# Homework

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1.1Units of Measurement

The Metric System - a “prefix-base unit” systems

 Measurement Metric Base Unit Length Meter       (   ) Volume (L) Mass (   ) Time Second     (   ) Temperature Celsius     (oC)

Scientific Notation -  a simplified expression of numbers, which contains a single digit number and 10 to an exponent

Procedure:

1.                 Leave one digit to the left of the decimal point and write “x 10" after the number.

2.       For numbers larger than one the power is positive, smaller than one it is negative .

3.       Count the number of places the decimal point moves, this is the power of the ten.

Examples:

25000                                               0.00125

1.2 Measured Numbers and Significant Figures

Exact Numbers - obtained by counting items or from a two units in the same system

Measured Numbers - there is always some uncertainty in every measurement

Significant Figures- an indication of the accuracy of a measurement

Procedure (for determining which Zeroes are Significant):

1) All zeros at the end of a number are placement zeros

e.g.: 2500 (____ placement zeros,  ____ S.F.)

2)  All zeros before a number are placement

e.g.: 0.00125 (____ placement zero, ____ S.F.)

3)  All zeros at the end of a number after the decimal are significant

e.g.:  0.0000430 (____ placement zero, ____ S.F.)

Example:

Write the following regular form numbers in scientific notation with proper significant figures:

4.50                      50                  1,250,000                        0.001050

Wayne McGowan

Chem 110G

1.3 Significant figures in Calculations

Multiplication and Division:

The answer cannot have more significant figures (sig. fig.s) than the least no. of sig. fig.s used in the calculation.

The answer should have the same number of decimal places as the measurement with the fewest decimal places.

1.4 Prefixes and Equalities

The special feature of the metric system is that a prefix can be attached to any base unit to increase or decrease its size by some factor of 10.

e.g.: a kilo anything is 1000 (103) anythings

1 kilometer (1 km) = 1000 meters (1000 m)

1 kiloliter (1 kL) = 1000 liters (1000 L)

1 kilogram (1 kg) = 1000 grams (1000 g)

Prefixes:

Prefix        Abbr          Numerical and Multiple

mega          ........ M                         1,000,000 = 1 x 106

kilo              ........ k                           1,000= 1 x 103

hecto           h.....                    100 = 1 x 102

deca             ........ da                         10 = 1 x 101

base unit     (no prefix)                      1

deci             ........ d                           0.1 = 1 x 10-1   = 1/101

centi            ........ c                           0.01 = 1 x 10-2   = 1/102

milli             ........ m                          0.001 = 1 x 10-3   = 1/103

micro            m.....                    0.000001 = 1 x 10-6   = 1/106

nano            ........ n                           0.000000001 = 1 x 10-9  = 1/109

1.5 Problem Solving Using Conversion Factors

Conversion Factors – the ratio of one unit to another

e.g.: 1 week = 7 days    Þ

Unit Analysis (Dimensional Analysis / the Factor-Label Method) -the method used to perform the majority of all of the calculations in this class

Plan on learning and mastering it as soon as possible in order to make future calculations much easier. If you have problems understanding this procedure see me during my office hours as soon as possible.

This method is extremely helpful for all problems involving units. That is to say, if the problem has units we can use dimensional analysis to set it up and solve it (almost all scientific questions, and "real life" questions have units).

No knowledge of equations is involved, and neither is knowledge of algebra.

Procedure:

1) Identify the unit(s) of your answer and write it (them) down on the far right side of your paper where your answer will appear.

2) Identify the piece(s) of information that you start with and write it (them) down on the far left of your paper.

The answer’s unit(s) will tell you what kind of unit(s) you need.  If you are starting with 2 separate units then the one on the top of the denominator line will lead you to the one on the top of your answers units, the one on the bottom leads you to the one on the bottom of your answers units.

3) Using available conversion factors cancel out of the beginning unit(s) and into a new unit that will be either the answers unit or a unit that gets you closer to the answers unit.

Most often there are several conversions that you will have to perform. Each conversion is done by putting the unit you want to convert out of on the opposite side of the denominator line and a unit that leads you to the desired unit on the same side of the denominator line.  Using this procedure you will "walk" toward your answer.

4) Put in the numbers and then multiply the numerators (tops) and divide the denominators (bottoms) to get the final answer.

NOTE:Once you are done with setting up the problem all the unwanted units should cancel out leaving only the desired (answers) unit(s); then all that is left to do is to punch in the numbers on your calculator to get the answer.

Examples:

How many seconds are in 10.0 years? (in this problem you are expected to know the conversions, if you wouldn't know them they would be given in the problem)

How many miles will you have driven if you drive 1.25 miles per hour for 0.75 hours?

How far can you drive on 50 dollars of gas? Gas costs 0.899 dollars per gallon, and your car gets 35 miles per gallon and holds 15 gallons of gas.

What is the maximum number of friends that can be fed pizza if each friend eats 5 slices and you only have 50 dollars? At the pizzeria that you selected each pizza costs 12 dollars and has 8 slices?

My puppy walked 106 yards in 1.00 minutes, how fast did he walk in miles per hour?

How many 24 ounce cans of tomato sauce do you need to make enchiladas for a party of 34 people? You plan for each person eating 3 enchiladas and the recipe makes 12 enchiladas from 16 ounces of sauce.

NOTE: When setting up a problem using unit analysis you should be thinking only about the units and NOT about the numbers. Sometimes it may be helpful to wait to put the numbers in after the set up is complete.

Metric Conversion

Procedure:

1.       write the unit to convert to on the far right

2.       write the number and unit to convert from on the far left

3a.     convert into the base unit write the equality to the base unit as a fraction with the base unit in the numerator and the given unit in the denominator

e.g.: to convert from mm to the base unit use

10-6

1 mm

3b.     convert from the base unit to the wanted unit write the equality to the base unit as a fraction with the base unit in the denominator and the wanted unit as the numerator

e.g.: to convert from the base unit to mm use

1 nm

10-9 m

4.       the equation is complete to compute quickly add all of the exponents in the numerator and subtract all of the exponents in the denominator

e.g.: convert 12,500 mm to nm

12,500 mm = 1.25x104 mm 10-6m   1 nm  = 1.25x10[4+(-6)-{-9}] =1.25x107nm

.......  1 mm   10-9m

Examples:

Convert 37,400 nanometers to micrometers.

Convert 0.0072 kilograms to milligrams.

Convert 0.0849 milliliters to microliters.

Convert 4,712 kilobytes to megabytes.

Wayne McGowan

Chem 110G

English System Conversions

English-English Equivalents:

Volume    ........                Length                            Mass      _

1 gal = ___   qt.... 1 mi = _____ ft              1 s.ton = _____ lb

= ___  pt         1 yd =  ____ ft                     1 lb      = ____ av.oz.

= ___ fl.oz.      1 ft  =  ____ in

Procedure:

1.       write the unit to convert to on the far right

2.       write the number and unit to convert from on the far left

3.       place the unit you are trying to get out of across the denominator line and

the answers unit or one that gets you closer to the answer in the numerator

Examples:

convert 1500 ft  into miles

1.                 .........                                                                     =                 miles

2.       1500 ft.......                                                                     =                 miles

3.       1500 ft              mile_                                                     =                 miles

ft

4.       1500 ft          1 mile_                                                      =  2.8x10-1  miles

5280 ft

How many inches are in 100 miles?

How many pints are in 32 fl.oz.?

English-Metric Intersystem Conversions

Metric-English Intersystem Equivalents:

Volume                               Length                                          Mass   _

1   L    = _____ qt                 1 m = _____  ft                       1   lb   =  _____ g

NOTE:It is best to only use the conversions given above. That way you get used to and remember only one for each measurement type, and the ones above are for the metric base unit, which you always change into first anyway.

Procedure:

1.       write the unit to convert to on the far right

2.       write the number to convert from on the far left

3a.     place the unit you are trying to get out of across the denominator line and the unit to be used in the intersystem conversion in the numerator

3b.     use the intersystem conversion factor to convert into the other system

3c.     convert from this unit to the wanted unit

4.       put the numbers in and calculate

Example:convert 12.59 kilograms to ave.oz.

1.                 .........                                                                    =               ave.oz.

2. 12.59 kg.........                                                                    =               ave.oz.

3. 12.59 kg        g                                                                             =               ave.oz.

kg

3. 12.59 kg       g       Lb                                                         =               ave.oz.

kg          g

3. 12.59 kg      g        Lb           ave.oz.                                   =                ave.oz.

kg           g          Lb

4. 12.59 kg103 g     1 Lb     16 ave.oz.                                    =  4.44x102 ave.oz.

1 kg     454 g       1 Lb

NOTE: The intersystem conversion terms have 3 significant figures.

How many pints are in 1725 mL?

How many centimeters are in 14 inches?

1.6 Density

Density is the ratio of a subject’s mass to its volume. Density units for the metric system are g/cc (cc = cubic centimeter) which is the standard for solids, liquids use g/mL and g/L are used for gasses.

To calculate the density of an object divide the mass by the volume and then convert by the normal method (dimensional analysis) into the units of choice.

Example:

127 lb of a rock sample displaces 18.5 qt of water, what is the rocks density?

What is the rocks density in g/ml ?

What is the density of a metal that displaces 1.17 L and has a mass of 16.82 kg?

43.2 Lb of a sample of sea water takes up 15.6 qt of space. What is the sample density?

Density as a conversion factor:

Any time you have a "dual unit" it can be used as a conversion factor.  The dual unit miles/hour will work to take you from hours to miles, or from miles to hours.  From density you have the conversion factor to take you from mass to volume, or from volume to mass.

e.g.  g/ml will take you from milliliters to grams, or from grams to milliliters

Example:

How many pounds does 2.50 quarts weigh if its density is 1.105 g/ml?

What volume in gallons does 1.24 av.oz of a gas occupy if it has a density of 1.055 g/L ?

How many kg does a 3.618 L sample of alcohol weigh if it has a density of 0.772 g/mL?

How many dL does a 162 mg sample of gas occupy if it has a density of 1.14 g/L?

1.7 Temperature

The temperature of an object tells us how hot or cold that object is.

Kelvin Temperature Scale

Absolute zero has the value of ___ Kelvin. The Kelvin scale has no negative numbers.

Temperature Conversions

oF to oC and oC to oF:

Fahrenheit has different size degrees from Celsius and the start point (0o) is different.

To convert from oC to oF first multiply by 180 oF/100 oC and then add 32 oF.

To convert from oF to oC first subtract 32 oF and then multiply by 100 oC/180 oF.

oC to K and K to oC

Centigrade/Celsius degrees are the same size as Kelvins, only 0 oC equals 273 Kelvins.

To convert from  oC  to K first multiply by 100 K/100 oC  then add 273.

To convert from K to oC  first subtract 273 and then multiply by 100 oC/100 K.

Examples:

34 oC =                .........                              =                 oF

55 oF =                 .........                              =                 oC

25 oC =                  ........                              =                 K       (read 298 Kelvins, no degrees)

315 K =                .........                              =                 oC      (read 42 degrees C)

What is the temperature in Celsius when it is 75 degrees Fahrenheit?

What is the temperature in Fahrenheit when it is 35 degrees Celsius?

What is the temperature in Kelvins when it is 35 degrees Celsius?

NOTE:  Keep in mind that only identical units can be added or subtracted.

Chem 110G

Wayne McGowan           HOMEWORK #1                                                                  ANSWER KEY

I.  Convert the following numbers into scientific notation.

1.  43,200                         4.32x104                                 2.  238,000                   2.38x105

3.  0.567                           5.67x10-1                                4.  0.00308                   3.08x10-3

II.  Write the following numbers into expanded form.

1.  2.40x101                        24.0                                       2.  2.8x10-3                   0.0028

3.  4.24x10-5                  0.0000424                                4.  7.923x105                792,300

VII.          Indicate the number of significant figures for each of the following numbers.

1.  0.00567                            3                                         2.  13,500                             3

3.  100.20                               5                                         4.  0.00100                          3

VIII.       Perform the following metric to metric conversion.  Express the final answer in scientific notation with proper significant figures.

1.  10.6 cm                                                                                                             = 1.06x10-1 m

2.  450,000 µg                                                                                                       =  4.5x10-1   g

3.  12 mL                                                                                                     =  1.2x100  cL

4.  50 Mton                                                                                                           =   5x107    ton

5.  4,600 nm                                                                                              =  4.6x100  µm

6.  0.058 kg                                                                                                   =  5.8x102  dg

7.  2460 µL                                                                                                  = 2.46x10-2 dL

8.  0.015 s                                                                                                               =   1.5x101   ms

9.  0.66 µm                                                                                                 =  6.6x10-4  mm

10.  0.062 g                                                                                                              =   6.2x100   cg

11.  12,500 cL                                                                                        =  2.65x102  pt

12.  61,500 nm                                                                                     =  2.42x10-3 in

13.  0.0520 kg                                                                                      =  1.83x100 oz

14.  9,000 µL                                                                                      =    3x10-1    oz

15.  1,989 mm                                                                                        = 2.17x100 yd

Chem 110G

Wayne McGowan           HOMEWORK #2                  Name ____________________________

1)   Write the symbols for the following elements:

a) oxygen -                                                     b) lithium -

c) sulfur -                                                        d) aluminum -

e) hydrogen -                                                            f) neon -

g) tin -                                                             h) gold -

2)   Write the name of the element for each symbol.

a) C -                        b) Cl -                            c) I -                      d) Hg -

e) F -                        f) Ar -                             g) Zn -                  h) Ni -

3)   Write the name of the element for each symbol.

a) He     -                  b) P -                              c) Na -                  d) Mg -

e) Ca -                                f) Br -                             g) Cd -                  h) Si -

4)   Is a proton, neutron, or electron described by the following?

a) has a mass about the same as a proton's

b) is found in the nucleus

c) is found in the largest part of the atom

d) carries a negative charge

5)   Write the names and symbols of the elements with the following atomic numbers:

a) 1 -                                  b) 11 -                                      c) 19 -

d) 26 -                                e) 35 -                                      f) 47 -

6)   Complete the following table for neutral atoms.

NAME           SYMBOL          A.N.                     M.N.          No.of P+    No.of N0    No.of e-

nitrogen                                                  15

20              22

88                                                   38

16             14

56            138

7)   What are the number of protons, neutrons and electrons in the following isotopes?

a.) 21H -                                             b.) 147N -

c.) 2614Si -                                              d.) 7030Zn -

8)   Write symbols for isotopes with the following:

Example: 2311Na

a.) an oxygen atom with 10 neutrons

b.) 4 protons and 5 neutrons

c.) 26 electrons and 30 neutrons

d.) a mass number of 24 and 13 neutrons

9)   Give the symbol of the element described by the following:

a.) an alkaline earth metal in Period 2

b.) Group 5A, Period 3

c.) a noble gas in Period 4

d.) a halogen in Period 5

e.) Group 4A, Period 4

10)Is each of the following a metal or nonmetal?

a.) in Group 2A

b.) good conductor of electricity

c.) chlorine

d.) iron

e.) is dull and brittle

11)Write the electron arrangements for each of the following atoms:

Example: sodium - 2,8,1

a)  phosphorus -                                   b) neon -                        c) 188O -

d)  an atom w/ A.N.18 -                        e) aluminum-                 f) silicon -

12)Identify the elements that have the following principal shell electron arrangements.

Energy Level            1                2                3

a.                                       2e-               5e-

b.                              2e-               8e-               6e-

c.                              2e-               4e-

d.                              2e-               8e-               8e-

e.                              2e-               8e-               3e-

13)The elements fluorine and chlorine are in the same group on the periodic table.

a.  Write the principal shell electron arrangements for F and Cl.

F (9) :                                     Cl (17):

b.  How many electrons are in each of their outer energy levels?

c.  What is their group number?

14)What is the number of electrons in the outer most shell and the group number for each of the following elements?

a.  magnesium                                      b.  chlorine

c.  oxygen                                            d.  nitrogen

15)What is the number of electrons in the outer most shell and the group number for each of the following elements?

a.  lithium                                            b.  silicon

c.  neon                                                d.  argon

16)    What would be the maximum number of electrons in the following?

a.  2p orbital                                        b.  2p subshell

c.  principal shell 2                              d.  3s orbital

17)Write the electron arrangement using subshell notation (1s2,2s2) for the following elements:

a.  magnesium

b.  phosphorus

c.  argon

d.  sulfur

e.  chlorine

18)Write the electron arrangement using subshell notation for the following elements:

a.  potassium

b.  sodium

c.  beryllium

d.  nitrogen

e.  carbon

19)    Identify the element that has the following electron arrangement:

a.  1s1

b.  1s2,2s2,2p3

c.  1s2,2s2,2p6,3s1

d.  1s2,2s2,2p6

20)Identify the element that has the following electron arrangement:

a.  1s2,2s2,2p2

b.  1s2,2s1

c.  1s2,2s2,2p6,3s2,3p4

d.  1s2,2s2,2p6,3s2,3p6,4s2

Wayne McGowan

Chem 110G

2.1  Elements and Symbols

Chemical Symbols - are one or two letter abbreviations for the names of the elements

The first letter of the symbol is capitalized; a second letter is lowercase.

Chemical Properties -  are characteristics that can be observed only when a chemical changes into a new chemical

e.g: decomposition

Physical Properties - are characteristics that can be observed or measured with out affecting the identity

e.g.: melting point, and boiling point

2.2 The Periodic Table

A Russian chemist, Dmitri Mendeleev, suggested that the elements could be arranged in groups or families according to their similar properties.  Today we call this arrangement of the elements the periodic table.

Groups - the vertical columns on the periodic table

Groups are sometimes called families as the elements in a group share similar chemical and physical properties.

Group 1A elements are the

Group 2A are the

Group 7A are the

Group 8A are the

Periods - the horizontal rows in the periodic table

Each row is counted from the top of the table down as Period 1, Period 2, and so on.

# Metals, Nonmetals, and Metalloids

Metals - those elements on the left of the zig-zag (except for hydrogen)

Metals are mostly shiny solids that can be shaped into wires (ductile) or hammered into a flat sheet (malleable) and are good conductors

Nonmetals - elements on the right of the zig-zag are dull and brittle

Metalloids - elements located along the zig-zag that exhibit some properties of the metals and the nonmetals

2.3 The Atom- is the smallest particle of an element that retains the characteristics of that element

## Atomic Theory:

1. All matter is made up of tiny particles called atoms.

2. All atoms of an element are similar to one another and different from atoms of other elements.

3. Atoms of two or more different elements combine to form compounds. Each compound is always made up of the same number and kinds of atoms.

4. A chemical reaction involves the rearrangement, separation, or combination of atoms. Atoms are never created or destroyed during a chemical reaction.

Atoms are the building blocks of everything we see around us; yet, we cannot see an atom or even a billion atoms with the naked eye.

## Parts of an Atom:

Atoms are composed of subatomic particles. There are three of interest to chemists, the proton, neutron, and electron.

The AMU – a mass unit defined as one-twelfth of the mass of the carbon

atom with 6 protons and 6 neutrons

## Structure of the Atom:

The dense nucleus, where the protons and neutrons are located, is extremely small compared to the electron orbits and contains most of the mass. Most of an atom is empty space, which is occupied only by fast moving electrons.

 Subatomic Particle Symbol Electrical Charge Approximate Mass Location in the atom Proton Neutron Electron 0.0005 amu outside nucleus

2.4 Atomic Number and Mass Number

Atomic Number -is the number of protons in the nucleus of an atom and is     used to identify each element

The number of protons is what determines which element it is (elements differ by their number of protons).

Usually an atom is electrically neutral. That means that the number of protons in an atom is equal to the number of electrons. This electrical balance gives an atom an overall charge of zero. Thus, in every atom, the atomic number also gives the number of electrons.

Mass Number - is the sum of the number of

2.5 Isotopes - are atoms of the same element that have a different number of neutrons

For example, all atoms of the element magnesium (Mg) have 12 protons. However, some magnesium atoms have 12 neutrons, others have 13 neutrons and still others have 14 neutrons.

Isotopic Symbol – shows the mass number as well as the atomic number for a particular isotope

eg: The isotopic symbol for an atom of Chlorine with 18 neutrons is:

Atomic Mass- the average mass of all the naturally occurring isotopes of that element

To determine the atomic mass of an element the masses and percent abundance of all of its isotopes must be known.

Most elements consist of several isotopes, and this is one reason atomic masses listed on the periodic table are seldom whole numbers.

Wayne McGowan

Chem 110G

2.6 Electron Energy Levels (e- shells)

In atoms e- exist only in the available e- shells. Generally the shells closest to the nucleus have the lowest energies.

The periodic table tells us how many e- per shell once we learn to use it. Shell 1 can hold up to ___ e-; shell 2 can hold up to ___ e-, shell 3 can take ___ e-, and shell 4 has room for ___ e-.

## The Principal Shell Electron Configuration:

### Element

Symbol

Atomic #

1

2

3

4

Nitrogen

Chlorine

Neon

The Quantum Mechanical Model - an e- in an atom has energy because of its motion and position(size and shape).

The shapes of e- orbitals correspond to the regions of space "filled" by the moving e-

Principal energy/shell - the different distances from the nucleus (the energy of position/size); related to the periods

Subshells (sublevels) -one or more energy states within the Principal shell (energy of motion/shape)

## Each orbital can hold a maximum of ___ electrons.

Electronic configuration (e- structure) - the e- arrangement of an atom; each element has a unique configuration made up of subshells distinguished by  principle numbers (the size of its orbital) and the letters s, p, d and f (the shapes of the orbitals).

e.g.: to designate the main energy levels (energy from size)for the s orbitals:
1  -   First energy level  - 1s2
2  -   Second energy level  - 2s2
3  -   Third energy level  - 3s2
4  -   Fourth energy level  - 4s2

Giving the main energy-level number followed by the letter corresponding to the subshell identifies a specific subshell.

Within the principle shell the subshell energies differ. The order of subshell energies is sublevel:

The sublevel structure of the first four principal shells is:
Principal shell        Subshells
First                     1s
Second
Third                    3s, 3p, 3d
Fourth

Each subshell consists of a certain number of electron orbitals.
An s sublevel has one orbital--the s orbital.
A p sublevel has three orbitals--the p orbitals.
A d sublevel has five orbitals--the d orbitals.
An f sublevel has seven orbitals--the f orbitals.

Orbitals within a given sublevel have the same electron energies.

Electron Spin:

Electrons have like charges and repel each other but they also have spin, which gives them magnetic properties and makes it possible for two e- to occupy an orbital. (thus a p subshell which has 3 orbitals can hold up to 3x2 = 6 e-)

Determining Electronconfigurations:

To determine which subshell an electron goes into use the periodic table. To do this we need to know only two things, where are these subshells in the periodic table and when we first encounter a subshell what is its principle quantum number.

The 1A and 2A groups make up the "s block", groups 3A through 8A make up the "p block", the transition metals make up the "d block" and the inner transition metals make up the "f block".

The first s orbital starts with the principle quantum number 1, the p with 2, d with 3 and f with 4 (1s, 2p, 3d, 4f).

Energy Overlap - a higher main energy level has lower energy than a sublevel of a

lower main energy level (e.g. 4s sublevel is of lower energy than the 3d sublevel)

The Electronic Configurations Of The Elements - written  1s2  2s2  2p6

The number on the left is the principal shell number and the letter is the subshell. The superscript is the number of electrons in that subshell.

Example:Give the electronic configuration for the following elements:

O   (8) =
P  (15) =
I   (53) =

## Pt (78) =

4.1 Valence Electrons - the e- in the outer shell, which determine the chemical properties

The group numbers are based on the number of valence e-. For main group elements the number of e- is equal to the group no. (groups 1A, 2A, 3A, 4A, 5A, 6A, 7A and 8A).

The Periodic Table and Electronic Configurations:

The elements having similar electronic configurations in their outermost energy levels are those with similar properties.

Parts Of The Periodic Table:

The elements are classified into four blocks, the s block elements, the p block elements, the d block elements and the f block elements. The s block is 2 elements wide, the p block  is 6 elements wide, the d block is 10 elements wide, and the f  block  is 14  elements wide.

Main group elements - elements that occupy the ___ and ___ blocks

Transition elements - ___ block elements

Inner transition elements / the rare earths - ___ block elements

Noble gases (Group VIIIA) - chemically inactive gases

The lack of chemical activity for the noble gases suggests that their electron configuration is quite stable. This stability tells us that having the outer shell filled is a desirable electronic configuration for an element.  We can now predict that each element bonds in such a way as to obtain an electronic configuration like its nearest noble gas neighbor.

Electron-Dot Structures - a convenient way to represent the valence e-

Valence e- are shown as dots placed on the 4 sides (the sides, top, and bottom) of the symbol for the element. Put a dot on each of four sides before doubling up.

e.g.:

Chem 110G

Wayne McGowan           HOMEWORK #2                                                                  ANSWER KEY

1)      Write the symbols for the following elements:

a) oxygen -O                 b) lithium - Li

c) sulfur - S                    d) aluminum -Al

e) hydrogen -H              f) neon - Ne

g) tin - Sn                       h) gold - Au

2)      Write the name of the element for each symbol.

a) C - carbon                 b) Cl - chlorine              c) I - iodine           d) Hg -mercury

e) F - fluorine                f) Ar - argon                  g) Zn - zinc           h) Ni - nickle

3)      Write the name of the element for each symbol.

a) He- helium               b) P - phosphorus                   c) Na - sodium     d) Mg magnesium

e) Ca - calcium              f) Br - bromine              g) Cd - cadmium  h) Si - silicon

4)      Is a proton, neutron, or electron described by the following?

a) has a mass about the same as a proton's                   neutron

b) is found in the nucleus                                               protons and neutrons

c) is found in the largest part of the atom            electrons

d) carries a negative charge                                  electrons

5)      Write the names and symbols of the elements with the following atomic numbers:

a) 1 - H, hydrogen                  b) 11 - Na, sodium                  c) 19 - K, potassium

d) 26 - Fe, iron              e) 35 - Br, bromine                 f) 47 - Ag, silver

6)      Complete the following table for neutral atoms.

NAME                   SYMBOL       A.N.                     M.N.          No.of P+    No.of N0    No.of e-

nitrogen                 N                 7                15                 7                8              7

calcium                  Ca              20               42                20              22             20

strontium               Sr               38               88                38              50             38

silicon                     Si                14               30                14              16             14

barium                   Ba               56             138               56              82             56

7)      What are the number of protons, neutrons and electrons in the following isotopes?

a.) 21H - 1 P+, 1 N0, 1 e-                     b.) 147N - 7 P+, 7 N0, 7 e-

c.) 2614Si - 14 P+, 12 N0, 14 e-            d.) 7030Zn -30 P+, 40 N0, 30 e-

8)      Write symbols for isotopes with the following:

Example: 2311Na

a.) an oxygen atom with 10 neutrons                                 188O

b.) 4 protons and 5 neutrons                                         94Be

c.) 26 electrons and 30 neutrons                                            5626Fe

d.) a mass number of 24 and 13 neutrons    2411Na

9)      Give the symbol of the element described by the following:

a.) an alkaline earth metal in Period 2                  Be

b.) Group 5A, Period 3                               P

c.) a noble gas in Period 4                           Kr

d.) a halogen in Period 5                                      I

e.) Group 4A, Period 4                               Ge

10)    Is each of the following a metal or nonmetal?

a.) in Group 2A                                          metal

b.) good conductor of electricity                 metal

c.) chlorine                                                  nonmetal

d.) iron                                                        metal

e.) is dull and brittle                                    nonmetal

11)    Write the electron arrangements for each of the following atoms:

Example: sodium - 2,8,1

a)  phosphorus - 2,8,5                      b) neon - 2,8                            c) 188O - 2,6

d)  an atom w/ A.N.18 - 2,8,8           e) aluminum-2,8,3                   f) silicon - 2,8,4

12)    Identify the elements that have the following principal shell electron arrangements.

Energy Level                  1                 2                3

a.                                    2e-               5e-                                  N - nitrogen

b.                                   2e-               8e-               6e-               S - sulfur

c.                                    2e-               4e-                                  C - carbon

d.                                   2e-               8e-               8e-               Ar- argon

e.                                    2e-               8e-               3e-               Al- aluminum

13)    The elements fluorine and chlorine are in the same group on the periodic table.

a.  Write the principal shell electron arrangements for F and Cl.

F (9) :2, 7                                Cl (17): 2, 8, 7

b.  How many electrons are in each of their outer energy levels?      7

c.  What is their group number?                                                       7

14)    What is the number of electrons in the outer most shell and the group number for each of the following elements?

a.  magnesium                2, 2A          b.  chlorine           7, 7A

c.  oxygen                      6, 6A          d.  nitrogen           5, 5A

15)    What is the number of electrons in the outer most shell and the group number for each of the following elements?

a.  lithium                      1, 1A          b.  silicon              4, 4A

c.  neon                          8, 8A          d.  argon               8, 8A

16)    What would be the maximum number of electrons in the following?

a.  2p orbital                            2                 b.  2p subshell      6

c.  principal shell 2                  8                 d.  3s orbital         2

17)    Write the electron arrangement using subshell notation (1s2,2s2) for the following elements:

a.  magnesium      1s2,2s2,2p6,3s2

b.  phosphorus     1s2,2s2,2p6,3s2,3p3

c.  argon               1s2,2s2,2p6,3s2,3p6

d.  sulfur               1s2,2s2,2p6,3s2,3p4

e.  chlorine            1s2,2s2,2p6,3s2,3p5

18)    Write the electron arrangement using subshell notation for the following elements:

a.  potassium                  1s2,2s2,2p6,3s2,3p6,4s1

b.  sodium            1s2,2s2,2p6,3s1

c.  beryllium                   1s2,2s2

d.  nitrogen           1s2,2s2,2p3

e.  carbon             1s2,2s2,2p2

19)    Identify the element that has the following electron arrangement:

a.  1s1                                                H - hydrogen

b.  1s2,2s2,2p3                          N - nitrogen

c.  1s2,2s2,2p6,3s1                     Na- sodium

d.  1s2,2s2,2p6                          Ne- neon

20)    Identify the element that has the following electron arrangement:

a.  1s2,2s2,2p2                          C - carbon

b.  1s2,2s1                                Li- lithium

c.  1s2,2s2,2p6,3s2,3p4              S - sulfur

d.  1s2,2s2,2p6,3s2,3p6,4s2                  Ca- calcium

Chem 110G

Wayne McGowan           HOMEWORK #3                  Name ____________________________

1)      Give a statement of the octet rule.

2)      Describe the ionic bond.  How does it differ from a covalent bond?

3)      What is an anion?  What is a cation?

4)      Give the formulas of the simple cations formed by the following elements.

a) H                      b)K                       c) Be                     d) Sr
e) Ga                     f) Fr                      g) Ca                     h) Rb

5)      Describe the kinds of simple ions formed by elements of the following groups.

a) group 1A                    b) group 2A                   c) group 3A
d) group 5A                   e) group 6A                   f) group 7A

6)      Give the formulas of the ionic compounds containing the following ions.

a)  Na+ and S2-                                   b)  Li+ and Br-

c)  Ca2+ and O2-                                 d)  Mg2+ and F-

e)  Al3+ and O2-                                               f)  Al3+ and Br-

7)      Give the formulas for the ionic compounds formed between the following ions.

a)  Na+ and SO42-                               b)  Al3+ and SO42-
c)  Ba2+ and NO3-                               d)  K+ and ClO3-
e)  Na+ and C2O42-                           f)  Cu+ and CO32-

8)      The formula for the ionic compound calcium phosphate is Ca3(PO4)2.  Explain why the parentheses are needed in this formula.

9)      Explain how simple positive and simple negative ions are named.

10)    Give the names of the following ions.

a)  Al3+                                                        b)  Br-
c)  CO32-                                                     d)  Ag+
e)  S2-                                                                         f)  Cl-
g)  CrO42-                                                      h) Cu+
i)   Ni2+                                                        j)  PO43-
k)  Cr2O72-                                                    l)  Fe2+
m) HSO4-                                                               n)  MnO4-
o)  HCO3-                                                   p) OH-
q)  Mg2+                                                      r)  NH4+
s)  Ca2+                                                       t)  K+

11)    Give the formulas for the following ions.

a)  hydrogen sulfate ion                                        b)  copper (II) ion
c)  chlorate ion                                            d)  fluoride ion
e)  nitrate ion                                              f)  sulfite ion
g)  iron (III) ion                                                    h)  acetate ion
i)  oxalate ion                                              j)  cyanide ion
k)  chlorite ion                                            l)  chromium (III)

## 12)    Name each of the following.

a) FePO4                                                      b) ZnC2O4
c) AgNO3                                                    d) SeF4
e) NH3                                                       f) Fe2O3
g) PbO2                                                       h) HgS
i)  KBr                                                        j) Ba3N2
k) Fe(OH)3                                                   l) Hg2I2

13)    Give the electron dot symbols for these elements

a) Be                 b) Ne                 c) H                       d) S                 e) Al             f) N

g) Te                       h) Rb                 i)  I                     j) Xe              k) C

14)    Give electron dot structures for the following.

a)  dichlorine, Cl2

b)  hydrogen iodide, HI

c)  water, H2O

d)  dihydrogen telluride, H2Te

e)  dichlorine oxide, Cl2O

f)  hydrazine,  N2H4

g)  phosphorus trichloride, PCl3

h)  boron trifluoride, BF3

i)  carbon tetrachloride, CCl4

j)  acetylene, a fuel, C2H2

k)  carbon dioxide CO2

l)  hydrogen cyanide, HCN

15)    Give electron dot structures and describe the shapes of the following.

a)  CCl2F2, dichlorodifluoromethane

b)  NF3, nitrogen trifluoride

c)  Cl2O, dichlorine oxide

d)  CF4, carbon tetrafluoride

e)  H2Se, dihydrogen selenide

f)  SiH4, silicon tetrahydride

16)    Define electronegativity.

17)    List the four most electronegative elements in descending order.

18)    What is a polar covalent bond?

19)    Describe a polar molecule and give an example.

20)    Are molecules of the following compounds polar or nonpolar?

a)  HF             b)  CS2         c)  CH4             d)  NCl3      e)  CH3Cl

Wayne McGowan

Chem 110G

4.2 The Octet Rule (Noble Gas Rule) and Ions

All elements want the electronic configuration of the nearest __________________ (atoms lose, gain, or share e- to get eight valence e-).

Bonding number - the number of e- lost or gained by an element (to attain the e- configuration of the nearest noble gas)

## Positive Ions (cations):

When metals react they lose their ____________________ in order to attain an e- arrangement like the prior noble gas.

## Negative Ions(anions):

Nonmetals form negative ions when they gain one or more e- to complete their valence shells.

## Ionic Charges from Group Numbers:

The elements in Groups 1A lose 1 e- for a 1+ charge, 2A lose 2 e- for a 2+, and 3A 3 e- for a 3+. Group 5A nonmetals gain 3 e- for a 3- charge, Group 6A gain 2 e- for a 2-, and Group 7A 1 e- for a 1-.

Formulas For Ions - represented by the symbol with the charge as a superscript

e.g.:

4.3 Ionic Compounds - consist of positive and negative ions with strong attractions between the oppositely charged ions (ionic bonds)

The e- lost by one ion are gained by another. This allows both elements to get an octet of e-.

## Charge Balance in Ionic Compounds -the total amount of positive charge is always equal to the total amount of negative charge

The formula of an ionic compound reflects the charge balance.

Determining Formulas of Compounds:

For ionic compounds the charge balance is the basis of determining the formula of the compound formed between any positive and negative ions.

Procedure:

1)      Identify each ion’s Bonding Number (the charge for ions) and write it above the elemental symbol.

2)      Determine the least common multiple of the two numbers and write this above the B.N.s (this is the total number of electrons exchanged).

3)      Divide each ions bonding number into the least common multiple, this is the number of each ion in the compound.

Examples:  What is the expected formula of the compound formed between:

sodium and chlorine             sodium loses 1 electron to be like Ne so it has a B.N. of 1
1                                     chlorine needs 1 electron to be like Ar so it has a B.N. of 1
1          1
Na + Cl Þ  NaCl                 1/1 = 1 sodium; 1/1 = 1 chlorine

magnesium and chlorine      magnesium loses 2 electrons to be like Ne, so it has a B.N. of 2
2                                     chlorine needs 1 electrons to be like Ar, so it has a B.N. of 1
2            1
Mg + Cl  Þ  MgCl2              2/2 = 1 magnesium; 2/1 = 2 chlorine

aluminum and sulfur

Pb+4 and O-2

4.4     Ion Names:

Multiple Charge Metals:

When two different ions are possible for a metal, a Roman numeral of the charge is placed in parentheses after the metal name.

e.g.: Fe2+ is                                        and Fe3+ is

Single Charge Metals:

Metal ions of Group 1A, 2A, Al, Zn, Ag, and Cd do not need Roman numerals.

e.g.: Mg2+ is                                       and  Li+ is

Nonmetals:

Negative nonmetal ions are named by using ide at the end of their element names.

e.g.: Cl- is                                 and O2- is

Example: Name the following ions:

S2-                         Ba2+                    Br-                        K+                         Cr+3                     P3-                         Sn2+

4.4 Naming Ionic Compounds – the cation name is followed by the anion name

Example:  Name the following:

K2O

MgCl2

Al2O3

CuCl2

Cu charge  +  2 Cl- charge      =  0

(?)               +  2(1-)                 =  0    Þ   Copper(II) chloride

(2+)             +  2-                     =  0

Fe2O3

4.5 Polyatomic Ions - a group of atoms with an electrical charge

Most polyatomic ions consist of a nonmetal bonded to oxygen atoms (oxoanions).

## Naming Polyatomic Ions:

The names of the most common polyatomic ions end in ate. An ite ending is used for the related ions that have one less oxygen atom.

e.g.:   CO3-2

NO3-                                                                   NO2-

PO4-3                                                                  PO3-3

SO4-2                                                                  SO3-2

More on Naming Oxoanions

Some nonmetals form more than one and oxoanion, so they are distinguished by use of different suffixes in their names. The different oxoanions of a given element differ in the number of combined oxygen atoms. When there are two oxoanions for an element, the one with the greater number of oxygen atoms has the -ate suffix and the one with the lesser number of oxygen atoms has the -ite suffix.

The easiest way to contend with all of the names of the oxoanions is to remember the formula for just one of the names, the one with the -ate ending, and then if there are more or less oxygens adjust the name accordingly.

example:

# of oxygens relative
ion                  to the -ate suffix                      name of oxoanion
ClO2-                one less                                    chlorite ion
ClO3-                has -ate suffix                          chlorate ion

To determine how many oxygens get the -ate suffix for an element we find the placement of the non-oxygen nonmetal in the periodic table. For elements in period 2 and group 7 the -ate suffix has 3 oxygens, for all of the rest it has 4 oxygens.

examples:
SO32- is the sulfite ion; sulfur is not in period 2 or group 7 so the -ate suffix has 4 oxygens, 3 is less than 4 so it gets the -ite suffix

PO33- is the phosphite ion; phosphorus is not in period 2 or group 7 so the -ate suffix has 4 oxygens, 3 is less than 4 so it gets the -ite suffix

NO2 is the nitrite ion; nitrogen is in period 2 so the -ate suffix has 3 oxygens, 2 is less than 3 so it gets the -ite suffix

Determining The Charge on Oxoanions

The negative charge on an oxoanion is equal to the non-oxygen nonmetal's bonding number for all nonmetals except carbon and nitrogen which have their bonding number minus two:

examples:
SO3-2 has 2 negatives; S has a bonding number of 2 and it is not carbon or nitrogen
ClO2-1 has 1 negative; Cl has a bonding number of 1 and it is not carbon or nitrogen
NO3-1has 1 negative; N has a bonding number of 3 and it is carbon or nitrogen, so 3-2=1

Other common Polyatomic Ions:

CrO42-         the chromate ion
Cr2O72-        the dichromate ion
MnO4-         the permanganate ion
C2O42-         the oxalate ion
C2H3O2-       the acetate ion
CN-             the cyanide ion

OH-             the hydroxide ion
NH4+           the ammonium ion  (the only common polyatomic cation)

## Writing Formulas for Compounds Containing Polyatomic Ions:

When more than one polyatomic ion is needed for charge balance parentheses are used to enclose the formula of the ion.

e.g.: What is the formula of the ionic compound containing the following:

magnesium and nitrate?

(Magnesium ion  Nitrate ion)

Mg2+           +           NO3-

To balance the positive charge of 2+, two nitrate ions are needed. The formula, including the parentheses around the nitrate ion, is as follows:

(2+)  +  2(1-)  =  0           Mg(NO3)2

Example:

Give the formulas of the compounds containing the following ions:

The barium ion and the nitride ion?

The cobalt(III) ion and the sulfate ion?

The lead(IV) ion and the oxide ion?

The gold(I) ion and the phosphate ion?

Name the following:

Ag2S

K3PO3

Pb3N4

Mn(NO3)2
WayneMcGowan

Chem 110G

4.6 Covalent Compounds (Molecules) - when atoms of two or more nonmetals share e-

The number of e- that an atom shares, and the number of covalent bonds it forms are usually equal to the Bonding Number.

For example, carbon has 4 valence electrons. Because carbon needs to acquire 4 more electrons for an octet, it forms 4 covalent bonds by sharing its 4 valence electrons.

The covalent bond containing a shared pair of e- is written as a line between the two atoms.

e.g.: H-H

Multiple Bonds - occurs when two or more pairs of e- are shared between atoms

## e.g.: O=C=O, H-CºC-H

Naming Covalent Compounds

The first nonmetal is named by its elemental name; the second nonmetal is named by its elemental name with the ending changed to “ide”. For two or more atoms of an element prefixes are placed in front of the name.

Prefixes:

mono      -        1-one (usually omitted)
-        2-two
tri            -        3-three
-          4-four
penta      -        5-five
-        6-six
hepta      -        7-seven
-         8-eight
nona       -        9-nine
-         10-ten

4.7 Bond Polarity

Electronegativity - the ability of an atom to attract additional electrons

The electronegativity of elements increase going across a period from left to right and within each group from bottom to top (the smallest elements have the greater electronegativities).

When atoms of different nonmetals share electrons, the more electronegative element has a stronger attraction for the e-. Then e- are shared unequally, resulting in a polar covalent bond.

This separation of charge, called a dipole, is represented by the symbols d+ and d- or as an arrow with a plus sign at the positive end (+®).

For example, in HCl the shared electron pair is pulled closer to the more electronegative chlorine atom.

e.g.:

4.8  Shapes and Polarity of Molecules

To determine the shape of a molecule we use the valence-shell electron-pair repulsion theory (VSEPR).

Most nonmetals have 4 regions of electron density around them (bonding and nonbonding e-) giving them the following shapes:

AX –

AX2 – bent/angular

AX3 –

AX4 – tetrahedral

Other Shapes:

## Polarity of Molecules

In a polar molecule one end of the molecule is more negatively charged than another end. Polarity in a molecule occurs when the polar bonds do not cancel each other. Dipoles in a molecule that cancel each other make the molecule nonpolar. For example, CO2 and CCl4 contain polar bonds; however, the symmetry of the polar bonds cancels the dipoles, leaving them nonpolar.

Often there are lone pairs around the central atom making the molecule nonsymmetrical. In H2O the dipoles do not cancel, which makes the molecule positive at one end and negative at the other end.

Determining Polar Molecules:

A polar molecule must satisfy two criteria.

1)  It must have one or more polar bonds.

2) It must have a shape that results in a net separation of positive and negative charge centers (nonsymmetrical).

NOTE:An easy way to determine if it is symmetric is to determine if it has only one type of  attachment,  with elements and nonbonding pairs being attachments. If there is only one type of attachment it is symmetric and thus nonpolar.

e.g.: water is a polar molecule (polar bonds and lack symmetry)

Example: Determine if the following are polar or nonpolar molecules.

I2               CS2                     NCl3                  CF4                  H2S                  CCl2F2

Chem 110G

Wayne McGowan           HOMEWORK #3                                                                  ANSWER KEY

1)      Give a statement of the octet rule.

The octet rule states that all elements want the electronic configuration of the nearest noble gas (having eight outer electrons).  This configuration can be achieved by some by gaining electrons and by others by losing electrons.

2)      Describe the ionic bond.  How does it differ from a covalent bond?

The ionic bond is a force of attraction due to oppositely charged particles (eg.Na+Cl-). The covalent bond is a force of attraction due to the sharing of a pair of electrons and differs from the ionic bond by forming discrete partners/pairs, which are molecules.

3)      What is an anion?  What is a cation?

An anion is a negatively charged ion (due to gaining additional electrons) and a cation is a positively charged ion (due to losing some of its electrons).

4)      Give the formulas of the simple cations formed by the following elements.

a) H     H+             b)K      K+            c) Be     Be2+         d) Sr     Sr2+
e) Ga   Ga3+          f) Fr     Fr+           g) Ca    Ca2+         h) Rb    Rb+

5)      Describe the kinds of simple ions formed by elements of the following groups.

a) group 1A +1              b) group 2A  +2             c) group 3A  +3
d) group 5A  -3                       e) group 6A  -2                        f) group 7A -1

6)      Give the formulas of the ionic compounds containing the following ions.

a)  Na+ and S2-      Na2S                     b)  Li+ and Br-           LiBr

c)  Ca2+ and O2-    CaO                     d)  Mg2+ and F-     MgF2

e)  Al3+ and O2-     Al2O3                                 f)  Al3+ and Br-     AlBr3

7)      Give the formulas for the ionic compounds formed between the following ions.

a)  Na+ and SO42-   Na2SO4                b)  Al3+ and SO42- Al2(SO4)3
c)  Ba2+ and NO3-   Ba(NO3)2              d)  K+ and ClO3-     KClO3
e)  Na+ and C2O42-Na2C2O4               f)  Cu+ and CO32-   Cu2CO3

8)      The formula for the ionic compound calcium phosphate is Ca3(PO4)2.  Explain why the parentheses are needed in this formula.

The parenthesis make clear that there are two phosphate ions in the compound; writing the formula Ca3P2O8 would not easily reveal the presence of the phosphate.

9)      Explain how simple positive and simple negative ions are named.

Simple positive ions are named as the element plus ion; if it is a multi-charge element then indicate the charge as a roman numeral following the name; eg. Mg2+ is magnesium ion, Fe2+ is iron II ion. Simple negative ions are named by using the root name of the element plus "ide";   eg. oxide ion, chloride ion.

10)    Give the names of the following ions.

a)  Al3+        aluminum                               b)  Br-         bromide
c)  CO32-      carbonate                                d)  Ag+        silver
e)  S2-                 sulfide                                     f)  Cl-          chloride
g)  CrO42-       chromate                                 h) Cu+         copper I
i)   Ni2+        nickle II                                   j)  PO43-           phosphate
k)  Cr2O72-     dichromate                             l)  Fe2+         iron II
m) HSO4-       hydrogen sulfate                    n) MnO4-        permanganate
o)  HCO3-   hydrogen carbonate               p) OH-              hydroxide
q)  Mg2+      magnesium                             r)  NH4+      ammonium
s)  Ca2+       calcium                                   t)  K+                    potassium

11)    Give the formulas for the following ions.

a)  hydrogen sulfate ion           HSO4-                   b)  copper (II) ion  Cu2+
c)  chlorate ion               ClO3-                    d)  fluoride ion     F-
e)  nitrate ion                 NO3-                     f)  sulfite ion         SO32-
g)  iron (III) ion              Fe3+                      h)  acetate ion       C2H3O2-
i)   oxalate ion                C2O42-                   j)  cyanide ion       CN-
k)  chlorite ion               ClO2-                    l)  chromium (III)           Cr3+

## 12)    Classify and name each of the following. C - covalent, I - ionic

a) FePO4               I - iron III phosphate             b) ZnC2O4    I - zinc oxalate
c) AgNO3              I - silver nitrate             d) SeF4        C- selenium tetrafluoride
e) NH3                  C - ammonia                f) H2SO3     C- sulfurous acid
g) PbO2                 I - lead IV oxide            h) HgS                  I - mercury II sulfide
i) HClO2               C - chlorous acid           j) Ba3N2           I - barium nitride
k) Fe(OH)3            I - iron III hydroxide    l) Hg2I2       I - mercury I iodide
m) KBr                 I - potasium bromide    n) Fe2O3     I - iron III oxide

13)    Give the electron dot symbols for these elements

. .                                           .                                    .                                 .
a) Be   Be .     b) Ne   :Ne :      c) H    . H       d) S   . S :     e) Al   . Al .    f) N   : N .
. .                                          . .                                          .

.                                                                             . .                                 . .                                .
g) Te   . Te :     h) Rb    Rb .       i)  I    : I  .      j) Xe  : Xe k) C   . C .
. .                                                                           . .                                 . .                               .

14)    Give electron dot structures for the following.

a)  dichlorine, Cl2                           Cl-Cl

b)  hydrogen iodide, HI                                             H-I

c)  water, H2O                                                         H-O
H
d)  dihydrogen telluride, H2Te        H-Te
H

e)  dichlorine oxide, Cl2O                                         Cl-O
Cl
f)  hydrazine,  H2H4                      H-N=N-H
Cl
g)  phosphorus trichloride, PCl3                                              Cl-P-Cl
F
h)  boron trifluoride, BF3                  F-B-F
Cl
i)  carbon tetrachloride, CCl4                                                                     Cl-C-Cl
Cl
j)  acetylene, a fuel, C2H2         H-C ºC-H

k)  carbon dioxide CO2                                                O=C=O

l)  hydrogen cyanide, HCN        H-C ºN

15)    Give electron dot structures and describe the shapes of the following.

F
a)  CCl2F2, dichlorodifluoromethane                     Cl-C-F          tetrahedral
F                         F
b)  NF3, nitrogen trifluoride         F-N-F     trigonal pyramid

c)  Cl2O, dichlorine oxide                                       Cl-O           bent/angular
F                         Cl
d)  CF4, carbon tetrafluoride         F-C-F     tetrahedral
F
e)  H2Se, dihydrogen selenide                                 H-Se           bent/angular
H
f)  SiH4, silicon tetrahydride       H-Si-H     tetrahedral
H

16)    Define electronegativity.

Electronegativity is the tendency of an element to attract additional electrons.

17)    List the four most electronegative elements in descending order.

F - fluorine, O - oxygen, Cl - chlorine, N - nitrogen

18)    What is a polar covalent bond?

A polar covalent bond is a covalent bond between two different elements, thus, one element attracts the shared electron pair closer to itself than the other.

19)    Describe a polar molecule and give an example.

Polar molecules contain polar covalent bonds and have no symmetry to cancel the over-all polarity.  eg. H2O, NH3, HCl, CCl2F2

20)    Are molecules of the following compounds polar or nonpolar?

a)  HF polar  b)  CS2 non-polar  c)  CH4 non-polar d)  NCl3polar  e)  CH3Cl polar

Chem 110G

Wayne McGowan           HOMEWORK #4                  Name ____________________________

1)      Balance the following equations.

a.     Zn +   O2 ®   ZnO                     b.     Cr +   O2 ®   Cr2O3
c.     KClO3 ®   KCl +   O2               d.     Li +   O2 ®   Li2O
e.     B +   F2 ®   BF3                        f.      C3H6 +   O2 ®   CO2 +    H2O
g.     Hg +   N2 ®   Hg3N2                 h.     N2 +   H2 ®   NH3
i.      WO3 +   H2 ®   W +   H2O       j.      C4H10 +   O2 ®   CO2 +   H2O
k.     Ca +   NH3 ®   CaH2 +   Ca3N2          l.      KNO3 +   C ®   K2CO3 +   CO +   N2

2)                If you had a mole of pennies and divided them among all of the 6 billion people on earth, how many dollars would each person have?

3)      Define or explain these terms.

a)  mole of an element

b)  molar mass

4)      Calculate the number of moles of atoms contained in each sample given below:

a)     24.8 g silver

b) 53.5 kg iron

c) 6.71 g sulfur

d) 0.0699 g of W

e) 131 mg silicon

5)      Calculate the mass of the following samples of elements:

a)     84 mol K

b)    7.60x10-3mol U

c)     0.0395 mol Li

d)    2.25x106 mol Ba

e)     4.69 mol B

6)      Calculate the number of atoms in each sample given below:

a) 3.18 g Cu

b) 0.171 g W

c) 12.5 g He

d) 453.6 g iron

e) 0.200 g carbon

7)      Determine the molar mass of the following compounds to four significant digits:

a)  isopropyl alcohol, C3H8O (rubbing alcohol)

b)  barium sulfate, BaSO4 (used for X rays)

c)  sodium nitrate, NaNO3 (found in the mineral Chile)

d)  potassium dichromate, K2Cr2O7

e)  titanium(IV) chloride (used in smokescreens), TiCl4

8)      Determine the number of moles of formula units in the following samples of compounds:

a)     311 g NaNO2

b)  75.0 g C16H30O

c)     2.88 x 105 g SO3

d)    2.24 x 10-7 g Au2S3

e)     9.10 g H2O2

9)      Acetylene, C2H2, burns in oxygen according to the following unbalanced equation:
C2H2(g) +   O2(g) ®   CO2(g) +   H2O(g)
Balance the equation.  Give the molar interpretation of the equation and give the molar ratios that relate the following pairs of reactants and products.

a.     C2H2 and O2

b. C2H2 and CO2

c. C2H2 and H2O

d. O2 and CO2

10)    Refer to the balanced equation in the previous question.  Find the number of moles of:

a. O2, to react with 0.570 mole of C2H2

b. C2H2, to produce 0.400 mole of CO2

c. H2O, formed when 17.6 moles of C2H2react

d. CO2, produced when 3.89 moles of O2 react with acetylene

11)    Nitrogen dioxide can form nitric acid by reaction with water as shown by the equation:
NO2 +   H2O ®   HNO3 +   NO
Balance the equation then find the mass in grams of:

a. HNO3, formed from 50 moles of NO2

b. H2O, to form 500 g of HNO3

c. NO2, to form 250 g of HNO3

d. HNO3, formed from 95.0 g of water

12)    An important reaction in a blast furnace used to make iron is:
Fe2O3 +   CO ®   Fe +   CO2

a. How many grams of Fe2O3 are needed to produce 887 g of Fe?

b. How many kilograms of CO are needed to produce 750 kg of Fe?

c. How many kilograms of Fe can be formed from 1.63 x 104 kg of Fe2O3?

d. How many kilograms of Fe can be formed from 779 kg of CO?

Wayne McGowan

Chem 110G

5.1 Chemical Changes

A chemical reaction always involves a chemical change because atoms of the reacting substances form new combinations with new properties.

5.2 Chemical Equations

A chemical equation tells us the materials we need and the products that will form in a chemical reaction.

Reactants                 Products

Equation:      C(s)   +   O2(g)    ®      CO2(g)

In an equation the formulas of the reactants are written on the left of the arrow and the formulas of the products on the right.

The bonds between the atoms of the reactants are broken and new bonds are formed to give the products.

Atoms cannot be gained, lost, or changed into other types of atoms during a reaction. A reaction must be written as a balanced equation, which shows the same number of atoms for each element on both sides of the arrow.

Balancing Chemical Equations:

1)       Write the formulas of the reactants and products.

2)   If the number of atoms of any element differs from side to side add necessary coefficients to the appropriate formulas.

(a)  Start with an element that appears in only one species on each side.
(b)  Work with one element at a time and leave diatomic elements until last.

(c)   Change coefficients as needed to maintain the balance.

NOTE: Since balancing is a trial-and-error procedure, don't expect to do it in your head or get it

right the first time.

Example: Balance the following equations:

Zn +     O2  ®     ZnO

Cr +     O2  ®     Cr2O3
KClO3  ®     KCl +     O2

B +     F2  ®     BF3
C3H6 +     O2  ®     CO2 +     H2O

## Additional  Information  given  in  Chemical  Equations:

It is sometimes useful to show required conditions and the physical states of the reactants and products; (g) gas, (l) liquid, (s) solid, and (aq) aqueous solution.

heat
HgO(s)    ®       Hg(l)  +     O2(g)

Solid HgO is heated and liquid mercury and oxygen gas form.

NOTE: Any special conditions are written above or below the arrow in an equation.

5.3  Types of Reactions

Combination (Synthesis) Reactions - two or more elements, or simpler compounds, bond together to form one product

Mg(s)  +    O2(g)   ®      MgO(s)

Decomposition Reactions - a reactant splits into two or more simpler products

CaCO3(s)   ®      CaO(s)  +     CO2(g)

Combustion Reactions - fuel and oxygen are reacted and produce carbon dioxide, water and heat (energy)

CH4(g)  +     O2(g)  ®    CO2(g)  +    H2O  +  heat

#### Replacement Reactions -elements in a compound are replaced by other elements

In a single replacement reaction, a reacting element switches place with an element in the other reacting compound.

Zn(s)  +     HCl(aq)  ®     ZnCl2(aq)  +    H2(g)

In a double replacement reaction, the positive ions in the reacting compounds switch places.

BaCl2(aq)  +    Na2SO4(aq)  ®     BaSO4(s)  +     NaCl(aq)

5.4 Oxidation-Reduction Reactions - have a loss and a gain of electrons among the reactants

In the oxidation part of the reaction, one substance loses electrons. Simultaneously, a reduction also occurs as another substance gains these electrons.

Ca  +    S  ®    CaS

oxidation

reduction

If we look at a single replacement reaction, we also find that it has two parts: an oxidation and a reduction.

Zn(s)  +    CuSO4(aq)  ®     ZnSO4(aq)  +    Cu(s)

We can rewrite the equation to show the ions of the compounds and the electrons exchanged.

Zn  +    Cu2+  +    SO42-  ®    Zn2+  +    SO42-  +    Cu

oxidation

reduction

# Oxidation and Reduction in Biological Systems

In the cells of the body oxidation of organic (carbon) compounds involves the transfer of hydrogen and oxygen atoms.

Oxidation always involves a loss of electrons, but it may also be seen as an addition of oxygen or the loss of hydrogen atoms. A reduction always involves a gain of electrons, and may also be seen as a loss of oxygen or the gain of hydrogen.

Wayne McGowan

Chem 110G

5.5 The Mole

In chemistry particles such as atoms, molecules and ions are counted by the mole, a unit that contains 6.022 x 1023 items. This very large number is called Avogadro’s number and for all practical purposes is just a number; just like a dozen is just a number.

One mole of any element (or anything) always contains Avogadro’s number of atoms (or anything).

1 mole of an element  =  6.022 x 1023 atoms of that element

As a conversion factor:                                                     =

One mole of a compound always contains Avogadro’s number of molecules or formula units. [Molecules are the groups of covalent compounds; formula units are the groups of ions given by the formula of the ionic compound.]

5.6 Molar Mass

For any element the quantity called its molar mass is the number of grams equal to the atomic mass of that element. There are 6.022 x 1023 atoms in any sample equal to the element’s atomic mass in grams. This is a matter of simply changing an elements weight unit from atomic-mass-units, amu, to grams.

NOTE: We use the periodic table to get the molar mass of an element.

# Calculations Using Molar Mass

Molar mass is used as a conversion factor to change from moles of a substance to grams of that substance, or from grams to moles.

e.g.:

1 mole Mg = 24.3 g Mg   Þ                                            Û

We use the first to change from moles to grams and the second to change from grams to moles.

Examples:

How many moles of gold are there in 455 grams of pure gold?

How many grams of potassium are there in 7.25 moles of potassium?

# Molar Mass of Compounds

To determine the molar mass of a compound multiply the molar mass of each element by its subscript in the formula and then add the results.

e.g.:

The molar mass of SO3 is obtained by adding the molar mass of 1 mole of S and 3 moles of O.

1S  = 1(         g/mol)

+ 3O = 3(         g/mol)

SO3 =             g/mol

# Example: Determine the Molar Mass of the following compounds:

H2O

NH3

NaCl

NaNO3

5.7 Mole Relationships in Chemical Equations

Participants in a chemical reaction relate through mole ratios equal to their coefficients in the balanced chemical equation.

e.g.: Silver reacts with sulfur to form silver sulfide.

Ag  +      S  ®

NOTE:  The number of silver atoms that react is two times the number of sulfur atoms. This

equation tells us that 2 moles of silver are needed for every 1 mole of sulfur that reacts.

Number of moles:

moles Ag  +       mole S  ®        mole

Using the molar mass of each substance we can also determine the mass in grams of reactants and products.

Number of grams:

2(107.9 g) Ag   +   32.1 g S     ®    247.9 g Ag2S

247.9 g reactants                     247.9 g product

# Mole-Mole Conversion Factors from an Equation

From the proportions given by the coefficients we can write mole-mole conversion factors between reactants and product.

The reaction  2Fe(s)  +  3S(s) ® Fe2S3(s)  gives the following conversion factors:

Fe and S:                                                     and

Fe and Fe2S3:                                               and

S and Fe2S3:                                                          and

# Calculating the number of Moles of Reactants and Products

If the number of moles of any participant (reactant or product) in a chemical reaction is given then the amount of any other participant (reactant or product) can be calculated using the balanced chemical equation.

Examples:

In the reaction of iron and sulfur how many moles of sulfur are needed to react with 6.0 moles of iron?

Fe  +    S  ®    Fe2S3

given               mole-mole ratio                                                       answer

In a combustion reaction propane reacts with oxygen. How many moles of CO2 can be produced when 6.0 moles of propane react?

C3H8  +     O2  ®    CO2  +    H2O

given                mole-mole ratio                                                      answer

Wayne McGowan

Chem 110G

5.8 Mass Calculations for Reactions

The mass of the reactants needed to produce a given amount of product can be calculated; as can the mass of product produced from a given amount of reactants.

Mass questions relating to the formation of ammonia:      N2 +     H2 Þ     NH3

1. How many grams of ammonia can be produced from specific masses of the reactants?

2. What mass of hydrogen or nitrogen are needed to produce a specific mass of ammonia?

3. What mass of hydrogen is needed to react with a specific mass of nitrogen?

Calculation Components:

The conversions between mass and moles (molar masses) and between moles and moles (molar ratios) are used together to determine the mass of a reactant or product.

Steps Involved for Finding the Masses of Substances in a Chemical Reaction:

Step 1. Convert the mass of substance A to moles of A using its molar mass.

Step 2. Convert the moles of A to moles of the desired substance B using the mole ratio derived

from the coefficients of the balanced equation.

Step 3. Convert the moles of B to grams of B using its molar mass.

e.g.: Calculate the mass of NO produced when 48.0 grams of O2 reacts in the following:

N2(g)  +    O2(g) ®   NO(g)

.

1st We convert the mass of O2 (the given) to moles of O2 using the molar mass of O2.

2nd We change the calculated moles of O2 to moles of NO using the mole-mole conversion factors derived from their coefficients in the balanced equation.

3rd  We convert from moles of NO to mass of NO using the molar mass of NO.

The calculation is normally conducted with all steps done together:

Example:

How many grams of O2 are required to react with 22.0 g of propane (C3H8) in the following reaction?

C3H8(g) +    O2(g) ®     CO2(g) +     H2O(g)

g C3H8           ® mole C3H8        ® mole O2           ® g O2

In the reaction of aluminum and oxygen how many grams of aluminum oxide are produced from 95.0 grams of aluminum?

Al  +     O2  ®     Al2O3

In a combustion reaction butane (C4H10) is reacted with excess oxygen. How many grams of butane are needed to react so that 625 grams of CO2 can be produced?

C4H10  +     02  ®    CO2  +     H2O

Hydrogen gas is produced in a reaction between sodium and water. How many grams of hydrogen gas can be produced from reacting 0.25 grams of sodium with excess water?

Na +     H2O  ®     NaOH +     H2

Chem 110

Wayne McGowan           HOMEWORK #4                                                                  ANSWER KEY

1)      Balance the following equations.

a. 2Zn + 1O2 ®2ZnO                    b. 4Cr + 3O2 ®2Cr2O3
c. 2KClO3 ®2KCl + 3O2               d. 4Li + 1O2 ®2Li2O
e. 2B + 3F2 ®2BF3                        f. 2C3H6 + 9O2 ®6CO2 + 6H2O
g. 3Hg + 1N2 ®1Hg3N2                           h. 1N2 + 3H2 ®2NH3
i. 1WO3 + 3H2 ®1W + 3H2O                  j. 2C4H10 + 13O2 ®8CO2 + 10H2O
k. 6Ca + 2NH3 ®3CaH2 + 1Ca3N2         l.  2KNO3 + 4C ®1K2CO3 + 3CO + 1N2

3)                If you had 1 mole of pennies and divided them among all of the 6 billion people on earth, how many dollars would each person have?

1 mol pennies    6.022 x 1023pennies      1 dollar                =  1.00 x 1012 dollars
6 x 109 persons            1 mol                100 pennies                                    person

3)      Define or explain these terms.

a)  mole of an element
A mole of an element has as many atoms as exactly 12 grams of carbon-12 (6.022x1023entities/mol).
b)  molar mass
Molar mass is the mass of one mole of your subject; for an elements it is equal to its atomic mass in grams.

4)      Calculate the number of moles of atoms contained in each sample given below:

a) 24.8 g silver           24.8 g Ag  1 mol Ag                            =  0.230 mol Ag
107.9 g

b) 53.5 kg iron           53.5 kg Fe  103 g      1 mol Fe                       =  957 mol Fe
1 kg         55.9 g

c) 6.71 g sulfur          6.71 g S   1 mol S                                          =  0.209 mol S
32.1 g

d) 0.0699 g of W       0.0699 g W   1 mol W                         =  3.80x10-4 mol W
183.9 g

e) 131 mg silicon       131 mg Si   10-3  1 mol Si                =  4.66x10-3 mol Si
1 mg      28.1 g

5)      Calculate the mass of the following samples of elements:

a)  84 mol K                84 mol K    39.1 g                                         =  3,284 g
1 mol K

b)  7.60x10-3mol U      7.60x10-3 mol U  238.0 g                             =  1.81 g
1 mol U

c)  0.0395 mol Li         0.0395 mol Li    6.90 g                                 =  0.273 g
1 mol Li

d)  2.25x106 mol Ba     2.25x106 mol Ba    137.3 g                =  3.09x108 g
1 mol Ba

e)  4.69 mol B             4.69 mol B    10.8 g                             =  50.7 g
1 mol B

6)      Calculate the number of atoms in each sample given below:

a) 3.18 g Cu          3.18 g Cu  1 mol Cu   6.022x1023 atoms   =  3.02x1022 atoms
63.5 g              1 mol

b) 0.171 g W          0.171 g W  1 mol W   6.022x1023 atoms   =  5.60x1020 atoms
183.9 g            1 mol

c) 12.5 g He           12.5 g He   1 mol He   6.022x1023 atoms   =  1.88x1024 atoms
4.0 g                1 mol

d) 453.6 g iron        453.6 g Fe  1 mol Fe   6.022x1023 atoms        =  4.90x1024 atoms
55.8 g              1 mol

e) 0.200 g carbon    0.200 g C  1 mol C   6.022x1023 atoms          =  1.00x1022 atoms
12.0 g             1 mol

7)      Determine the molar mass of the following compounds to four significant digits:

a)  isopropyl alcohol, C3H8O (rubbing alcohol)                       60.00 g/mol

b)  barium sulfate, BaSO4 (used for X rays)                                     233.4 g/mol

c)  sodium nitrate, NaNO3 (found in the mineral Chile)           85.00 g/mol

d)  potassium dichromate, K2Cr2O7                                                  294.2 g/mol

e)  titanium(IV) chloride (used in smokescreens), TiCl4           189.7 g/mol

8)      Determine the number of moles of formula units in the following samples of compounds:

a)  311 g NaNO2          311 g NaNO2  1 mol NaNO2              = 4.51 mol NaNO2

69.00 g

b)  75.0 g  C16H30O     75.0 g C16H301 mol C16H30O                   = 0.315 mol C16H30O

238 g

c)  2.88 x 105 g of SO3  2.88 x 105 g SO3  1 mol SO3              = 3.60 x 103 mol SO3
80.1 g

d)  2.24 x 10-7 gAu2S3   2.24 x 10-7gAu2S3 1 mol Au2S3                   = 4.57x10-10molAu2S3
490.3 g

e)  9.10 g H2O2           9.10 g H2O2  1 mol H2O2                    = 0.268 mol H2O2
34.0 g

9)      Acetylene, C2H2, burns in oxygen according to the following unbalanced equation:
2C2H2(g) + 5O2(g) ®4CO2(g) + 2H2O(g)
Balance the equation.  Give the molar interpretation of the equation and give the molar ratios that relate the following pairs of reactants and products.

a. C2H2 and O2               2 mol C2H2          Û        5 mol O2

5 mol O2                          2 mol C2H2

b. C2H2 and CO2            2 mol C2H2          Û       4 mol CO2

4 mol CO2                      2 mol C2H2

c. C2H2 and H2O            2 mol C2H2          Û       2 mol H2O

2 mol H2O                       2 mol C2H2

d. O2 and CO2                 5 mol O2                       Û       4 mol CO2

4 mol CO2                        5 mol O2

10)    Refer to the balanced equation in the previous question.  Find the number of moles of:

a. O2, to react with 0.570 mole of C2H2

0.570 mol C2H2     5 mol O2                                                  =  1.43 mol C2H2
2 mol C2H2

b. C2H2, to produce 0.400 mole of CO2

O.400 mol CO2    2 mol C2H2                                                =  0.200 mol C2H2
4 mol CO2

c. H2O, formed when 17.6 moles of C2H2react

17.6 mol C2H2       2 mol H2O                                                =  17.6 mol H2O
2 mol C2H2

d. CO2, produced when 3.89 moles of O2 react with acetylene

3.89 mol O2         4 mol CO2                                                 =  3.11 mol CO2
5 mol O2

11)    Nitrogen dioxide can form nitric acid by reaction with water as shown by the equation:
3NO2 + 1H2O ®2HNO3 + 1NO
Balance the equation then find the mass in grams of:

a. HNO3, formed from 50 moles of NO2

50 mol NO2 2 mol HNO3   63 g HNO3                                 =  2,100 g HNO3
3 mol NO2     1 mol HNO3

b. H2O, to form 500 g of HNO3

500 g HNO3 1 mol HNO3   1 mol H2    18 g H2                     =  71.4 g H2O
63 g HNO3     2 mol HNO3 1 mol H2O

c. NO2, to form 250 g of HNO3

250 g HNO3 1 mol HNO3     3 mol NO     46 g NO2                     =  274 g NO2
63 g HNO3     2 mol HNO3   1 mol NO2

d. HNO3, formed from 95.0 g of water

95.0 g H21 mol H2  2 mol HNO3   63 g HNO3                      = 665 g HNO3
18 g H2O      1 mol H2O     1 mol HNO3

12)    An important reaction in a blast furnace used to make iron is:
Fe2O3 + 3CO ® 2Fe + 3CO2

a. How many grams of Fe2O3 are needed to produce 887 g of Fe?

887 g Fe  1 mol Fe   1 mol Fe2O3   159.6 g Fe2O3               =  1269 g Fe2O3
55.8 g Fe       2 mol Fe        1 mol Fe2O3

b. How many kilograms of CO are needed to produce 750 kg of Fe?

750 kg Fe  103 g    1 mol Fe   3 mol CO   28 g CO    1 kg          =  565 kg CO
1 kg       55.8 g       2 mol Fe    1 mol CO   103 g

c. How many kilograms of Fe can be formed from 1.63 x 104 kg of Fe2O3?

1.63x104 kg Fe2O3  1 mol Fe2O3      2 mol Fe     55.8 g Fe            =  1.14x104 kg Fe
159.6 g       1 mol Fe2O3   1 mol Fe

d. How many kilograms of Fe can be formed from 779 kg of CO?

779 kg CO  103 1 mol CO  2 mol Fe   55.8 g Fe    1 kg          =  1.03x103 kg Fe
1 kg     28 g CO   3 mol CO   1 mol Fe    103 g

Chem 110G

Wayne McGowan           HOMEWORK #5                  Name ____________________________

1)      Explain how polar molecules attract one another.  What is a hydrogen bond?

2)      Describe what happens, in terms of the interaction between water molecules and ions, when the ionic compound potassium chloride, KCl, dissolves in water.

3)      a. Draw a water molecule and describe its polarity.

b. Show how hydrogen bonding occurs between two water molecules.

c. Why does HF also exhibit hydrogen bonding?

d. Would you expect CH4 molecules to hydrogen bond? Explain.

4)      Using Le Chatelier's principle, predict any shift in equilibrium in the reaction:

N2(g) + 3H2(g) <====> 2NH3(g) + energy

when the following occur.

a. the temp. of the equilibrium system is increased

b. the conc. of H2 is increased

c. the conc. of N2 is decreased

d. the temp. of the equilibrium system is decreased

5)      Why would you expect a solution of a soluble ionic compound to conduct electricity?

6)      Indicate the type of electrolyte represented in the following equations:

H2O

a. CH3OH ----> CH3OH(aq)

H2O

b. MgCl2 ----> Mg2+ + 2Cl-

H2O

c. HClO <----> H+ + ClO-

7)      What is the difference between a 10% (v/v) methyl alcohol (CH3OH) solution and a 10% (m/m) methyl alcohol solution?

8)      Calculate the mass percent of the solute in each of the following solutions:

a. 8.0 g NaBr in 100.0 g H2

b. 75.0 g KCl in 500.0 g of solution

9)      Write two conversion factors for the following solution concentrations:

a. 4% (m/v) KCl

b. 1.0% (m/m) K2CO3

c. 15% (v/v) isopropyl

10)    Calculate the amount of solute (g or mL) needed to prepare the following solutions:

a. 75 g of a 25% (m/m) NaOH solution

b. 150 mL of a 40.0% (m/v) LiNO3 solution

c. 450 mL of a 2.0% (m/v) KCl solution

11)    Calculate the amount of solution (g or mL) that holds the following amounts of solute?

a. 10.0 g HCl from a 1.0% (m/m) HCl solution

b. 5.0 g LiNO3 from a 25% (m/v) LiNO3 solution

c. 40.0 g KOH from a 10.0% (m/v) KOH solution

12)    A sample of vinegar is found to contain 18.9 g of acetic acid, HC2H3O2 in 34l mL

of the solution. Calculate the molarity of acetic acid in the vinegar.

13)    A 325-mL sample of beer is found to contain 9.75 grams of ethyl alcohol, C2H6O.

Calculate the molarity of ethyl alcohol in the beer.

14)    Determine the number milliliters of solution needed to provide the following:

a.  0.475 mole of HC2H3O2 from a 0.789  M acetic acid (HC2H3O2) solution?

b.  16.3 g of MgCl2 from a 1.50 M solution?

c.  137.3 g of FeCl3 from a 1.250 M solution?

15)    Determine the number of grams of solute needed to prepare the following solutions.

a.  235 mL of a 4.95 x 10-2 M HC2H3O2 (acetic acid) solution

b.  8.30 L of a 0.396 M FeCl3 solution

c.  740 mL of a 0.200 M NaOH solution

16)    To what volume should 150 mL of 5.0 M NaOH be diluted to give a 0.10 M solution?

17)    How much of a 12 M NH3 is needed to make 75 mL of 1.0 M NH3 by dilution?

18)    What kind of mixture, solution, colloidal dispersion, or suspension:

a) can be separated into its components by filtration?

b) has the smallest particles of all kinds?

e) is likeliest to be the least stable at rest over time?

19)    In general terms, how does an osmotic membrane differ from a dialyzing membrane?

Wayne McGowan

Chem 110G

8.1Solutions

Solutions are mixtures in which one substance, called the solute, is uniformly dispersed in another substance, called the solvent.

In solutions the solute and the solvent can be mixed in varying proportions (have varying concentrations).

Water as a Solvent

In the H2O molecule an oxygen atom shares e- with two hydrogen atoms. Because the oxygen atom is much more electronegative the O—H bonds are polar. Water is also asymmetric and hence is a polar molecule.

Hydrogen Bonding - occur between polar molecules where a partially positive hydrogen is attracted to the strongly electronegative atoms of ___, ___, ___ or ___ in another molecule

In water, hydrogen bonds form between the oxygen atom of one water molecule and a hydrogen atom in another water molecule.

Hydrogen bonds are much weaker than covalent or ionic bonds but there are many of them linking molecules together. As a result hydrogen bonding plays an important role in the properties of water and biological compounds.

8.2 Formation of Solutions

When a solid ionic compound dissolves in water the negatively charged oxygen atom attracts the positive ions and the positively charge hydrogen atoms attract the negative ions.

H                                           H

d-¬O ®d+     ¬Cl- Na+®      d-¬O ®d+

H                                           H

As the water pull the ions into solution the dissolved Na+ and Cl- ions are surrounded by water molecules forming a hydration sphere.

Like Dissolves Like - is a way of saying that the polarities of a solute and a solvent must be similar in order to form a solution

For anything to dissolve there must be an attraction between the solute particles and the solvent particles.

Compounds containing non-polar molecules do not dissolve in water because the water is polar. Non-polar solutes require non-polar solvents for a solution to form.

8.3 Solubility and Saturated Solutions

Solubility – is a term used to describe the amount of a solute that can dissolve in a given amount of solvent

Solubility is the maximum amount of solute that can be dissolved in a given amount of solvent at a certain temperature.

Saturated Solution - when a solution contains all the solute that can dissolve

Additional solute will remain undissolved on the bottom of the container.

When a solution becomes saturated equilibrium exists between the forward reaction that dissolves the solute and the reverse reaction of crystallization.

solute undissoved  Û solute dissolved

When equilibrium is reached there is no further change in the amount of dissolved solute.

Le Chatelier’s Principle:

An equilibrium responds to a change by shifting in a direction that absorbs/reduces the change (adding more of one of the participants shifts the equilibrium to the other side).

From the equation:

soluteundissolved + heat  Û solutedissolved

we find that increasing temperature                               the solubility of a solute.

From the equation:

gasundissolved  Û gasdissolved + heat

we find that gas solubility                             with temperature and                           with the partial pressure of the gas.

Effect of Temperature:

The solubility of most solids in water increases as temperature increases. The solubility of a gas in a liquid is decreased as temperature increases.

Effect of Pressure:

The solubility of a gas in a liquid increases as the pressure of that gas above the liquid increases. When a gas is at higher pressures there are more gas molecules available to enter and dissolve in the liquid.

8.4 Electrolytes

Electrolytes produce aqueous solutions that conduct electricity (due to the mobile ions of acids, bases, and salts).

Classifications:

Strong electrolyte – a high percentage of ionization in water

eg:

Weak electrolyte –  a  low percentage of ionization in water

eg:

Non-electrolyte – does not ionize in water

eg:

8.5 Percent Concentration

Solute Concentrations - specify the amount of solute in a given amount of solution

Concentration  =

of a solution

Mass Percent  (m/m %)  - the percent by mass of solute in a solution

Mass %  =       mass of solute      x  100%  (same units, usually grams)

mass of solution

Volume Percent (v/v %) - the percent by volume of solute in a solution

Volume %  =   volume of solute    x  100%         (same units, usually milliliters)

volume of solution

Mass/Volume Percent (m/v %)

mass/volume %  =     grams of solute     x  100%(always grams and milliliters)

milliliters of solution

Example:

A solution is made with 20 g Na2SO4 in 200 g of solution. What is the m/m%?

A solution is made from 25 g NaCl  and 100 g H20 what is the m/m %?

A beer is found to have 10 ml of alcohol in 200 ml of the beer. What is the v/v%?

Wayne McGowan

Chem 110G

Percent Concentrations as Conversion Factors

In the preparation of solutions we often need to calculate the amount of solution containing a needed amount of solute or the amount of solute in a given amount of solution. When given the concentration of the solution we have the conversion term needed for these types of calculations.

Percent concentrations can be used as conversion factors to convert between the amount of solute and the amount of solution.

e.g. 18% m/m KCl  Þ

Example:

How much glucose must be added to make 500 ml of 10% m/v glucose solution?

How much of a 15% m/m NaI solution can be prepared from 60 g NaI ?

How much NaHCO3 is needed to make 2.0 Liters of a 20.0% m/v solution?

8.6 Molarity (M)

Molarity is the concentration term that states the number of moles of solute per 1 liter of solution.

Molarity (M)  =

Solutes in solutions take part in chemical reactions and chemists are interested in the number of reacting particles. For this purpose, chemists use molarity.

e.g.: If 1.0 mole of NaCl were dissolved in enough water to prepare 1.0 L of solution the resulting NaCl solution has a molarity of 1.0 M.

M    =                                               =

Example:

What is the molarity (M) for 60.0 g of NaOH in 0.250 L of solution?

A sample of vinegar is found to contain 18.9 g of Acetic Acid (AA), HC2H3O2 in 34l mL of the solution. Calculate the molarity of acetic acid in the vinegar.

Molarity as a Conversion Factor

When using molarity in a calculation include the units so that molarity becomes a conversion factor.

Molarity can be used in two ways:

1. To find the number of moles in a given volume of solution.
2.   To find the volume of solution containing a specific number of moles of solute.

Examples:
What volume of 1.5 M HCl solution is needed to provide 6.0 moles of HCl?

How many milliliters of a 0.789  M acetic acid (HC2H3O2) solution are needed to provide 0.475 mole of acetic acid?

How many moles of acetic acid are in 602 ml of a 0.789 M acetic acid solution?

How many grams of KCl would you need to prepare 0.250 L of a 2.00 M KCl solution?

Dilution

Dilution occurs when additional solvent is added to an existing solution.

When more solvent is added the solution volume increases which causes a decrease in the concentration.

NOTE: The amount of solute does not change.

The number of moles of solute is the product of the volume and the molarity.
n  =  VcMc                     (n = no. of moles; V = volume; M = molarity; c = concentrated)

Adding water to the sample dilutes it.
n  =  VdMd                    (d refers to diluted)

The number of moles is unchanged by dilution.
n  =  VcMc   and   n  =  VdM

With "n" being the same for both this gives

The General Dilution Formula:
VcMc  =  VdMd

VcCc  =  VdCd             (when other conc. units are used)

Examples:
To what volume should 150 mL of 5.0 M acetic acid be diluted to give a 0.100 M solution?

How many milliliters of 12 M NH3 are needed to prepare 75 mL of 1.0 M NH3 by dilution?

A NaCl solution is 2.00 M.  Find the molarity when a 10.0-mL sample is diluted with water to a volume of 40.0 mL

NOTE: The terms "of" and "is" links information together, either both are concentrate values or dilute values.

Wayne McGowan

Chem 110G

8.7 Colloids and Suspensions

Colloids

The particles in colloidal dispersions, or colloids, are much larger than solute particles in a solution. Colloids are homogeneous mixtures that do not separate or settle out. Colloidal particles are small enough to pass through filters but too large to pass through semi-permeable membranes.

Suspensions

Suspensions are heterogeneous, non-uniform mixtures that are very different from solutions or colloids. The particles of a suspension are so large that they can often be seen with the naked eye. They are trapped by filters and semi-permeable membranes.

 Type of mixture Type of particle Settling Separation Solution Small particles such as atoms, ions, or small molecules Particles do not settle Particles cannot be separated by filters or semi-permeable membranes Colloid Larger molecules or groups of molecules or ions Particles do not settle Particles can be separated by semi-permeable membranes but not by filters Suspension Very large particles that may be visible Particles do settle Particles can be separated by filters

8.8 Osmosis and Dialysis

Osmosis - a process where water moves through a very selective semi-permeable membrane from a solution that has a lower concentration of solute into a solution of higher concentration

This movement of water happens in a direction that equalizes (or attempts to equalize) the concentrations of both sides of the membrane.

Osmotic Pressure - the pressure that prevents the flow of additional water into the more concentrated solution

The osmotic pressure of a solution depends on the number of solute particles in the solution. The greater number of particles dissolved in a solution the higher its osmotic pressure. Pure water has no osmotic pressure.

If a pressure greater than the osmotic pressure is applied to a solution then osmosis is reversed and solvent flows out of the solution and the level in the solvent compartment increases.

Isotonic Solutions - exert the same osmotic pressure as body fluids

Dialysis

Dialysis is a process that is similar to osmosis. In dialysis a semi-permeable membrane, called a dialyzing membrane, permits small solute molecules and ions as well as solvent water molecules to pass through but retains large particles such as colloids. Dialysis is a way to separate solution particles from colloids.

9.1 Arrhenius Acids and Bases

Acids - substances that produce hydrogen ions, H+, in water

The hydrogen ions, H+, give acids a sour taste, change blue litmus to red, and corrode some metals.

H2O

HCl(g)    ®       H+(aq) + Cl-(aq)

# polar covalent              ionization

compound                    in water

## Naming Acids

Binary Acids - compounds involving hydrogen and a nonmetal

When these compounds dissolve in water the resulting solutions display acidic properties. In water they are named:  hydro + root of name + ic acid

e.g.:

FORMULA OF              NAME OF                               NAME OF

##### COMPOUND                                     COMPOUND                                                 WATER SOLUTION

HF                                                                                 Hydrofluoric acid

HCl                                Hydrogen chloride

HBr                                Hydrogen bromide                  Hydrobromic add

Oxoacids – compounds involving hydrogen, a nonmetal and one or more oxygens

We name an oxoacid by taking the name of the corresponding oxoanion, changing the suffix, and adding the word acid. The -ate ending of the ion name is changed to -ic for the acid name, and the -ite ending of the ion name is changed to -ous for the acid name.

(see Chapter 4, for more on naming)

e.g.:

ion          oxoanion name             acid             oxoacid name
ClO2-         chlorite ion                   HClO2         chlorous acid
ClO3-         chlorate ion                  HClO3             chloric acid

Bases - compounds that form hydroxide ions, OH-, in water

Most bases are ionic compounds (metal hydroxides).

H2O

NaOH(s)     ®      Na+(aq) + OH-(aq)

# ionic compound          ionization in water

Ammonia is a common molecular base:

NH3(g)  +   H2O     Û       NH4+(aq)  +  OH-(aq)

# covalent compound                   ionization in water

# Naming Bases

Ionic bases are named according to the ionic nomenclature rules as metal hydroxides.

e.g.:   NaOH  -

KOH  -

Molecular bases are named by normal covalent compound naming rules.

e.g.:   NH3 -  nitrogen trihydride, or ammonia

9.2 Bronsted-Lowry Acids and Bases

Acids - proton (H+) donors

Bases - proton (H+) acceptors

NOTE: Arrhenius acids and bases are not different from Bronsted-Lowry acids and bases, they

are just defined differently.

It is worth noting that a proton (H+) does not actually exist by itself. Its attraction to polar water molecules is so strong that the proton bonds to the water molecule and forms a hydronium ion, H3O+.

e.g.: In the formation of a hydrochloric acid solution there is a transfer of a proton from HCl to water. By donating a proton HCl is acting as an acid and by accepting a proton water is acting as a base.

HCl(g)    +   H2O     ®       H3O+(aq)  +  Cl-(aq)

(H+ donor)      (H+ acceptor)                              (acidic solution)

In another reaction ammonia reacts with water, where water acts as an acid by donating a proton.

NH3(g)   +   H2O      Û      NH4+(aq)    +  OH-(aq)

(H+ acceptor)        (H+ donor)                             (basic solution)

Chem 110G

Wayne McGowan           HOMEWORK #5                                                                  ANSWER KEY

1)      Explain how polar molecules attract one another.  What is a hydrogen bond?

Polar molecules attract each other through their regions of opposite charges.

A hydrogen bond is a special polar bonding that involves hydrogen and a highly electronegative element.

2)      Describe what happens, in terms of the interaction between water molecules and ions, when the ionic compound potassium chloride, KCl, dissolves in water.

When KCl is subjected to the presence of water molecules the water molecules positive region (the hydrogen side) pulls/attracts the negative chloride ion, while the water molecules negative region (the oxygen side) pulls/attracts the positive potassium ions.  This pulls the ions into solution where they are surrounded by a hydration sphere.

3)      a. Draw a water molecule and describe its polarity.

H

d-  ¬O ®  d+

### H

b. Show how hydrogen bonding occurs between two water molecules.

H               H

d-¬O®d+ d-¬O ®d+

### H               H

c. Why does HF also exhibit hydrogen bonding?

Hydrogen bonding requires hydrogen and a highly electronegative element (so that the bond is very polar), which fluorine is.

d. Would you expect CH4 molecules to hydrogen bond? Explain.

No.  Carbon has about the same electronegativity as hydrogen, thus, the bond is not sufficiently polar.

4)      Using Le Chatelier's principle, predict any shift in equilibrium in the reaction:

N2(g) + 3H2(g) <====> 2NH3(g) + energy

when the following occur.

a. the temp. of the equilibrium system is increased                 shift left

b. the conc. of H2 is increased                                        shift right

c. the conc. of N2 is decreased                                        shift left

d. the temp. of the equilibrium system is decreased                shift right

5)      Why would you expect a solution of a soluble ionic compound to conduct electricity?

For a solution to conduct electricity it must have mobile ions to act as carriers of electrical charge.  A solution of a soluble ionic compound would dissolve/dissociate into mobile ions.

6)      Indicate the type of electrolyte represented in the following equations:

H2O

a. CH3OH ----> CH3OH(aq)                 non electrolyte

H2O

b. MgCl2 ----> Mg2+ + 2Cl-                strong electrolyte

H2O

c. HClO <----> H+ + ClO-                            weak electrolyte

7)      What is the difference between a 10% (v/v) methyl alcohol (CH3OH) solution and a 10% (m/m) methyl alcohol solution?

The 10% m/m concentration is based on the relative masses of the alcohol and solvent while the 10% v/v concentration is based on the relative volumes of the alcohol and solvent.

8)      Calculate the mass percent of the solute in each of the following solutions:

a. 8.0 g NaBr in 100.0 g H2

8.0g NaBr             x 100%    =  7.4% m/m NaBr

100.0g H2O + 8.0g NaBr

b. 75.0 g KCl in 500.0 g of solution

75.0g KCl       x 100%                 =  15.0% m/m KCl

500.0g solution

9)      Write two conversion factors for the following solution concentrations:

a. 4% (m/v) KCl                   4 g KCl          100 ml solution

100 ml solution        4 g KCl

b. 1.0% (m/m) K2CO3      1.0g K2CO3        100 g solution

100 g solution        1.0 g K2CO3

c. 15% (v/v) isopropyl   15 ml isopropyl   100 ml solution

100 ml solution    15 ml isopropyl

10)    Calculate the amount of solute (g or mL) needed to prepare the following solutions:

a. 75 g of a 25% (m/m) NaOH solution

75 g solution    25 g NaOH                                                              =  18.75  NaOH

100 g solution

b. 150 mL of a 40.0% (m/v) LiNO3 solution

150 ml solution    40.0 g LiNO3                                              =  60.0 g LiNO3

100 ml solution

c. 450 mL of a 2.0% (m/v) KCl solution

450 ml solution       2.0 g KCl                                                =  9.0 g KCl

100 ml solution

11)    Calculate the amount of solution (g or mL) that holds the following amounts of solute?

a. 10.0 g HCl from a 1.0% (m/m) HCl solution

10.0 g HCl100 g solution                                                      = 1,000 g solution

1.0 g HCl

b. 5.0 g LiNO3 from a 25% (m/v) LiNO3 solution

5.0 g LiNO3         100 ml solution                                                   = 20 ml solution

25 g LiNO3

c. 40.0 g KOH from a 10.0% (m/v) KOH solution

40.0 g KOH           100 ml solution                                                  = 400 ml solution

10.0 g KOH

12)    A sample of wine vinegar is found to contain 18.9 g of acetic acid, HC2H3O2 in 34l mL

of the solution. Calculate the molarity of acetic acid in the vinegar.

18.9 g AA 1 mol AA
60 g AA                                                              =  0.924 mol
341 ml sol’n 10-3 L                                                                             L
1 ml

13)    A 325-mL sample of beer is found to contain 9.75 grams of ethyl alcohol, C2H6O.

Calculate the molarity of ethyl alcohol in the beer.

9.75 g ethyl 1 mol ethyl
46 g ethyl                                                           =  0.652 mol
325 ml sol'n 10-3 L                                                                             L
1 ml

14)    Determine the number milliliters of solution needed to provide the following:

a.  0.475 mole of HC2H3O2 from a 0.789  M acetic acid (HC2H3O2) solution?

0.475 mol HC2H3O2        1 L        1 ml                                   =  602 ml
0.789 mol   10-3 L

b.  16.3 g of MgCl2 from a 1.50 M solution?

16.3 g MgCl2 1mol MgCl2         1 L       1 ml                        =  114 ml
95.3 g          1.50 mol   10-3 L

c.  137.3 g of FeCl3 from a 1.250 M solution?

137.3 g MgCl2 1mol FeCl3         1 L         1 ml                               =  677.2 ml
162.2 g        1.250 mol   10-3 L

15)    Determine the number of grams of solute needed to prepare the following solutions.

a.  235 mL of a 4.95 x 10-2 M HC2H3O2 (acetic acid) solution

235 mL  10-3  4.95x10-2mol   60.0 g                                          =  0.698 g
1 mL           1 L            1 mol

b.  8.30 L of a 0.396 M FeCl3 solution

8.30 L  0.396 mol   162.2 g                                                            =  533 g
1 L          1 mol

c.  740 mL of a 0.200 M NaOH solution

740 mL  10-3  0.200 mol    40.0 g                                    =  5.92 g
1 mL        1 L           1 mol

16)    To what volume should 150 mL of 5.0 M NaOH be diluted to give a 0.10 M solution?

Vd  =  Vc  Mc   =  150 mL  5.0 M                                                 =  7500 mL
Md               0.100 M

17)    How much of a 12 M NH3 is needed to make 75 mL of 1.0 M NH3 by dilution?

Vc  =  Vd  Md   =  75 mL  1.0 M                                                   =  6.25 mL
Mc                 12 M

18)    What kind of mixture, solution, colloidal dispersion, or suspension:

a) can be separated into its components by filtration?  stirred suspensions

b) has the smallest particles of all kinds?                      Solutions

e) is likeliest to be the least stable at rest over time?      unstirred suspensions

19)    In general terms, how does an osmotic membrane differ from a dialyzing membrane?

Osmotic membranes have pores so small only the water molecules can get through, whereas Dialyzing membranes have pores large enough to let small ions and small molecules through.

Chem 110G

Wayne McGowan           HOMEWORK #6                  Name ____________________________

1)      Tell whether each of the following solutions is acidic, basic, or neutral.

a) [H+] = 7.9 x 10-6 mol/L and [OH-] = 1.3 10-9 mol/L

b) [H+] = 1.0 x 10-7 mol/L and [OH-] = 1.0 x 10-7 mol/L

c) [H+] = 7.6 x 10-8 mol/L and [OH-] = 1.2 x 10-7 mol/L

2)      What are acidic and basic solutions in terms of the hydronium ion concentration and the hydroxide ion concentration?  What are they in terms of pH?

3)      What is the essential difference between an aqueous solution of a strong acid and that of a weak acid?

4)      Write a balanced equation for the ionization of each of the following in water:

a. HCN, a weak acid

b. Ba(OH)2, soluble

c. NH3, a weak base

d. HNO3, a strong acid

5)      What are the formulas of the conjugate acids of the following?

a) HSO3-

b) Br-

c) H2O

d) CH3CO2

6)      Write the formulas of the conjugate bases of the following:

a) NH3

b) HNO2

c) HSO3-

d) H2SO3

7)      Which of the following is a neutralization reaction?

a. 2HCl + Ca(OH)2 ----> CaCl2 + 2H2O

b. H2SO4 + Pb(NO3)2 ----> PbSO4 + 2HNO3

c. Mg + 2HCl ----> MgCl2 + H2

d. HClO2 + NaOH ----> NaClO2 + H2O

8)      Balance each of the following neutralization reactions:

a. __ HNO3          +  _  Ba(OH)2 ---->   _ Ba__(NO3)__          + __ H2O

b. __ H2SO4 + __ Al(OH)3 ---->  _  Al__(SO4)__           + __ H2O

9)      How many moles of sodium hydroxide can react with 0.256 mol of H2SO4 ,assuming that both H+ in H2SO4 are neutralized? (Hint: this is a stoichiometry problem)

10)    How many grams of sodium carbonate will neutralize 5.24 g of HCl?

11)    How many grams of calcium metal can react with 118 mL of 0.375 M hydrochloric acid if the reaction is

Ca(s) + 2HCl(aq) ----> CaCl2(s) + H2(g)

12)    What is Kw and what does it represent?

13)    What are acidic and basic solutions as described in terms of the hydronium ion concentration and the hydroxide ion concentration? What are they in terms of pH?

14)    Calculate the pH of the following solutions and indicate whether it is acidic or basic.

a. an ammonia solution that has 8.4 x 10-9 M H3O+

b. a sample of saliva that has 3.5 x 10-7 M H3O+

c. a sample of rain that has 1.8 x 10-7 M H3O+

d. an apple juice that has 9.3 x 10-4 M H3O+

15)    What is the pH of 0.001 M HCl(aq), assuming 100% ionization?

16)    What is the pH of 0.01 M NaOH(aq), assuming 100% ionization?

17)    Describe a simple buffer solution and explain how a buffer system functions.

18)    Use the following general equation for an acid in equilibrium with its conjugate base to explain how a buffer solution functions upon the addition of small amounts of acid or base.                     HA + H2O <==> H3O+ + A-

Wayne McGowan

Chem 110G

## Conjugate Acid-Base Pairs

When the loss or gain of one H+ relates compounds they make up a conjugate acid-base pair. Because protons are transferred in both a forward and reverse reaction each acid-base reaction contains two conjugate acid-base pairs.

e.g.: In this general reaction the acid HA has a conjugate base A- and the base B has a conjugate acid BH+.

___  __ Conjugate acid-base pair___  __

|                                                               |

HA        +       B            Û         A-         +          BH+

H+ donor         H+ acceptor               H+ acceptor             H+ donor

|_________________________________|

Conjugate acid-base pair

The relationship between a conjugate pair is like the relationship between two people playing catch: whoever has the ball is the pitcher and who ever is waiting for the ball is the catcher; whichever has the proton is the acid and whichever is waiting for the proton is the base.

9.3  Strengths of Acids and Bases

Strong acids give up protons easily, whereas in a weak acid most of the molecules keep their protons and just a few give them up.

Strong bases have a strong attraction for protons, whereas weak bases have little attraction for protons.

The six strong acids dissociate nearly completely (»100%) to give H3O+ ions and anions.

e.g.:     H2O

HCl(g)  ®  H3O+(aq) + Cl-(aq)   (shown by the arrow pointing only one way)

Formulas

#### Perchloric acid

Sulfuric acid

Hydroiodic acid

HI

Hydrobromic acid

HBr

Hydrochloric acid

HCl

Nitric acid

Other acids are weak acids and even at high concentrations they produce low concentrations of hydronium ions.

e.g.:           H2O

HNO2(aq)  Û  H3O+(aq) + NO2-(aq)   (shown by the reversible arrow)

# Strong and Weak Bases

LiOH, KOH, NaOH, and Ca(OH)2 are strong bases that dissociate completely (100%).

e.g.:       H2O

KOH(s)  ®                                                 (shown by the arrow pointing only one way)

The weak bases only partially dissociate to produce the hydroxide ion.

e.g.:

NH3(g)  +  H2O  Û                                     (shown by the reversible arrow)

9.4  The Self-Ionization of Water

In water one water molecule can donate a proton to another to produce H3O+ and OH-, which means that water can behave as both an acid and a base.

Conjugate acid-base pair

|                                        |

H2O  +  H2O   Û   H3O+(aq)  +  OH-(aq)

|______________________|

Conjugate acid-base pair

In pure water the transfer of a proton between two water molecules produces equal number of H3O+ and OH-.

[H3O+] = [OH-] = 1.0 x 10-7M                   (pure and/or neutral water)

Square brackets around the symbols indicate their concentrations in moles per liter (M).

The Water Dissociation Constant (Kw)

When we multiply the [H3O+] and [OH-] concentrations it forms the ion-product constant for water, Kw, which is 1.0 x 10-14 M2.

Kw

=  (1.0 x 10-7 M) (1.0 x 10-7 M)  = 1.0 x 10-14 M2

NOTE: The ion product ([H3O+] x [OH-]) is a constant value (Kw) which means that an increase in the concentration of one of the ions will cause an equilibrium shift that decreases the other ion.

Acidic Solutions

In an acidic solution the  [H3O+] is greater than the [OH-].

Basic Solutions

In a basic solution the [OH-] is greater than the [H3O+].

Example:

If the [H3O+] in a solution is 1.0 x 10-4 M then what is the [OH-] and is this an acidic or basic solution?

If the [OH­-] in a solution is 1.0 x 10-6 M then what is the [H3O+] and is this an acidic or basic solution?

The product of the ions [H3O+] and [OH-] is always equal to 1.0 x 10-14 M2 at 25oC. Therefore, we can use the Kw to calculate the [H3O+] or [OH­-] if the other is known.

9.5 The pH Scale

The pH scale is a scale of acidity that is essentially equivalent to “the power of H+” when [H3O+] is represented numerically as 10 raised to an exponent (power).

# Calculating the pH of Solutions

The pH scale is a logarithmic scale that corresponds to the [H3O+] of aqueous solutions. Mathematically pH is the negative logarithm of the [H3O+].

# Logarithms

Any number can be expressed in terms of 10 raised to a power. Examples are 100 expressed as 102 and 0.001 as 10-3. Any number can be expressed as a power of 10, for example 2 is 100.301 and 9.5 is 100.978.

The logarithm of a number is the power to which 10 is raised to give the number.  Thus, the log of 102 is 2, the log of 10-3 is -3, the log of 2 is 0.301 and the log of 9.5 is 0.978.

Since the logarithms of hydronium and hydroxide ion concentrations are typically negative (i.e., log 10-7 = -7), the definitions of pH is the negative logarithm. e.g. -log 10-7 = -(-7) = 7.

# Acidity

Solutions with a pH value less than 7 are acidic, whereas solutions with a pH greater than 7 are basic. Keep in mind that higher values of pH correspond to lower concentrations of hydronium ions. This is a consequence of the negative logarithm in the definition of pH.

#### NOTE: In the pH scale that there is a 10 fold difference in acidity between pH units.

Wayne McGowan

Chem 110G

9.6 Neutralization Reactions of Acids and Bases

# Acids reacting with Metals

Acids react with and are neutralized by certain metals, known as active metals, to produce hydrogen gas (H2) and a salt of that metal.

Mg(s)    +     HCl(aq)   ®   MgCl2(aq)   +   H2(g)

Zn(s)  +    HCl(aq)  ®  ZnCl2(aq)  +  H2(g)

# Acids reacting with Hydroxide

A reaction between an acid and a hydroxide (base) neutralize each other to form a salt and water.

HCl  +  NaOH  ®  NaCl  +  H2O

In the neutralization the H+ reacts with OH- to form water, leaving Na+ and Cl- ions in solution.

# Acids reacting with Carbonates and Bicarbonates

When acids are neutralized by addition to a carbonate or bicarbonate solution the reaction produces bubbles of carbon dioxide gas, a salt and water. In the reaction H+ is transferred to the carbonate to give carbonic acid, H2CO3, which breaks down rapidly to CO2 and H2O.

Bicarbonate (hydrogen carbonate):

H+(aq)  +  HCO3-(aq)  ®  CO2(g)  +  H2O

Carbonate:

H+(aq)  +  CO32-(aq)  ®  CO2(g)  +  H2O

# Balancing Acid-Hydroxide Neutralization Reactions

For complete neutralization all H+ must react with a OH- and vise versa. To balance a neutralization reaction make sure there are as many H+ (from the acid formula) as there are OH-(from the base formula).

Example:

Balance the following neutralization reaction.

?H2CrO4    + ? Fe(OH)3 Þ? Fe?(CrO4)? + ? H2O

To balance the H+ and OH- we must find the lowest common multiple that 2 (H2CrO4) and 3 (Fe(OH)3) go into; this would be 6. Then each has the coefficient equal to how many times it goes into the least common multiple.

H2CrO4    +    Fe(OH)3 Þ? Fe?(CrO4)? + ? H2O

The salt will always have a coefficient of 1 and have the subscript for the cation equal to the coefficient for the base ( the source of the cation) and have a subscript for the anion equal to the coefficient for the acid (the anion source). The number of waters will always be equal to the least common multiple, which equals the total number of H+ and OH- (since we get 1 H2O from every H+ and OH- neutralized).

_  H2CrO4 + _   Fe(OH)3 Þ _  Fe__(CrO4)__ +  _  H2O

Balance the following neutralization reactions:

H2SO4     +       NaOH    Þ    Na__(SO4)___   +       H2O

H3PO3     +       Ca(OH )2   Þ    Ca__(PO3)___   +       H2O

H3AsO4     +      Pb(OH)4    Þ    Pb__(AsO4)___   +       H2O

9.7 Buffers

Buffers resists a change in pH when small amounts of acid or base are added.

When a small amount of a strong acid or base is added to pure water the pH changes drastically. However, if a solution is buffered then there is little change in pH.

A buffer must contain an acid to react with any OH that is added and it must have a base to react with any added H3O+. However, that acid and base must not be able to neutralize each other. Therefore, a combination of an acid-base conjugate pair is used in buffers.

Most buffer solutions consist of nearly equal concentrations of a weak acid and a salt containing its conjugate base (the salt is the primary source of the conjugate base, which is needed to neutralize any incoming acid). The buffering ability (the amount of base or acid that can be neutralized) depends on the amount of the weak acid and the amount of the weak base present.

For example, a typical buffer contains acetic acid (HC2H3O2) and a salt such as sodium acetate (NaC2H3O2). Being a weak acid acetic acid dissociates only slightly in water to form H3O+ and a very small amount of acetate ions. The presence of the salt provides a much larger concentration of acetate ion, which is necessary for its buffering capability.

HC2H3O2(aq)  +  H2O(l)  Û H3O+(aq)  +  C2H3O2-(aq)

When a small amount of acid is added it will be neutralized by the acetate ion and the equilibrium shifts left to produce more of the acetic acid. (Adding acid is adding H3O+ to the right, which shifts to the left.)

If a small amount of base is added it is neutralized by the acetic acid and the equilibrium shifts right to produce more acetate ions. (Adding a base removes H3O+ from the right, which shifts the equilibrium back to the right.)

NOTE: [H3O+] and [OH-] do not change very much during buffering, hence there is little change in pH.

Chem 110G

Wayne McGowan           HOMEWORK #6                                                                  ANSWER KEY

1)      Tell whether each of the following solutions is acidic, basic, or neutral.

a) [H+] = 7.9 x 10-6 mol/L and [OH-] = 1.3 10-9 mol/L             acidic

b) [H+] = 1.0 x 10-7 mol/L and [OH-] = 1.0 x 10-7 mol/L                    neutral

c) [H+] = 7.6 x 10-8 mol/L and [OH-] = 1.2 x 10-7 mol/L          basic

2)      What are acidic and basic solutions in terms of the hydronium ion concentration and the hydroxide ion concentration?  What are they in terms of pH?

Acidic solutions have higher hydronium concentrations than hydroxide concentrations.

Basic solutions have higher hydroxide concentrations than hydronium concentrations.

3)      What is the essential difference between an aqueous solution of a strong acid and that of a weak acid?

In a solution of a strong acid, the solute is near fully ionized. In a solution of weak acid, the percentage ionization is small.

4)      Write a balanced equation for the ionization of each of the following in water:

H2O

a. HCN, a weak acid                HCN  ®¬ H+ + CN-

H2O

b. Ba(OH)2, soluble                 Ba(OH)2  ®Ba2+ + 2OH-

H2O

c. NH3, a weak base                 NH3 + H2®¬NH4+ + OH-

H2O

d. HNO3, a strong acid            HNO3  ®H+ + NO3-

5)      What are the formulas of the conjugate acids of the following?

a) HSO3-               H2SO3

b) Br-                                 HBr

c) H2O                  H3O+

d) CH3CO2          CH3CO2H

6)      Write the formulas of the conjugate bases of the following:

a) NH3                              NH2-

b) HNO2               NO2-

c) HSO3-                         SO32-

d) H2SO3                        HSO3-

7)      Which of the following is a neutralization reaction?

a. 2HCl + Ca(OH)2 ----> CaCl2 + 2H2O                yes

b. H2SO4 + Pb(NO3)2 ----> PbSO4 + 2HNO3                  no

c. Mg + 2HCl ----> MgCl2 + H2                                      yes

d. HClO2 + NaOH ----> NaClO2 + H2O                yes

8)      Balance each of the following neutralization reactions:

a. 2 HNO3  + 1 Ba(OH)2 ----> 1 Ba(NO3)2           + 2 H2O

b. 3 H2SO4 + 2 Al(OH)3 ----> 1 Al2(SO4)3 + 6 H2O

9)      How many moles of sodium hydroxide can react with 0.256 mol of H2SO4 ,assuming that both H+ in H2SO4 are neutralized? (Hint: this is a stoichiometry problem)

2.56 mol H2SO4  2 mol NaOH                                                         = 5.12 mol NaOH

1 mol H2SO4

10)    How many grams of sodium carbonate will neutralize 5.24 g of HCl?

5.24 g HCl  1 mol HCl  1 mol Na2CO3   106 g Na2CO3                  = 7.61 g Na2CO3

36.5 g HCl    2 mol HCl       1 mol Na2CO3

11)    How many grams of calcium metal can react with 118 mL of 0.375 M hydrochloric acid if the reaction is

Ca(s) + 2HCl(aq) ----> CaCl2(s) + H2(g)

118 mL  10-3L    0.375 mol HCl    1 mol Ca     40.1 g                    = 0.887 g Ca

1 mL            1 L           2 mol HCl  1 mol Ca

12)    What is Kw and what does it represent?

Kw is the water dissociation constant, which is the equilibrium constant of the self ionization of water H2O + H2O <==> H3O+ + OH-and is equivalent to the product of the concentrations of hydronium and hydroxide ions. [H3O+][OH-] = 1 x 10-14 M2 (H2O is a liquid of constant concentration so it does not appear in the equation)

13)    What are acidic and basic solutions as described in terms of the hydronium ion concentration and the hydroxide ion concentration? What are they in terms of pH?

Acidic solutions have higher hydronium concentrations than hydroxide concentrations and have pH values less than seven. Basic solutions have higher hydroxide concentrations than hydronium concentrations and have pH values above seven.

14)    Calculate the pH of the following solutions and indicate whether it is acidic or basic.

a. an ammonia solution that has 8.4 x 10-9 M H3O+
pH = -log [H3O+] = 8.08 basic

b. a sample of saliva that has 3.5 x 10-7 M H3O+
pH = -log [H3O+] = 6.46 acidic

c. a sample of rain that has 1.8 x 10-7 M H3O+
pH = -log [H3O+] = 6.74 acidic

d. an apple juice that has 9.3 x 10-4 M H3O+
pH = -log [H3O+] = 3.03 acidic

15)    What is the pH of 0.001 M HCl(aq), assuming 100% ionization?

0.001 M HCl(aq) = 0.001 M H3O+(aq) = 1 x 10-3 M H3O+(aq)

pH = -log [H3O+] = 3.00 acidic

16)    What is the pH of 0.01 M NaOH(aq), assuming 100% ionization?

0.01 M NaOH(aq) = 0.01 M OH-(aq) = 1 x 10-2 M OH-(aq)

[H3O+] = Kw / [OH-] = 1.0 x 10-14 M2 / 1 x 10-2 M = 1 x 10-12 M H3O+(aq)

pH = -log [H3O+] = 12.0 basic

17)    Describe a simple buffer solution and explain how a buffer system functions.

A buffer solution is a solution that resists change in pH upon the addition of small amounts of acid or base and consists of a weak acid or base and a salt of that acid or base, which provides its conjugate. When acid is added its proton is absorbed by the weak base or the conjugate of the weak acid (depending on if a weak acid or weak base system is used), when base is added it removes the proton from the weak acid or the conjugate of the weak base. In both of these systems the addition of acid or base just shifts the equilibrium.

18)    Use the following general equation for an acid in equilibrium with its conjugate base to explain how a buffer solution functions upon the addition of small amounts of acid or base.                     HA + H2O <==> H3O+ + A-

If acid is added the conjugate base, A-, absorbs the proton and the equilibrium shifts to the left. If base is added the weak acid, HA, donates its proton and the equilibrium shifts right.

Chem 110G

Wayne McGowan           HOMEWORK #7  Name ____________________________

1)      What kind of bond between atoms predominates among organic compounds?

2)      Which of the following structures are possible, given the numbers of bonds that various atoms can form and explain your answer if not possible?

a) CH3CH2CH2OCH3

b) CH2CH2CH2CH3

c)  CH3 == CHCH2CH3

d) CH3CH==CHCH2CH3

e) NH2CH2CH2CH3

3)      Write full (expanded) skeletal structures for each of the following molecular formulas. In some structures you will have to use double or triple bonds.

a) CH3OH

b) CH2Cl2

## c) N2H4

d) C2H6

e) CH2O

f) CHO2H

g) NH2OH

h) C2H2

4)      Draw the condensed structural formula for each of the following alkanes:

a. butane                                                     b. 2,3 – dimethylpentane

c. 2,2,4 - trimethyloctane                                     d. 3,3 - dimethylhexane

e. 4 - ethyl - 2 – methylheptane

5)      Indicate whether each of the following pairs of alkanes are structural isomers:

a. 2-methylpentane and 3-methylpentane

b. 3-methylpentane and 2,2,3-trimethylbutane

c. 3-methylhexane and 3,3-dimethylpentane

d. 3-ethylhexane and 2,2,3-trimethylpentane

6)      Draw a structure for each of the following cycloalkanes, using the geometric formula:

a. cyclobutane                                   b. methylcyclopropane

c. 1,1 – dimethylcyclopentane           d. 1 - ethyl - 4 - methylcyclohexane

7)      Write the condensed structural formula for each of the following names:

a.  2-chloropropane                                    b.  1-bromo-2-chloropropane

c.  1,2-difluorocyclopentane              d.  1,1,3,5-tetrachlorocyclohexane

8)      Write the condensed structural formula for each of the following names:

a.  1-chlorobutane                                      b.  1-bromo-2-chlorocyclopropane

c.  3,3-dibromo-2-methylpentane      d.  2,4-dichloro-3-methylhexane

9)      The isomers of C2H4Cl2 are used as industrial solvents.  Write the condensed structural formulas for the two isomers and give their IUPAC names.

10)    Write the four isomers of C3H6F2 and give the IUPAC name for each.

11)    Write the balanced equations for the complete combustion of each of the following:

a.  cyclohexane

b.  butane

c.  octane

12)    Some haloalkanes used as refrigerants have the following names.  Draw their structures.

Freon 14               tetrafluoromethane

Freon 114             1,2-dichloro-1,1,2,2-tetrafluoroethane

Freon C318          octafluorocyclobutane

Wayne McGowan

Chem 110G

10.1 Organic Compounds

Organic compound are covalent molecules that contain carbon hydrogen bonds and normally form non-polar molecules.

The attractions between non-polar molecules are weak, which accounts for their low melting and boiling points.

10.2  Bonding in Organic Compounds

The simplest organic compounds are the hydrocarbons, which consist of only carbon and hydrogen. In other organic compounds oxygen, sulfur, nitrogen and the halogens are bonded to carbon. In all organic molecules every carbon atom always has four bonds.

10.3 Alkanes

Alkanes are the family of hydrocarbons that contain only single bonds between carbons. Having only carbon-carbon single bonds Alkanes are referred to as “Saturated” as they are saturated with hydrogen (the minimum amount of carbon bonds means they have the maximum amount of hydrogen bonds).

Names – The IUPAC system uses prefixes to indicate the number of carbon atoms in the chain (see the next page) and the suffix ane indicates the alkane family

In an alkane the groups attached to each carbon rotate freely about the bond connecting the two carbon atoms.

terminal carbons                                  central carbons

H                                                                   H

|                                                                     |

H – C—       =      CH3—                                 — C —         =       — CH2

|                                                            |

H                                                                   H

expanded          condensed                   expanded                condensed

Molecular Formulas - all alkanes have the formula CnH2n+2

Example:

What is the formula of propane (3 carbons)?

Cycloalkanes

Cycloalkanes result when alkanes have cyclic (ring) structures.

The use of geometric figures is the conventional way to show cyclic structures. When using geometric figures each corner represents a carbon atom.

C3H6                         C4H8                  C5H10              C6H12

Names - cycloalkanes have the prefix cyclo in front of the name of the alkane

Molecular Formulas – all cycloalkanes have the formula CnH2n due to having no terminal carbons

10.4 Branched Alkanes

When an alkane has four or more carbon atoms the atoms can be arranged so that a side group, called a branch or substituent, is attached to the main (parent) carbon chain.

Naming Parent Chains and Alkyl Groups

The branches off of the main, or parent, carbon chain are called alkyl groups. The alkyl group is named by replacing the ane ending of the corresponding alkane name with yl. Alkyl groups do not exist on their own; they must be attached to a parent carbon chain.

Alkane

Name of

Alkane

Corresponding

Alkyl group

## Alkyl group

One carbon

CH4

(1)             ane

CH3

Methyl

Two carbons

CH3-CH3

(2) Ethane

CH3-CH2

yl

Three carbons

CH3-CH2-CH3

(3)             ane

CH3-CH2-CH2

|

CH3-CH-CH3

Propyl

Isopropyl

Four carbons

CH3-(CH2)2-CH3

(4) Butane

CH3-(CH2)2-CH2

yl

Five carbons

CH3-(CH2)3-CH3

(5)             ane

CH3-(CH2)3-CH2

Pentyl

Six carbons

CH3-(CH2)4-CH3

(6) Hexane

CH3-(CH2)4-CH2

yl

Seven carbons

CH3-(CH2)5-CH3

(7)             ane

CH3-(CH2)5-CH2

Heptyl

Eight carbons

CH3-(CH2)6-CH3

(8)             ane

CH3-(CH2)6-CH2

yl

Naming Alkanes with Alkyl Groups:

Step 1: Find the longest chain of carbon atoms; this is the parent.

CH3

|

CH3 – CH – CH2 – CH2 – CH3

Step 2: Number the carbons in the parent starting from the end nearest a substituent.

CH3

|

CH3 – CH – CH2 – CH2 – CH3

Step 3: Give the location and name of each alkyl group, these go in front of the parent name. Use a prefix (di, tri, tetra, etc.) to indicate a group that appears more than once. Hyphens separate numbers from words and commas separate numbers. When there are multiple substituents that allow the numbering from both ends use the direction that gives the lowest numbers. All ties go to the alphabetically first substituent. (When a single substituent is attached to a cycloalkane no number is needed.)

CH3

|

CH3 – CH – CH2 – CH2 – CH3

CH3  CH3

|        |

CH3 – CH2 – CH – CH– CH3

CH3              CH3

|                    |

CH3 – CH – CH2 – C– CH3

|

CH3

Step 4: List the substituents in alphabetical order (discounting di, tri etc.).

CH3 – CH2   CH3

|        |

CH3—CH2—CH2 – CH—CH—CH3

NOTE: Be sure that you find the longest continuous carbon chain.

Example: Name the following alkane:

CH3

|

CH2

|

CH2             CH2– CH3

|                                  |

CH3 – CH – CH2 – C – CH3

|

CH2

|

CH2

|

CH3

#### The IUPAC name gives all information needed to draw the structure. The parent name gives the number of carbon atoms. The first part of the names gives the substituents and where they are.

To draw a condensed structural formula:

Step 1: Draw the parent backbone and number it.

2, 3 – dimethylhexane

Step 2: Place the substituents on the parent.

2, 3 dimethylhexane

Step 3: Add the correct number of hydrogen atoms to give four bonds to each carbon atom.

2, 3 dimethylhexane

NOTE: We will soon omit the hydrogens . We do this to make the picture less “messy”.

C  C

|    |

e.g.:  C–C–C–C–C–C

The naming rules can seem very complicated at first glance and hence can be intimidating. Try to relax and just practice going from names to structures and then from structures to names. You should find that with a bit of practice it gets much easier. The rules will soon start to make since. All organic molecules use similar rules and hence you will get a lot of practice.

Wayne McGowan

Chem 110G

Examples:  Name the following:

CH3

|

CH2             CH2 – CH3                     CH3              CH2 – CH3

|                                  |                                                                       |                                     |

CH3 – CH – CH2 – CH – CH3                     CH2 – CH2 – C – CH3

|

CH2

|

CH3

CH3            CH3                                      CH3             CH2 – CH3

|                                |                                                                         |                                   |

CH3 – CH – CH – CH – CH3            CH3 – CH – CH2 – C – CH3

|                                                                                                                           |

CH2                                                           CH3

|

CH3

CH2 – CH3  CH3                                         CH3          CH2 – CH3

|                                  |                                                                                           |                            |

CH3 – CH – CH2 – CH – CH3           CH3 – CH2 – C – CH – CH – CH3

|                                                                                                               |            |

CH2                                                   CHCH3

|

CH3

Structural Isomers

Structural Isomers result when two compounds have the same molecular formula but a different arrangement of atoms.

Example:

What are the three structural isomers of C5H12.

Are the following isomers, the same or neither:

CH3                                                             CH3

|                                                                   |

CH3 – CH – CH2 – CH2 – CH3          CH3 – CH2 – CH – CH2 – CH3

CH3                                                                       CH3

|                                                                            |

CH3 – CH – CH2 – CH2 – CH3          CH3 – CH2 – CH2 – CH– CH3

CH3                                                              CH3

|                                                                   |

CH3 – CH – CH2 – CH2 – CH3          CH3 – CH2 – CH– CH3

CH3                                         CH3 –CH–CH3

|                                                        |

CH3 – CH – CH2 – CH2 – CH3                    CH2 – CH2 – CH3

If you are given the names to tell if they are isomers then look to see if they have the same number of carbons and are all straight chain alkanes. If they do have the same number of carbons and are both straight chain alkanes then they have the same number of hydrogens as well and hence are isomers.

10.5 Haloalkanes

Haloalkanes are the hydrocarbons of an alkane with one or more halogens attached.

Halogens are named as substituents: fluorine is fluoro, chlorine is chloro, bromine is bromo, and iodine is iodo. The halo-substituents use the same rules as the alkyl groups.                                                                                                                                                            F

|

CH3 – Cl                    CH3 – CH2 – Br             CH3 – CH – CH3

10.6 Properties of Alkanes

Alkanes are nonpolar, which makes them insoluble in water and have the lowest melting and boiling points of all the organic compounds.

###### Reactivity:

Alkanes are the least reactive family of organic compounds. Their main reaction is to undergo combustion when they react with oxygen.

Alkane + O2 ® CO2 + H2O + energy

Example:

Show the combustion of methane.

10.7 Functional Groups

Functional groups are a feature, usually a substituent, which determines the reactivity of the compound.

Alkenes and Alkynes

Alkenes contain a double bond functional group between two carbons; Alkynes contain a triple bond.

H    H                                       H   H

|      |                                         |      |

H – C – C – H                         H – C = C – H                          H – C º C – H

|      |

H    H

Condensed structural formulas:

CH3 – CH3                                CH2 = CH2                                 HC º CH

#### Alcohols and Ethers

The functional group in alcohols is the hydroxyl (–OH) group. In ethers the functional group is an oxygen bridging two carbons.

– O – H                                                            – O –

e.g.:

H    H                                                             H           H

|      |                                                             |             |

H – C – C – O – H                                                    H – C – O – C – H

|      |                                                             |             |

H    H                                                            H           H

Condensed structural formulas:

CH3 – CH2 – OH                                               CH3 – O – CH3

Aldehydes and Ketones

Both aldehydes and ketones contain a carbonyl group (C = O)

In an aldehyde the carbonyl group is on a terminal carbon. In a ketone the carbonyl group is on a central carbon.

O                                                         O

||                                                         ||

– C – H                                                         – C –

e.g.:

O                                                      O

||                                                       ||

CH3 – C – H                                               H3C – C – CH3

Carboxylic Acids

Carboxylic acids are organic compounds containing the carboxyl functional group.

#### O

||

– C – O – H             or      –COOH                or         –CO2H

e.g.:

O

||

CH3 – C – O – H                 or      CH3COOH           or       CH3CO2H

#### Esters are similar to a carboxylic acid except the oxygen bridges to a carbon and not to hydrogen.

O

||

– C – O –                  or        – COO –             or        – CO2

e.g.:

O

||

CH3 – C – O – CH3 or      CH3COOCH3       or    CH3CO2CH3

Amines

Amines are derivatives of ammonia, NH3, in which carbon atoms replace one, two, or three

of the hydrogen atoms.

NH3                            CH3 – NH2            CH3 – NH             CH3 – N – CH3

|                            |

CH3                     CH3

Chem 110G

Wayne McGowan           HOMEWORK #7                                                  ANSWER KEY

1)      What kind of bond between atoms predominates among organic compounds?

Covalent

2)      Which of the following structures are possible, given the numbers of bonds that various atoms can form and explain your answer if not possible?

a) CH3CH2CH2OCH3

Possible

b) CH2CH2CH2CH3

Impossible, the first carbon has only 3 bonds

c)  CH3 == CHCH2CH3

Impossible, the first carbon has 5 bonds

d) CH3CH==CHCH2CH3

Possible

e) NH2CH2CH2CH3

Possible

3)      Write full (expanded) skeletal structures for each of the following molecular formulas. In some structures you will have to use double or triple bonds.

a) CH3OH       H

|

H-C-O-H

|

H

b)CH2Cl2        Cl

|

### H-C-H

|

Cl

c) N2H4           H   H

|     |

## H-N-N-H

d) C2H6             H    H

|      |

H–C—C–H

|      |

H    H

e) CH2O           O

||

H-C-H

f) CHO2H        O

||

H-C-O-H

g) NH2OH       H

|

H-N-O-H

h) C2H2     H—CºC—H

4)      Draw the condensed structural formula for each of the following alkanes:

a. butane                                                     b. 2,3 – dimethylpentane

CH3-CH2-CH2-CH3                                 CH3-CH - CH-CH2-CH3

CH3  CH3

c. 2,2,4 - trimethyloctane                                     d. 3,3 - dimethylhexane

CH3                                                                CH3

CH3-C -CH2-CH-CH2-CH2-CH2-CH3     CH3-CH2-C -CH2-CH2-CH3

CH3     CH3                                                     CH3

e. 4 - ethyl - 2 – methylheptane

CH3-CH-CH2-CH -CH2-CH2-CH3

CH3        CH2-CH3

5)      Indicate whether each of the following pairs of alkanes are structural isomers:

a. 2-methylpentane and 3-methylpentane                      yes

b. 3-methylpentane and 2,2,3-trimethylbutane              no

c. 3-methylhexane and 3,3-dimethylpentane                           yes

d. 3-ethylhexane and 2,2,3-trimethylpentane                          yes

6)      Draw a structure for each of the following cycloalkanes, using the geometric formula:

a. cyclobutane                                   b. methylcyclopropane

-C

c. 1,1 – dimethylcyclopentane           d. 1 - ethyl - 4 - methylcyclohexane

C

C-               -C-C

C

7)      Write the condensed structural formula for each of the following names:

a.  2-chloropropane                                    b.  1-bromo-2-chloropropane

Cl                                                      Cl

C-C-C                                               C-C-C-Br

c.  1,2-difluorocyclopentane              d.  1,1,3,5-tetrachlorocyclohexane

F                                   Cl                -Cl

Cl                -Cl

F

8)      Write the condensed structural formula for each of the following names:

a.  1-chlorobutane                                      b.  1-bromo-2-chlorocyclopropane

Cl

C-C-C-C

Br-           -Cl

c.  3,3-dibromo-2-methylpentane               d.  2,4-dichloro-3-methylhexane

C  Br                                                Cl    Cl

C-C-C-C-C                                       C-C-C-C-C-C

Br                                                     C

9)      The isomers of C2H4Cl2 are used as industrial solvents. Write the condensed structural formulas for the two isomers and give their IUPAC names.

Cl                                            Cl

C-C                                         C-C

Cl                                        Cl

1,2-dichloroethane                1,1-dichloroethane

10)    Write the four isomers of C3H6F2 and give the IUPAC name for each.

F                          F                                    F                               F

C-C-C                  C-C-C                            C-C-C                  C-C-C

F                               F                                         F                     F

1,1-difluoropropane                                  1,3-difluoropropane

1,2-difluoropropane                                  2,2-difluoropropane

11)    Write the balanced equations for the complete combustion of each of the following:

a.  cyclohexane     1 C6H12 + 9 O2  ---->  6 CO2 + 6 H2O

b.  butane             2 C4H10 + 13 O2  ---->  8 CO2 + 10 H2O

c.  octane              2 C8H18 + 25 O2  ---->  16 CO2 + 18 H2O

12)    Some haloalkanes used as refrigerants have the following names.  Draw their structures.

F

Freon 14               tetrafluoromethane                  F-C-F

F

Cl  Cl

Freon 114             1,2-dichloro-1,1,2,2-tetrafluoroethane        F-C-C-F

F   F

Freon C318          octafluorocyclobutane        F         F

F-            -F

F-            -F

F        F

Chem 110G

Wayne McGowan           HOMEWORK #8                  Name ____________________________

1)      Draw the condensed structural formulas for each of the following compounds:

a.  3-methyl-1-hexene                        b.  1-methylcyclopentene

c.  1,1-dichloro-2-butene                             d.  1,3-dimethyleyclobutene

2)      Draw condensed structural formulas for each of the following compounds:

a.  cis-2-pentene                                b.  cis-1-bromo-2-chloroethene

c.  trans-2,3-dibromo-2-butene                   d.  cis-3-hexene

3)      Name the following compounds:

a.       C          C-C                                       b. Cl          C

C=C                                            C=C

C –C                                               Br          Br

c.   Br                                                d.  C-C          C-C

C=C                                                    C=C

Br        C        -C                                                       C-C

4)      Draw the condensed structural formulas for each of the following alkynes:

a.  3-chloro-1-propyne                      b.  3-ethyl-1-pentyne

c.  1-butyne                                       d. 1-bromo-4-methyl-2-hexyne

5)      Write an equation for the following reactions:

a.  hydration of 1-methylcyclobutene

b.  hydrogenation of 3-hexene

c.  addition of hydrogen bromide to 2-methyl-2-butene

d.  addition of chlorine to 2,3-dimethyl-2-pentene

6)      Draw the condensed structural formulas for each of the following:

a.  ethylbenzene                                 b.  1,3,5-trichlorobenzene

c.  meta-chlorotoluene                       d.  ortho-diethylbenzene

7)      Give the names of the following:

a.                C-C-C                                      b.             Cl

–OH                               Cl-

Cl

c.               Cl                                    d.              C-C

C–           –C-C                                               –NH2

Wayne McGowan

Chem 110G

11.1 Alkenes and Alkynes

Alkenes and Alkynes are families of hydrocarbons that contain double and triple bonds.

They are called unsaturated hydrocarbons (alkane carbons are “saturated” with hydrogens).

An alkene contains at least one double bond between carbons and has a flat geometry at the double bond.

A             X

C = C

B             Y

A, B, X, Y and the two Carbons all lie flat in the same plane.

11.2 Names

1. Identify the longest carbon chain that contains the double or triple bond. Use the suffix –ene for an alkene and –yne for an alkyne.

2. Number the parent chain from the end nearest the double or triple bond. Indicate it’s position with the number of the first carbon (for 3 or less carbons no number is needed).

CH3–CH2–CH=CH2       CH3–CH=CH–CH3       CH3–CºC–CH3

3. Place the number and names of substituents in front of the parent name.

CH3                  H3C   CH3                    CH3

|                            |     |                           |

CH2=CH–CH2–CH–CH3              CH3–C=C–CH3        CH3–CH–C º CH

4. For cycloalkenes the double bond is between carbons 1 and 2.

Example:

Name the following:

C2H5                  H3C              C2H5

|                                |            |                                                CH3

CH3–CH=C–CH–CH3    CH3–CH–CH2–CH–C º CH

11.3 Cis-Trans Isomers

There is no rotation of the rigid double bond. Any groups attached remain fixed on one side. In a cis isomer the parent is on the same side of the double bond. In the trans isomer the parent is on opposite sides.

A          X

C = C

B          Y

A and X are cis to each other, as are B and Y. A and Y are trans to each other, as are B and X.

The cis/trans given is based upon the parent chain. If the parent ends at the double bond then the largest substituent determines the cis/trans isomer. If one of the carbons has identical groups then it does not have cis-trans isomers.

same side                                      opposite sides

 CH2 – CH3
 CH3 – CH2
 CH2 – CH3

H

C = C                                            C = C

 CH3 – CH2

H                   H                                                    H

Addition of reactants occurs because double and triple bonds are easily broken, which provides electrons for new single bonds.

The general equation for the addition of a reactant A – B to an alkene can be written as follows.

A    B

\        /                     addition      |      |

C = C    +    A – B      ®       – C – C –

/        \                                        |      |

alkene

Hydrogenation – adds hydrogen across the double bond

H   H

\        /                     addition       |      |

C = C    +    H – H      ®        – C – C –

/        \                                         |      |

double bond                           single bond

Halogenation – adds a halogen across the double bond

Cl   Cl

\        /                      addition       |      |

C = C    +    Cl – Cl      ®       – C – C –

/        \                                          |      |

alkene                                 haloalkane

Hydro-Halogenation – adds a hydrogen halide across the double bond

H   Cl

\        /                      addition       |     |

C = C    +    H – Cl      ®        – C – C –

/        \                                          |     |

alkene                                  haloalkane

Hydration/Hydrolysis – adds water across the double bond

H   OH

\        /                       addition       |      |

C = C    +    H – OH      ®       – C – C –

/        \                                           |      |

alkene                                         alcohol

Markovnikov’s Rule:

When water or a hydrogen halide adds to a double bond the H attaches to the carbon with the most H atoms. (think of hydrogens as money and the saying that “the rich get richer”; later we will see where “the poor get poorer”)

OH    H

CH3 – CH = CH2   +   H – OH        ®           CH3 – CH – CH2

major product:

11.6 Aromatic Compounds

The simplest aromatic compound is benzene. The carbon atoms were arranged in a flat ring with alternating single and double bonds between the carbon atoms.

Û                         Þ

resonance structures                   most common representation

Names

These compounds are usually named as benzene derivatives, however, many use their common names, such as toluene, aniline and phenol.

–CH3                              –NH2                            –OH

Toluene                            Aniline                           Phenol

The carbon containing these special substituents is always carbon number 1 with the number assumed and not part of the name.

When a benzene ring is a substituent, C6H5– , it is named as a phenyl group.

When there are two or more substituents on a benzene ring the ring is numbered to give the lowest numbering (rules like cyclic alkanes).

When there are only two substituents on a benzene ring the prefixes ortho (o) for 1, 2, meta (m) for 1, 3, and para (p) for 1, 4 are often used to show their relative position to one another.

Example:

Name the following:

–Cl              C2H5–         –C2H5           Br–         –OH

Draw 4-ethyl-2,2-dimethyl-3-hexene:

Draw 3-ethyl-6-methyl-4-propyl-3-heptene

Name the following:

 CH3             CH2 – CH3  |                   | CH = CH  – C – CH3                      |                    CH3
CH3          CH2 – CH3

|                |

CH ºCH2 – C  – CH – CH – CH3

|       |

CH3 CH3

Chem 110G

Wayne McGowan           HOMEWORK #8                                                                  ANSWER KEY

1)      Draw the condensed structural formulas for each of the following compounds:

a.  3-methyl-1-hexene                        b.  1-methylcyclopentene

#### C=C-C-C-C-C                                    C—

c.  1,1-dichloro-2-butene                             d.  1,3-dimethyleyclobutene

Cl                                                             C-

C-C=C-C

Cl                                                                             -C

2)      Draw condensed structural formulas for each of the following compounds:

a.  cis-2-pentene                                b.  cis-1-bromo-2-chloroethene

C         C-C                                       Br        Cl

C=C                                                   C=C

c.  trans-2,3-dibromo-2-butene                   d.  cis-3-hexene

C         Br                                          C-C          C-C

C=C                                                       C=C

Br       C

3)      Name the following compounds:

a.       C         C-C                              b. Cl          C

C=C                                            C=C

C –C                                               Br          Br

trans-3-methyl-3-hexene                  trans-1,2-dibromo-1-chloro propene

c.   Br                                                d.  C-C          C-C

C=C                                                    C=C

Br        C        -C                                                       C-C

1,1-dibromo-1-butene                      3-ethyl-3-hexene

4)      Draw the condensed structural formulas for each of the following alkynes:

a.  3-chloro-1-propyne                      b.  3-ethyl-1-pentyne

Cl                                                         C-C

C ºC-C                                             C ºC-C-C-C

c.  1-butyne                                       d. 1-bromo-4-methyl-2-hexyne

##### C ºC-C-C                                                      Br             C

C-C ºC-C-C-C

5)      Write an equation for the following reactions:

a.  hydration of 1-methylcyclobutene                   OH

C-                          H+      C-          -H

+  H2O     ®

b.  hydrogenation of 3-hexene

Pt

C-C-C=C-C-C  +  H2  ®  C-C-C-C-C-C

c.  addition of hydrogen bromide to 2-methyl-2-butene

C                                  C

C-C=C-C  +  HBr   ®  C-C-C-C

BrH

d.  addition of chlorine to 2,3-dimethyl-2-pentene

C  C                            C  C

C-C=C-C-C  +  Cl2  ®  C-C-C-C-C

Cl Cl

6)      Draw the condensed structural formulas for each of the following:

a.  ethylbenzene                                 b.  1,3,5-trichlorobenzene

Cl

-C2H5                                       Cl-

Cl

c.  meta-chlorotoluene                       d.  ortho-diethylbenzene

Cl                                    C-C         C-C

C-

7)      Give the names of the following:

a.                C-C-C                                      b.             Cl

–OH                               Cl-

Cl

ortho-propyl phenol                                  1,3,5-trichloro benzene

c.               Cl                                    d.              C-C

C–           –C-C                                               –NH2

3-chloro-4-ethyl toluene                  ortho-ethyl aniline

Chem 110G

Wayne McGowan           HOMEWORK #9                Name ____________________________

1)      Write the skeletal formula of each of the following alcohols:

a.  ethanol

b.  3-methyl-1-butanol

c.  2,4-diclorocyclohexanol

d.  1-propanol

e.  2,2,4-trimethyl-3-hexanol

2)      Write the skeletal formula of each of the following phenols:

a.  ortho-ethylphenol                        b.  2,4-dichlorophenol

c.  2,4-dimethylphenol             d.  2-ethyl-5-methylphenol

3)      Why is propanol more soluble in water than hexanol?

4)      Write the skeletal formula for each of the following ethers:

a.  methyl propyl ether                               b.  ethyl cyclopropyl ether

c.  methoxycyclopentane                                      d.  1-ethoxy-2-methylbutane

5)      Write the condensed structural formula for each of the following compounds:

a.  propanal                                                          b.  butanal

c.  3-methyl-2-pentanone                                     d.  2 - chloro pentanal

6)      Draw the skeletal formula of the alcohol needed for the following oxidation products:

a.  ethanal                                                   b.  4-methyl-2-pentanone

c.  cyclohexanone                                       d.  propanal

e.  3-methylbutanal

7)      Give the IUPAC names of the following compounds.

a) CH3CH2OH                                   b) HOCH2CH2OH

OH                                                         OH

c) HOCH2CHCH2OH                        d) CH3CHCH3

8)      What are the IUPAC names of the following compounds?

a) CH3CH2CH2OH                            b) HOCH2CH2CH2CH3

CH3                                                         CH3

c)  HOCCH3                                                d) HOCH2CH

CH3                                                         CH3

9)      Draw a figure that illustrates a hydrogen bond between two molecules of ethanol.

Use a dotted line to represent this bond, and write in the d+ and d- symbols where they belong.

10)    Write the structures of the alkenes that form when the following alcohols undergo

dehydration. Where more than one alkene is possible, identify which forms in the greatest relative amount. Write the names of the reactant alcohols and product alkenes.

CH3                                                    OH

a) CH3CHCH2OH                                       b) CH3CHCH2CH3

c)                     CH3                                     d)            OH

CH3CH2CH2CCH3                             CH3CHCHCH3

OH                                             CH2CH3

11)    Write the structure of any alcohol that could be used to prepare each of the following

compounds by an oxidation. Name each alcohol used.

a)       O                                                       b)                O

||                                                                           ||

HOCCH2CH2CH3                          CH3CH2CH2CCH(CH3)2

12)    Which of the following compounds contain the ether functional group?

a) CH3CH2OCH2CH2CH2OCH3                  b)          O

||

CH3COCH3

c)          O                                          d)                      O

||                                                                              ||

CH3OCCH2CH3                                CH3OCH2CH2CCH3

Wayne McGowan

Chem 110G

12.1  Alcohols, Phenols, and Thiols

# Phenols -have a hydroxyl group (-OH) as a substituent on an benzene ring

Thiols - similar to alcohols with a thiol group (-SH) as a substituent on the parent

# Names of Alcohols and Phenols

1) Number the main chain starting at the end closest to the –OH group.

2) The alcohol is indicated by an ol ending and uses a number to show the location of the hydroxyl group.

OH

|

CH3CH2CH2 -OH                                 CH3CHCH3

For Phenols carbon 1 has the –OH group and the ring is numbered to give other substituents the lowest numbers.

OH                                   OH                                         OH

Br

Cl

Example:                                                                                                             OH

Name the following alcohols:

CH3CH2CH–OH                               CH3CHCH2CH2CH2CH3

|                                                  |

CH3CHCH3                                 CH3CHCH2 –OH                           C-C

Classification of Alcohols - by the number of carbon groups attached to the carbon with the hydroxyl (-OH)

A primary (1o) alcohol has one alkyl group, a secondary (2o) alcohol has two alkyl groups and a tertiary (3o) alcohol has three alkyl groups.

# H                                     R2                                   R2

|                                                |                                     |

R1 – C – OH                              R1 – C – OH                  R1 – C – OH

|                                                |                                     |

H                                     H                                    R3

Physical Properties  -are primarily related to hydrogen bonding

The -OH groups cause increases in melting and boiling points and solubility in water.

As the size of an alcohol increases (longer carbon chains) it becomes more alkane-like.

The more -OH groups there are per molecule the more polar regions there are and hence the more soluble it is in water.

12.2  Some Common and Important Alcohols and Phenols

OH                       OH                                          OH

|                             |                                             |

CH3                                      CH2 – CH3                    CH3– CH – CH3

wood alcohol                liquor alcohol                 isopropyl alcohol

OH    OH             OH    OH    OH                                 OH

|          |                |        |        |

CH2 – CH2            CH2– CH – CH2

OH

ethylene glycol     glycerol (glycerin)                      resorcinol

12.3 Ethers – have an oxygen bridging 2 carbon groups (-O-)

# Names

For the common name write each alkyl group attached to the oxygen atom in alphabetical order followed by ether.

CH3 – O – CH2 – CH2 – CH3

For the IUPAC name the smallest alkyl group attached to the oxygen along with the oxygen is named as an alkoxy substituent.

CH3 – O –            CH3 – CH2 – O –           CH3 – CH2 – CH2 – O –

Example:

Name the following ethers using common names and the IUPAC names:

CH3 – O – CH2 – CH2 – CH3            CH3 – CH2 – O – CH2– CH3

CH3                                                                                                                          CH3

|                                                                                    |

CH3 – CH – O– CH3                                          CH3 – CH2 – O – CH2– CH– CH3

12.4 Properties of Alcohols, Phenols, and Ethers

The oxygen and hydrogen in the hydroxyl group participate in hydrogen bonding. Ethers do not hydrogen bond with each other but they do hydrogen bond with water.

# Boiling Points

Higher temperatures are required to provide the energy needed to break the hydrogen bonds between alcohol molecules. Thus, alcohols have higher boiling points than alkanes and ethers of similar mass.

# Solubility in Water

The oxygen atom in alcohols and ethers influences their solubility in water. In alcohols, the polar –OH group can hydrogen bond with water, which makes alcohols with one to four carbon atoms very soluble in water.

Ethers are more soluble in water than alkanes of similar mass but are not as soluble as alcohols.

12.5 Reactions of alcohols and Thiols

# Dehydration of alcohols to Form Alkenes

Alcohols lose a water molecule when they are heated with an acid catalyst and a double bond forms.

H   OH

|     |

C – C      ®

For 2o and 3o alcohols the hydrogen is taken from the carbon with the fewest hydrogens (the poor get poorer).

H    OH  H

|      |      |

C – C – C – C       ®

# Oxidation of Alcohols

In organic chemistry it is convenient to think of oxidation as a loss of hydrogen atoms or the addition of oxygen atoms.

The oxidation of a primary alcohol produces an aldehyde.

O – H

|                    [O]

R – C – H             ®

|

H

Aldehydes can oxidize further to form a carboxylic acids.

O

||                             [O]

R – C – H            ®

In the oxidation of secondary alcohols the products are ketones.

O – H

|                             [O]

R1 – C – R2           ®

|

H

# Oxidation of Thiols

Thiols undergo oxidation to yield a disulfide.

[O]

R – S – H + H – S – R      ®

Wayne McGowan

Chem 110G

12.6 Aldehydes and Ketones

Aldehydes and Ketones contain the carbonyl group. In an aldehyde the carbonyl group is terminal, in a ketone the carbonyl group is central.

O                                                      O

||                                                        ||

R1 – C – H                                                 R1 – C – R2

# Naming Ketones

1) Name the longest carbon chain containing the carbonyl group with the suffix one.

2) Number the main chain starting from the end nearest the carbonyl group. Use the carbonyl carbon number to indicate its position. Ketones with fewer than 5 carbons do not need, nor use, a number.

# Naming Aldehydes

1) Name the longest carbon chain containing the carbonyl group with the suffix al. No number is needed for the aldehyde group.

The aldehyde of benzene is benzaldehyde.

H         O

C

Example:

Name the following:

O            CH3                 CH3  CH3          O                           CH3         O

||              |                        |         |                 ||                           |               ||

H – C – CH2– CH– CH3                  CH2– CH– CH2 – CH            CH3– CH– CH2–C– CH3

12.7 Some Important Aldehydes and Ketones

O                                                       O

||                                                        ||

H – C – H                                      CH3 – C – CH3

12.8 Properties of Aldehydes and Ketones

The polar carbonyl group has an influence on the boiling points and the solubility of aldehydes and ketones in water.

# Boiling Points

The polar carbonyl group gives aldehydes and ketones higher boiling points than alkanes and ethers of similar mass. However, because there is no hydrogen on the oxygen atom, aldehydes and ketones cannot form bonds with each other. Thus they have boiling points that are lower than alcohols.

Solubility in Water

Carbonyl compounds with one to four carbons are very soluble in water.

# Oxidation and Reduction

Oxidation is the gain of oxygen or the loss of hydrogen. Reduction is the opposite; the gain of hydrogen or loss of oxygen.

Aldehydes oxidize readily to carboxylic acids. In contrast, ketones do not undergo further oxidation.

O                                  O

||          [O]                   ||

CH3 – C – H             ®           CH3 – C – OH

12.9 Chiral Molecules

Another group of isomers, called stereoisomers, have the atoms bonded in the same sequence but differ in the way they are arranged in space.

The left hand is the mirror image of the right hand. Your hands are not superimposable. Objects such as hands that have nonsuperimposable mirror images are chiral.

Molecules in nature also have mirror images and often the “left-handed” stereoisomer has a different biological effect than does the “right-handed” one.

# Chiral Carbon Atoms

A compound is chiral if it has at least one carbon atom bonded to four different atoms or groups. There are two different ways that it can bond to four atoms or groups of atoms. The resulting structures are nonsuperimposable mirror images of each other. When stereoisomers cannot be superimposed they are called enantiomers.

# Drawing Fischer Projections                     (used for carbohydrates)

A simplified system for drawing stereoisomers that shows the arrangements of the atoms. The horizontal lines represent the bonds that come forward in the three-dimensional structure and the vertical lines represent the bonds that point away.

The carbonyl group is always written at the top and the letter L assigned to the stereoisomer which has the –OH group on the left of the chiral carbon. The letter D is assigned to the structure where the –OH is on the right of the chiral carbon.

H         O

C

|

HO – C – H

|

R

L – isomer

H         O

C

|

H – C – OH

|

R

#### D – isomer

Examples:

Write the structure of the carbonyl produced from the following alcohols by an oxidation. Name each alcohol used and carbonyl produced.

OH

|                                                                 [O]

CH2CH2CH2CH3                               ®

OH

|                                              [O]

CH3CH2CHCHCH3                                     ®

|

CH2CH2CH3

Write the structure of any alcohol that could be used to prepare each of the following compounds by an oxidation. Name each alcohol used and carbonyl produced.

CH3      O

[O]                               |            ||

®                         CH3CHCH2CH

O

[O]                                             ||

®                                  CH3CH2CH2CCH2CH3

Chem 110G

Wayne McGowan           HOMEWORK #9                                                                ANSWER KEY

1)      Write the skeletal formula of each of the following alcohols:

a.  ethanol                      C-C-OH

OH  C

b.  3-methyl-1-butanol                                         C-C-C-C

c.  2,4-diclorocyclohexanol        Cl -                    - OH

Cl

d.  1-propanol                                                      C-C-C-OH

C      C

e.  2,2,4-trimethyl-3-hexanol                                C-C-C-C-C-C

C  OH

2)      Write the skeletal formula of each of the following phenols:

a.  o-ethylphenol                      b.  2,4-dichlorophenol

C2H5                                                Cl

- OH                             Cl -              - OH

c.  2,4-dimethylphenol             d.  2-ethyl-5-methylphenol

CH3                                                 C2H5l

CH3 -            - OH                                                         - OH

CH3

3)      Why is propanol more soluble in water than hexanol?

Because it is a shorter chain alcohol and short chain alcohols have a greater solubility in polar solvents than longer chain alcohols do.

4)      Write the skeletal formula for each of the following ethers:

a.  methyl propyl ether                               b.  ethyl cyclopropyl ether

C-O-C-C

-         O -C-C

c.  methoxycyclopentane                                      d.  1-ethoxy-2-methylbutane

C

- O -C                                         C-C-O-C-C-C-C

5)      Write the condensed structural formula for each of the following compounds:

a.  propanal                                                          b.  butanal

O                                                              O

C-C-C                                                      C-C-C-C

c.  3-methyl-2-pentanone                                     d.  2 - chloro pentanal

O C                                                      O Cl

##### C-C-C-C-C                                                              C-C-C-C-C

6)      Draw the skeletal formula of the alcohol needed for the following oxidation products:

a.  ethanal                                                   b.  4-methyl-2-pentanone

OH                                                                            OH  C

C-C                                                            C-C-C-C-C

c.  cyclohexanone                                       d.  propanal

OH

- OH                                             C-C-C

e.  3-methylbutanal

OH  C

C-C-C-C

7)      Give the IUPAC names of the following compounds.

a) CH3CH2OH                                   b) HOCH2CH2OH

ethanol                                             1,2-ethanediol

OH                                              OH

c) HOCH2CHCH2OH                        d) CH3CHCH3

1,2,3-propanetriol                            2-propanol

8)      What are the IUPAC names of the following compounds?

a) CH3CH2CH2OH                            b) HOCH2CH2CH2CH3

1-propanol                                       1-butanol

CH3                                                         CH3

c)  HOCCH3                                                d) HOCH2CH

CH3                                                         CH3

2-methyl-2-propanol                        2-methyl-1-propanol

9)      Draw a figure that illustrates a hydrogen bond between two molecules of ethanol.

Use a dotted line to represent this bond, and write in the d+ and d- symbols where they belong.                       CH2CH3

d+   d-  /

H¼¼¼: O :

/               \

CH3CH2 – O :              H

. .

10)    Write the structures of the alkenes that form when the following alcohols undergo

dehydration. Where more than one alkene is possible, identify which forms in the greatest relative amount. Write the names of the reactant alcohols and product alkenes.

CH3                                                    OH

a) CH3CHCH2OH                                       b) CH3CHCH2CH3

2-methyl-1-propanol                        2-butanol

CH3                                                CH3CH = CHCH3

CH3C = CH2                                2-butene

2-methyl propene

c)                     CH3                                     d)            OH

CH3CH2CH2CCH3                             CH3CHCHCH3

OH                                             CH2CH3

2-methyl-2-pentanol                        3-methyl-2-pentanol

CH3                            CH3CH = CCH3

CH3CH2CH = CCH3                                          CH2CH3

2-methyl-2-pentene                          3-methyl-2-pentene

11)    Write the structure of any alcohol that cold be used to prepare each of the following

compounds by an oxidation. Name each alcohol used.

a)       O                                                       b)                O

||                                                                 ||

HOCCH2CH2CH3                          CH3CH2CH2CCH(CH3)2

OH                                                                     OH

CH2CH2CH2CH3                                     CH3CH2CH2CHCH(CH3)2

1-butanol                                          2-methyl-3-hexanol

12)    Which of the following compounds contain the ether functional group?

a) CH3CH2OCH2CH2CH2OCH3                  b)          O

||

CH3COCH3

ether                                                 not ether (ester; due to double bonded oxygen on the same carbon)

c)          O                                          d)                      O

||                                                                              ||

CH3OCCH2CH3                                CH3OCH2CH2CCH3

not ether (ester; due to double         ether (and ketone)

bonded oxygen on same carbon)

Chem 110 G

Wayne McGowan           HOMEWORK #10              Name ____________________________

1)      Draw the condensed structural formulas of each of the following carboxylic acids:

a.  butanoic acid                      b.  para-ethylbenzoic acid

c.  2-methyl propanoic acid     d.  2,4-dibromobutanoic acid

e.  m-methyl benzoic acid                  f.  4,4-dibromohexanoic acid

2)      In the following pairs, which compound would be more soluble in water?  Explain.

a.  propanoic acid or 1methoxy propane

b.  butanoic acid or ethanoic acid

c.  sodium pentanoate  or pentanoic acid

3)      Write the skeletal structures of the esters formed by reacting each of the following carboxylic acids with ethanol:

a.  methanoic acid                    b.  propanoic acid                   c.  2-methylpentanoic acid

4)      When acetic acid (ethanoic acid) reacts with the following alcohols, what are the

structures of the esters that form?

a) methanol                                       b) 2-propanol                c) phenol

5)      Give the names of the carboxylic acids used to produce the following esters:

O                                                        O

a.  C-C-C-O-C-C                               b.  C-O-C-C-C-C-C-C

O C                                                    O

c.  C-O-C-C-C-C                               d.  C-C-C-O-C-C-C-C

6)      Give the IUPAC names of the esters in problem #5.

a.                                                       b.

c.                                                       d.

7)      Draw the condensed structural formulas of each of the following esters:

a.  hexyl ethanoate

b.  propyl propanoate

c.  ethyl 2-ethyl butanoate

d.  methyl benzoate

8)      Write the equation for the hydrolysis of each compound. If no reaction occurs, write

“no reaction.”

a)       O         CH3

||           |

CH3COCH2CHCH3

b)                  O

||

CH3CH2OC

c)      O

||

CH3CCH2OCH3

d) CH3CH2OCH2CH2CH3

9)      Give the IUPAC names for each of the following amines:

a.  CH3CH2-NH2

b.  CH3-NH-CH2CH2CH3

CH2CH3

c.  CH3CH2-N-CH2CH3

d.                 - NHCH3

10)    There are four amine isomers with the molecular formula C3H9N.  Draw their condensed structural formulas.  Name and classify each as a primary, secondary, or tertiary amine.

a.                          b.                          c.                          d.

11)    Draw the condensed structural formulas of the amides formed in each of the following reactions:

O

a.  C-C-C-C-C-OH  +  NH3       ®

C      O

b. C-C-C-C-OH   +  N-C-C-C  ®

O

c.  C-C-C-OH   +   N-              ®

Wayne McGowan

Chem 110G

13.1 Carboxylic Acids

In a carboxylic acid a hydroxyl group is attached with a carbonyl group forming a carboxyl group.

O

| |

-C – OH

# Names

1) Identify the longest carbon chain containing the carboxyl group and replace the e of the alkane name by oic acid.

2) Number the carbon chain beginning with the carboxyl carbon as carbon 1.

The aromatic carboxylic acid is called benzoic acid. The carboxyl carbon resides off of the carbon ring and is bonded to carbon 1. The ring is numbered in the direction that gives substituents the smallest possible numbers.

CO2H                               CO2H                                   CO2H

C2H5

Cl

13.2 Properties

Carboxylic acids are among the most polar organic compounds because the functional group consists of two polar groups: a hydroxyl (-OH) group and a carbonyl (C = O) group.

Carboxylic acids form hydrogen bonds with other carboxylic acid molecules and water. This ability to form hydrogen bonds has a major influence on both their boiling points and solubility in water.

Melting and Boiling Points

Carboxylic acids have higher boiling points that alcohols, ketones, and aldehydes of similar mass.

# Solubility in Water

Carboxylic acids with one to four carbons are very soluble in water because the carboxyl group forms hydrogen bonds with several water molecules.

13.3 Acidity of Carboxylic Acids

One of the most important properties of carboxylic acids is their ionization in water as weak acids. In the ionization a carboxylic acid donates a proton to a water molecule to produce a carboxylate ion and a hydronium ion. Carboxylic acids are more acidic than other organic compounds but only a small percentage (< 1%) of the carboxylic acid molecules are ionized.

# Neutralization of Carboxylic Acids

Carboxylic acids are weak acids and are completely neutralized by strong bases. The products are water and a salt, which consists of a carboxylate ion and the metal ion from the base.

O                                                 O

| |                                                       | |

R  - C -  OH    +     MOH     ®        R -  C  -  O-  M+     +     H2O

13.4 Esters

Esters are the derivatives of carboxylic acids in which an alkoxy group (-OR) can be thought of as replacing the hydroxyl (-OH) group in the carboxylic acid.

In a reaction called esterification a carboxylic acid reacts with an alcohol. In the reaction water is produced from the  – OH removed from the carboxylic acid and a – H lost by the alcohol.

O                                                         O

| |                                                              | |

# Names

The name of an ester consists of two parts taken from the names of the alcohol (1st part) and the acid (2nd part). The first part is the name of the alkyl group from the alcohol. The second part is the name of the carboxylic acid with the suffix “oate”.

 O  || C – CH3 CH3  H CH3CH3
 CH3 – O

H   +  H – O

Ester name

## +                                                          =

Example:

Name the following esters and give the names of the alcohol and carboxylic acid from which they were formed:

O

| |

C-C-C-C-O-C-C

ester:

alcohol:

acid:

O

| |

C-C-C-O-C

ester:

alcohol:

acid:

C     O  C

|      | |   |

C-C-O-C-C-C

ester:

alcohol:

acid:

O          O- C2H5

C

ester:

alcohol:

acid

13.5 Chemical Properties of Esters

# Acid Hydrolysis of Esters

In hydrolysis esters are split apart with water when heated in the presence of a strong acid, usually H2SO4 or HCl. The products are the carboxylic acid and the alcohol from which it was made. Therefore, hydrolysis is the reverse of the esterification reaction.

O                                                         O

| |                                                     | |

R – C – O – R’  +  H – OH    Û    R – C – O – H  +  R’ –OH

ester                   water              carboxylic acid       alcohol

Base Hydrolysis of Esters

When an ester undergoes hydrolysis with a strong base the products are the carboxylic acid salt and the corresponding alcohol.

The base hydrolysis reaction is also called saponification, which refers to the reaction of a long-chain fatty acid to make soap. The carboxylic acid, which is produced in acid hydrolysis, is converted in an irreversible reaction to its carboxylate ion by the strong base.

O                                                            O

| |                                                       | |

R – C – O – R    +    NaOH     Þ     R – C – O-  Na+   +   R –OH

ester          sodium hydroxide     carboxylic acid salt      alcohol

Wayne McGowan

Chem 110G

13.6 Amines

Amines are considered as derivatives of ammonia (NH3) in which one or more hydrogen atoms is replaced with carbon groups.

# Classification of Amines

Amines are classified in a similar way to alcohols, by the number of carbon atoms directly bonded to the nitrogen atom. In a primary (1o) amine one carbon is bonded to a nitrogen atom. In a secondary (2o) amine two carbons are bonded to the nitrogen. A tertiary (3o) amine has three carbons bonded to the nitrogen.

# Naming Amines

1) Find the longest chain containing the amine and name it with the suffix amine.

CH3 – NH2                    CH3 – CH2 – NH2

2) Number from the end nearest the amine and include the carbon number to indicate its position (for 3 or more carbons).

NH2

|

CH3 - CH2 - CH2 - NH2            CH3 - CH - CH3

3) In secondary and tertiary amines the largest group attached to the nitrogen is the parent. The smaller alkyl groups are named with the prefix N- followed by the alkyl name and listed alphabetically.

CH3                                             CH3

|                                                    |

CH3-CH2-CH2 - N - CH3          CH3 - CH2 - CH2 - N - CH2 - CH3

The aromatic amines use the name aniline.

NH2                                 NH2                                     NH2

Br

Cl

13.7 Properties of Amines

Because amines contain a polar N – H bond they form hydrogen bonds. However, the nitrogen atom in amines is not as electronegative as the oxygen in alcohols, which makes the hydrogen bonds weaker in amines.

Melting and Boiling Points

Amines have higher boiling points than hydrocarbons of similar mass but lower than the alcohols.

In primary amines -NH2 can form more hydrogen bonds, which gives them higher boiling points than the secondary amines of the same mass. It is not possible for tertiary amines to hydrogen bond with each other (no N-H bonds), which makes their boiling points much lower and similar to those of alkanes and ethers.

# Solubility in Water

Like alcohols, the smaller amines, including tertiary ones, are soluble in water because they form hydrogen bonds with water. However, in amines with more than six carbon atoms the effect of hydrogen bonding is diminished and their solubility in water decreases.

# Amines React as Bases

In water amines act as bases because the lone electron pair on the nitrogen atom can accept a proton from water and produces hydroxide ions.

CH3 – NH2   +    H2O     Û     CH3 – NH3+         +        OH-

In a neutralization reaction an amine acts as a base and reacts with an acid to form an amine salt.

As ionic compounds, amine salts are solids at room temperature, colorless, and soluble in water and body fluids. For this reason, amines used as drugs are converted to their amine salts.

When an amine salt reacts with a strong base it is converted back to the amine, which is also called the free amine or free base.

CH3  - NH3+  Cl-  +  NaOH  ®  CH3 – NH2  +  NaCl  +  H2O

13.8 Amides

The amides are derivatives of carboxylic acids in which a nitrogen group replaces the hydroxyl group. An amide is produced during amidization when a carboxylic acid reacts with ammonia or an amine.   A molecule of water is eliminated and the fragments of the carboxylic acid and amine molecules join to form the amide, much like the formation of an ester.

O                                 H                        O    H

| |                          |                          | |     |

R – C – OH   +   H – N – H   ®    R – C – N – H   +   H2O

# Naming Amides

Primary amides are named by dropping the oic acid from the carboxylic acid names and adding the suffix amide.  For secondary and tertiary amides use N- before the alkyl group name to show its attachment.

O                                             O                                          O

| |                                             | |                                           | |

H – C – NH2                  CH3C NH2            CH3CH2CH2C – NH – CH3

# Physical Properties of Amides

For primary amides, the –NH2 group can form several hydrogen bonds, which gives primary amides high melting points. The melting points of the secondary amides are lower because the number of hydrogen bonds decreases. Tertiary amides have even lower melting points because they cannot form hydrogen bonds with other tertiary amides.

The amides with one to five carbon atoms are soluble in water because they can hydrogen bonds with water molecules.

# Hydrolysis of Amides

Amides undergo hydrolysis when water is added back to the amide bond to split the molecule. When an acid is used the hydrolysis products are the carboxylic acid and the ammonium salt. In base hydrolysis the amide produces the salt of the carboxylic acid and ammonia or amine.

# Acid Hydrolysis of Amides

O                                                                O

| |                                                           | |

R – C – NH2  +  HOH +  HCl     ®      R – C – OH       +      NH4+Cl-

# Base Hydrolysis of Amides

O                                                            O

| |                                                      | |

R – C – NH – R  +  NaOH     ®      R – C – O-Na+    +    R – NH2

Chem 110 G

Wayne McGowan           HOMEWORK #10                                                              ANSWER KEY

1)      Draw the condensed structural formulas of each of the following carboxylic acids:

a.  butanoic acid                      b.  para-ethylbenzoic acid

O                                                      O

HO-C-C-C-C                                    HO-C-          -C-C

c.  2-methyl propanoic acid     d.  2,4-dibromobutanoic acid

O C                                                   O Br     Br

HO-C-C-C                                        HO-C-C-C-C

e.  m-methyl benzoic acid                  f.  4,4-dibromohexanoic acid

C                                  O                  Br

HO-C-                                     HO-C-C-C-C-C-C

Br

2)      In the following pairs, which compound would be more soluble in water?  Explain.

a.  propanoic acid or 1methoxy propane

Acids are more soluble than ethers of the same length.

b.  butanoic acid or ethanoic acid

The shorter chain acid.

c.  sodium pentanoate  or pentanoic acid

The sodium salt is much more soluble due to it already having a charge.

3)      Write the skeletal structures of the esters formed by reacting each of the following carboxylic acids with ethanol:

a.  methanoic acid                    b.  propanoic acid                   c.  2-methylpentanoic acid

O                                   O                                             O C

C-C-O-C                       C-C-O-C-C-C                         C-C-O-C-C-C-C-C

4)      When acetic acid (ethanoic acid) reacts with the following alcohols, what are the

structures of the esters that form?

a) methanol                                       b) 2-propanol                c) phenol

O                                             O      C                           O

C-C-O-C                                 C-C-O-C-C                             C-C-O-

5)      Give the names of the carboxylic acids used to produce the following esters:

O                                                        O

a.  C-C-C-O-C-C                               b.  C-O-C-C-C-C-C-C

propanoic acid                                 hexanoic acid

O C                                                             O

c.  C-O-C-C-C-C                               d.  C-C-C-O-C-C-C-C

2-methyl butanoic acid                    propanoic acid

6)      Give the IUPAC names of the esters in problem #5.

a.  ethyl propanoate                         b.  methyl hexanoate

c.  methyl-2-methyl butanoate                  d.  butyl propanoate

7)      Draw the condensed structural formulas of each of the following esters:

O

a.  hexyl ethanoate                            C-C-C-C-C-C-O-C-C

O

b.  propyl propanoate             C-C-C-O-C-C-C

O  C-C

c.  ethyl 2-ethyl butanoate                 C-C-O-C-C-C-C

O

d.  methyl benzoate                           C-O-C-

8)      Write the equation for the hydrolysis of each compound. If no reaction occurs, write

“no reaction.”

a)       O         CH3                                            O                          CH3

||           |               ®              ||                            |

CH3COCH2CHCH3              CH3COH  +  HOCH2CHCH3

b)                  O                                                       O

||               ®                                     ||

CH3CH2OC                        CH3CH2OH  +  HOCH

c)      O

||                                    no reaction

CH3CCH2OCH3

d) CH3CH2OCH2CH2CH3        no reaction

9)      Give the IUPAC names for each of the following amines:

a.  CH3CH2-NH2                      ethanamine

b.  CH3-NH-CH2CH2CH3                  N-methyl-1-propanamine

CH2CH3

c.  CH3CH2-N-CH2CH3            N,N-diethyl ethanamine

d.                 - NHCH3              N-methyl aniline

10)    There are four amine isomers with the molecular formula C3H9N.  Draw their condensed structural formulas.  Name and classify each as a primary, secondary, or tertiary amine.

N                              N                                                      C

a.  C-C-C             b.  C-C-C             c.  C-C-N-C                   d.  C-N-C

1- propanamine                               N-methyl ethanamine

2- propanamine                               N,N-dimethyl methanamine

11)    Draw the condensed structural formulas of the amides formed in each of the following reactions:

O                                                 O

a.  C-C-C-C-C-OH  +  NH3       ®   C-C-C-C-C- NH2

C      O                                                    C      O

b. C-C-C-C-OH   +  N-C-C-C  ®     C-C-C-C-N-C-C-C

O

c.  C-C-C-OH   +   N-              ®      C-C-C-N-

Chem 110G

Wayne McGowan           HOMEWORK #11              Name ____________________________

1)      What functional groups are found in all monosaccharides?

2)      What is the difference between an aldose and a ketose?

3)      What are the functional groups and number of carbons in an aldohexose?

4)      Classify each of the following monosaccharides according to the carbonyl, number of carbons and the L or D enantiomer.

a)                          b)                          c)                          d)                          e)

CH2OH                CHO                CH2OH                       CHO                CHO

C=O               H-C-OH              C=O                       H-C-OH            H-C-OH

HO-C-H               H-C-OH              CH2OH                   H-C-OH         HO-C-H

H-C-OH             H-C-OH                                           HO-C-H           HO-C-H

H-C-OH                  CH2OH                                               CH2OH                  H-C-OH

CH2OH                                 Dihydroxyacetone                                      CH2OH

Fructose               Ribose                                                Xylose            Galactose

5)      Classify each of the following monosaccharides according to the carbonyl, number of

carbons and the L or D enantiomer.

a)                          b)                          c)                          d)

CH2OH                  CH2O                    CHO                     CHO

C=O                      C=O                  H-C-OH             HO-C-H

H-C-OH              HO-C-H               HO-C-OH             HO-C-H

H-C-OH                 H-C-OH                H-C-OH             HO-C-H

CH2OH            HO-C-H                  H-C-OH                    CH2OH

CH2OH                          CH2OH

Ribulose              Sorbose                 Glucose              Ribose

6)         Stearic acid and linoleic acid both have 18 carbon atoms.  Why does stearic acid melt

at 70*C, but linoleic acid melts at -5*C?

7)      A container of margarine says that it contains partially hydrogenated corn oil.

a.  How has the liquid corn oil been changed?

b.  Why is the margarine product solid?

8)      Describe the differences between the following pairs:

a.  fat and oil

b. a fat and a phosphoglyceride

9)      How does the polarity of the phospholipids contribute to their function in cell

membranes?

10)    Write the structure of the following fatty acids:

a.  stearic acid                                   b.  linoleic acid

11)    Which of the following fatty acids are saturated, and which are unsaturated?

a.  linolenic acid                                b.  palmitic acid

c.  myristic acid                                 d.  oleic acid

12)    A mixed triglyceride contains two palmitic acid molecules to every one oleic acid

molecule.  Write two possible structures (isomers) for the compound.

13)    Draw the structure of a phospholipid that contains choline and palmitic acids.

14)    How are phospholipids similar to soaps?

15)    Draw the structure for the steroid nucleus.

Wayne McGowan

Chem 110G

14.1 Carbohydrates

Carbohydrates such as table sugar, lactose in milk, and cellulose are all made of carbon, hydrogen, and oxygen.

In photosynthesis energy from the sun is used to combine CO2 and water into the carbohydrate glucose.

6CO2 + 6H2O + energy       Û        C6H12O6 + 6O2

In our body tissues glucose is oxidized in a series of metabolic reactions known as respiration, which releases chemical energy to do work in the cells.

# Types of Carbohydrates

Monosaccharides-the simplest carbohydrates

Disaccharides - consist of two monosaccharide units joined together

C6H12O6 + C6H12O6        ®      C12H22O11 + H2O

Polysaccharides- carbohydrate polymers containing many monosaccharide units.

14.2 Classifications of Monosaccharides

Monosaccharides are simple sugars that have an unbranched chain of three to eight carbons; one of them has a carbonyl group and the rest have hydroxyl groups.

1st  – By the enantiomer: The enantiomer is distinguished by the chiral carbon furthest from the carbonyl using the Fischer Projection. The “L” isomer has the hydroxyl on the left and the “D” isomer has it on the right.

2nd – By carbonyl: In an aldose the carbonyl group is on the first carbon and in a ketose the carbonyl group is on the second carbon.

3rd  – By number of carbons: A monosaccharide with three carbons is a triose, one with four is a tetrose, a pentose has five, and a hexose contains six carbons.

14.3 D and L notations from Fischer Projections

In the Fischer projection the carbon chain is written vertically with the carbonyl group at the top.

The letter L is assigned to the stereoisomer if the –OH group is on the left of the chiral carbon. In a D stereoisomer the –OH is on the right of the chiral carbon. The very bottom carbon atom is not chiral because it does not have four different groups bonded to it.

CHO                               CHO

HO – H                              H – OH

CH2OH                          CH2OH

-Glyceraldeyde                -Glyceraldeyde

Most of the carbohydrates have five or six carbon atoms, which means they have several chiral carbons. Then we use the chiral carbon furthest from the carbonyl group to determine the D or L isomer.

CHO         CHO             CHO           CHO

HO – H        H – OH        HO – H          H – OH

HO – H        H – OH           H – OH     HO – H

HO – H        H – OH        HO – H          H – OH

CH2OH     CH2OH      HO – H      H – OH

CH2OH       CH2OH

-Ribose        -Ribose

-Glucose      -Glucose

Example:

Classify the following carbohydrates according to stereoisomer, carbonyl and number of carbons:

CHO                   CH2OH                 CH2OH                         CHO

HO – H                      = O                  O =                                  H – OH

HO – H                H – OH                  H – OH                                      HO – H

H – OH                           CH2OH             HO – H                                                        H – OH

CH2OH                                                      HO – H                             H – OH

CH2OH                          CH2OH

14.5 Cyclic Structures of Monosaccharides

Monosaccharides exist as ring structures formed when a carbonyl group and a hydroxyl group in the same molecule react.

In the cyclic structure, carbon 1 is bonded to the ring by its carbonyl reacting with the carbon 5 hydroxyl (taking its hydrogen as the oxygen bonds with carbon 1) and a new –OH group is formed from the carbonyl oxygen. There are two ways to place the new –OH, either up or down, which gives two isomers called anomers. The –OH group on carbon 1 is down in the a (alpha) anomer and up in the b (beta) anomer.

14.7 Disaccharides

A disaccharide is composed of two monosaccharides linked together. The most common disaccharides are maltose, lactose, and sucrose. The hydrolysis gives the following monosaccharides.

Maltose + H2O  ®

Lactose + H2O  ®

Sucrose + H2O  ®

14.8 Polysaccharides

Three biologically important polysaccharides – starch, cellulose, and glycogen – are all polymers of D-glucose that differ only in the type of glycosidic bonds and the amount of branching in the molecule.

Plant Starch: Amylose and Amylopectin

Starch, a storage form of glucose in plants, is found as insoluble granules.

Starches hydrolyze easily in water and acid to give smaller saccharides called dextrins, which then hydrolyze to maltose and finally glucose.

# Animal Starch: Glycogen

Glycogen is a polymer of glucose that is stored in the liver and muscle of animals. It is hydrolyzed in our cells at a rate that maintains the blood level of glucose and provides energy between meals.

# Structural Polysaccharide: Cellulose

Cellulose is the major structural material of wood and plants. In cellulose glucose molecules form a long unbranched chain.

15.1 Lipids

Lipids are a family of nonpolar biomolecules that are soluble in organic (nonpolar) solvents but not in water.

15.2 Fatty Acids

Fatty acids are the simplest type of lipids and are also found as components in several of the more complex lipids.

All fatty acids are long carbon chain (12-18 carbons) carboxylic acids.

Saturated fatty acids contain only single bonds between carbons. Monounsaturated have one double bond and polyunsaturated have two or more double bonds.

15.3 Waxes - esters of a saturated fatty acid and a long-chain alcohol.

Ester bond

# Fatty acid – Long-chain alcohol

Wax

## Fats and Oils: Triacylglycerols (triglycerides)

In the body fatty acids are stored as fats and oils. These substances, also called triglycerides or triacylglycerols, are triesters of glycerol (1,2,3-propanetriol) and fatty acids.

O                                                          O

| |                                                          | |

CH2O– H + HO –C–(CH2)16CH3                CH2 – O – C – (CH2)16CH3

O                                                          O

| |                                                          | |

CHO – H + HO –C–(CH2)16CH3       ®      CH  – O – C – (CH2)16CH3      +      3 H2O

O                                                          O

| |                                                                 | |

CH2O– H + HO –C–(CH2)16CH3                 CH2 – O – C – (CH2)16CH3

Most fats and oils are mixed triacylglycerols that contain two or three different fatty acids.

# Melting Points of Fats and Oils

A fat is a triacylglycerol that is solid at room temperature.

An oil is a triacylglycerol that is liquid at room temperature.

Saturated fatty acids have higher melting points than unsaturated fatty acids because they pack together more tightly. Animal fats usually contain more saturated fatty acids than do vegetable oils.

15.4 Chemical Properties of Triacylglycerols

The chemical reactions of the triacylglycerols are the same as we discussed for alkenes and carboxylic acids and esters.

# Hydrogenation

The addition of hydrogen to an unsaturated fat to make it saturated.

# Hydrolysis

The products of hydrolysis of the ester bonds are glycerol and three fatty acids.

# Saponification

When a fat is heated with a strong base saponification of the fat gives glycerol and the salts of the fatty acids, which are soaps.

Fat or oil + strong base ® glycerol + salts of fatty acids (soaps)

Wayne McGowan

Chem 110G

15.5 Glycerophospholipids (phospholipids)

Glycerophospholipids are a family of lipids similar to triacylglycerols except that one fatty acid is replaced by a phosphate group with an amino alcohol

O

| |

CH2 – O – C – R1

## O

| |

CH  – O – C – R2

O

| |

CH2 – O – P – O – CH2CH2+N(CH3)3

|

O-

Phospholipids contain both polar and nonpolar regions, which allow them to interact with both polar and nonpolar substances. Phospholipds are the most abundant lips in cell membranes, where they play an important role in cellular permeability.

15.6 Steroids: Cholesterol and Steroid Hormones

Steroids are compounds containing the steroid nucleus, which consists of three cyclohexane rings and one cyclopentane ring fused together.

#### Cholesterol -one of the most important and abundant steroids in the body

Steroid Hormones- are closely related in structure to cholesterol and depend on

cholesterol for their synthesis.

15.7 Cell Membranes/The Fluid Mosaic Model

In a cell membrane two rows of phospholipids are arranged into a lipid bilayer, like a sandwich. Their nonpolar tails, which are hydrophobic, move to the center while their polar heads, which are hydrophilic, align on the outer edges.

Chem 110G

Wayne McGowan           HOMEWORK #12         Name ______________________________

1)      Classify each of the following proteins according to its function:

a.  hemoglobin, oxygen carrier in the blood.

b.  collagen, a major component of tendons and cartilage

c.  keratin, a protein found in hair

d.  amylase, an enzyme that hydrolyzes starch

2)      Describe the functional groups found in all amino acids.

3)      Write the general structure of the zwitterion of an amino acid:

4)      Write the ionized form of an amino acid at high pH.

5)      What is the primary structure of a protein?

6)      State whether the following statements apply to primary, secondary, tertiary or quaternary protein structure:

a.  Side groups interact to form disulfide bonds or ionic bonds.

b.  Peptide bonds join amino acids in a polypeptide chain.

c.  Several polypeptides are held together by hydrogen bonds between adjacent           chains.

d.  Hydrogen bonding between carbonyl oxygen atoms and nitrogen atoms of   amide groups causes a polypeptide to coil.

e.  Hydrophobic side chains seeking a nonpolar environment move toward the inside of the folded protein.

f.  Protein chains of collagen form a triple-helix.

g.  an active protein contains four tertiary subunits.

7)      How does denaturation of a protein differ from its hydrolysis?

8)      Match the terms:  1.  enzyme - substrate complex         2.  enzyme  3.  substrate

with the following phrases:

a.  has a tertiary structure that recognizes the substrate

b.  the combination of an enzyme with the substrate

9)      Match the terms:  1.  active site 2.  lock-and-key model          3.  induced-fit model

with the following phrases:

a.  the portion of an enzyme where catalytic activity occurs

b.  an active site that adapts to the shape of a substrate

c.  an active site having a rigid shape

10)    Indicate if the following describe a competitive or a noncompetitive enzyme inhibitor:

a.  The inhibitor has a structure similar to the substrate.

b.  The effect of the inhibitor cannot be reversed by adding more substrate.

c. The addition of more substrate reverses the inhibition.

Chem 110G

Wayne McGowan

16.1 Functions of Proteins

Class of Protein

Function in the Body

#### Examples

Provide structural components

Collagen in tendons and cartilage. Keratin in hair, skin, wool, and nails.

# Movement of muscles

Myosin and Actin contract muscle fibers

Carry essential substances throughout the body

Hemoglobin transports oxygen. Lipoproteins transport lipids

# Store nutrients

Casein stores protein in milk. Ferritin stores iron in the spleen and liver

Regulate body metabolism and nervous system

Insulin regulates blood glucose level. Growth hormone regulates body growth

Catalyze biochemical reactions in the cells

Sucrase catalyzes the hydrolysis of sucrose. Trypsin catalyzes the hydrolysis of proteins.

Recognize and destroy foreign substances

Immunoglobulins stimulate immune responses.

16.2 Amino Acids

Proteins are polymers of amino acids. Every amino acid contains two functional groups, an amino group and a carboxyl group.  Amino acids differ by the identity of their side group (R).

R

H                                O

N – C – C

|

H            H             OH

At the pH of most body fluids, the carboxyl group loses a H+ and the amino group accepts it forming the dipolar form of an amino acid, called a zwitterion.

R      O                          R     O

|       | |                    +     |       | |      -

H2N – CH – C – OH ® H3N – CH – C – O

zwitterion

# Classification of Amino Acids        (See Table 16.2)

Nonpolar - contain an alkyl or aromatic side chain

Polar - have side chains that contain polar groups

Acidic - have side chains that contain a carboxyl group

Basic - contain an amine group

# Amino Acid Stereoisomers

All of the amino acids except glycine are chiral since the a carbon is attached to four different atoms or groups.

NOTE: In biological systems only L amino acids are incorporated into proteins.

16.3 Amino Acids as Acids and Bases

At the pH known as the isoelectric point (pI) the amino acid is in the zwitterion form. However, when the pH is different from the pI the zwitterion accepts or donates H+.

Acidic solution

R       O                                               R       O

+      |        | |                                       +        |        | |

H3N – CH – C – O- + H3O+      ®      H3N – CH – C – OH + H2O

Zwitterion

Basic Solution

R       O                                            R       O

+      |        | |                                              |        | |

H3N – CH – C – O- + OH-      ®     H2N – CH - C – O - + H2O

Zwitterion

16.4 Formation of Peptides

A peptide bond is an amide bond that forms when the carboxyl group of one amino acid reacts with the amine group of the next amino acid.

O                                O                                     O                              O

+             | |                   +            | |                     +              | |                              | |

H3N – CH – C – O- + H3N – CH – C – O-  ®  H3N – CH – C – NH – CH – C – O- + H2O

|                                   |                                      |                          |

R1                               R2                                  R1                      R2

amino acid 1              amino acid 2

Chem 110G

Wayne McGowan           HOMEWORK #11                                                              ANSWER KEY

1)      What functional groups are found in all monosaccharides?

All monosaccharides have multiple hydroxy functional groups and a carbonyl fuctional group.

2)      What is the difference between an aldose and a ketose?

An aldose is a sugar that has an aldehyde functional group (carbonyl is terminal) while a ketose is a sugar that has a ketone functional group (carbonyl is central).

3)      What are the functional groups and number of carbons in an aldohexose?

An aldohexose has the aldehyde functional group (aldo), multiple hydroxy functional groups and has six carbons (hexose).

4)      Classify each of the following monosaccharides according to the carbonyl, number of carbons and the L or D enantiomer.

a)                          b)                          c)                          d)                          e)

CH2OH                CHO                CH2OH                       CHO                CHO

C=O               H-C-OH              C=O                       H-C-OH            H-C-OH

HO-C-H               H-C-OH              CH2OH                   H-C-OH         HO-C-H

H-C-OH             H-C-OH                                           HO-C-H           HO-C-H

H-C-OH                  CH2OH                                               CH2OH        H-C-OH

CH2OH                                 Dihydroxyacetone                                      CH2OH

Fructose               Ribose                                                Xylose            Galactose

D-ketohexose       D-aldopentose     ketotriose         L-aldopentose  D-aldohexose

5)      Classify each of the following monosaccharides according to the carbonyl, number of

carbons and the L or D enantiomer.

a)                          b)                          c)                          d)

CH2OH                  CH2O                   CHO                     CHO

C=O                      C=O                  H-C-OH             HO-C-H

H-C-OH              HO-C-H               HO-C-OH             HO-C-H

H-C-OH                 H-C-OH                H-C-OH             HO-C-H

CH2OH            HO-C-H                  H-C-OH                    CH2OH

CH2OH                 CH2OH

Ribulose               Sorbose                Glucose                Ribose

D-ketopentose      L-ketohexose       D-aldohexose       L-aldopentose

6)         Stearic acid and linoleic acid both have 18 carbon atoms.  Why does stearic acid melt at 70*C, but linoleic acid melts at -5*C?

Stearic acid is saturated fatty acid while linoleic acid is a polyunsaturated fatty acid.  The unsaturated fatty acids have "kinks" in them due to the double bonds, which means less intermolecular attractions and thus a lower melting point.

7)      A container of margarine says that it contains partially hydrogenated corn oil.

a.  How has the liquid corn oil been changed?

The double bonds between carbons have been converted to single bonds.

b.  Why is the margarine product solid?

Saturated fats have a higher melting point than unsaturated fats.

8)      Describe the differences between the following pairs:

a.  fat and oil

Fats and oils are both triglycerides, the differences are that fats are solid at room temperature and oils remain liquid at room temperature.

b. a fat and a phosphoglyceride

A fat is a triglyceride having three fatty acids, while a phosphoglyceride has two fatty acids and a phosphate group.

9)      How does the polarity of the phospholipids contribute to their function in cell membranes?

Having a polar and a non-polar end allows the phospholipid to have both a hydrophobic and a hydrophilic portion.  These different portions allow it to interact with both polar (aqueous) and non-polar molecules.  In the cell membrane the phospholipids make up a bilayer (two layers) with the hydrophilic portions pointing outward (toward the aqueous environments) and the hydrophobic portions pointing inward (toward a non-polar environment).

10)    Write the structure of the following fatty acids:

a.  stearic acid                                   b.  linoleic acid

CH3(CH2)16COOH                           CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH

or                        O

|  |

C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-OH

O

|  |

C-C-C-C-C-C=C-C-C=C-C-C-C-C-C-C-C-C-OH

11)    Which of the following fatty acids are saturated, and which are unsaturated?

a.  linolenic acid                                b.  palmitic acid

POLYUNSATURATED                           SATURATED

c.  myristic acid                                 d.  oleic acid

SATURATED                                  MONOUNSATURATED

12)    A mixed triglyceride contains two palmitic acid molecules to every one oleic acid molecule.  Write two possible structures (isomers) for the compound.

C-O-PALMITIC                                       C-O-OLEIC

|                                                                 |

C-O-OLEIC                                               C-O-PALMITIC

|                                                                 |

C-O-PALMITIC                                       C-O-PALMITIC

13)    Draw the structure of a phospholipid that contains choline and palmitic acids.

O

|  |

C-O-C(CH2)14CH3

O

|  |

C-O-C(CH2)14CH3

O

|  |

C-O-P-O-CH2CH2N+(CH3)3

|

O-

14)    How are phospholipids similar to soaps?

They both have a polar end/portion and a non-polar end/portion; which allows interaction with both polar (water) and non-polar (grease, oil etc.) solvents and solutes.

15)    Draw the structure for the steroid nucleus.

Wayne McGowan

Chem 110G

16.5 Levels of Protein Structure

Primary Structure of Proteins

The primary structure is

# Secondary Structure

The secondary structure of a protein describes the way the amino acids are arranged in space. The three most common are:

# The Alpha Helix

The corkscrew shape of an a helix is held in place by hydrogen bonds between each N – H group and the C = O group in the next turn of the helix, four amino acids down the chain.

Beta-Pleated Sheet

In a b-pleated sheet polypeptide chains are held together side by side by hydrogen bonds between the peptide chains.

# Tertiary Structure

The tertiary structure of a protein involves interactions between the side groups of the amino acids. Segments of the chain twist and bend until the protein acquires a specific three-dimensional shape.

Cross-Links in Tertiary Structures

1.     Hydrophobic interactions occur between nonpolar side groups to form a hydrophobic center at the                            of the protein.

2.     Hydrophilic interactions occur between the external aqueous environment and amino acids that have                                         side groups.

3.     Salt bridges are ionic bonds between side groups of basic and acidic amino acids, which have positive and negative charges.

4.     Hydrogen bonds form between

5.     Disulfide bonds (-S-S-) are covalent bonds that form between the –SH groups of cysteines

Quaternary Structure:

A biologically active protein consists of two or more tertiary subunits. The subunits are held together in the quaternary structure by the same interactions that stabilize the tertiary structure.

# Denaturation of Proteins

Denaturation of a protein occurs when there is a disruption of any of the bonds that stabilize the secondary, tertiary, or quaternary structure. With the loss of its overall shape the protein is no longer biologically active.

Denaturing agents include:

16.6 Enzymes

Enzymes catalyze nearly all the chemical reactions that take place in the body. As a catalyst an enzyme increases the rate of a reaction by changing the way a reaction takes place, but is itself not changed at the end of the reaction.

As catalysts enzymes lower the activation energy of the reaction so that less energy is required to accomplish the reaction.

16.7 Enzyme Action

Each enzyme has a unique three-dimensional shape that recognizes and binds a small group of reactants (                                ).

Active Site

Within its large tertiary structure there is a region called the                                       where the enzyme binds the substrates and catalyzes the reaction. The active site of a particular enzyme fits the shape of only a few types of substrates, which makes enzymes very specific.

Lock and Key model

In the lock-and-key model the active site has a

Induced Fit model

In the induced-fit model the active site                                      the shape of the substrate.

# Enzyme Catalyzed Reaction

The substrate within the active site forms an enzyme-substrate (ES) complex. As soon as the reaction is complete the products are released from the enzyme.

Step 1                                                        Û

## Step 2                                                                                               ®

### Û                                   ®

##### Enzyme + substrate

16.8 Factors Affecting Enzyme Activity (how fast an enzyme catalyzes the reaction)

# Temperature

At higher temperatures, enzyme activity increases. Enzymes are most active at their optimum temperature.

pH

Enzymes are most active at their optimum pH, the pH that maintains the proper tertiary structure of the protein.

# Substrate Concentration

Increasing the substrate concentration increases the rate of the catalyzed reaction as long as there are more enzymes present than substrate.

# Enzyme Inhibition

Inhibitors cause enzymes to lose catalytic activity.

Competitive Inhibition

A competitive inhibitor has a structure                        to the substrate and competes for the

# Noncompetitive Inhibition

When the noncompetitive inhibitor is bonded to the enzyme the                          of the enzyme is

16.9 Enzyme Cofactors

Many enzymes require small molecules or                           called cofactors to catalyze properly.

Chem 110G

Wayne McGowan           HOMEWORK #12                                                              ANSWER KEY

1)      Classify each of the following proteins according to its function:

a.  hemoglobin, oxygen carrier in the blood.

TRANSPORT PROTEIN

b.  collagen, a major component of tendons and cartilage

STRUCTURAL

c.  keratin, a protein found in hair

STRUCTURAL

d.  amylase, an enzyme that hydrolyzes starch

ENZYME

2)      Describe the functional groups found in all amino acids.

All amino acids contain an amine and a carboxylic acid.  (they also all have an alpha carbon and a unique side group, which could have additional functional groups)

3)      Write the general structure of the zwitterion of an amino acid:

R     O

+      |       | |       -

H3N – CH – C – O

4)      Write the ionized form of an amino acid at high pH.

R      O

|       | |       -

H2N – CH – C – O

5)      What is the primary structure of a protein?

The primary structure of a protein is the sequence of its amino acids, this sequence determines all of the other levels of structure.

6)      State whether the following statements apply to primary, secondary, tertiary or quaternary protein structure:

a.  Side groups interact to form disulfide bonds or ionic bonds.

TERTIARY

b.  Peptide bonds join amino acids in a polypeptide chain.

PRIMARY

c.  Several polypeptides are held together by hydrogen bonds between adjacent           chains.

SECONDARY (beta pleated sheet)

d.  Hydrogen bonding between carbonyl oxygen atoms and nitrogen atoms of   amide groups causes a polypeptide to coil.

SECONDARY (alpha helix)

e.  Hydrophobic side chains seeking a nonpolar environment move toward the inside of the folded protein.

TERTIARY

f.  Protein chains of collagen form a triple-helix.

SECONDARY

g.  an active protein contains four tertiary subunits.

QUATERNARY

7)      How does denaturation of a protein differ from its hydrolysis?

Denaturation occurs by disrupting all levels of structure, while hydrolysis (the splitting of amino acids from one another by the addition of water) disrupts only the primary level of structure.

8)      Match the terms:  1.  enzyme - substrate complex         2.  enzyme  3.  substrate

with the following phrases:

a.  has a tertiary structure that recognizes the substrate

ENZYME

b.  the combination of an enzyme with the substrate

ENZYME - SUBSTRATE COMPLEX

9)      Match the terms:  1.  active site 2.  lock-and-key model          3.  induced-fit model

with the following phrases:

a.  the portion of an enzyme where catalytic activity occurs

ACTIVE SITE

b.  an active site that adapts to the shape of a substrate

INDUCED-FIT MODEL

c.  an active site having a rigid shape

LOCK-AND-KEY MODEL

10)    Indicate if the following describe a competitive or a noncompetitive enzyme inhibitor:

a.  The inhibitor has a structure similar to the substrate.

COMPETITIVE

b.  The effect of the inhibitor cannot be reversed by adding more substrate.

NONCOMPETITIVE

c. The addition of more substrate reverses the inhibition.

COMPETITIVE