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Summer Review. Unit I: Chemical Foundations Unit 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

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Summer review

Summer Review

Unit I: Chemical Foundations

Unit II: Atoms, Molecules, Ions

Unit III: Stoichiometry


What you are already expected to know

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

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

AP Chemistry Unit Layouts

Semester 1

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

  • 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)


What will we be reviewing in class

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

The Fundamental SI Units


Types of error

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

Rules for Counting Significant Figures - Details

  • Exact numbershave an infinite number of significant figures. Can come from counting or definition.

  • 15 atoms

  • 1 inch = 2.54cm, exactly


Rules for significant figures in mathematical operations

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 operations1

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)


Summer review

Universe

Matter

Energy

Physical Change

Homogeneous

Heterogeneous

PotentialEnergy

KineticEnergy

Solution

Mixture

Pure Substance

Position

Composition

Chemical Change

Element

Compound

Gravitational

Electrostatic

Electron Levels

Nucleus

Protons

Neutrons

Electrons


Separation of mixtures

SEPARATION OF MIXTURES

  • - mixtures can be separated into pure substances by physical means.

    • distillation

    • filtration

    • centrifuging

    • magnet

    • evaporation

    • chromatography


Summer review

Simple laboratory distillation apparatus.


Summer review

CENTRIFUGE


Summer review

Paper Chromatography

Chromatography has two phases of matter: a stationary

phase (the paper) and a mobile phase ( the liquid).


Law of conservation of mass

Law of Conservation of Mass

  • Discovered by Antoine Lavoisier

  • Mass is neither created nor destroyed

  • Combustion involves oxygen, not phlogiston


Other fundamental chemical laws

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 laws1

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

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

The Mass and Charge of the Electron, Proton, and Neutron


The chemists shorthand atomic symbols

The Chemists’ Shorthand: Atomic Symbols

39

Mass number 

K

 Element Symbol

19

Atomic number 


The chemists shorthand formulas

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


Summer review

Ions

  • Cation: A positive ion

    Mg2+, NH4+

  • Anion: A negative ion

    Cl, SO42

  • 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 & Anions

  • Cations are positive ions.

  • Na ----> Na+ + e-

  • Anions are negative ions.

  • Cl2 + 2e- ----> 2Cl-


Periodic table

Periodic Table

  • Elements classified by:

  • Properties

    -atomic number

  • Groups (vertical)

    1A = alkali metals

    2A = alkaline earth metals

    7A = halogens

    8A = noble gases

  • Periods (horizontal)


Summer review

The periodic table.


Chemical symbols

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

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

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

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 symbols1

The Chemists’ Shorthand: Atomic Symbols

39

1+

Mass number 

 Ion charge

K

 Element Symbol

19

Atomic number 


Charges on common ions

Charges on Common Ions

-4

-3

-2

-1

+1

+2

+3


Common names

Common Names

  • sugar of lead

  • blue vitriol

  • quicklime

  • Epsom salts

  • milk of magnesia

  • gypsum

  • laughing gas

lead(II) acetate

copper(II) sulfate

calcium oxide

magnesium sulfate

magnesium hydroxide

calcium sulfate

dinitrogen monoxide


Naming compounds

Naming Compounds

1. Cation first, then anion

2. Monatomic cation = name of the element

  • Ca2+ = calciumion

    3. Monatomic anion = root + -ide

  • Cl = chloride

  • CaCl2 = calcium chloride

Binary Ionic Compounds:


Naming compounds continued

Naming Compounds(continued)

Binary Ionic Compounds (Type II):

-metal forms more than one cation

-use Roman numeral in name

  • PbCl2

  • Pb2+is cation

  • PbCl2 = lead (II) chloride

  • plumbous chloride


Naming compounds continued1

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 = diphosphoruspentoxide

Binary compounds (Type III):


Summer review

NOMENCLATURE OF COMPOUNDS

Binary -- 2 elements

Ternary -- (3 elements) - Ionic

(metal ion + polyatomic ion)

Type I - Ionic

(Type I metal + nonmetal)

Group I, II, Al+3, Ag1+,

Cd2+, & Zn2+

NaCl -- Sodium Chloride

Ca3(PO4)2 -- calcium phosphate

FeSO4 -- iron (II) sulfate

-- ferrous sulfate

Type II - Ionic

(Type II metal + nonmetal)

All other metals

Fe2S3 -- iron (III) sulfide

-- ferric sulfide

Type III - covalent

(2 nonmetals)

CO2 -- carbon dioxide


Chemical nomenclature

Chemical Nomenclature

  • Name each of the following:

  • CuCl

  • HgO

  • Fe2O3

  • MnO2

  • PbCl2

  • CrCl3

copper(I) chloride cuprous chloride

mercury(II) oxide mercuric oxide

iron(III) oxide ferric oxide

manganese(IV) oxide manganic oxide

lead(II) chloride plumbous chloride

chromium(III) chloride chromic chloride


Chemical nomenclature1

Chemical Nomenclature

  • Name each of the following:

  • P4O10

  • N2O5

  • Li2O2

  • Ti(NO3)4

  • SO3

  • SF6

  • O2F2

tetraphosphorusdecoxide

dinitrogenpentoxide

lithium peroxide

titanium(IV) nitrate

sulfur trioxide

sulfur hexafluoride

dioxygendifluoride


Common nomenclature mistakes

Compounds:

SO3 --Sulfur trioxide

NO2 -- Nitrogen dioxide

NO3 -- Nitrogen trioxide

Polyatomic ions:

SO32- -- Sulfite ion

NO21- -- Nitrite ion

NO31- -- Nitrate ion

Common Nomenclature Mistakes


Summer review

A flow chart for naming acids. An acid is best considered as

one or more H+ ions attached to an anion.


Binary acids

Binary Acids

  • made up of two elements -- hydrogen and a nonmetal

  • named by using:

  • prefix hydro + root of nonmetal + ic + acid

    HCl-- hydrochloric acid

    H2Se -- hydroselenic acid


Ternary acids oxyacids

Ternary Acids (oxyacids)

  • contain three elements -- hydrogen, nonmetal, and oxygen.

  • most oxygenper + root of nonmetal + ic + acid

  • less oxygenroot of nonmetal + ic + acid

  • less oxygenroot of nonmetal + ous + acid

  • least oxygenhypo + root of nonmetal + ous + acid


Ternary acids continued

Ternary Acids(continued)

  • HBrO4 perbromic acid

  • HBrO3 bromic acid

  • HBrO2 bromous acid

  • HBrOhypobromous acid

  • H3PO4 phosphoric acid

  • H3PO3 phosphorous acid

  • H3PO2 hypophosphorus acid


Salt nomenclature ionic compounds

Salt Nomenclature (Ionic compounds)

  • Binary salts (metal and nonmetal)

  • name of positive ion + root of nonmetal + ide

    NaCl -- sodium chloride

    K2S -- potassium sulfide


Salt nomenclature continued

Salt Nomenclature (continued)

  • Ternary salts ( metal and polyatomic ion)

  • name of positive ion + root of nonmetal + ate or ite

  • If the salt comes from anicacid, changeictoate.

  • H2CO3carbonicacid Na2CO3 sodiumcarbonate

  • H3PO4phosphoricacid K3PO4potassiumphosphate

  • If the salt comes from anousacid, changeoustoite.

  • H2SO3sulfurousacid Li2SO3 lithiumsulfite

  • HClOhypochlorousacid NaClO sodiumhypochlorite


Avogadro s number equals 6 022 10 23 units

Avogadro’s number equals6.022  1023 units


Calculating moles number of atoms

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

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 Equations

Chemical change involves a reorganization of the atoms in one or more substances.


Physical states

Physical States

  • solid (s)

  • liquid (l)

  • gas (g)

  • aqueous (aq)


Important equation symbols

Important Equation Symbols

  • yields ------>

    cat.

  • catalyst ------->

    H2SO4

  • catalyst ------>

  • heat ------->

  • light

  • light -------->

  • elect.

  • electricity ------>


Four steps in balancing equations

Four Steps in Balancing Equations

1. 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 caution

Balancing Equations Caution

The 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

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

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

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 gram1

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

Percent Composition

  • Mass percent of an element:

  • For iron in iron (III) oxide, (Fe2O3)


Composition

% 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

% 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

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

Empirical Formula – the lowest whole number proportion of a compound

Molecular Formula – the actual formula of a compound


Summer review

Example:

C2H4O2 – Molecular formula

CH2O – Empirical Formula


Steps to determine 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 Formula

Step 4: Write Empirical Formula


Summer review

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

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


Summer review

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

Limiting Reactant

The limiting reactant is the reactant that is consumed first,limiting the amounts of products formed.


Solving a stoichiometry problem

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

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 problem1

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


Summer review

  • 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 = % yield

      Theoretical 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

% Yield

  • Values calculated using stoichiometry are always theoretical yields!

  • Values determined experimentally in the laboratory are actual yields!


Limiting reactant yield

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 yield1

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 yield2

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?

% Yield = 52.1 %


Molarity calculations for stock solutions

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

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


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