Physical Properties

A trait we can observe w/o changing the identity of the substance. EXAMPLES:ColorShape Dimensions (size)TextureTasteTemperatureMelting PointDensity SmellFreezing PointMass

Chemical Properties

A trait we can observe by changing the identity of the substance. EXAMPLES: FlammabilityDecomposableDigestible ToxicityAbility to Oxidization Reactivity/Inert

Physical Change

A change that affects only physical properties and does not change the identity of the substance. EXAMPLES:RipCrumbleWetThrowDrawFoldKickMix

Chemical Change

A change that alters the identity of the substance. EXAMPLES:Eating/Digesting Burning React w/ acid or anythingDecompose Oxidization

Elements, Compounds, and Mixtures

Element: substances that cannot be separated by chemical or physical means.

Atom: smallest unit of an element.Compound: substance made of 2 or more elements. It can only be separated by chemical means.

Molecule: smallest unit of a compound.Mixture: A combination of substances that are not chemically combined. Can be separated by physical means.

Homogeneous: looks the same throughout. Heterogeneous: composed of different parts.

Density

A ratio of mass to volume.

D= m/vD=Density g/ml=g/cm^3V=Volume milliliters=cm^3M=Mass grams

Example Problem: Density

A rock has a mass of 8g. When Bob puts it in a graduated cylinder it displaces 4L of water.

What is the density?

ANSWER

D=M/VD=8g/4LD=2L

Scientific Notation

A way of writing numbers that are too big or too small to be conveniently written in decimal form.

EXAMPLES:300 = 3×102

6,720,000,000 = 6.72×109

Sig Figs

Multiplication and Division Rules:The number of sig figs in your answer is the

same as the least number of sig figs in your measurements.

EX: find the area from these measurements. 505cm long = 3 sig figs 10cm wide = 1 sig

figA = L x WA= 505 x 10 = 5050cm2

A= 5,000cm2

Extra Practice

Find the rate of measurement. 250.0mL 0.25s

ANSWER

Find the rate of measurements. 250.0 mL = 3 sig figs 0.25s = 2 sig figs

250.0/0.25 = 1000mL/s 1.0 x 103

Sig Figs Continued

Addition and Subtraction Rules:The estimated number in your answer is located in the same place value column as

the estimated number in the measurement’s that is farthest to the left.

EX: 101.0g+ 10.5g111.5g 112 g

Extra Practice

212.25C and 12.3C

ANSWER

212.25C+ 12.3C199.95C 200.0C

Accuracy/Precision

Accuracy/Precision Continued

the accuracy of a measurement system is the degree of closeness of measurements of a quantity to that quantity's actual value. The precision of a measurement system, also called reproducibility or repeatability, is the degree to which repeated measurements under unchanged conditions show the same results.

Quantitative/Qualitative

Quantitative: Is expressed or related to in terms of quantity as in numbers.

Qualitative: Related to or involving quality or kind.

Ionic Compound Occurs between One Metal and Non-Metal Crystilanial Solids (Made of Ions) High Melting and Boiling points Conduct with electricity when melted No Definite shape

Covalent Compound Occurs Between Two Metals Gases, Liquids, Soft Solids (Made of Molecules) Low Melting and Boiling points Poor electrical conductors Define Shape

Periodic Table To the left of the red line are non-metal To the right of the red line are metals

Example…is which type of bond?1.NaBr2.HCL3.AgCL4.C2S

Answers1. Ionic Bond- Na is a non-metal, Br is a Metal2. Ionic Bond- H is a non-metal, Cl is a metal3. Ionic Bond- Ag is a non-metal, Cl is a metal4. Covalent Bond- C is a metal, S is a metal

To Tell The Difference Look at periodic table Locate your elements Look for the stair case in the periodic table Under elements B,SI,AS,TE,AT

Isotopes An isotope is an atoms with the same number of

protons, but differing numbers of neutrons Isotopes are different forms of a single element Different Atomic Mass

How To Calculate Average Atomic Mass 1. Convert the percentages into decimals. (This

percentage is known as its relative abundance or percent abundance)

2. Multiply the percentage of each isotope by its respective mass

3. Add the numbers from step two together

Examplemass number exact weight percent abundance

12 12.000000 98.90

13 13.003355 1.10

Answer This is the solution for carbon:

(12.000000) (0.9890) + (13.003355) (0.0110) = 12.011 amu

Basics Find your element on the Period Table element's atomic number and atomic weight.

atomic number is the number in the upper left corner atomic weight is the number on the bottom

Protons and Electrons The atomic number is the number of protons in

an atom of an element. krypton's atomic number is 36, so an atom of

krypton has 36 protons Electrons have the same quantity as protons

Neutrons Mass Number = (Number of Protons) + (Number

of Neutrons) For krypton, this equation becomes: 84 = (Number of Protons) + (Number of Neutrons) 84 = 36 + (Number of Neutrons) NUETRONS = 48

Summary For any element: Number of Protons = Atomic Number Number of Electrons = Number of Protons =

Atomic Number Number of Neutrons = Mass Number - Atomic

Number

Ions Ions are formed when an atom gains or loses

electrons and gets charged If an atom gains electrons, number of electrons are

higher than protons an atom gets negatively charged

Similarly if an atom loses electrons the total number of protons become more than the number of electrons atom becomes positively charged

What is an ion?

Ion electrically charged atom or atom group: an atom or group of atoms that has acquired an

electric charge by losing or gaining one or more electrons

Knowing the charge

So… Determine the charge of each element when present in an ionic

compound.Use the table above to determine these charges. For example, O = -2, Rb = +1.

Use the appropriate number of each ions such that: The sum of all charges adds up to zero. The simplest ratio of ions is used.

For example, if magnesium (Mg) and bromine (Br) are mixed: Metal (Mg) + Non-metal (Br) IS an ionic compound. Mg ⇒ Mg+2 and Br ⇒ Br-1 For the final steps:

One +2 ion is exactly balanced by two -1 ions. 2:1 is the simplest possible ratio.

Thus, the formula of the ionic compound formed is MgBr2

Alpha There are three primary types of radiation:

Alpha - these are fast moving helium atoms. They have high energy, but due to their large mass, they are stopped by just a few inches of air, or a piece of paper.

Beta These are fast moving electrons. Since electrons

are lighter than helium atoms, they are able to penetrate further, through several feet of air, or several millimeters of plastic or less of very light metals.

Gamma Gamma - these are photons, just like light, except

of much higher energy, X-Rays and gamma rays are really the same thing, the difference is how they were produced. Depending on their energy, they can be stopped by a thin piece of aluminum foil, or they can penetrate several inches of lead

Review A nuclear reaction can be described by an equation,

which must be balanced. The symbol for an atom or atomic particle includes the

symbol of the element, the mass number, and the atomic number.

The mass number, which describes the number of protons and neutrons, is attached at the upper right of the symbol.

The atomic number, which describes the number of protons in the nucleus, is attached at the lower left of the symbol

Electron ConfigurationsDiagram way/ Orbital notation

1.List all of the orbitals with boxes

2.Find out how many electrons are in the element

3. Electrons fill the lowest energy levels first (follow diagonal rule)

4.Reminder* Only 2 electrons fit into a degenerate orbital

Long way/List way

1.Write out the orbitals

2.Depending on the orbital write the amount of electrons in the exponent

3.Do this until your configuration is complete

4.Reminder * S orbital- up to 2 electrons, P orbital- up to 6,

D orbital- up to 10, F orbital- up to 14

Noble gas notation

1.Refer to noble gas before element

2.Write noble gas in brackets

3.Figure out next orbital and fill like usual

Sample Problem:

Be- Beryllium

Write the:Orbital NotationList wayNoble gas notation

Answers to Sample Problem:Orbital Notation:

2s

1s

Electron Configuration:1s^2 2s^2

Noble Gas: [He]2s^2

Wavelength, Energy & Frequency CalculationsWavelength- distance from crest to crest of a wave, one full cycle of

a wave-- measured in metersFrequency- how often a wave passes per second-- measured in 1/s

per secondAs wavelength gets smaller, wave frequency gets largerEquation: c= speed of light= 3.0x 10^8 m/s

Energy of a wave is related to its frequency: E= energy in joulesh= plank's constant= 6.626 x 10^-34 Joules/second

Sample Problem:

Ex. A light at a stoplight has a wavelength of 5m

1. What is the frequency of light?2. What is the energy of light?

•

Answers to Sample Problem:Answers1. wavelength= speed/frequency5 m= 3.0 x 10^-8 m/s ___________frequencyfrequency= 3.08 x 10^-8/ 5frequency= 6.16 x 10^-9 1/s

2. E= hvh= 6.6 x 10^-34 J x SE= (6.6 x 10^-34)(6.16 x 10^-9)E= 4.07 x 10^-42 J

TrendsElectronegativity-chemical property describing

tendency of an atom to attract electrons to itself, which is affected by both its atomic number and the distance that its valence electrons reside from the charged nucleus.

Ionization Energy- describes the amount of energy required to remove an electron from the atom or molecule.

Atomic Radius- measure of the size of an element's atoms.

Electronegativity, Ionization Energy & Atomic Radius Trends

Periodic Table Families and Periods

Periods- go across the table; there are 7Groups- go vertically down the table; 8 are classifiedFamilies- represent elements of the same nature ex. group 8 is called the Noble Gas family because all of these elements have something in common, they all have complete valence shells

Valence Electron trends on periodic table

The Valence Electron trends show the amount of valence electrons in each element. Each column has a different number of valence electrons.

Transition Metals

Ion Trends on the Periodic TableThe Ion trends on the periodic table show

the charge of each element on the periodic table. Each column has a different charge.

Ion Trends

Naming Covalent CompoundsNaming Covalent Compounds

1) Name the 1st element written in the formula2) Use a prefix to tell how many we have

1) Mono2) Di3) Tri4) Tetra5) Penta6) Hexa7) Hecta8) Octa9) Nona10) Deca

Naming Covalent Compounds Naming Covalent Compounds Cont.Cont.

3) Name the 2nd element, change the ending to -ide4) Add a prefix to tell how many

◦CO- Carbon Monoxide◦CO₂- Carbon Dioxide◦SO₃- Sulfur Trioxide ◦H₂O- Dihydrogen Oxide◦S₆F₂- Hexasulfur Diflouride

Naming Ionic CompoundsNaming Ionic Compounds

1) Name the cation (First element in compound)

2) Is the cation a transition metal1) If yes, use a roman numeral to tell the change2) If no, move on

3) Name the anion4) Change the ending to –ide unless it’s a

polyatomic ionEx. 1) Cu₂O₃Copper III Oxide Ex. 2) Na₃PSodium Phosphide

Naming AcidsNaming Acids

Hydrogen ChloridePut anion root in formula Hydro_ic acidHydrochloric Acid

Hydrogen ChloratePut anion root into formula ___ic acid Chloric Acid

Hydrogen Chlorite Put anion root into formula ___ous acid Chlorous acid

Naming Hydrated CrystalsNaming Hydrated Crystals

CuCl₂•4H₂OCuCl₂ Name normal4H₂O Name the Prefix Copper II Chlorine • Tetra Hydrate

5 types of chemical reactions

• Synthesis: 2Cu+O2 2CuO• The two elements combine

• Decomposition: MgCl2Mg+Cl2• The Compound decomposes into two elements

• Single replacement: MgCl2+CuCuCl2+Mg• The single element is switched with the “ “ from the compound

• Double replacement: MgCl2+CuOMgO+CuCl2• One element from each compound is switched

• Combustion: CH3OH+O2CO2+H2O• Hydrocarbon + oxygen = Carbon dioxide+ water

Predicting Products

• If there are two individual elements as the reactants, The reaction will be synthesis.– The product will be a compound.

• If the reactant is a compound, the reaction will be decomposition.– The product will be two separate elements.

• If the reactants are a compound and a single element, the element will switch with one of the elements in the compound. The reaction will be single replacement.

– The product will be a new element and a new compound

• If the reactants are two compounds (respectively) an element from each compound will switch. The reaction will be double replacement.

– The product will be two new compounds.

• If the reactants are a hydrocarbon and an oxygen, the reaction will be combustion.

– The product will be carbon dioxide and water.

Law of Conservation of Matter

• The law of conservation of matter states that matter cannot be created or destroyed, only transferred.– This is why we must balance equations.

Balancing Equations

• When balancing equations, you must have the same amount of each element on either side of the reaction.– __Mg+__O2__MgO

• There is too much O and too little Mg and MgO, so this equation must be balanced to follow the law of conservation of matter

• Add coefficients where needed

– 2 Mg+O22MgO

Balancing Equations

• There are 7 diatonic elements.

• When listed alone, these elements always occur with the subscript of 2

• The elements are Br2, I2, N2, Cl2, H2, O2 and F2.– These can be remembered using the word

BrINClHOF.

Net Ionic Equations

• First start by writing an ionic equation– This includes aqueous substances that

dissociate into ions– Zn(s)+CuSO4(aq)ZnSO4(aq)+Cu(s)

• The aqueous substances dissociate

– Zn(s)+Cu+2 (aq) +SO4-2(aq)Zn=2 (aq) +SO4-2(aq)

+Cu(s)

• (aq)=aqueous, (s)=solid, (l )=liquid• Remember to add charges

Net Ionic Equations

Factor Label Method

• The factor-label method was made to keep track of units in conversion problems. In the method, conversion factors are set up in fraction form. To get the unit you want, you put it on the top of the fraction next to it. You put the unit on the bottom that you are trying to get rid of.

Factor Label Method Formulas

Factor Label Method Examples

Molar Mass

Molar Mass Example

Empirical Formula

• An empirical formula is the lowest term ratio of elements

• Its not always the true formula• When given the % composition, you can find

the empirical formula• Step 1: % to mass• Step 2: Mass to Moles• Step 3: Divide by small• Step 4: Multiply till whole

Empirical Formula Examples

Molecular Formulas

• The ‘true’ formulas• Not in lowest terms1. Get atomic mass2. Multiply by number of subscript 3. Add numbers together4. Divide by molecular mass

Molecular Formula Example

Percent Composition

Mole Ratios

• The ratio between the amounts in moles of any two compounds involved in a chemical reaction.

• Used as conversion factors between products and reactants in many chemistry problems.

• Example: 2 H2(g) + O2(g) → 2 H2O(g)

The mole ratio between O2 and H2O is 1:2. For every 1 mole of O2 used, 2 moles of H2O are formed.

The mole ratio between H2 and H2O is 1:1. For every two moles of H2 used, 2 moles of H2O is formed.

Stoichiometry• The part of chemistry that studies amounts of substances that are involved in

reactions• Helps you figure out how much of a compound you will need, or how much you

started with. • To solve:

1.Balance the equation.2.Convert units of a given substance to moles.3.Using the mole ratio, calculate the moles of substance yielded by the reaction.4.Convert moles of wanted substance to desired units.

• Ex:2Mg + O2 2MgOIf you have 1.2 moles of Mg, how many moles of O2 will you get? 1.2 mol Mg x O2/2Mg = .6 mol O2

Limiting and Excess Reactants

• If you have more than needed, you have excess.• If you need more than you have, you have limiting.• Example:

2Al2O3+3C 3CO2+4Al

If 2.3 g AlO3 and 3.2 g of C mix, which is the limiting and which is the excess?2.3 g AlO3 x 1 mol/101.96 g = .02 mol Al2O3 (have)3.2 g C x 1 mol/12.01 g = .27 mol C.02 x 3C/2Al2O3 = .03 mol C (need)

• Excess : C• Limiting: Al2O3

Theoretical Yield

• The quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction.

• Commonly expressed in grams or moles• To calculate:

1. Use the molar mass of your reactant to convert grams of your reactant to moles.2. Use the mole ratio between reactant and product to convert moles reactant to moles product.3. Use the molar mass of the product to convert moles product to grams of product.

Theoretical yield (cont.)

• Na2S(aq) + 2 AgNO3(aq) → Ag2S(s) + 2 NaNO3(aq)• How many grams of Ag2S will form when 3.94 g of AgNO3 and

an excess of Na2S are reacted together?Atomic weight of Ag = 107.87 gAtomic weight of N = 14 gAtomic weight of O = 16 gAtomic weight of S = 32.01 gAtomic weight of AgNO3 = (107.87 g) + (14.01 g) + 3(16.00 g)Atomic weight of AgNO3 = 107.87 g + 14.01 g + 48.00 gAtomic weight of AgNO3 = 169.88 gAtomic weight of Ag2S = 2(107.87 g) + 32.01 gAtomic weight of Ag2S = 215.74 g + 32.01 gAtomic weight of Ag2S = 247.75 g

• Two moles of AgNO3 is needed to produce one mole of Ag2S.

Theoretical yield (cont.)

• The excess of Na2S means all of the 3.94 g of AgNO3 will be used to complete the reaction.

grams Ag2S = 3.94 g AgNO3 x 1 mol AgNO3/169.88 g AgNO3

x 1 mol Ag2S/2 mol AgNO3 x 247.75 g Ag2S/1 mol Ag2S • 2.87 g of Ag2S will be produced from 3.94 g of AgNO3.

Percent yield

• To find percent yield:1. Balance the chemical equation 2.  Find the limiting reagent3.  Find the theoretical yield 4.  Find the actual yield5.  Find the percentage yield 

• Percent yield = mass of actual yield/mass of theoretical yield x

100         

KINETIC MOLECULAR THEORY (COLLISION THEORY)

Gas particles exhibit constant random motion Collisions happen All collisions are perfectly elastic (can bounce off each

other)What Collisions Mean in the Real World- Collisions between molecules of gas are what creates

air pressure The more collisions that occur, the more pressure is

excited All gas particles move in straight lines, but they

bounce off anything they collide with As the temperature increases the particles move

faster, and more collisions occur, that increases the pressure

As you increases the number of particles, the number of collisions increases, which increases the pressure

ABSOLUTE ZERO AND STP

Absolute Zero- Absolute zero is the point where no more heat

can be removed from a system, according to the absolute or thermodynamic temperature scale.

0 K or -273.15°C.  STP-

0˚C 1 atm 273.15 K 101,325 Pa

PRESSURE CONVERSIONS Pressure is Usually Measured in:

Atmosphere (atm) Bar (Bar) Pascals (Pa) Millimeters of Mercury (mmHg) Torr (torr)

Conversions: 1 atm=101325 Pa 1 bar= 100025 Pa 1 torr= 133.32 Pa 1mmHg= 133.32 Pa 1mmHg= 1 torr

Practice: 14.7KPa= _______ atm

14.7KPa*1000Pa/1KPa=14700Pa

BOYLE’S GAS LAW A relationship between pressure and volume Inverse relationship:

When pressure goes up, volume goes down. When volume goes up, pressure goes down PV=P2V2

Practice problem: A balloon with a volume of 2.0 L is filled with a gas at 3

atmospheres. If the pressure is reduced to 0.5 atmospheres without a change in temperature, what would be the volume of the balloon?

V1 = 2.0 LP1 = 3 atmP2 = 0.5 atmV2 = (2.0 L)(3 atm)/(0.5 atm)

Answer= 12L

CHARLES’S GAS LAW

A relationship between volume and temperature

When temperature goes up, volume goes up When temperature goes down, volume goes

down Direct relationship V/T=V2/T2 Practice problem:

Given 300.0 mL of a gas at 17.0 °C. What is its volume at 10.0 °C?

(300.0) / (290.0 K) = (x) / (283.0 K) 292.76mL

GAY-LUSSAC GAS LAW

Relates pressure to temperature When temperature goes up, pressure goes up When temperature goes down, pressure goes

down P/T=P2/T2 Practice problem:

Find the final pressure of gas at 150 K, if the pressure of gas is 210 kPa at 120 K.

P1= 210 kPa T1= 120 K P2= ? T2= 150 K P2= (210 kPa) (150 K) / 120 K

P2= 262.5 kPa

COMBINED GAS LAW

Combines Boyles, Charles and Gay-Lussac’s gas laws into one equation.

PV/T=P2V2/T2 Practice Problem:

Oxygen occupies a fixed container of 5.5L at STP. What will happen to the pressure ifthe temperature rises to 300K?

P2 = (5.5 L) (101.3 Kpa) (300 K) = 111.3 KPa (5.5 L) (273 K)

IDEAL GAS LAW Ideal Gases DO NOT EXIST in the real world But we pretend real gases act like ideal gases so we can predict

what will happen Properties

Infintetly small (no volume) No attraction to other particles (No intermolecular force) Don’t condense to liquids

PV=nRT P is the pressure of the gas (in atmospheres, ATM) V is the

volume of the container (in liters, L) n is the number of moles of gas in the container (in moles, mol) R is Universal Gas Constant (which is 0.0820574587 L · ATM · K-1 · mol-1) T is the temperature of the gas (in Kelvin)

Practice: 6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How

many moles of this gas are present? n = PV / RT n = ( 3.0 atm x 6.2 L ) / ( 0.08 L atm /mol K x 310 K)

n = 0.75 mol

DALTON’S GAS LAW At a particular temperature, the total pressure

of a mixture of two or more non-interacting (ideal) gases is equal to the sum of partial pressures of the individual gases

P1+P2+P3… Practice:

The pressure of a mixture of nitrogen, carbon dioxide, and oxygen is 150 kPa. What is the partial pressure of oxygen if the partial pressures of the nitrogen and carbon dioxide are 100 kPA and 24 kPa, respectively?

P = Pnitrogen + Pcarbon dioxide + Poxygen

150 kPa = 100 kPa + 24 kPa + Poxygen

Poxygen = 150 kPa - 100 kPa - 24 kPa Poxygen = 26 kPa

GRAHAM’S GAS LAW

States of Matter

Parts of Solutions

Solution- a homogeneous mixture (same throughout)

Solute- part of solution that gets dissolvedSolvent- part that surrounds the solute to

make it dissolve

Phase Diagram

Shows what temperature and pressure combinations result in each state of matter for a particular chemical

Critical point- the last pressure and temperature combination where liquids and gases exist separately

Heating Curve

Shows the temperature at which the state changes occur and describes how the substances uses heat

Cooling Curve

Opposite phases of the heating curve diagram, work backwards and they decrease

Gas-condensation-liquid-freezing-solid

Solubility Curve

The relationship between temperature and solubility

Photos were not working, see notes(class handout), know how to interpret

Molarity Calculations

M= m/VMolarity= moles/litersUse unit (M)Example: You make a solution

using 47.9 g of HCl. The volume of the solution is 753 mL. What is the Molarity of the solution?

47.9g*1 mole/36.46g=1.31mol753mL*1L/1000mL=.75LM= 1.31mol/.75LM=1.75M

Intermolecular Forces

Dipole-dipole forces: the attraction found between two polar molecules

Vanderwaals forces- attraction between two nonpolar molecules

Hydrogen bonding- a strong force of attraction, not considered a bond. Requires a very electronegative element (Oxygen, Nitrogen, Fluorine) bonded to a Hydrogen

Intermolecular Forces

Cohesive forces- attracted to other molecules of the same type Surface tension Pepper flakes float in a glass of

water until a spoon is inserted into the cup and they sink

Adhesive forces- molecules are attracted to other molecules of a different type Capillary action Pulls the liquid up the side of a

tube creating a meniscus

Specific heat calculationsThe specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius•Delta, Δ or D= change of a variable.•M=Mass of the sample•Q= amount of heat that must be added.•T= temperature of the substance.•C= specific heat

Formulas:c = Q/mΔt ΔQ = mcΔt

Properties of Acids and Bases

Acids• Sour• Burns/Stings• Reacts with metal• Electrolyte (conducts

electricity)• CorrosiveExamples: citric acid, vinegar

Bases• Bitter• Slippery• Non-reactive with metals• Electrolyte (conducts

electricity)• CorrosiveExamples: ammonia, baking

soda

pH/OH Calculations

• pH=-log[H+]• [H+]=10-pH

• pOH=-log[OH-]• [OH-]=10-pOH

• pH+pOH=14