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

01_13

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

02_33

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