Summer Review
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Transcript of Summer Review
Summer Review
Unit I: Chemical FoundationsUnit II: Atoms, Molecules, Ions
Unit III: Stoichiometry
What you are already expected to know:
• Sig Figs• Metric Conversions• Dimensional Analysis• Changing between units of temperature• Classification of matter (compound, element, homogeneous,
heterogeneous)• Atomic Structure & Electron Configuration• Mole-molecules-gram calculations• Basic Bonding Theory & Assigning Oxidation #s• Naming compounds• Types of Rxns & Balancing Eqns• Basic Solution Chemistry
What if you don’t know these things?
• All of these concepts were part of your summer assignment packet
• If you cannot complete some of the questions, then you need to come before/after school to receive additional assistance
AP Chemistry Unit Layouts
Semester 1• Unit 1: Chemical
Foundations (2 wks)• Unit 2: Stoichiometry,
Reactions in Solutions (5 wks)
• Unit 3: Atomic Structure & Periodicity (2 wks)
• Unit 4: Bonding & Intermolecular Forces (5 wks)
• Unit 5: Gases (2 wks)
Semester 2• Unit 6: Kinetics (3 wks)• Unit 7: Equilibrium (6 wks)• Unit 8: Thermodynamics &
Free Energy (4 wks)• Unit 9: Electrochemistry (1
wks)• AP REVIEW• AP TEST: MONDAY, MAY 5th• End of Year Project
What will we be reviewing in class?
UNIT 1: Chemical Foundations• Challenging dimensional analysis problems• Basic Bonding Theory• Naming Compounds• Limiting Reactants & Stoichiometry• Empirical & Molecular Formulas• Molarity Calculations• Separation Techniques
The Fundamental SI Units
Physical Quantity Name AbbreviationMass kilogram kgLength meter mTime second sTemperature Kelvin KElectric Current Ampere AAmount of Substance mole molLuminous Intensity candela cd
Types of Error
• Random Error (Indeterminate Error) - measurement has an equal probability of being high or low.
• Systematic Error (Determinate Error) - Occurs in the same direction each time (high or low), often resulting from poor technique.
Rules for Counting Significant Figures - Details
• Exact numbers have an infinite number of significant figures. Can come from counting or definition.
• 15 atoms• 1 inch = 2.54 cm, exactly
Rules for Significant Figures in Mathematical Operations
• Multiplication and Division: # sig figs in the result equals the number in the least precise measurement used in the calculation.
• 6.38 2.0 =• 12.76 13 (2 sig figs)
Rules for Significant Figures in Mathematical Operations
• Addition and Subtraction: # sig figs in the result equals the number of decimal places in the least precise measurement.
• 6.8 + 11.934 =• 18.734 18.7 (3 sig figs)
Universe
MatterEnergy
HomogeneousPhysical Change Heterogeneous
Pure Substance Solution Mixture
Element CompoundChemical Change
Electron Levels Nucleus
Electrons Protons Neutrons
Potential Energy
Kinetic Energy
Position Composition
Gravitational Electrostatic
SEPARATION OF MIXTURES
• - mixtures can be separated into pure substances by physical means. – distillation– filtration– centrifuging– magnet– evaporation– chromatography
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Thermometer
Vapors
Distillingflask
Burner
Condenser
Receivingflask
Distillate
Water out Coolwater in
Simple laboratory distillation apparatus.
CENTRIFUGE
Paper Chromatography
Chromatography has two phases of matter: a stationaryphase (the paper) and a mobile phase ( the liquid).
Law of Conservation of Mass
- Discovered by Antoine Lavoisier
- Mass is neither created nor destroyed
- Combustion involves oxygen, not phlogiston
Other Fundamental Chemical Laws
- A given compound always contains exactly the same proportion of elements by mass.
- Carbon tetrachloride is always 1 atom carbon per 4 atoms chlorine.
Law of Definite Proportion -- Joseph Proust
Other Fundamental Chemical Laws
- When two elements form a series of compounds, the ratios of the masses of the second element that combine with 1 gram of the first element can always be reduced to small whole numbers.
- The ratio of the masses of oxygen in H2O and H2O2 will be a small whole number (“2”).
Law of Multiple Proportions--John Dalton
The Modern View of Atomic Structure
- electrons- protons: found in the nucleus, they have a
positive charge equal in magnitude to the electron’s negative charge.
- neutrons: found in the nucleus, virtually same mass as a proton but no charge.
The atom contains:
The Mass and Charge of the Electron, Proton, and Neutron
Particle Mass (kg) Charge
Electron 9.11 10 31 1
Proton 1.67 10 27 1+
Neutron 1.67 10 27 0
The Chemists’ Shorthand: Atomic Symbols
K Element Symbol39
19
Mass number
Atomic number
The Chemists’ Shorthand:Formulas
• Chemical Formula:• Symbols = types of atoms• Subscripts = relative numbers of atoms
CO2
• Structural Formula:• Individual bonds are shown by lines.
O=C=O
Ions• Cation: A positive ion
Mg2+, NH4+
• Anion: A negative ionCl, SO4
2
• Polyatomic: an ion containing a number of covalently bonded atoms acting as a single unit.
• Ionic Bonding: Force of attraction between oppositely charged ions.
Cations & Anions
• Cations are positive ions.• Na ----> Na+ + e-
• Anions are negative ions.• Cl2 + 2e- ----> 2Cl-
Periodic Table• Elements classified by:- Properties- atomic number
• Groups (vertical) 1A = alkali metals 2A = alkaline earth metals 7A = halogens 8A = noble gases
• Periods (horizontal)
02_29
1H
3Li
11Na
19K
37Rb
55Cs
87Fr
4Be
12Mg
20Ca
38Sr
56Ba
88Ra
21Sc
39Y
57La*
89Ac†
22Ti
40Zr
72Hf
104Unq
23V
41Nb
73Ta
105Unp
24Cr
42Mo
74W
106Unh
25Mn
43Tc
75Re
107Uns
26Fe
44Ru
76Os
108Uno
27Co
45Rh
77Ir
109Une
110Uun
111Uuu
28Ni
46Pd
78Pt
29Cu
47Ag
79Au
30Zn
3 4 5 6 7 8 9 10 11 12
48Cd
80Hg
31Ga
49In
81Tl
5B
13Al
32Ge
50Sn
82Pb
6C
14Si
33As
51Sb
83Bi
7N
15P
34Se
52Te
84Po
8O
16S
9F
17Cl
35Br
53I
85At
10Ne
18Ar
36Kr
54Xe
86Rn
2He
58Ce
90Th
59Pr
91Pa
60Nd
92U
61Pm
93Np
62Sm
94Pu
63Eu
95Am
64Gd
96Cm
65Tb
97Bk
66Dy
98Cf
67Ho
99Es
68Er
100Fm
69Tm
101Md
70Yb
102No
71Lu
103Lr
1A
2A
Transition metals
3A 4A 5A 6A 7A
8A1
2 13 14 15 16 17
18A
lkal
i met
als
Alkalineearth metals Halogens
Noblegases
*Lanthanides
† Actinides
The periodic table.
Chemical Symbols• Symbols commonly missed.• A -- Al, Ar, As, Au, & Ag.• B -- Ba, Bi, B, Br, & Be.• C -- C, Ca, Cd, Cl, Cr, Co, Cs, & Cu.• M -- Mg, Mn, & Mo.• S -- S, Sb, Si, Sr, & Sn.• Latin -- Fe, Au, Ag, Sb, Pb, Na, K, Hg, & Cu.• German -- W
Physical Properties of Metals
Metals are:1. efficient conductors of heat and electricity.2. malleable (Can be hammered into thin
sheets).3. ductile (Can be pulled into wires).4. lustrous (shiny). 5. tend to lose electrons and form cations.• Examples are: Na, Cu, Au, Ag, & Fe.
Metalloids
• substances with the properties of both metals and nonmetals.
• also called semimetals• Lie along the zigzag line between metals
and nonmetals• The six metalloids are:• B, Si, Ge, As, Sb, and Te.
Physical Properties of Nonmetals
Nonmetals are:1. nonconductors of heat and electricity
(insulators).2. not malleable, but are brittle.3. not ductile.4. dull and without a luster.5. tend to gain electrons to form anions.• Examples are: H, He, N, O, S, & P.
The Chemists’ Shorthand: Atomic Symbols
K Element Symbol39
19
Mass number
Atomic number
1+ Ion charge
Charges on Common Ions
+1
+2 +3
-4 -3 -2 -1
Common Names
• sugar of lead• blue vitriol• quicklime• Epsom salts• milk of magnesia• gypsum• laughing gas
lead(II) acetatecopper(II) sulfatecalcium oxidemagnesium sulfatemagnesium hydroxidecalcium sulfatedinitrogen monoxide
Naming Compounds
1. Cation first, then anion
2. Monatomic cation = name of the element• Ca2+ = calcium ion
3. Monatomic anion = root + -ide• Cl = chloride
• CaCl2 = calcium chloride
Binary Ionic Compounds:
Naming Compounds(continued)
- metal forms more than one cation- use Roman numeral in name
• PbCl2
• Pb2+ is cation
• PbCl2 = lead (II) chloride
• plumbous chloride
Binary Ionic Compounds (Type II):
Naming Compounds(continued)
- Compounds between two nonmetals- First element in the formula is named first.- Second element is named as if it were an anion.- Use prefixes- Never use mono-
• P2O5 = diphosphorus pentoxide
Binary compounds (Type III):
NOMENCLATURE OF COMPOUNDS Binary -- 2 elements Ternary -- (3 elements) - Ionic
(metal ion + polyatomic ion)
Ca3(PO4)2 -- calcium phosphate FeSO4 -- iron (II) sulfate
-- ferrous sulfate
Type I - Ionic(Type I metal + nonmetal)Group I, II, Al+3, Ag1+,Cd2+, & Zn2+
NaCl -- Sodium Chloride
Type II - Ionic(Type II metal + nonmetal)All other metalsFe2S3 -- iron (III) sulfide -- ferric sulfide Type III - covalent
(2 nonmetals)CO2 -- carbon dioxide
Chemical Nomenclature• Name each of the following:• CuCl • HgO• Fe2O3
• MnO2
• PbCl2
• CrCl3
copper(I) chloride cuprous chloridemercury(II) oxide mercuric oxideiron(III) oxide ferric oxidemanganese(IV) oxide manganic oxidelead(II) chloride plumbous chloridechromium(III) chloride chromic chloride
Chemical Nomenclature
• Name each of the following:• P4O10
• N2O5
• Li2O2
• Ti(NO3)4
• SO3
• SF6
• O2F2
tetraphosphorus decoxidedinitrogen pentoxidelithium peroxidetitanium(IV) nitratesulfur trioxidesulfur hexafluoridedioxygen difluoride
Common Nomenclature Mistakes
•Compounds:
•SO3 --Sulfur trioxide
•NO2 -- Nitrogen dioxide
•NO3 -- Nitrogen trioxide
•Polyatomic ions:
•SO32- -- Sulfite ion
•NO21- -- Nitrite ion
•NO31- -- Nitrate ion
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Does the anioncontain oxygen?
No Yes
hydro -+ anion root+ -ichydro (anion root)ic acid
Check the ending of the anion.
-ate
anion or element root+ -ous(root)ous acid
-ite
anion or element root+ -ic(root)ic acid
A flow chart for naming acids. An acid is best considered asone or more H+ ions attached to an anion.
Binary Acids
• made up of two elements -- hydrogen and a nonmetal
• named by using:• prefix hydro + root of nonmetal + ic +
acid
HCl -- hydrochloric acidH2Se -- hydroselenic acid
Ternary Acids (oxyacids)
• contain three elements -- hydrogen, nonmetal, and oxygen.
• most oxygen per + root of nonmetal + ic + acid
• less oxygen root of nonmetal + ic + acid• less oxygen root of nonmetal + ous + acid• least oxygen hypo + root of nonmetal +
ous + acid
Ternary Acids(continued)
• HBrO4 perbromic acid
• HBrO3 bromic acid
• HBrO2 bromous acid
• HBrO hypobromous acid
• H3PO4 phosphoric acid
• H3PO3 phosphorous acid
• H3PO2 hypophosphorus acid
Salt Nomenclature (Ionic compounds)
• Binary salts (metal and nonmetal)
• name of positive ion + root of nonmetal + ide
NaCl -- sodium chlorideK2S -- potassium sulfide
Salt Nomenclature (continued)
• Ternary salts ( metal and polyatomic ion)• name of positive ion + root of nonmetal + ate or ite• If the salt comes from an ic acid, change ic to ate.• H2CO3 carbonic acid Na2CO3 sodium carbonate• H3PO4 phosphoric acid K3PO4 potassium phosphate
• If the salt comes from an ous acid, change ous to ite. • H2SO3 sulfurous acid Li2SO3 lithium sulfite • HClO hypochlorous acid NaClO sodium hypochlorite
Avogadro’s number equals
6.022 1023 units
Calculating Moles & Number of Atoms• 1 mol Co = 58.93 g• (5.00 x 1020 atoms Co)(1mol/6.022 x 1023 atoms)
= 8.30 x 10-4 mol Co
(8.30 x 10-4 mol)(58.93g/1 mol) = 0.0489 g Co
Moles are the doorway grams <---> moles <---> atoms
•
Calculating Mass from Moles
• CaCO3
• 1 Ca = 1 (40.08 g) = 40.08 g• 1 C = 1 (12.01 g) = 12.01 g • 3 O = 3 (16.00 g) = 48.00 g 100.09 g
• (4.86 molCaCO3)(100.09 g/1 mol) = 486 g CaCO3
Chemical Equations
Chemical change involves a reorganization of the atoms in one or
more substances.
Physical States
• solid (s)• liquid (l)• gas (g)• aqueous (aq)
Important Equation Symbols
• yields ------> cat.
• catalyst ------->
H2SO4
• catalyst ------>
∆
• heat ------->
light
• light -------->
elect.
• electricity ------>
Four Steps in Balancing Equations1. Get the facts down. (WRITE THE REACTION)2. Check for diatomic molecules (subscripts).3. Balance charges on compounds containing a metal, ammonium
compounds, and acids (subscripts).4. Balance the number of atoms (coefficients).
a. Balance most complicated molecule first.b. Balance other elements.c. Balance hydrogen next to last.d. Balance oxygen last.
**It may be helpful to write water has H(OH) if used in a single or double replacement rxn!
Balancing Equations CautionThe identities (formulas) of the compounds must never be changed in balancing a
chemical equation!Only coefficients can be used
to balance the equation-subscripts will not change!
Balancing Equations
When solid ammonium dichromate decomposes, it produce solid chromium(III) oxide, nitrogen gas, and water vapor.
(NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) + HOH(g)
(NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) + 4HOH(g)
Calculating Masses of Reactants and Products
1. Balance the equation.2. Convert mass to moles.3. Set up mole ratios.4. Use mole ratios to calculate moles of
desired substituent.5. Convert moles to grams, if necessary.
Gram to Mole & Gram to Gram
__Al(s) + __I2(s) ---> __AlI3(s)
2Al(s) + 3I2(s) ---> 2AlI3(s)
How many moles and how many grams of aluminum iodide can be produce from 35.0 g of aluminum?
Gram to Mole & Gram to Gram
• 2Al(s) + 3I2(s) ---> 2AlI3(s)
• (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al) = 1.30 mol AlI3
• (35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al)(407.68 g/1 mol) = 529 g AlI3
Percent Composition
• Mass percent of an element:
• For iron in iron (III) oxide, (Fe2O3)
mass Fe% ..
. 111 69159 69
100% 69 94%
mass mass of element in compoundmass of compound
% 100%
% Composition
• CuSO4. 5 H2O
• 1 Cu = 1 (63.55 g) = 63.55 g• 1 S = 1 (32.06 g) = 32.06 g• 4 O = 4 (16.00 g) = 64.00 g• 5 H2O = 5 (18.02 g) = 90.10 g
249.71 g
% Composition (Continued)
• % Cu = 63.55 g/249.71g (100 %) = 25.45 % Cu• % S = 32.06 g/249.71 g (100 %) = 12.84 % S• % O = 64.00 g/249.71 g (100 %) = 25.63 % O• % H2O = 90.10 g/249.71 g (100 %) = 36.08 % H2O
Check: Total percentages. Should be equal to 100 % plus or minus 0.01 %.
Formulas
• molecular formula = (empirical formula)n
[n = integer]
• molecular formula = C6H6 = (CH)6
• empirical formula = CH
Empirical Formula – the lowest whole number
proportion of a compound
Molecular Formula – the actual formula of a compound
Example:C2H4O2 – Molecular formulaCH2O – Empirical Formula
Steps to Determine Empirical Formula
Step 1: Assume you have 100 g of compound, convert each % to g
Step 2: Use atomic mass for each element to calculate mass in moles
Step 3: Divide each mole value by smallest # of mols present (will give a whole number)
Step 4: Write Empirical FormulaStep 4: Write Empirical Formula
Example Problem: Determine the empirical and molecular formulas for a compound that gives
the following percentages: 71.65% Cl, 24.27% C, and 4.07% H the molar mass is known to be
98.96 g/mol
Step 1: Assume you have 100 g of compound, convert each % to g
Step 2: Use atomic mass for each element to calculate mass in moles
Step 3: Divide each mole value by smallest # of mols present (will give a whole number)
Step 4: Write Empirical Formula
Steps to Determine Molecular Formula
Step 1: Obtain Empirical Formula
Step 2: Compute Molar Mass for Empirical Formula
Step 3: Calculate Ratio… Molar Mass = whole
Empirical Formula Mass number Step 4: Write Molecular Formula by multiplying integer by each
element in the compound
Example Problem: Determine the empirical and molecular formulas for a compound that gives the following percentages: 71.65% Cl, 24.27% C, and
4.07% H the molar mass is known to be 98.96 g/mol
Step 1: Obtain Empirical Formula
Step 2: Compute Molar Mass for Empirical Formula
Step 3: Calculate Ratio… Molar Mass = whole
Empirical Formula Mass number Step 4: Write Molecular Formula by multiplying integer
by each element in the compound
Limiting Reactant
The limiting reactant is the reactant that is consumed first, limiting the
amounts of products formed.
Solving a Stoichiometry Problem
1. Balance the equation.2. Convert masses to moles.3. Determine which reactant is limiting.4. Use moles of limiting reactant and mole
ratios to find moles of desired product.5. Convert from moles to grams.
Limiting Reactant Problem
If 56.0 g of Li reacts with 56.0 g of N2, how many grams of Li3N can be produced?
__Li(s) + __N2(g) ---> __Li3N(s)
6 Li(s) + N2(g) ---> 2 Li3N(s)
(56.0 g Li) (1 mol/6.94g)(1 mol N2/6 mol Li) (28.0 g/1 mol) = 37.7 g N2
Since there were 56.0 g of N2 and only 37.7 g used, N2 is the excess and Li is the Limiting Reactant.
Limiting Reactant Problem
6 Li(s) + N2(g) ---> 2 Li3N(s)
(56.0 g Li)(1 mol/6.94g)(2 mol LiN3/6 mol Li) (34.8 g/1 mol) = 93.6 g Li3N
• How many grams of nitrogen are left?56.0g N2 given - 37.7 g used = 18.3 g excessN2
• Calculating percent yield– Actual yield – what you got by actually doing the
experiment– Theoretical yield – what stoichiometric calculations say
the reaction should have produced.
Actual yield (data table) x 100 = % yieldTheoretical yield (Stoich)
5 g of Zn reacts with excess hydrochloric acid. 9.4 g of zinc chloride are formed. What is the percent yield for this reaction?
% Yield
• Values calculated using stoichiometry are always theoretical yields!
• Values determined experimentally in the laboratory are actual yields!
Limiting Reactant & % Yield
If 68.5 kg of CO(g) is reacted with 8.60 kg of H2(g), what is the theoretical yield of methanol that can be produced?
__H2(g) + __CO(g) ---> __CH3OH(l)
2 H2(g) + CO(g) ---> CH3OH(l)
(68.5 kg CO)(1 mol/28.0 g)(2 mol H2/1 mol CO)(2.02 g/1mol) = 9.88 kg H2
Limiting Reactant & % Yield
• 2 H2(g) + CO(g) ---> CH3OH(l)
• Since only 8.60 kg of H2 were provided, the H2 is the limiting reactant, and the CO is in excess.
(8.60 kg H2)(1000 g/1 kg)(1 mol/2.02 g)(1 mol CH3OH/2 mol H2)(32.0 g/1 mol) = 6.85 x 104 g CH3OH
Limiting Reactant & % Yield• 2 H2(g) + CO(g) ---> CH3OH(l)
• If in the laboratory only 3.57 x 104 g of CH3OH is produced, what is the % yield?
%100_
_%yieldltheoretica
yieldactualYield
%1001085.61057.3% 4
4
gxgxYield
% Yield = 52.1 %
Molarity Calculations for Stock Solutions
• Molarity = (moles)÷Liters• You will have to create your own stock solns
for lab• Basic Chart to help:
To dilute solutions
• Use this formula:
Molarity1xVolume1=Molarity2xVolume2
WATCH: Volume units should be the same Volume 1 & 2 mean FINAL volumes Usually you will be solving for Volume 1 in dilutions