balance a Chemical Equation
Step 1: write the unbalanced equation
Step 2: list the atoms involved
list each atom type below the equation, indicating how many of each are on both sides
Step 3: balance one element at a time
start with elements that appear in only one reactant and one product. Leave H and O last if
possible
Step 4: use coefficients
place numbers in the front of compounds to balance the number of atoms
Step 5: double-check
ensure all elements are balanced and verify that coefficients are in the lowest whole numbers

CH4 + O2  CO2 + 2H2O

-2Al+1O2 Al2O3
Al = 1  Al=2
O=2 O=3
Al=2
O=3
HW:
Fe2O3+C  Fe+CO2
C2H6+O2 CO2+H2O

HOMEWORK: C4H10 + O2  CO2 +H2O

Left Right
C-4 C-1
H-10 H-2
O-2 O-3
C4H10 + O2  CO2 +H2O
4CO2+H2O
Balance carbon atoms

left right
C4 C4
H10 H8
O2 O9

Balance hydrogen atoms

4CO2+5H2O

left right
C4 C4
H10 H10
O2 O14

Balance oxygen atoms

C4H10 + 7O2  4CO2+5H2O

left right
C4 C4
H10 H10
O14 O14
Mole Concept and Molar Mass
Mole: The mole is a fundamental concept in chemistry that represents a specific quantity of
particles, similar to how a “dozen” represents 12 items.
 A mole is defined as 6.022 * 10^23 particles (atoms, molecules, ions, etc.), which is
known as Avogadro’s number
Molar Mass: the molar mass of a substance is the mass of one mole of that substance. It is
expressed in grams per mole (g/mol)
H2O  H2 + O
1*2+16
2*16=18 g/moL
HCL  H + CL
1 + 35.5 = 36.5 g/moL
NaCl  Na + CL
23 + 35.5 = 58.5 g/moL
H2SO4 H2+S+O4
1*2 + 32+ 16*4
2 + 32 + 64 = 98 g/moL
KMnO4 K + Mn + O
39.1 + 55 + 16(4)
39.1+55+64 = 158.1 g/moL
Fe2O3  Fe + O
56(2)+16(3)
112+48 = 16 g/moL

HOMEWORK
C6H8O6
Molar mass= (6×12.01)+(8×1.008)+(6×16.00)
Molar mass= 72.06+8.064+96.00=176.124 g/mol
CH3COOH
Molar mass= (2×12.01)+(4×1.008)+(2×16.00)
Molar mass= 24.02+4.032+32.00=60.052 g/mol

Stoichiometry
Avogadro’s number = 6.022 * 10^23 = 1 mole
1) Mass to moles: moles = (g)/molar mass
2) Moles to mass: (g)=moles*molar mass
3) Moles to particles: number of particles = moles * 6.022 * 1023
4) Particles to moles: moles=number of particles/6.022*1023

1. calculate the number of moles in 36 grams of water (H2O)

-mole=mass/molar mass
-molar mass = H2O = 1*2 + 16 = 18 g/mol
-mole= 36g / 18g/mol
2mol  1 mol X X = 6.022 * 10^23 * 2mol

2. Calculate the number of moles in 2 grams of H2SO4

Formula: mass(g)/molar mass(g/mol)
Molar mass of hydrogen: 98.086 g/mol
2g/98.086 g/mol ≈ 0.0204 moles

3. How many grams are in 2 moles of carbon

Mass = moles * molar mass
= 2 moles * 12g/mol
= 24g

4. How many grams are in 0.75 moles of sodium chloride (NaCl)

Mass = Moles * Molar mass
=0.75moles * 58.5 g/mol
=43.875g

5. How many grams are in 3.5 moles of water (H2O)

Mass = moles * molar mass
=3.5moles * 18.016g/mol
=63.056g

6. How many molecules are in 2 moles of water (H2O)

Particles = Moles * Avogadro’s Number
2 mol * 6.022 * 10^23
=1.024*10^24

7. How many molecules are in 0.5 moles of carbon dioxide (CO2)

Particles = moles * Avogadro’s number
0.5 mol * 6.022 * 10^23
3.011 * 10^23

8. How many atoms are in 2.5 moles of oxygen (CO2)

2.5 * 6.022 * 10^23
1.505 * 10^24

9. How many moles are in 3.011 * 10^23 molecules of water (H2O)

Moles = particles / Avogadro’s number
= 3.011 * 10^23 / 6.022 * 10^23
= 0.5
10. How many moles are in 9.022 * 10^23 of oxygen (O)
 = 9.022 * 10^23 / 6.022 * 10^23
=1.5

Limiting Reactants and Percentage Yield

Steps to Solve Limiting Reactant Problems:

1. Write the balanced equation
2. Convert the given quantities of reactants to moles using molar masses
3. Use the mole ratio from the balanced equation to calculate how much product each
reactant can produce
4. Determine the limiting reactant (the reactant that produces the least amount of
product)
5. Use the limiting reactant to calculate the amount of product
-If you start with 4 moles of H2 and 3 moles of O2, which reactant is limiting, and how many
moles of H2O would be produced?
Mole of Product/Mole of Reactant * No of moles given
H2 = 2 / 2 * 4 = 4 moles
O = 2 / 1 * 3 = 6 moles
H2 can produce 4 moles of water (H2O) O2 can produce 6 moles of H2O
The limiting reactant is the one that produces the fewer amount of product (in this case it is H2)

-N2+H2=N2H4
If you have 6 moles of nitrogen gas (N2) and 7 moles of hydrogen gas (H2), how many moles of
hydrazine (N2H4) can be produced, and which reactant is limiting?
Balanced equation= N2+2H2→N2H4
N2 = 2/1 * 6 = 12
H2 = 2/2 * 7 = 7
Limiting reactant is 7

1. Write balanced equation: 2NaOH +H2SO4→Na2SO4+2H2O

2. Find the molar masses of the reactants:

-sodium hydroxide (NaOH)
 Na: 23
 O: 16
 H: 1
Molar mass = 40 g/mol
-sulfuric acid (H2SO4):
 H: 1
 S: 32
 O: 16
Molar mass = 2(1)+32+4(16) = 98 g/mol
3. Calculate the number of moles of each reactant:
Moles of NaOH= Mass of NaOH/Molar mass of NaOH

Moles of H2SO4= Mass of H2SO4/Molar mass of H2SO4

4. Determine limiting reactant

-2 moles of NaOH gives us 1 mole of Na2SO4

-limiting reactant formula= mole of product/ mole of reactant * mole given

NaOH: 1 mole/ 2 * 0.25 = 1.125mol
 H2SO4: 1mol/1mol * 0.102 = 0.102
-H2SO4 (0.102) is the limiting reactant because it would be completely used up first

5. Calculate how much Na2SO4 is produced

-molar mass= 2(23) + 32 + 4(16) = 142 g/mol
-mass = 0.102mol (limiting reactant) * 142g/mol = 14.484 g

Percentage Yield: compares the actual amount of product obtained in an experiment to the
theoretical amount predicted by stoichiometry
Percentage Yield = (Actual Yield / Theoretical Yield) * 100
Example:
1. If 10 grams of sodium (Na) react with chlorine (Cl2) to form 15 grams of sodium chloride
(NaCl), what is the percent yield?
2Na + Cl2  2NaCl
Theoretical yield: molar mass of Na = 23g/mol
mass/molar mass = 10g/23g/mol = 0.42mol
2 mol of Na produces 2 moles of NaCl (2:2)
0.43mol of Na produces
2mol of Na * * = 2 moles of NaCl * 0.43

x = 2moles * 0.43 / 2 = 0.43

Theoritical yield = 0.43mol * 58.44g/mol = 25.129
Percentage yield = 15g / 25mol = 0.6
0.6*100=60%

1. Balance
 H2O + O2 = H2O
2. Calculate moles of hydrogen: moles of H2 = mass of H2 / molar mass of H2
-molar mass of hydrogen: 2 g/mol
5g
= 2.5 moles of H2
2 g ∕ mol
2. Calculate the theoretical yield of water (H2O)
1 mole of hydrogen produces 1 mole of water (1:1) = 2.5 moles of hydrogen produce 2.5
moles of water
 Molar mass of H2O: 18 g/mol
To calculate the theoretical yield: Mass of H2O = moles of H2O * molar mass H2O
2.5 ×18=45 moles of H2O

3. Calculate the percentage yield

Actual yield: 8 grams
Formula: (actual yield/theoretical yield) * 100

Concentration and Volume

Concentration refers to the amount of solute present in a given quantity of solvent of solution.
It helps quantify how much of substance is dissolved in a particular amount of liquid. The
concentration solution can be expressed in various ways, including molarity, molality, and
percent concentration.
Volume in STP and Volume in RTP
 STP is standard Temperature and Pressure
 RTP is Room Temperature and Pressure

STP: refers to a standard condition of 0 degrees celsius (273.15 K) and 1 atm pressure. At STP,
the volume of one mole of an ideal gas is 22.4 L
Molar volume at STPL V=22.4 L/mol (for an ideal gas)
RTP: This refers to to conditions of 25 degrees celcius (298.15 K) and 1 atm pressure. At RTP the
volume of 1 mole of an ideal gas is 24 L.
Molar volume at RTP: V-24 L/mol city

Dilution:
 C1V1 = C2V2
 C1 = Initial Contraction
 V1 = Initial Volume
 C2L: Final Concentration
 V2: Final Volume

 You have 100 cm^3 of 2 mol/dm^3 HCl and need to dilute it to 0.5 mol/dm^3. What will
the final volume be?
-C1V1= C2V2
-C1= 2mol/dm^3
-V1= 100cm^3
-C2= 0.5 mol/dm^3
-V2= y
 A laboratory technician has 1.5 mol/dm^3 solution of HCl and needs to prepare 500
cm^3 of 0.2 mol/dm^3 HCl. Calculate the volume of the concentrated solution and water
required.