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Lab equipment

Lab equipment practice

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Laboratory spatula: Scoops stuff, moves it, applies it

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Erlenmeyer flask: Beakers with thin necks

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Test tube rack: Holds test tubes

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Graduated cylinder: Holds, measures liquids

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Watch glass: Holds liquid to evaporate it, holds solids for weighing, or covers beaker

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Crucible: Holds hot objects

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Laboratory iron rings: Holds items above work surface

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Evaporation dish: Evaporates

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Funnel:

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Test tube: Holds materials

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Ring stand (with attached titration device): Holds iron rings

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Spring clamps: Holds test tubes

Click!

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Wire gauze: Holds hot materials, diffuses heat

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Beakers: Holds liquids for stirring, heating, etc.

Click!

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Pipeclay triangle: Supports a crucible over a Bunsen burner

Click!

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Test tube brush: Cleans test tubes

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Florence flask: Beaker that ensures uniform heating, stirring, with thin neck

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Pasteur pipet (dropper): Moves liquid; large mouth; ungraduated

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Graduated pipet (dropper): Moves liquid, in measured amounts

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Bunsen burner: Emits sootless open gas flame

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Disposable weighing boat: Used for weighing

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Wash bottle: Used for dispensing fluid

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Wash bottle: Used for dispensing fluid

Click!

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Crucible tongs: Picks up hot stuff

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Forceps: Picks up small stuff

Click!

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Beaker tongs: Picks up beakers

Scientific method

General terms

Observations
1. Quantitative
2. Qualitative
Hypotheses: theoretical explanation for observations
Experiments: obtaining new observations, testing hypotheses
Theory/Law meaning set 1
1. Theory: set of tested hypotheses
2. Law: theory, applied to broad range
Theory/Law meaning set 2
1. Law: what happens
2. Theory: why happens

Measurement

Precision vs accuracy: [1]
1. Precision: close to other measurement
2. Accuracy: close to "true" measurement
Types of error:
1. Random/Indeterminate error: equal probability of decreasing or increasing result
2. Systematic/Determinate error: consistently skewed probability; decreases accuracy, not precision
Measurement: number & unit [2]
1. Mass: Gram (g)
2. Length: Meter (m)
3. Time: Second (s)
4. Temperature: Kelvin (K) [3]
5. Electric current: Ampere (A)
6. Amount of substance: Mole (mol)
7. Luminous intensity: Candela (cd)
Prefixes:
1. Mega: 10⁶
2. Kilo: 10³
3. none: 10¹
4. Deci: 10^-1
5. Centi: 10^-2
6. Milli: 10^-3
7. Micro: 10^-6
8. Nano: 10^-9
9. Pico: 10^-12

Significant figures

Simplify to scientific notation
Zeroes: [4]
1. If placeholder, insignificant
2. If in-number, significant
3. If after number and after decimal point, significant
Multiplication and division: [5]
1. Least number of significant digits
Addition and subtraction:
1. Round to least significant digit after

History

400 BCE Demokritos, Leucippus use term "atomos"
Alchemy
1500s:
1. Georg Bauer: Systematic metallurgy
2. Paracelsus: Medicinal application of minerals
1600s:
1. Robert Boyle: The Skeptical Chemist: Quantitative experimentation, identification of elements
1700s:
1. Georg Stahl: Phlogiston theory
2. Joseph Priestly: Discovery of oxygen
3. Antoine Lavosier: Role of oxygen in combustion, law of conservation of mass, modern chemistry textbook
1800s:
1. Joseph Proust: Law of definite composition
2. John Dalton: Atomic Theory, law of multiple proportions
3. Joseph Gay-Lussac: Combining volumes of gases, existence of diatomic molecules
4. Amadeo Avogadro: Molar volumes of gases
5. Jons Jakob Berzelius: Relative atomic masses, modern symbols for the elements
6. Dmitri Mendeleyev: Periodic table
7. J.J. Thomson: Discovery of the electron
8. Henri Becquerel: Discovery of radioactivity
1900s:
1. Robert Millikan: Charge and mass of the electron
2. Ernest Rutherford: Existence of the nucleus, its relative size
3. Meitner & Fermi: Sustained nuclear fission
4. Ernest Lawrence: The cyclotron and trans-uranium elements

Dalton

Atomic Theory (1808)
1. All matter is atoms (extremely small)
2. Atoms in element are identical in size, mass, and other properties; atoms of other elements differ in all
3. Atoms cannot be subdivided, created, destroyed
4. Atoms of diff elements combine in whole-number ratios to form chemical compounds
5. In chemical reactions, atoms are rearranged
Changes:
1. none
2. Atoms in element have characteristic average mass, unique to that element.
3. Atoms cannot be subdivided, created, or destroyed in ordinary chemical reactions; nuclear reactions can cause these reactions
4. none
5. none

Thomson

Cathode ray tube (1897) [6]
From classical mechanics:
1. ΔX₍₊₎ ∝ e₍₊₎/m₍₊₎
2. ΔX_(-) ∝ e_(-)/m_(-)
From the fact that the cathode was H₂:
1. |e₍₊₎| = |e_(-)| (since H is neutral; theoretically, mult electrons could have instead existed, making this incorrect)
From the proportions and the balanced equation:
1. ΔX₍₊₎/ΔX_(-) = (|e_(-)|/m_(-))/(|e₍₊₎|/m₍₊₎) = m_(-)/m₍₊₎
Proposed "plum pudding model"
1. Positive pudding
2. Negative plums

Millikan

Oil drop [7]
1. Suspended oil with electricity
2. Electron mass: 9.109 x 10^-31 kg
3. Electron charge: 1.5924(17)×10−19 C

Rutherford

Gold-foil experiment [8]
1. RaBr (radium bromide) source of α particles
2. Experiment 1: (control)
  1. RaBr
  2. Detection screen opposite RaBr
  3. Results:
    1. 132,000 α particles/min
    2. Expected
3. Experiment 2: (first gold-sheet experiment)
  1. RaBr
  2. Gold sheet (thinner than human hair)
  3. Detection screen opposite RaBr
  4. Results:
    1. 132,000 α particles/min
    2. Appears to confirm plum-pudding model
4. Experiment 3: (backscatter)
  1. RaBr
  2. Gold sheet (thinner than human hair)
  3. Detection screen opposite RaBr
  4. Detection screen on same side as RaBr
  5. Results:
    1. 20 α particles/min on same-side detector
    2. Couldn't be explained as background noise
    3. Proves non-zero amount of backscatter (P = count rate backscattered particles / count rate of incident particles = 20/132000 = 2e-4)
    4. "As if you'd fired a 15" shell at tissue paper, and it came back and hit you!"
Conclusions:
1. Plum-pudding model false
2. Au atoms must be mostly empty
3. Very small, heavy, positively charged area
4. Diameter of nucleus: 1*e-14 meters
5. From the balanced nature of particle: Charge of nucelus: +Ze (atomic number * absolute value of electron charge)

General principles, 1900s

Cathode rays have identical properties regardless of element used
Atoms are neutral
1. Positive particles must balance electrons
2. Electrons are small; a 3rd particle must exist

Matter

Properties

Hierarchy

Matter
1. Mixtures
  1. Homogenuous (Solutions)
  2. Heterogenuous (Solutions)
2. Pure substances
  1. Compounds
  2. Elements
    1. Atoms
      1. Electrons (is lepton) [9]
      2. Protons (3 quarks)
      3. Neutrons (3 quarks)

Phase/state differences

Solid
1. Definite volume
2. Definite shape
3. Fixed particle positions
4. Fixed particle distances (small)
5. Molecular bonds present
Liquid
1. Definite volume
2. Indefinite shape
3. Unfixed particle positions
4. Unfixed particle distances (small)
5. Molecular bonds present
Gas
1. Indefinite volume
2. Indefinite shape
3. Unfixed particle positions
4. Unfixed particle distances (large)
5. Molecular bonds present
Plasma
1. Indefinite volume
2. Indefinite shape
3. Unfixed particle positions
4. Unfixed particle distances
5. Molecular bonds dissasociated

Extensive vs intensive

Extensive (dependent on amount of matter)
1. Volume
2. Mass
3. Energy content
Intensive (independent of amount of matter)
1. Melting point
2. Freezing point
3. Boiling point
4. Condensation point
5. Density

Separation

Mixture
1. Chromatography
2. Filtration
3. Evaporation
4. Fractional distillation (different boiling points)
Compound (must be chemically separated)
1. Electrolysis

Atomic structure

Abnormal:
1. Isotope: Abnormal neutron count
2. Ion: Unbalanced electron/proton count
Atomic mass: average natural isotope mass

Nuclear symbol

Mass number (A): number of protons and neutrons (p⁺ + n⁰)	235	U	Element symbol
Atomic number (Z): number of protons (p⁺)	92

Particulate symbols, charges, and mass

particle	symbol	charge (C)	mass (kg)
electron	e^-	-1.6e-19	9.11e-31
proton	p⁺	+1.6e-19	1.673e-27
neutron	n⁰	0	1.675e-27

Particulate composition

Protons (1e)
1. Two up quarks (+2/3e)
2. One down quark (-1/3e)
Neutrons (0)
1. One up quarks (+2/3e)
2. Two down quarks (-1/3e)
Gluons hold together particles
Electron (-1e)

Molecules

Molecule vs compound

Molecule:
1. 2+ atoms joined chemically
Compound:
1. Molecule
2. 2+ different elements joined chemically

Bonding types

Two atoms close enough for atomic orbital to mix. Electronegativity values are:
1. Very different: Ionic bonding.
2. Similar: The atoms are classified as:
  1. Metals: Metallic bonding.
  2. Nonmetals: Covalent bonding. Electronegativity values are:
    1. Very close: Nonpolar covalent bonding. (Electron is equally shared; no polar region of molecule.)
    2. Different: Polar covalent bonding. (Electron is unequally shared; polar region of molecule.)
Discrete/indiscreet:
1. Discrete: Represented by formula
2. Indiscrete: Covalent Network Structures, no formula

Nuclear reactions

Mass defect (ΔE = Δmc²)
1. ΔE: change in energy
2. m: mass defect
3. c²: speed of light squared (really big; 8.98755179 × 10¹⁶ m2 / s2)

Fission

Splitting heavy nucleus into two lighter nuclei (plus neutrons)
Uranium fission: ₀¹n + ₉₂²³⁵U = ₅₆¹⁴²Ba + ₃₆⁹¹U + 3₀¹n (with variation on the fission fragments' mass numbers) [10][11][12]
Chain reaction: (self-sustaining fission)

Event	Neutrons causing fission	Result
subcritical	<1	reaction stops
critical	=1	reaction continues
supercritical	>1	reaction expands

Ensuring self-sustaining reaction:
1. Increasing size increases likelihood neutrons will collide with additional matter
2. Encasing sphere in neutron reflector (eg tungsten carbide) increases likelihood neutrons will collide with additional matter
Fission bomb design:
1. Little Boy: conventional explosive "gun" mechanism shoved two subcritical uranium 235 pieces together
2. Fat Man: conventional explosive compressed plutonium core
Fission reactor [13]

Fusion

Combining two lighter nuclei into a heavier nucleus
Deuterium-tritium: ₂³He + ₁¹H = ₂⁴He + ₁⁰n [14]
Star fusion:
1. 1.3xSun mass or less: almost all energy from proton-proton fusion [15][16]
2. 1.3xSun mass or more: almost all energy from CNO cycle [17][18]

Balancing

Σin = Σout
1. ΣA_reactants = ΣA_products
2. ΣZ_reactants = ΣZ_products

A:	1 +	235 =	142 +	91 +	3(1)
Equation:	₀¹n +	₉₂²³⁵U =	₅₆¹⁴²Ba +	₃₆⁹¹U +	3₀¹n
Z:	0 +	92 =	56 +	36 +	0

Nuclear balancing practice

₈₈²²⁶Ra = ₂⁴α + X

Answer
₈₆²²²Rn

₀¹n + ₉₂²³⁵U = ₅₃¹³⁹I + 2₀¹n + X

Answer
₃₉⁹⁵Y

Decay

Alpha decay
1. Alpha particle emitted
2. Alpha particle is Helium nucleus (₂⁴α = ₂⁴He)
3. Limited to very large nuclei
4. Example: ₉₂²³⁸U = ₂⁴α + ₉₀²³⁴Th
Beta decay
1. Beta particle emitted & neutron converted to proton
2. Beta particle is electron (_-1⁰β = _-1⁰e)
3. Example: ₉₀²³⁴U = ₉₁²³⁴Pa + _-1⁰e
Gamma emission
1. Gamma particle emitted during other decay; nucleus unchanged
2. Gamma particle is high-energy light
3. Example: ₉₂²³⁸U = ₂⁴α + ₉₀²³⁴Th + 2₀⁰γ
Positron emission
1. Positron emitted & proton converted to neutron
2. Positron is anti-electron (₁⁰e)
3. Example: ₁₁²²Na = ₁⁰e + ₁₀²²Ne
Electron capture
1. Inner-orbital electron captured by nucleus
2. Electron converts proton to neutron
3. Example: ₈₀²⁰¹Hg + _-1⁰e = ₇₉²⁰¹Au + 2₀⁰γ
Decay reasons: return to stability (proper A and Z)
Decay series [19]
1. Shit gets bad
2. Shit gets better
ln(N/N₀)=-kt
1. N₀ = initial nucleides
2. N = remaining nucleides
3. k = rate constant
4. t = elapsed time
t_0.5 = ln(2)/2 = 0.693/k
1. k = rate constant
2. t = elapsed time

Ions

Terms

Cation: +
Anion: -
Ionic bonding: attraction between + and -
Predicting charges:
1. Alkali metals: Lose one electron, become 1+
2. Alkali earth metals: Lose two electrons, become 2+
3. Transition metals: no rules!!!!! (except balance electrons, yadda yadda)
4. Boron group: Loses 3 electrons, become 3+
5. Carbon group: Loses or gains 4 electrons
6. Pnictogens: Gains 3 electrons, become 3-
7. Chalcogens: Gain 2 electrons, become 2-
8. Halogens: Gain 1 electron, become 1-
9. Noble gases: No ions

Ionic equation balancing practice

Barium nitrate: Ba²⁺ + NO₃^-

Answer

2+ + 1- (unbalanced)

Ba²⁺(NO₃^-)₂

2+ + 2- (balanced)

Ammonium sulfate: NH₄⁺ | SO₄^2-

Answer

1+ | 2- (unbalanced)

(NH₄⁺)₂SO₄^2-

2+ | 2- (balanced)

Iron(III) chloride: Fe³⁺ | Cl^-

Answer

3+ | 1- (unbalanced)

Fe³⁺Cl₃^-

3+ | 3- (balanced)

Aluminum sulfide: Al³⁺ | S^2-

Answer

3+ | 2- (unbalanced)

Al³⁺₂S₃^2-

6+ | 6- (balanced)

Magnesium carbonate: Mg²⁺ | CO₃^2-

Answer
2+ \| 2- (balanced)

Zinc hydroxide: Zn²⁺ | OH^-

Answer

2+ | 1- (unbalanced)

Zn²⁺ | (OH^-)₂

2+ | 2- (balanced)

Magnesium carbonate: Mg³⁺ | CO₄^3-

Answer
3+ \| 3- (balanced)

Naming ionic compounds

Cation
1. Monoatomic: name of element (eg, calcium)
2. Transition metals with multiple oxidation states: use Roman numerals in name (eg, II)
Anion:
1. Monoatomic: root + ide (eg, chlor-ide)
Binary molecular compounds: (between two nonmetals)
1. First element named first with element name; prefix if subscript
2. Second element named second with root-ide; always prefixed
Prefixes:
1. 1 = mon(o)
2. 2 = di
3. 3 = tri
4. 4 = tetra
5. 5 = penta
6. 6 = hexa
7. 7 = hepta
8. 8 = octa
9. 9 = nona
10. 10 = deka

Naming ionic compounds practice

Formula to words

H₂O:

Answer
Hydrogen monoxide

P₂O₅:

Answer
Diphosphorus pentoxide

CO₂:

Answer
Carbon dioxide

P₂O₅:

Answer
Carbon monoxide

P₂O₅:

Answer
Dinitrogen monoxide

N₂O₄:

Answer
Nitrogen tetroxide

SO₃:

Answer
Sulfur trioxide

NO:

Answer
Nitrogen monoxide

NO₂:

Answer
Nitrogen dioxide

As₂O₅:

Answer
Diarsenic pentoxide

PCl₃:

Answer
Phosphorous trichloride

CCl₄:

Answer
Carbon tetrachloride

SeF₆:

Answer
Selenium hexafluoride

Words to formula

Diphosphorus pentoxide:

Answer
P₂O₅

Carbon dioxide:

Answer
CO₂

Carbon monoxide:

Answer
P₂O₅

Dinitrogen monoxide:

Answer
P₂O₅

Silicon dioxide:

Answer
SiO₂

Carbon tetrabromide:

Answer
CBr₄

Sulfur dioxide:

Answer
SO₂

Phosphorus pentabromide:

Answer
PBr₅

Iodine trichloride:

Answer
ICl₃

Nitrogen triiodide:

Answer
NI₃

Dinitrogen trioxide:

Answer
N₂O₃

Solutions

Matter; Can be separated by physical means:
1. Yes: Pure substances; Can it be decomposed by ordinary chemical means
  1. Yes: Compounds (water, NaCl, sucrose)
  2. No: Elements (H, He, Li)
2. No: Mixtures; Is the composition uniform?
  1. Yes: Homogenous mixtures (air, sugar in water, stainless steel)
  2. No: Heterogenous mixtures (granite, wood, blood)
Solute: dissolved substance in a solution
Solvent: dissolving medium in a solution
Best dissolution:
1. Nonpolar solutes dissolve best in nonpolar solvents (fats, steroids, waxes, benzene, hexane, toulene)
2. Polar and ionic solutes dissolve best in polar solvents (inorganic salts, sugars, water, small alcohols, acetic acid)
Factors affecting solubility:
1. Solids:
  1. Increases with temperature (generally)
  2. Increases with increasing surface area of the solid (generally)
  3. Increases with stirring
2. Gases:
  1. Decreases with increases in temperature (generally)
  2. Increases with pressure above the solution (generally)
Saturation:
1. A solution that contains the maximum amount of solute that may be dissolved under existing conditions is saturated.
2. A solution that contains less solute than a saturated solution under existing conditions is unsaturated.
3. A solution that contains more dissolved solute than a saturated solution under the same conditions is supersaturated.
Electrolytes:
1. Electrolyte: A substance whose aqueous solution conducts an electric current.
2. Nonelectrolyte: A substance whose aqueous solution does not conduct an electric current.
3. Can be tested by running current through solution and ammeter; no current means nonelectrolyte
Ionization:
1. Ionic compounds disassociate (eg NaCl(s) = Na+(aq) + Cl-(aq))
  1. Ions prefer positioning to conduct a current rather than forming a solid.
  2. Polar water molecule pulls positive ions to the negative end (oxygen) and negative ions to the positive end (hydrogen).
2. Some covalent compounds ionized: hydrogen chloride molecules, which are polar, give up their hydrogens to water, forming chloride ions (Cl-) and hydronium ions (H3O+).
3. Acids:
  1. Strong acids are totally ionized
  2. Weak acids (often containing carboxyl group) rarely (5%) ionized

Solution concentration

Mole fraction – the ratio of moles of solute to total moles of solution
1. Mole fraction of A = x_A = n_A / (n_A + n_B)
Molarity is the ratio of moles of solute to liters of solution
1. Molarity = M = moles of solute / Liters of solution

Solution concentration practice

Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?

Answer
1.5 L * (0.5 mol / 1 L) * (58.44 g / 1 mol) = 43.8 g NaCl

Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution?

Answer
M_stockV_stock = M_diluteV_dilute (11.6 mol / L)(x L) = (3.0 mol / L)(0.250 L) x = (3.0 mol) * (0.250 L) / (11.6 L) x = 0.065 L

How many grams of NaOH are contained in 270.0 mL of a 0.450 M sodium hydroxide solution?

How much of each starting material would you use to prepare 2.00 L of each of the following solutions? (a) 0.250 M NaOH from solid NaOH (b) 0.160 M NaOH from 1.00 M NaOH stock solution (c) 0.150 M K2CrO4 from solid K2CrO4 (d) 0.210 M K2CrO4 from 1.75 M K2CrO4 stock solution

Moles

General

1 mole = 6.022e23 (6.022 * 10²³) atoms = Avogadro's number
Exactly 12 grams of carbon-12 in 1 mole of carbon-12
Formulas:
1. Empirical formula: lowest whole-number ratio of atoms in a compound (eg, CH)
2. Molecular formula: actual ratio of atoms in compound; is always multiple of empirical formula (eg, C₆H₆ = (CH)₆)
3. Always empirical for ionic compounds, such as NaCl, MgCl₂, Al₂(SO₄)₃, K₂CO₃
4. Possibly empirical for molecular compounds, such as H₂O, C₆H₁₂O₆, C₁₂H₂₂O₁₁

Moles practice

How many grams of lithium are in 3.50 moles of lithium?

Answer
(3.50 moles Li / 1) * (6.94 g Li / 1 mol Li) = 45.1 g Li

How many moles of lithium are in 18.2 grams of lithium?

Answer
(18.2 grams Li / 1) * (1 mol Li / 6.94 g Li) = 2.62 mol Li

How many atoms of lithium are in 3.50 moles of lithium?

Answer
(3.50 moles Li / 1) * (6.022e23 atoms Li / 1 mol Li) = 2.11e24 atoms Li

How many atoms of lithium are in 18.2 grams of lithium?

Answer
(18.2 grams Li / 1) * (1 mol Li / 6.94 g Li) * (6.022e23 atoms Li / 1 mol Li) = 1.58e24 atoms Li

Calculate each of the following quantities. (a) Mass (g) of 0.59 mol of MnSO4 (b) Amount (mol) of compound in 13.3 kg of Fe(ClO4)3 (c) Number of N atoms in 78.2 mg of NH4NO2

Calculate each of the following quantities. (a) Mass (g) of 0.66 mol of KMnO4 (b) Amount (mol) of O atoms in 8.85 g of Ba(NO3)2 (c) Number of O atoms in 7.8 10-3 g of CaSO4 · 2 H2O

Formula Mass

Add up each element's atomic weight, multiplied by the number of times the element appears.

eg: Calculate the formula mass of sodium chloride.

sodium chloride = NaCl
Sodium atomic weight: 22.99 g Na * 1
Chloride atomic weight: 35.45 g Cl * 1
Sodium chloride formula mass: 22.99 g Na + 35.45 g Cl = 58.44 g NaCl

Divide each element mass in the formula by total mass and multiply by 100% to find percent composition. Add all percentages to check correctness; should equal approximately 100%.

eg: Calculate the percentage composition of sodium chloride.

Sodium percentage composition: (22.99 g Na / 58.44 g NaCl) * 100% = 39.34%
Chloride percentage composition: (35.45 g Cl / 58.44 g NaCl) * 100% = 60.66%
Total: 39.34% + 60.66% = 100% (+/- a tiny bit is fine)

To determine empirical formula:

Assume 100g of compound.
Determine moles of each element per 100g.
Divide each molesofelement by the smallest molesofelement.
If fractional, multiply by reciprocal of smallest fraction until all are whole numbers.

Empirical formula practice

Calculate the formula mass and percentage composition of magnesium carbonate.

Answer
magnesium carbonate = MgCO₃ Formula mass: 24.31 g Mg + 12.01 g C + 3(16.00 g O) = 84.32 g MgCO₃ Percentage composition: Magnesium: (24.31 g/84.32 g) * 100% = 28.83% Carbon: (12.01 g/84.32 g) * 100% = 14.24% Oxygen: (48 g/84.32 g) * 100% = 56.93% Total: 28.83% + 14.24% + 56.93% = 100%

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

Answer

Pretend 100g.
Find mass:
1. Carbon: 49.32% C * 100g = 49.32g C
2. Oxygen: 43.84% C * 100g = 43.84g O
3. Hydrogen: 6.85% C * 100g = 6.85g H
Find moles:
1. Carbon: 49.32g C/(12.01g C/mol C) = 4.107 mol C
2. Oxygen: 43.84g O/(16.00g O/mol O) = 2.740 mol O
3. Hydrogen: 6.85g H/(1.01g H/mol H) = 6.782 mol H
Divide moles:
1. Carbon: 4.107 mol C/2.740 mol O = 1.499 C/O
2. Oxygen: 2.740 mol O/2.740 mol O = 1 O/O
3. Hydrogen: 6.782 mol H/2.740 mol O = 2.475 H/O
Multiply:
1. Carbon: 1.499 C/O * 2 O ≈ 3
2. Oxygen: 1 O/O * 2 O ≈ 2
3. Hydrogen: 2.475 H/O * 2 O ≈ 5
Consolidate: C₃H₅O₂

The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

Answer
Add masses: Carbon: (12.01g C/mol C) * 3 mol C = 36.03 g C Oxygen: (16.00g O/mol O) * 2 mol O = 32.00 g O Hydrogen: (1.01g H/mol H) * 5 mol H = 5.05 g H Total: 73.08 g Divide given mass by total: 146 g/73.08 g = 2 Multiply empirical formula by number: (C₃H₅O₂) * 2 = (C₃H₅O₂)₂ = C₆H₁₀O₄

A chloride of silicon contains 79.1 mass % Cl. If the molar mass is 269 g/mol, what is the molecular formula?

Answer
Pretend 100g. Find mass: 79.1% Cl * 100g = 79.1 g Cl 79.1 g / (35.453 g/mol) = 2.23 mol Cl 100 g (Cl + Si) - 79.1 g Cl = 20.9 g Si 20.9 g / (28.09 g/mol) = 0.744 mol Si Find ratios: 2.23 Cl / 0.744 = ca. 3 Cl 0.744 Cl / 0.744 = ca. 1 Si Empirical formula: SiCl₃ Molar mass: 134.449 g/mol 269 g/mol / 134.449 g/mol = 2 Molecular formula: Si₂Cl₆

A compound contains only carbon, hydrogen, and oxygen. Combustion of 8.544 mg of the compound yields 12.81 mg CO2 and 3.50 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?

Answer

8.544 mg C_xH_yO_z + ∞ mg O₂ = 12.81 mg CO₂ + 3.50 mg H₂O
Molar masses:
1. C_xH_yO_z: 176.1 g/mol
2. CO₂: 44.01 g/mol
3. H₂O: 18.016 g/mol
Product milligrams:
1. 12.81 mg CO₂ = 12.01 (element molar mass) / 44.01 (compound molar mass) * 12.81 = 3.496 mg C
2. 3.50 mg H₂O = 2.016 (element molar mass) / 18.016 (compound molar mass) * 3.50 = 0.3917 mg H
3. O: 8.544 - (3.496 + 0.3917) = 4.656 mg O
Moles:
1. C: 3.496 / 12.011 = 0.2911
2. H: 0.3917 / 1.008 = 0.3886
3. O: 4.656 / 16.00 = 0.2911
Ratio:
1. C: 0.2911 / 0.2911 = 1 ; *3 = 3
2. H: 0.3886 / 0.2911 = 1.33 ; * 3 = 4
3. O: 0.2911 / 0.2911 = 1 ; * 3 = 3
Empirical formula: C₃H₄O₃
Empirical formula mass: 88.06 g
171.6 / 88.06 = 2
Chemical formula: C₆H₈O₆

A chloride of silicon contains 79.1 mass % Cl. (a) What is the empirical formula of the chloride? (b) If the molar mass is 269 g/mol, what is the molecular formula?

A 0.370-mol sample of a metal oxide (M2O3) weighs 55.4 g. (a) How many moles of O are in the sample? (b) How many grams of M are in the sample? (c) What element is represented by the symbol M?

Menthol (script M = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen, and oxygen. When 0.1595 g menthol was burned in a combustion apparatus, 0.449 g of CO2 and 0.184 g of H2O formed. What is menthol's molecular formula?

Mass Spectrometry

https://en.wikipedia.org/wiki/Mass_spectrometry

Equipment: [20]
1. Vacuum pump: 10^-5 to 10^-8
Process: [21]
1. Sample added
2. Sample vaporized
3. Sample ionized
4. Sample accelerated into magnetic field
5. Sample path pulled up/down by electromagnet
Effect:
1. Produces spectra of masses from the molecules in a sample of material, and fragments of the molecules.
Purpose:
1. Determines elemental composition of a sample.
2. Determines the masses of particles and of molecules.
3. Determines potential chemical structures of molecules by analyzing the fragments.
4. Determines the identity of unknown compounds by determining mass and matching to known spectra.
5. Determines the isotopic composition of elements in a molecule.

Processes table

Ionization method	Typical Analytes	Sample Introduction	Mass Range
Electron Impact (EI)	Relatively small, volatile	gas chromatography or liquid/solid probe	to 1,000 Daltons
Chemical Ionization (CI)	relatively small, volatile	gas chromatography or liquid/solid probe	to 1,000 Daltons
Electrospray (ESI)	peptides, proteins, nonvolatile	liquid chromatography or syringe	to 200,000 Daltons
Fast Atom Bombardment (FAB)	carbohydrates, organometallics, peptides, nonvolatile	Sample mixed in viscous matrix	to 6,000 Daltons
Matrix Assisted Laser Desorption (MALDI)	peptides, proteins, nucleotides	sample mixed in solid matrix	to 500,000 Daltons

Charts

Y axis: Amount of each ion detected as compared to the most-detected ion.
X axis: m/z (mass (unified atomic mass unit) to charge (of the ion) ratio)

Carbon dioxide (CO₂) has a single molecular ion peak at 44 m/z ([CO₂]⁺) and three fragment peaks at 12 m/z ([C]⁺), 16 m/z ([O]⁺), and 28 m/z ([CO]⁺).

Bromine has two isotopes 50.69% ⁷⁹Br and 49.31% ⁸¹Br. This causes it to form three molecular ion peaks at 158 m/z ([⁷⁹Br⁷⁹Br]⁺) 160 m/z ([⁷⁹Br⁸¹Br]⁺) and 162 m/z ([⁸¹Br⁸¹Br]⁺) and two fragment peaks at 79 m/z ([⁷⁹Br]⁺) 81 m/z ([⁸¹Br]⁺).

Mass spectrometry charts practice

See: http://www.sciencegeek.net/APchemistry/APtaters/MassSpec.htm

Methylene Chloride (CH₃Br). Br is 50.69% ⁷⁹Br and 49.31% ⁸¹Br.

Answer

Methylene Chloride (CH₂Cl₂). Cl is 75.77% ³⁵Cl and 24.23% ³⁷Cl.

Answer

Vinyl Chloride (CH₂CHCl). Cl is 75.77% ³⁵Cl and 24.23% ³⁷Cl.

Answer

Stoichiometry

When the equation is balanced it has quantitative significance:
1. C₂H₅OH + 3O₂ = 2CO₂ + 3H₂O
2. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water
Calculating masses of reactants and products
1. Balance the equation.
2. Convert mass or volume to moles, if necessary.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of desired substituent.
5. Convert moles to mass or volume, if necessary.

Stoichiometry practice

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

Answer
Aluminum oxide: Al_x⁺³O_y^-2 Balance: Al₂O₃ Equation: XAl + YO₂ = ZAl₂O₃ Balance: 4Al + 3O₂ = 2Al₂O₃ Al₂O₃ molar mass = 26.98 * 2 + 16.00 * 3 = 101.96 g/mol Convert: 6.50 g Al * (1 mol Al / 26.98 g Al) * (2 mol Al₂O₃ / 4 mol Al) * (101.96 g / 1 mol Al₂O₃) 6.50 / 26.98 x 2 / 4 x 101.96 = 12.3 g

Bornite (Cu3FeS3) is a copper ore used in the production of copper. When heated, the following reaction occurs.

2 Cu₃FeS₃(s) + 7 O₂(g) = 6 Cu(s) + 2 FeO(s) + 6 SO₂(g)

If 3.98 metric tons of bornite is reacted with excess O2 and the process has an 76.9% yield of copper, how much copper is produced?

Answer
3.98 metric tons = 3.98e6g 3.98e6g (2 Cu₃FeS₃(s)) + ∞(7 O₂(g)) = x(6 Cu(s)) Molar mass Cu₃FeS₃: 342.681 g/mol (3.98e6) * (1/342.681) * (6/2) * (63.546) * (1/1e6) * (0.769) = 1.70 metric tons Cu

Chromium(III) oxide reacts with hydrogen sulfide (H2S) gas to form chromium(III) sulfide and water. Cr2O3(s) + 3 H2S(g) → Cr2S3(s) + 3 H2O(l) (a) To produce 484 g of Cr2S3, how many moles of Cr2O3 are required? (b) How many grams of Cr2O3 are required?

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example, SrH2(s) + 2 H2O(l) → Sr(OH)2(s) + 2 H2(g). You wish to calculate the mass (g) of hydrogen gas that can be prepared from 5.99 g of SrH2 and 4.37 g of H2O. (a) What amount (mol) of H2 can be produced from the given mass of SrH2? (b) What amount (mol) of H2 can be produced from the given mass of H2O? (c) Which is the limiting reactant? (d) How many grams of H2 can be produced?

Diborane (B2H6), a useful reactant in organic synthesis, may be prepared by the following reaction, which occurs in a nonaqueous solvent. NaBH4(s) + BF3(g) B2H6(g) + NaBF4(s) [unbalanced]

If the reaction has a 75.0% yield of diborane, how many grams of NaBH4 are needed to make 25.0 g B2H6?

Bacterial digestion is an economical method of sewage treatment. 5CO2(g) + 55NH4+(aq) + 76O2(g) reaction arrow identifying the presence of bacteria C5H7O2N(s) + 54NO2-(aq) + 52H2O(l) + 109H+(aq) bacterial tissue

The above reaction is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. How much bacterial tissue is produced in a treatment plant for every 1.0 ✕ 104 kg of wastewater containing 3.0% NH4+ ions by mass?

Limiting reactant

Limiting reactant practice

What is the limiting reactant in the synthesis of ammonia (NH₃) if you have 5.0 moles of N₂ and 10.0 moles of H₂(g) in the reaction vessel?

N₂(g) + 3H₂(g) = 2NH₃(g)

Answer
N₂ as limiting: (N₂)_5.0 + (3H₂(g))_∞ = (2NH₃(g))_5/2 H₂ as limiting: (N₂)_∞ + (3H₂(g))_10/3 = (2NH₃(g))_10/6 H₂ is limiting.

How many milliliters of 0.383 M HCl are needed to react with 7.0 g of CaCO3? 2 HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)

How many grams of NaH2PO4 are needed to react with 38.23 mL of 0.295 M NaOH? NaH2PO4(s) + 2 NaOH(aq) → Na3PO4(aq) + 2 H2O(l)

A sample of impure magnesium was analyzed by allowing it to react with excess HCl solution: Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) After 1.62 g of the impure metal was treated with 0.100 L of 0.774 M HCl, 0.0125 mol HCl remained. Assuming the impurities do not react, what is the mass % of Mg in the sample?

Precipitation Reactions

Double Replacement:
1. AX + BY = AY + BX
2. The ions of two compounds exchange places in an aqueous solution to form two new compounds.
3. One of the compounds formed is usually a precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

Lead(II) nitrate + potassium iodide = lead(II) iodide + potassium nitrate
Double replacement equation: Pb(NO₃)₂(aq) + 2KI(aq) = PbI₂(s) + 2KNO₃(aq)
Compounds as aqueous ions: Pb2+(aq) + 2 NO₃-(aq) + 2 K+(aq) +2 I-(aq) = PbI₂(s) + 2K+(aq) + 2 NO₃-(aq)
Eliminate spectator ions:Pb2+(aq) + 2 K+(aq) +2 I-(aq) = PbI₂(s) + 2K+(aq)

Solubility rules

ON AP EXAM

All sodium, potassium, ammonium, and nitrate salts are soluble in water.

NOT ON AP EXAM

Ion	Solubility	Exceptions
CO32-	Insoluble	Group IA and NH4+
PO₄3-	Insoluble	Group IA and NH₄+
OH-	Insoluble	Group IA and Ca2+, Ba2+, Sr2+
S₂-	Insoluble	Groups IA, IIA, and NH₄+
NO₃-	Soluble	None
ClO₄-	Soluble	None
Na+	Soluble	None
K+	Soluble	None
NH₄+	Soluble	None
Cl-, I-	Soluble	Pb2+, Ag+, Hg₂2+
SO₄2-	Soluble	Ca2+, Ba2+, Sr2+, Pb2+, Ag+, Hg2+

Salt chart

Gas

[16:40:28] Yixuan Liu: 1.7 atm Xe and 8.0 atm F2 gas is in a sealed metal container at constant volume. When the Xe is used up, there is 4.6 atm remaining of F2 and a solid in the container. What is the empirical formula of the solid? [16:40:43] Yixuan Liu: and then all the answer choices are XeF with a different subscript for F

Quantum mechanics

Failure of the classical model

Coulomb force: F(r) = (-e)(e)/4πε₀r² = -e²/4πε₀r²
1. Doesn't work with F = ma!
2. e = absolute value of an electron's charge
3. r = distance between two charges
4. ε₀ = permittivity constant of a vacuum = 8.854e-12 C²J^-1m^-1
Photoelectric effect:
1. Electron release:
  1. ν₀ = threshold frequency
  2. If ν < ν₀: no electrons released
  3. If ν > ν₀: electrons released
  4. Above ν₀, no additional electrons released (until a significantly higher v)
2. Kinetic energy vs frequency:
  1. Predicted: ν has no relation on KE
  2. Observed: above ν₀, ν has linear relationship with KE
3. Kinetic energy vs intensity (I):
  1. Predicted: I exponentially increases KE
  2. Observed: I has no relation on KE
    1. Reason: unless individual particle has energy > ν₀, doesn't emit electron
4. Number of electrons emitted vs intensity (I):
  1. Predicted: I has no effect on #e
  2. Observed: I has linear relationship with #e
Einstein plotted KE against ν for several metals
1. All had same slope: 6.626e-34 Js (Planck's constant, "h")
2. All had same y intercept: (6.626e-34)ν₀
3. KE = hν - hν₀
  1. Both sides are energy units!
  2. Energy of photon proportional to frequency
  3. Quantized packets of light
4. hν = E_i = energy of incident photon
5. hν₀ = φ = workfunction
6. Therefore:
  1. KE = E_i - φ
  2. E_i = KE + φ

Photoelectric effect practice

http://www.sciencegeek.net/Activities/Ehf.html

If a beam of light with energy 4.0 eV (1 eV = 1.602e-19 J) strikes a gold surface (φ = 5.1 eV), what is the maximum kinetic energy of the ejected electrons?

Answer
KE_max = E_i - φ E_i = 4.0 eV φ = 5.1 eV KE_max = 4.0 eV - 5.1 eV KE_max = -1.1 eV = 0

Electromagnetic waves

E(x₁t) = Acos(2πx/λ - 2πνt)
1. E = electric field
2. x = position of the wave
3. t = time
4. ν = frequency
5. A = amplitude
E(x₁t) = Acos(2πx/λ)
1. only if time constant
2. A max at intervals of λ
E(x₁t) = Acos(2πνt)
1. only if position constant
2. A max at intervals of int/λ
Speed = λν m/sec
c = 2.9979e8

Zinc experiment

φ Zn = 6.9e-19 J
λ UV lamp = 254 nm
λ red laser pointer = 700 nm
Calculate energy:
1. Equation (for light):
  1. E = hν
  2. ν = c/λ
  3. E = hc/λ
2. UV lamp: E = (6.626e-34)(2.998e8)/(254e-9) Js*m/s/m = 7.82e10-19 J
Red laser pointer: E = (6.626e-34)(2.998e8)/(700e-9) Js*m/s/m = 2.84e10-19 J
Ejects electrons?
1. UVL: yes
2. RLP: no
Calculate photons:
1. UVL: (1e-3J/s) * (photon/2.84e-19 J) * 60s = 2.1e17 photons
It works!

Single-electron atoms

Schroedinger equation: ĤΨ = EΨ
1. Ĥ = Hamiltonian operator
2. Ψ = wavefunction (description of the orbital)
3. E = binding energy

Hydrogen orbital change

Wavefunctions: http://quantum.phys.cmu.edu/CQT/chaps/cqt02.pdf

For a hydrogen atom:
1. ĤΨ = ΨE
2. ĤΨ = Ψ * (-1/n²) * (me⁴/8ε₀²h²)
  1. m = m_e = mass of electron
  2. e = charge on the e-
  3. ε₀ = permittivity constant of a vacuum = 8.854e-12 C²J^-1m^-1
  4. h = 6.626e-34 Js
  5. n = principle quantum number (interger)

Rydberg constant in binding energy in general

Rydberg constant (R):
1. ĤΨ = Ψ * -R_H/n²
2. R_H = (me⁴/8ε₀²h²) = 2.18e-18 J
Binding energy (E): E = -R_H/n² (Always negative)
Ionization energy (IE): IE_n = -E_n (Always positive)

States of excitement in general

"Excited state"
1. n₁ = ground
2. n₂ = first excited state
3. n₃ = second excited state
4. n₄ = third excited state
E_n = -Z²R_H / n² (ONLY IF 1 ELECTRON; EITHER HYDROGEN OR 1+ IONS)
1. He¹⁺ 1e^- atom: Z = 2
2. Li²⁺ 1e^- atom: Z = 3
3. Tb⁶⁴⁺ 1e^- atom: Z = 65

Series of hydrogen orbital changes

Series:
1. n_f = 1: Lyman series, UV range
2. n_f = 2: Balmer series, UV range
3. n_f = 3: Paschen series, UV range
4. n_f = 4: Brackett series, UV range

Photon emission in hydrogen orbital change

Setup
1. Evacuated glass tube, filled with H₂
2. Negative electrode, positive electrode
3. Disperse emission
4. Analyse wavelength
ν = ΔE/h = (E_i - E_f)/h
Four wavelengths:
1. Red: n=3 to n=2; 656 nm; longest wavelength, lowest frequency, lowest energy
2. Green: n=4 to n=2; 486 nm
3. Purple: n=5 to n=2; 434 nm
4. Violet: n=6 to n=2 (shortest wavelength, highest frequency, highest energy); 410 nm

Frequency in hydrogen orbital change

General:
1. Combine
  1. ν = (E_i - E_f)/h
  2. E_n = -R_H / n²
  3. ν = ((-R_H / n_i²) - (-R_H / n_f²))/h
2. Simplify
  1. ν = (-R_H/h) * ((1/n_i²) - (1/n_f²))/h
  2. ν = (R_H/h) * ((1/n_f²) - (1/n_i²))
For n_f=2:
1. ν = (R_H/h) * (1/4 - (1/n_i²))

Rydberg constant in frequency in general

Rydberg constant Mk. II:
1. R_H/h = ℜ = 3.29e15 s^-1
2. R∞ = 1.097373157e7 s^-1

Energy levels practice in general

Calculate the wavelength of radiation emitted by a hydrogen atom when an electron makes a transition from the n=3 to the n=2 energy level. (E_n=3 to E_n=2).

Answer
Frequency: ν = (R_H/h) * ((1/n_f²) - (1/n_i²)) ν = 2.18e-18/6.626e-34 * ((1/(2^2)) - (1/(3^2)) ν = 4.5695409e+14 s^-1 Wavelength: λ = c/ν = 2.998e8/4.5695409e+14 = 6.56e-7 m = 656 nm

1e^- atom

Frequency:
1. For n_i > n_f: ν = (R_HZ²/h) * ((1/n_f²) - (1/n_i²))
2. For n_f > n_i: ν = (R_HZ²/h) * ((1/n_i²) - (1/n_f²))
Binding energy:
1. Hydrogen: E_n = -R_H/h
2. All 1-electron: E_n = -Z²R_H/h

Quantum numbers

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

Ψ_nlm(r,θ,φ)
n
1. Name: Principle quantum number
2. Describes: Total binding energy of electron (potential + kinetic)
3. Describes: Shell (and number of subshells)
4. Rules: Any positive integer above 1
l
1. Name: Angular momentum quantum number
2. Describes: Angular energy of the electron
3. Describes: Subshell (l = 0 (s), l = 1 (p), l = 2 (d), l = 3 (f))
4. Rules: Any positive integer above 0 up to l = n-1
m (or m_l)
1. Name: Magnetic quantum number
2. Describes: Shape of orbital / how electron behaves in magnetic field / z component of angular momentum
3. Describes: Subshell description (m = -1 (x), m = 0 (z), m = +1 (y))
4. Rules: Any integer above 0 from m = -l to m = l
Alternate description:
1. Ψ₁₀₀(r,θ,φ) = 1s orbital
Degenerate orbitals: for n orbitals, n² orbitals are degenerate (have the same energy)
For 1-electron atoms, all subshells have the same energy

	State label	Wavefunction	Orbital	E_s	E_s[J]
n = 1 l = 0 m = 0	100	Ψ₁₀₀	1s	-R_H/1²	-2.9e-18 J
n = 2 l = 0 m = 0	200	Ψ₂₀₀	2s	-R_H/2²	-5.45e-19 J
n = 2 l = 1 m = +1	210	Ψ₂₁₁	2p_x or 2p_y (opp of 5th)	-R_H/2²	-5.45e-19 J
n = 2 l = 1 m = 0	210	Ψ₂₁₀	2p_z	-R_H/2²	-5.45e-19 J
n = 2 l = 1 m = -1	21-1	Ψ_21-1	2p_x or 2p_y (opp of 3rd)	-R_H/2²	-5.45e-19 J

Wavefunction physical interpretation

https://youtu.be/Pj2fkkZ6Gto?t=1630

Max Born
1. [Ψ_nlm(r,θ,φ)]²
2. Probability/Volume
3. Probability of finding an electron in a given volume
4. Never reaches 0 through space
Components
1. Ψ_nlm(r,θ,φ) = R_nl(r) x Y_lm(θ,φ)
2. Wavefunction = radial wavefunction x angular wavefunction
3. For all s orbitals, Y is a constant
Areas of 0 probability: Nodes
1. Only occurs between orbitals
2. Number of nodes = n - l - 1

Radial probability distribution

For s orbitals: RPD = Ψ² * 4πr²Ψ²dr
1. Probability = Probability/Volume * Volume
2. Most probable : r_mp = a₀ = Bohr radius = 0.529 Ångstroms

p orbitals

l = 1
3 orbitals
1. for m = +/-1, p_x/p_y
θ and φ (angular) dependence
Total of 6 lobes
1. Each plane: Two lobes, separated by nodal plane
Ψ²_{2p_y} =
1. Highest probability: Along Y axis
2. Positive Ψ: Where y is positive
3. Nodal plane: xz plane (φ = 0 degrees)
Nodes:
1. Total: n - 1
2. Angular: l
3. Radial: n - l - 1
As l increases, r_mp decreases

Spin magnetic quantum number

m_s = +1/2 (spin up) or -1/2 (spin down)
Independent of orbital, only describes electron
Discovery:
1. Emission spectrum of sodium
2. Would expect 1 line at given frequency
3. 2 lines very slightly above and below given frequency (doublet)

Pauli exclusion principle

No two electrons in same atom can have same quantum numbers
Distinction: orbital has 3 quantum numbers, electron 4
Limits to 2 electrons per orbital

Multi-electron atoms

How many electrons can exist in a 2p orbital? 6. 2p has 3 complete orbitals, 2px, 2py, 2pz; each complete orbital can have 2 electrons.

Shroedinger equation for multiple electrons

Base: ĤΨ = EΨ

H (1 electron): ĤΨ(r₁θ₁φ₁) = EΨ(r₁θ₁φ₁)
He (2 electrons): ĤΨ(r₁θ₁φ₁r₂θ₂φ₂) = EΨ(r₁θ₁φ₁r₂θ₂φ₂)
Li (3 electrons): ĤΨ(r₁θ₁φ₁r₂θ₂φ₂r₃θ₃φ₃) = EΨ(r₁θ₁φ₁r₂θ₂φ₂r₃θ₃φ₃)
At high enough levels, it is mathematically impossible to solve the Shroedinger equation; instead, an approximation is used.

Hartree orbitals

Approximation: Treat the wavefunction of the many orbitals as the product of a one electron approximation for each.

He (2 electrons):
1. Ψ(r₁θ₁φ₁r₂θ₂φ₂) = Ψ(r₁θ₁φ₁) * Ψ(r₂θ₂φ₂)
2. Ψ(e^-#1, e^-#2) = Ψ(e^-#1) * Ψ(e^-#2)
3. Ψ_100-1/2 * Ψ_100+1/2
4. 1s(1) * 1s(2)
Li (3 electrons):
1. Ψ(r₁θ₁φ₁r₂θ₂φ₂r₃θ₃φ₃) = Ψ(r₁θ₁φ₁) * Ψ(r₂θ₂φ₂) * Ψ(r₃θ₃φ₃)
2. Ψ(e^-#1, e^-#2, e^-#3) = Ψ(e^-#1) * Ψ(e^-#2) * Ψ(e^-#3)
3. Ψ_100-1/2 * Ψ_100+1/2 * Ψ_200-1/2
4. 1s(1) * 1s(2) * 2s(1)

Shorthand

H (1 electron): 1s¹
He (2 electrons): 1s²
Li (3 electrons): 1s²2s²
Be (4 electrons): 1s²2s¹
B (5 electrons): 1s²2s²2p¹

Multi-electron vs hydrogen atom wavefunctions

Argon: 1s²2s²2p⁶3s²3p⁶

Similarities:
1. Similar in shape
2. Identical Radial Probability Distribution and nodes
Differences:
1. Each multi-electron orbital is smaller than the corresponding hydrogen atom orbital, because the nucleus is more positively charged.
2. Orbital energy depends on both the shell (n) and the subshell (l) or angular momentum quantum number.
3. Binding orbital energy in multi-electron atoms is lower (more negative) than in corresponding H-atom orbitals.
4. In hydrogen, all orbitals for a given n have same energy level; in a multi-electron atom, orbitals increase in energy for both n and l
  1. For 1-electron: E_n = -IE_n = -Z²R_H/n²
  2. For multi-electron: E_n = -IE_n = -(Z^nl_eff)²R_H/n²
  3. Z != Z_effective
Shielding and binding energy:
1. Scenario: Helium (He) nucleus has 2 protons (z = 2). Electron 1 is very close to nucleus. Electron 2 is far enough out that it doesn't affect electron 1. Electron 1 is thus "no shielded" and electron 2 is "totally shielded".
2. E_He = -IE_He = -(Z_eff)²R_H/n² = -(+2)²R_H/1² = 8.72 x 10^-18 J, Z_eff = 1
3. E_He = -IE_He = -(Z_eff)²R_H/n² = -(+1)²R_H/1² = 2.18 x 10^-18 J, Z_eff = 2
4. Reality: IE_He = 3.94 x 10^-18 J
Finding Z_eff:
1. Z_eff = [n²(IE/R_H)]^1/2
2. He: [1²(3.94 x 10^-18 J/2.18 x 10^-18 J)]^1/2 = 1.34
Shielding
1. S is less shielded than P, because over RPD, average Z_eff of 2p is less than Z_eff of 2s (etc.)
2. RPD of orbitals

Electron configurations

Aufbau principle:
1. Pauli Exclusion Principle: Only 1 spin in any given suborbital
2. Hund's rule: At the same energy level, a single e^- enters state before a second enters said state
Oxygen: https://youtu.be/f7RRqxv2pzg?t=2198
1. O: 1s²2s²2p⁴
2. O_ml: 1s²2s²2px²2px¹2px¹
Sodium:
1. Na: [Ne]3s¹
2. Na_ml: 1s²2s²2px²2px²2px²3s¹

Exceptions

1: Half-filled d orbitals are more stable than half-filled s
1. V: [Ar]4s²3d³
2. Cr: [Ar]4s¹3d⁵
3. Mn: [Ar]4s²3d⁵
2: Filled d orbitals are more stable than half-filled s
1. Ni: [Ar]4s²3d⁸
2. Cu: [Ar]4s¹3d¹⁰
3. Zn: [Ar]4s²3d¹⁰

Ion electron configurations

A d orbital with 2+ electrons has less energy (is more stable) than a s orbital. When removing the highest-energy atoms (making something ionic) an S

Vi: [Ar]4s²3d³
1. Vi (reordered from lowest to highest energy orbital): [Ar]3d³4s²
Vi^-: [Ar]4s¹3d³

Practice

http://www.sciencegeek.net/Chemistry/taters/Unit2ElectronNotations.htm http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/electronconfigpractice.html

Photoelectric spectroscopy (PES)

https://en.wikipedia.org/wiki/Spectroscopy

X-rays are commonly used because they have sufficient energy; UV rays may be, but are sometimes insufficient.
Neon:
1. 1s²2s²2p⁶ -> 1s²2s²2p⁵ + e^- with most KE
2. 1s²2s²2p⁶ -> 1s²2s¹2p⁶ + e^- with middle KE
3. 1s²2s²2p⁶ -> 1s¹2s²2p⁶ + e^- with least KE
IE = E_i + KE
1. IE = Ionization energy
2. E_i = Energy of photon
3. KE = Kinetic energy of electron
Neon IE (with E_i = 1254)
1. 1s²2s²2p⁶ ; KE = 1232 eV ; IE_2p = 22 eV (1254 - 1232)
2. 1s²2s²2p⁶ ; KE = 1206 eV ; IE_2p = 48 eV (1254 - 1206)
3. 1s²2s²2p⁶ ; KE = 384 eV ; IE_2p = 870 eV (1254 - 384)
4. Each would be represented as a bar with x = KE in order 1s, 2s, 2p
Practice: If PES produces 5 lines, which elements could it be?
1. 5 lines, 5 orbitals: 1s2s2p3s3p: Al, Si, P, S, Cl, Ar
Practice: PES would reveal how many lines in hafnium (z = 72)?
1. 1s2s2p3s3p3d4s4p4d4f5s5p5d6s (or 1s2s2p3s3p4s3d4p5s4d5p6s4f5d)
2. 14

https://youtu.be/LPh2Ut7D4WA?t=746

Periodic trends

Ionization energy

First (smallest) ionization energy
1. Implied whenever "ionization energy" is stated
2. IE = -E_nl
3. Boron:
  1. B(1s²2s²2p¹) -> B⁺(1s²2s²) + e^-
  2. ΔE = E_p - E_r = IE = -E_2p
Second ionization energy
1. FILLER
2. Boron:
  1. B⁺(1s²2s²) -> B²⁺(1s²2s¹) + 2e^-
  2. ΔE = IE₂ = -E_2s in B₊
Third ionization energy
1. FILLER
2. Boron:
  1. B²⁺(1s²2s¹) -> B³⁺(1s²) + 3e^-
  2. ΔE = IE₃ = -E_2s in B₊
Periodic table:
1. [22], [23]
2. Moving right in a row, Z (charge) while n (shell) remains constant; thus, Z_eff (effective charge) increases, so IE increases
3. Moving down in column, Z (charge) and n (shell) increases; increasing n overpowers increasing Z, so Z_eff decreases and IE decreases
4. Exceptions in Lithium - Neon row: http://staff.norman.k12.ok.us/~cyohn/index_files/ionizationenergynotes_files/image001.gif
  1. Beryllium (1s²2s² to Boron 1s²2s²2p¹: Energy required to add new orbital is greater than the increased distance from the nucleus
  2. Nitrogen (1s²2s²2p³ to Boron 1s²2s²2p⁴: Oxygen has 4 electrons in a 6 electron-orbital (3 subshells of 2 each) which forces 2 electrons to pair up

Which element has smaller IE (and why): Al or P?

Al (lower Z_eff)
P (lower Z_eff)
Al (higher Z_eff)
P (higher Z_eff)

Electron affinity

Electron affinity (EA or E_ea) = -ΔE
Chlorine:
1. Cl + e^- -> Cl^- ; ΔE = -349 kJ/mol ; EA = 349 kJ/mol
2. Energy is released, so negative ion is more stable than atom.
Nitrogen:
1. N + e^- -> N^- ; ΔE = +7 kJ/mol ; EA = -7 kJ/mol
2. Energy is added, so atom is more stable than negative ion.
Periodic table:
1. Moving right in a row, EA increases.
2. Moving down in a column, EA decreases.

Electronegativity

Electronegativity (χ) is proportional to (0.5)(EA + IE)
Periodic table:
1. Upper right: high χ (electron expector)
2. Bottom left: low χ (electron donor)

Atomic radius

Atomic radius: value of r which encompasses 90% of electron density
Atomic radius: value of r which is 0.5 of distance between two atoms in a compound
Both are very similar values
Periodic table:
1. Moving right in a row, r decreases (because Z_Eff decreases).
2. Moving down in a column, r increases (because new orbitals are larger).
Ions:
1. F^- is larger than F (more shielding, less Z_Eff))
2. Na⁺ is smaller than Na (less shielding, more Z_Eff))

Isoelectronic

Neon: 1s²2s²2p⁶
Flourine:
1. F: 1s²2s²2p⁵
2. F^-: 1s²2s²2p⁶
Sodium:
1. Na: 1s²2s²2p⁶3s¹
2. Na⁺: 1s²2s²2p⁶
Which atom is isoelectronic with Krypton (Z=36)? Selenium^2- (Z=34).

Covalent bonds

Chemical bonds: rearrangement of the nuclei and electrons of the bonded atoms results in a lower energy than separate atoms.
Covalent bond: a pair of electrons shared between 2 atoms.
Terms:
1. Internuclear distance: r
2. Energy: E = nuclear-nuclear repulsion + electron-nuclear attraction + electron-electron repulsion
3. Disassociation energy: ΔE_d = energy necessary to break covalent bond = bond strength = distance from E_{H + H} to E_H₂
4. Energy vs interatomic distance [24][25]
Example with H:
1. H⁺ + e^- = 0 kJ/mol
2. 2H (unbonded): E = 2 * -1312 kJ/mol = -2624 kJ/mol
3. H₂ (bonded): E = -3048 kJ/mol
4. H₂ is lower energy, so 2H bonds
5. ΔE_d = -2624 kJ/mol - (-3048 kJ/mol) = 424 kJ/mol
N₂ vs H₂: N₂ is stronger bond, because its ΔE_d is lower than H₂'s, which means that the bond is more stable.

Quantitative chemical shit

Finding the stuff you've got in a pile of shit

Gravimetric analysis

Obtain sample (analyte) with some
Form precipitate: Add solute to form insoluble precipitate containing sample and some known (eg, AgNO₃
Isolate precipitate:
1. Filter the precipitate from the solvent (eg, using sintered glass filter and vacuum pump)
2. Precipitate has water removed (eg, drying/heating/etc.)
Calculate mass of sample
1. Precipitate is weighed
2. Proportion of mass attributable to the precipitating agent is subtracted
3. Remainder is mass of analyte

Later

molecular geometry https://www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&q=molcular%20geometry&oq=molcular%20geometry&aqs=chrome..69i57j0l5.2521j0j7

Shroedinger: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/scheq.html

TITRATION!!!

Reorganize

General naming rules, move ionic naming rules

Practice photoelectric effect.

Gas laws

Solubility rules (NO3, NH4, and Group I)

http://ocw.mit.edu/courses/chemistry/5-111-principles-of-chemical-science-fall-2008/exams/

Dtown

Redox

http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions http://www.webqc.org/balance.php?reaction=Ag%2BHNO3%3DAgNO3%2BH2O%2BNO2 http://www.chemteam.info/Redox/Balance-Redox-Acid.html http://www.sciencegeek.net/APchemistry/APtaters/Redox/redox1.htm

Lab equipment

Lab equipment practice

Scientific method

General terms

Measurement

Significant figures

History

Dalton

Thomson

Millikan

Rutherford

General principles, 1900s

Matter

Properties

Hierarchy

Phase/state differences

Extensive vs intensive

Separation

Atomic structure

Nuclear symbol

Particulate symbols, charges, and mass

Particulate composition

Molecules

Molecule vs compound

Bonding types

Nuclear reactions

Fission

Fusion

Balancing

Nuclear balancing practice

Decay

Ions

Terms

Ionic equation balancing practice

Naming ionic compounds

Naming ionic compounds practice

Solutions

Solution concentration

Solution concentration practice

Moles

General

Moles practice

Formula Mass

Empirical formula practice

Mass Spectrometry

Processes table

Charts

Mass spectrometry charts practice

Stoichiometry

Stoichiometry practice

Limiting reactant

Limiting reactant practice

Precipitation Reactions

Solubility rules

Gas

Quantum mechanics

Failure of the classical model

Photoelectric effect practice

Electromagnetic waves

Zinc experiment

Single-electron atoms

Hydrogen orbital change

Rydberg constant in binding energy in general

States of excitement in general

Series of hydrogen orbital changes

Photon emission in hydrogen orbital change

Frequency in hydrogen orbital change

Rydberg constant in frequency in general

Energy levels practice in general

1e- atom

Quantum numbers

Wavefunction physical interpretation

Radial probability distribution

p orbitals

Spin magnetic quantum number

Pauli exclusion principle

Multi-electron atoms

Shroedinger equation for multiple electrons

Hartree orbitals

Shorthand

1e^- atom