C HEM 167 F INAL R EVIEW Part 1. E LECTROLYTES VS. N ONELECTROLYTES Eletrolyte: Can break apart into ions Examples: salts – NaCl, acids – HF, bases –

CHEM 167 FINAL REVIEWPart 1

ELECTROLYTES VS. NONELECTROLYTES

Eletrolyte: Can break apart into ionsExamples: salts – NaCl, acids – HF, bases – NaOHTypes: Strong – complete dissociation into ions

Weak – not complete dissociation Nonelectrolyte: does not break apart into

ionsExamples: ethanol, sugars (sucrose, fructose)

MASS SPECTROMETER

VIADDVaporization: heat, laser, needs to be in gas phaseIonization: electron shower/magnetic plateAcceleration: ions are sped through spectrometerDeflection: Magnetic field sorts larger/smaller ionsDetection: how many of which ion are present Gives a Mass:Charge ratio Data displayed on graph with peaks Helps to find average mass of an element

FRACTIONAL ABUNDANCE OF AN ISOTOPE

Multiply fraction by the mass of the isotopeExample: 24Mg 23.99 amu – 78.99%

25Mg 24.99 amu – 10.00% 26Mg 25.99 amu – 11.01%

Work: (.7899x23.99) + (.1x24.99) + (.1101x25.99)= 24.31 amu which is value on periodic table Can also be asked to find the fractional

abundance

FIND THE FRACTIONAL ABUNDANCE

Problem: Bromine has two naturally occurring isotopes 79Br 78.918 amu and 81Br 80.916 amu. The average mass for bromine is 79.904 amu. What is the fractional abundance of 79Br?

FINDING FRACTIONAL ABUNDANCE

Work: X + Y = 1 Y = 1-X X= 79Br,Y=81Br78.918(X) + 80.916(1-X) = 79.90478.918X + 80.916 – 80.916X = 79.90478.918X-80.916X = 79.904-80.916-1.998X = -1.012X = 0.51 = 79Br

PERCENT BY MASS/EMPIRICAL FORMULA

Assume a 100g sample %mass mass of element moles mole

ratioProblem: A compound is 54.05% Ca, 43.24% O, and 2.71 % H. What is its empirical formula?

ANSWER

54.05%54.05g Ca/40.1g/mol1.348 mol Ca43.24%43.24g O/16g/mol2.703 mol O2.71%2.71g H/1.01g/mol2.683 mol HCa has the smallest number of moles1.348mol/1.3481Ca2.703mol/1.3482O2.683mol/1.3482HAnswer: CaO2H2Ca(OH)2

LINE DRAWINGS AND THEIR FORMULAS

If no element is written assume is C at every joint and end. Fill in H’s until number of bonds is met

Examples: drawn on board/paper

NAMING COMPOUNDS

Know group charges and specific element charges, such as B, Al, Zn, and Ag

Know polyatomic ions Carbon-carbon bonds: single –ane, double –

ene, triple –yne Transition metals need oxidation state in

parentheses Covalent bonding needs prefixes Examples: a) Fe2O3 iron (III) oxide

b) Cl2O7 dichlorine heptoxide

c) (NH4)2S ammonium sulfide

SOLUBILITY VS. INSOLUBILITY

Review or write down solubility rulesExamples: are the following salts soluble or insoluble?a) KBr - solubleb) AgCl - insoluble c) Al(OH)3 - insoluble

d) BaSO4 - soluble

e) Ca3(PO4)2 - insoluble

BREAKDOWN OF AN ELEMENT

An element contains a specific number of protons, neutrons, and electrons

If the element is neutral then the number of electrons and protons is equal

The number of protons and neutrons is equal to the mass of the element (A)

The number of protons is equal to the atomic number of the element (Z)

TYPES OF BONDING

Ionic bonding: between a metal and nonmetal; is a transfer of electrons

Covalent bonding: between nonmetals; is a sharing of electrons

Metallic bonding: between metals; electrons are free flowing and shared between many nuclei

BALANCING EQUATIONS

Matter cannot be created or destroyed, need the same amount of each element on both sides of the arrow.

Balance these equations:a) C4H10 + O2 CO2 + H2O

b) NaClO3 NaCl + O2

c) CaCl2 + NaOH NaCl + Ca(OH)2

d) C + SO2 CS2 + CO

ANSWERS

a) 2,13,8,10b) 2,2,3c) 1,2,2,1d) 5,2,1,4

MOLES, MOLARITY, AND DILUTION

Moles = mass/molar mass; this is true for elements and compounds

Molarity = moles/liters Dilution = M1V1 = M2V2Example Dilution problem:What volume (mL) of 6M NaOH must be diluted to create a solution of 200mL and 1.5M?

ANSWER

Use M1V1=M2V2Assign M1=6M V1=? M2=1.5M V2=200mLV1=(M2V2)/M1V1=50mL

TOTAL IONIC, MOLECULAR, NET IONIC

Total ionic equation: write out all the ions on both sides of arrow

Molecular equation: combines ions and includes states; aq, and l

Net ionic equation: eliminate spectator ions from the total ionic equation

Write the net ionic equation of:AgNO3 + NaCl *look for formed insoluble salts

ANSWER

AgNO3 + NaCl AgCl + NaNO3 AgCl is insoluble

Total ionic:Ag+ + NO3

- + Na+ + Cl- AgCl + Na+ + NO3-

Spectator ions: Na+ and NO3

-

Net ionic:Ag+

+ Cl- AgCl

GRAPHS TO KNOW

Coulomb’s Law: repulsion on top half (same ions), attraction on lower half (different ions). As radius (r) increases force (F) decreases.

Maxwell-Boltzmann distribution: explains how the speed of molecules in a gas increases with higher temperature, and the speed also increases with smaller mass

EXAM TWO MATERIALTYPES OF LIGHTS Incandescent bulbs: a wire inside is heated up by

electricity, excites the electrons in tungsten (W) which then emits excited electrons to create light.

Fluorescent bulbs: an arc/plasma is sent through bulb, excites the electrons of the gas in the bulb (gas varies), then emitted excited electrons interact with phosphorous coating to create light.

LEDs (light emitting diodes): promote electron from ground to excited state, emits light to fall back to ground state. Has a narrow wavelength and frequency distribution, single color, high efficiency

Laser (light amplification by stimulated emission of radiation): emits light same as LED. Need population inversion. All emitted photons are in same phase (coherent light). Single color with narrow wavelength and frequency distribution

ATOMIC MODELS AND PRINCIPLES

Bohr’s: states that 1) electrons exist in well defined orbits 2) each orbit has specific energy associated with it and 3) energy is released or absorbed when electrons change energy levels

Heisenberg uncertainty principle: proves Bohr’s model wrong because it states that it is impossible to pin down where an electron is at exactly.

Another thing to prove Bohr’s model wrong is that electrons diffract through slits so they can also behave like waves

ORBITALS Orbitals have a set of quantum numbers that

describe them: n, l, ml, and ms . N: principle quantum level, gives number of energy

shell L: secondary quantum number; determines shape of

orbital. Is equal to n-1or less. 0=s, 1=p, 2=d, 3=f Ml: magnetic quantum number; tells how many

orbitals are in a subshell. Is equal to any number from –l to +l

Ms: spin quantum number. Can only be +1/2 or -1/2. Each orbital can only have 2 electrons and they spin in opposite directions.

Pauli exclusion principle: each electron in an atom has a unique set of quantum numbers

Hund’s rule: single electrons first fill empty orbitals before they pair up in an occupied one.

SHAPE OF AN ORBITAL

S orbital: shaped like a sphere, groups 1&2 P orbial: has two lobes, groups 3-8 D orbital: has 4 lobes, transition metals F orbital: has 6 lobes, lanthanides &actinides Orbitals also have nodes (areas where no

electrons exist) the number of nodes is equal to n-1. Nodes can be planar or spherical.

Examples:Which orbital is this? n=2, l=1, ml=0, ms=+1/2

Which quantum numbers describe a 3p orbital?

ANSWER

2p because n=2 and l=1p n= 3, l=1, ml=-1,0,1, ms= +1/2 or -1/2

PHOTOELECTRIC EFFECT

Energy cost to remove one electron from a surface

E = h*v Energy cost to remove one electron using a

certain wavelengthE=(h*c)/l and c=lv Example: What is the wavelength in meters of a wave with a photon energy of 4.5*10^-28J? Know photoelectric graphs, there are 4

ANSWER

Use E=(h*c)/l = (l h*c)/Eh= Planck’s constant = 6.626*10^-34J*sc= speed of light = 3*10^8m/sl= 442m

ELECTRON CONFIGURATIONS

Fill in the orbitals across the periods of the periodic table

There are full electron configurations and condensed configurations that use noble gases

Examples:a) Write the full configuration of Ga.b) Write the condensed state of Ga.

ANSWER

1s22s2 2p63s23p64s23d104p1

[Ar] 4s23d104p1

GROUND STATE AND EXCITED STATE

Electrons can either be in their ground state or their excited state. This can be demonstrated in electron configurations.

Examples:1) Write the stable configuration of nitrogen

(N)2) Which is an excited state?

a) 1s22s22p63s23p64s1

b) 1s22s22p64s1

c) 1s22s22p63s1

ANSWER

1s22s2 2p3

b

PERIODIC PROPERTIES

Zeff: increases as move left to right because Zeff = Z-S. S= shielding (core) and remains the same, Z=atomic number and it increases.

Radius, diameter, and volume: increase as move left and down towards francium (Fr).

Ionization energy: increases as move right and up towards fluorine (F). Is a positive value

Electron affinity: increases in magnitude as move right and up towards fluorine (F). Is a negative value. Noble gases are extremely positive and do not follow trend.

LATTICE ENERGY

The energy cost to break an ionic species into cations and anions

Larger energy is characterized by larger ionic charges and smaller distance apart (smaller r)

Example:a) Which has the smallest radius?

O2-, F-, Ne, Na+, Mg2+

b) Which will have the largest lattice energy?NaCl, MgCl2, MgO, Na2O

In general cations are smaller than anions because of the missing electron(s) which occupy a lot of space by repulsion.

GAS LAWS

P1V1=P2V2, P1/T1=P2/T2, V1/T1=V2/T2 Combines into ideal gas law PV=nRT where

R=0.08206 L*atm/mol*K Conditions where any gas behaves most

ideally are at high temperatures and low pressures

Van der Waals equation helps account for not ideal conditions (P+(n2/V2)a) (V-nb) = nRT

Example: What is the V occupied by 2 moles of N2 at P=1atm and T=273K?

ANSWER

Use ideal gas lawV=(nRT)/PV=(2*0.08206*273)/1=44.8L

DALTON’S LAW AND MOLE FRACTION

Dalton’s Law: total pressure is equal to the sum of the partial pressures.

Mole fraction: equal to partial moles divided by total moles is symbolized by Xi

Partial pressure can be found by applying mole fraction and Dalton’s law. Pi=Xi * PT

CALCULATING THE EXTENT OF A REACTION

Need to make sure the equation is balanced and then use the stoichiometry to decide if there is a limiting reactant.

Example:a) N2 + H2 NH3 *balance it

What mass of N2 is needed to fully react with 5.11g of H2?

b) O2 + 2H2 2H2O

What is the yield of H2O if 7.9g O2 and 2.9g H2 react to completion?

ANSWERS

N2 + 3H2 2NH3

5.11g H2/2.02 2.53 mol H2

H2 and N2 in 1:3 ratio; 2.53mol/30.843 mol N2

0.843mol * 28g/mol = 23.6g N2

7.9g O2/32 0.247mol O2

2.9g H2/2.021.435mol H2

Limiting reactant is O2 because of mole ratio

0.247mol O20.494mol H2O from stoichiometry

0.494mol * 18g/mol = 8.89g H2O

REACTIONS WITH SOLIDS AND GASES

Can use gaseous reactants in reactions to calculate product yields by using the ideal gas law.

Examples:a) O2(g) + 2H2(g) 2H2O(g)

What is the V of H2 needed to make 36g H2O at T=30C and P=0.995 atm?

b) CaO(s) + CO2 (g) CaCO3 (s)

What V of CO2 can be reacted exactly with 112g CaO at T=800C and P=1.25 atm?

ANSWERS

First find moles of H2O created36g/18g/mol 2mol H2O2mol H2 by reaction stoichiometryUse number of moles in PV=nRTV=(nRT)/PV=(2*0.08206*303)/0.995 = 50L First find moles of CaO used112g/56.1g/mol2mol CaO2mol CO2Use number of moles in V=(nRT)/PV=(2*0.08206*1073)/1.25 = 141L

PRESSURE MEASURING DEVICES

Capacitance manometer: uses a flexible diaphragm and measures the distance change (capacitance) between the diaphragm and a metal plate.

Ionization gauge: measures the current created by ions, an increase in P makes the current increase

Thermocouple gauge: as pressure decreases the measured temperature of the filament increases

Mass spectrometer: measures partial pressures of the sample