Alkenes & Alkynes

A. Carbon can form double and triple bonds with carbon atoms.

 B. Naming rules are almost the same as with Alkanes (single bond chains)

                a. Double bonds end in "-ene"

                b. Triple bonds end in "-yne"

                c. When numbering the base chain, double/ triple bonds always get the lowest number

 ex. 

Name: 4 Methyl 2 Pentene

ex.

Name: 4, 4 Dimethyl 4 Heptyne

ex. 

Draw 2, 3, 4 Trimethyl 1 Octene

Trans & Cis Butene

A. If 2 adjacent carbons are bonded by a double bond AND have side chains on them there are 2 possible compounds. 

     a. Cis: The side chains are on the same side as the base chain

Cis 2 Butene

     b. Trans: The side chains are on opposite sides of the base chain

Trans 2 Butene

Multiple Double Bonds

A. More than 1 double bond can exist in a molecule

B. Use the same multipliers inside the base chain name

ex. 

 1,3 Butadiene

ex. Draw 2 methyl 1,4 pentadiene

Polar and Non-Polar Solvents Lab

Objective: To determine if glycerin is Polar or Non-polar

Materials:

·         Test tubes

·         Test tube stoppers

·         Test tube rack

·         Scupula

·         Safety goggles and apron

·         Sodium chloride

·         Sucrose

·         Iodine crystals

·         Paint thinner (Turpentine)

·         Glycerine

Part 1

Procedure:

1.       Set up 6 clean, dry test tubes. Divide the test tubes into 2 groups of 3. Label a piece of paper in front of each pair A, B, and C.

2.       Fill each test tube from the first set half way with water. Fill each test tube from the second set half full with paint thinner.

3.       In the first set, carefully add enough salt crystals to cover the bottom of the test tube.

4.       Stopper the test tube and carefully shake it several times. Examine the test tube and look for any crystals that may be stuck against the walls. Repeat this process with the first test tube in the second set.

5.       Repeat steps 3 and 4 with crystals of sugar in test tube B from each set.

6.       Repeat steps 3 and 4 with crystals of iodine in test tube C from each set.

7.       Dispose of the test tube solutions in the Waste Chemical container.

Observations:

Solvents	Solute	Solute	Solute
	Table Salt	Sugar	Iodine
Water	Dissolved	Dissolved	Did not dissolve
Paint thinner	Did not dissolve	Did not dissolve	Dissolved

Part 2

Procedure:

1.       Fill a clean test tube a quarter way full of water and then add twice as much paint thinner to the test tube.

2.       Stopper the test tube and shake it gently.

3.       Examine what has happened to the solutions.

4.       Using a second test tube, repeat steps 1 through 3 using Glycerin in place of paint thinner.

Analysis:

1) Based on the results from part 1, what conclusions can you draw about the types of solutes that can dissolve in polar and non-polar substances?

Polar substances dissolve polar solutes.

Non-polar solutes dissolve in non-polar substances.

2) Using the conclusion you made in question 1, predict whether or not Glycerin is polar or non-polar.

Glycerin is polar because it did not dissolve.

3) What would you predict to happen if you combine water, glycerin and paint thinner all in one test tube.

The glycerin will dissolve in the water and the paint thinner will float on top.

Organic Chemistry

I. Organic chemistry is the study of carbon compounds

                A. Carbon forms multiple covalent bonds

                B. Carbon compounds can form chains, rings, or branches

                                a. There are less than 100 000 non-organic compounds

                                b. organic compounds make up 400 000 compounds

                C. The simplest organic compounds are made of carbon and hydrogen

Expanded structural diagram

Condensed structural formula:

1 3-dichlorocyclopentane

III. Saturated compounds

                A. Have no double/ triple bonds

                B. Compounds with only single bonds are called alkalines and always end in "-ane"

IV.  Nomenclature

                A. There are 3 categories of organic compounds

                                                a. straight chains

                                                b. cyclic chains

                                                c. aromatics

                B. Straight chains

                                a. Rules for naming straight chain compounds

                                                1. Circle the longest continuous chain of carbons. This is your base chain.

       2. Name this base chain using the following chart and adding the ending "-ane"

Number of Carbon Atoms	Stem Name	Side Chain Name	Multiplier
1	Meth	Methyl
2	Eth	Ethyl	Di
3	Prop	Propyl	Tri
4	But	Butyl	Tetra
5	Pent	Pentyl	Penta
6	Hex	Hexyl	Hexa
7	Hept	Heptyl	Hepta
8	Oct	Octyl	Octa
9	Non	Nonyl	Nona
10	Dec	Decyl	Deca

In this case the name of the base chain is: Butane

      3. Number the base chain so that the side chains have the lowest possible number    

           4. Name the side chain using the appropriate stem name and the ending "-yl"

The side chain in this example is called: Methyl

          5. Give the side chain the appropriate number

The side chain in this example is now called: 2 Methyl

          6. If there is more than 1 side chain repeat steps 4 and 5 and list the side chains                                                                 in alphabetical order

THE COMPLETE NAME FOR THIS STRAIGHT CHAIN IS: 2 METHYL PROPANE

Types of Bonds

A. Intramolecular: bonds that exist within a molecule

                -ionic, covalent

B. Intermolecular: bonds that exist between molecules

                -the stronger the intermolecular bonds the higher the boiling point or melting point

                -2 types of intermolecular bonds: Van der  Waasl bonds and hydrogen bonds

I. Van der Waals Bonds

                A. Based on the electron distribution

                                a. 2 categories

                B. Dipole-dipole bonds

                                a. If a molecule is polar the positive end of one molecule will be attracted to the negative end of another molecule

II. London Depression Forces (LDF)

                A. present in all molecules

                B. creates weakest bonds

                C. If a substance is non-polar, dipole-dipole bonds do not exist, only LDFs

                D. Electrons are free to move around and will randomly be grouped on one side of the molecule

                E. This creates a temporary dipole and can cause a weak bond to form

                F. The more electrons in the molecule the stringer the LDF can be

IV. Hydrogen Bonding

                A. If hydrogen is bonded to certain elements (F, O or N) the bond is highly polar

                B. This forms a very strong intermolecular bond 

Polar Molecules

I. Polar Molecules
                A. Have an overall charge separation

                B. Unsymmetrical molecules are usually polar

                C. Molecular dipoles are the result of unequal sharing of electrons

II. Predicting Polarity

                A. If a molecule is symmetrical, the pull of an electron is usually unbalanced

                B. Molecules can be unsymmetrical in 2 ways

                                a. different atoms

                                b. different number of atoms

Bonding

I. Types of Bonds

                A. There are 3 main types of bonds

                                a. Ionic (metal and non-metal)

                                                i. electrons are transferred from metal to non-metal

                                b. covalent (non-metal and non-metal)

                                                i. electrons are shared between non-metals

                                c. metallic (metal and metal)

                                                i.holds pure metals together by electrostatic attraction

II. Electronegativity

                A. Electronegativity (en)* is a measure of an atom's attraction for electrons in a bond

                B. Atoms with a bigger en attract/want electrons more

                C. Polar covalent bonds form from an unequal sharing of electrons

                D. Non-polar covalent bonds form from unequal sharing of bonds

III. Bonds

                A. The type of bond formed can be predicted by looking at the difference in electronegativity of elements

                                a. en > 1.7 = ionic bond

                                b. en < 1.7 = polar covalent bond

                                c. en = 0 = non-polar covalent

bond

Ion Concentrations

I. Dissociation

                A. Ionic compounds are made up of 2 parts

                                a. Cation: positively charged particle

                                b. Anion: negatively charged particle

                B. When ionic compounds are dissolved in water the cation and anion separate from each other. This process is called dissociation

                C. When writing dissociation equation the atoms and charges must balance

                                a. The dissociation of sodium chloride is:

                                NaCl ---> Na⁺ + Cl^-

Ex. Complete the following dissociations

1. Fe(OH)₂ ---> Fe²⁺ + 2OH^-

2. Na₃PO₄ ---> 3Na⁺ + PO₄^3-

3. Fe₃(PO₄)₂ ---> 3Fe²⁺ + 2PO₄^3-

                        D. If the volume does not change then the concentration of individual ions depends on the balanced coefficients in the dissociation equation

Ex. Determine the [Na⁺] and [PO₄^3-] in a 1.5M solution on Na₃PO₄

Na₃PO₄ -->  3Na⁺ + PO₄^3-

1.5M x (1/1) = 1.5M [PO₄^3-]

1.5M x (3/1) = 4.5M [Na⁺]

Ex. Determine both ion concentrations when a 2.5M solution of Lithium sulphate dissociates.

Li₂SO₄ ---> 2Li⁺ + SO₄^2-

2.5M x (2/1) = 5.0M [Li^-]

2.5M x (1/1) = 2.5M [SO₄^2-]

Dilutions

I. Diluting Solutions

                A. When 2 solutions are mixed the concentration chenges

                B.  Dilution: process of decreasing the concentration of a solution by adding solvent (usually water)

                                a. The amount of solute doesn't change, only the volume changes

                                                i. v₁=v₂

                C. Because concentration is [mol/L] we can write:

                                a. C= n/v              or           n=C/v   so C₁v₁=C₂v₂

Ex. Determine the concentration when 100mL of 10.0M HCL is diluted to a final volume of 400mL

C₁= 0.1M

V₁=100mL

C₂=  ?

V₂= 400mL

C₁v₂=C₂v₂

C₁v₁ / v₂= C₂

(0.1M)(100mL) / (400mL) = 0.025 mol/ mL

Ex. How much water must be added to 10.0mL of 10.0M Na₂SO₄ to give a solution with a concentration of 0.50M?

C₁=10.0M

V₁= 10.0mL

C₂=0.50M

V₂=?

C₁V₁=C₂V₂

C₁V₁ / C₂= V₂

(10.0M)(10.0mL) / (0.5M) = 200mL

ΔV = V_f - V_i

ΔV = 200 - 10

ΔV = 190 mL of water needs to be added

Ex. How much water must be evaporated from 2.00L of 0.250M KCl solution for the final concentration to be 2.75M?

C₁=0.250M

V₁= 2.00L

C₂=2.75M

V₂=?

C₁V₁=C₂V₂

C₁V₁ / C₂= V₂

(0.250M)(2.00L) / (2.75M) = 0.182L

ΔV = V_f - V_i

ΔV = 0.182 - 2.00

ΔV = 0.182 - 200

ΔV= -1.82

Ex. [100 mL of 0.250M]_a sodium nitrate is mixed with [200mL of 0.100M]_b sodium nitrate. Find the concentration.

-Find the mol and volume of each solution separately

Solution a.  (0.100L) x (0.250mol / 1L) = 0.025 mol

Solution b. (0.200L) x (0.100mol / 1L) = 0.0200 mol

-Add together

mol_tot = 0.025mol + 0.0200mol = 0.0450mol

volume_tot= 100mL + 200mL = 300mL

(mol_tot) / (Volume_tot) = Molarity_tot

(0.0450mol) / (0.300L) = 0.150M

Ex. What mass of silver nitrate is needed to make 75.0mL of a 0.130M solution?

(0.130mol / 1L) x (0.075L) x (169.9g / 1mol) = 1.66g

Titrations

I. A titration is an experimental technique used to determine the concentration of an unknown solution

II. Terms & Equipment

                A. Buret - contains the known solution. Used to measure how much was added

                B.  Stopcock - valve used to control the flow of solution from the buret

                C. Pipet - used to accurately measure the volume of the unknown solution

                D. Erlenmeyer flask - container for unknown solution

                E. Indicator - Used to identify the end point of the titration

                F. Stock solution - known solution

Solution Stoichiometry

I.  SOLUTIONS

                A. Solutions are homogeneous mixtures composed of a solute and a solvent

                                a. Solute: The chemical present in lesser amount

                                                1. Whatever is being dissolved

                                b. Solvent: The chemical present in greater amount

                                                2. Whatever does the dissolving

                B. Chemicals dissolved in water are aqueous (aq)

II. MOLARITY

                A. Concentration can be expressed in many different ways

                                a. Ex. g/L, mL/L, % by volume, % by mass, mol/L

                                b. The most common is mol/L (M), which is also called molarity

                                                1. mol/L = M

                                                2. [HCl] = find the concentration of HCl

EX.  What is the concentration of 0.118 mol of water in 2.50 L?

0.11 x (1 mol/ 2.50 L) = 0.047M

EX. A solution of Pb(NO₃)₂ has a concentration of 0.330M. How many moles of solute are contained in 100 mL?

(0.330 mol/L) x (0.100L/ 1) = 0.0330 mol

EX. Determine the concentration of 12.5g of PbCl₂ in 30mL of water

12.5g x (1 mol/ 278.2g) x(1/ 0.030L) = 1.5M

EX. What volume of 12.7M  HNO₃ would be needed to produce  25.0g of HNO₃?

(1L/ 12.7 mol) x (25g/ 1) x (1 mol/ 63 g) = 0.031 L

-Nicole

Limiting Reactants

I. In chemical reactions, usually one chemical gets used up first.

                A. This is known as the limiting reactant

                B. Once the L.R. is used up, the reaction stops

                C. The L.R. determines the quantity of products formed

                D. To find the L.R. assume one reactant is used up. Determine how much of this reactant is used up

Ex.

2.5 mol of CuSO₄ reacts with 2.5 mol of NaNO₃. A) Determine the L.R. B)Calculate how many moles of NaNO₃ are formed.

CuSo₄ + 2NaNO₃ --> Cu(NO₃)₂ + Na₂SO₄

A) 2.5 mol x (2/1*) = 5.0 mol of NaNO₃

*What you need over what you have

Since you do not have 5.0 mol of NaNO₃ and you only have 2.5 mol of it, the L.R. is NaNo₃

B) 2.5 mol x (1/2) = 1.25 mol of NaNO₃

Ex.

Determine the L.R. when 1.4 mol of hydrogen reacts with 0.09 mol of oxygen.

2H₂ + O₂ --> 2H₂O

1.4 mol of hydrogen x (1/2) = 0.70 mol of O₂

Therefore the L.R. is H₂

Ex.

What is the L.R. when 125g of P₄ reacts with 320g of Cl₂ to form phosphorus trichloride?

P₄ + 6Cl₂ --> 4PCl₃

125g x (1 mol/ 124g*) x (6/1) x (71.0g/ 1 mol**) = 429g of Cl₂

* Molar mass of P

**Molar mass of Cl

Therefore Cl₂ is the L.R.

Ex. In the formation of water 50g of oxygen react with 60g of hydrogen. Determine the L.R. and theoretical yield.

2H₂ + O₂ --> 2H₂O

50g of O₂ x (1 mol/ 32g) x (2/1) x (2.0g/ 1 mol) = 6.25 of H₂

Therefore the L.R. is O₂

Energy and Percent Yield

I. Enthalpy is the energy stored in chemical bonds

                A. The symbol of enthalpy is H

                                a. units of joules (J)

                B. Change in enthalpy is ΔH

                C. In exothermic reactions enthalpy decreases

                D. In endothermic reactions enthalpy increases

II. Calorimetry

                A. To experimentally determine the heat released we need to know 3 things

                                a. temperature change (ΔT)

                                b.  mass (m)

                                c. specific heat capacity (C)

These are related by the equation:

ΔH = mCΔT

Calculate the heat required to warm a cup of 723 g of water (c=6.98J/g˚C) from 30.0˚C to 50.0˚C
C = 6.98J/g˚C
M = 723g
T = 20.0˚C

▲H = mC▲T
= (723g)(6.98 J/g˚C)(20.0˚C)
= 100930.8
= 1.01×10⁴

Ex. 9500 J of heat are added to a 778 g glass of water initially at 56.0˚C calculate the final temperature of the water (c = 6.21 J/g˚C)
▲H = 9500 J
m = 778 g
C = 6.21 J/g˚C
Ti = 56.0˚C

▲H = mC▲T
= mC (Tf-Ti)

95000J = (778g)(6.21J/g˚C)(Tf-56.0˚C)
95000J = (4831J/˚C)(Tf-56.0˚C)
95000J / 4831J/˚C = (4831J/˚C )(Tf-56.0˚C) / (4831J/˚C )
19.7˚C = Tf-56.0˚C
Tf = 75.7˚C

III. Percent Yield

                A.  The theoretical yield of a reaction is the amount of products that should be formed.

                B. The actual amount depends on the experiment

                c. The percent yield is like a measure of success

                                a. How close is the actual amount to the predicted amount?

Ex. The production of Octane (C₈H₁₈) is given by the equation:
C₈H₁₈ → 8C + 9H₂

If 38.0g of Octane are produced, determine the theoretical (predicted) yield of Carbon.

If 38.0g of Octane are produced, determine the theoretical (predicted) yield of Carbon.

 What is the percent yield of Carbon if 11.0 g is produced?
(11.0 g / 32 g) x 100 = 34%