Chemistry Course Pack, Winter 2014

Table of Contents:

Topic Page Topic PageIntroduction to Chemistry Nuclear ChemistryUnit Conversions Intro to Organic ChemistryAtomic Structure Hydrocarbon DerivativesThe Periodic TableBondingBasic Chemical ReactionsGas LawsSolutionsAcid-Base ChemistryIntroduction to Thermodynamics Appendix A: AlgebraChemical Equilibrium Appendix B: Useful Constants

About the Authors:

Peace Corps Volunteers from LR-3 and LR-4 worked on this course pack, with primary authorship by Dan Frechtling. Frechtling has a degree in Cellular and Molecular Biology from the University of Michigan, and taught chemistry to 10th, 11th, and 12th grades at Gorblee Central High School in Gorzohn, Grand Bassa County.

A Note on Using this Course Pack:

This course pack is intended to supplement other resources and lessons; if taken alone, it provides a brief introduction to chemistry, and some practice. For students, it is suggested that they use other textbooks and notes from teachers in conjunction with this guide.

Following each section are practice problems. Solutions are provided at the back of the text; students are encouraged to attempt the practice problems before checking their answers against key.

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Section 1: Introduction to Chemistry

Chemistry is the science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.

Put simply, chemistry explains what everything is made of, and how it can change.

The curriculum for chemistry in Liberia recognizes three branches of chemistry: Inorganic chemistry, physical chemistry, and organic chemistry. Inorganic chemistry is the branch of chemistry that describes the structure and properties of non-living things. Physical chemistry explains the behavior of matter under different conditions. Organic chemistry is the study of complex carbon compounds. Each of these branches of chemistry will be treated in detail in subsequent chapters.

1.1 States of Matter

It is useful to remember the states of matter as learned in previous classes: Solid, liquid, gas, and plasma. Matter can exist in each of those states depending on temperature and pressure conditions; matter in different states will behave in different ways. Think – liquid water and ice are very different in appearance and behavior, but are chemically the same.

Solids are the state of matter in which matter is most densely packed, and unable to move very far. As a result, solids are very hard and don’t change shape – they are said to have definite shape and definite volume.

In the liquid state, matter is able to move around; its shape can change. In this state, shape is indefinite, but volume is definite.

In the gas phase, matter is free to move around, and does. The shape of a gas can change, and so can the volume. Gases are said to have indefinite shape and indefinite volume.

1.2 Chemical Change and Physical Change

You should have learned in a previous course about the difference between chemical change and physical change. Physical change is when a substance changes state or shape, without becoming a different substance – think of ice melting to become water, or boiling to become water vapor.

Chemical change is a change of substance into a different substance.

Practice:

1. Tell whether the following is a physical change or a chemical change:a. A piece of paper is burned.

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b. Cassava is pounded to make GBc. A piece of paper is folded.d. Dry ice (solid carbon dioxide) changes to carbon dioxide gas

Section 2: Unit Conversions

In science there is an accepted system of units called the International System of Units, or the SI System, and it defines the units of different quantities like time, length, and mass.

The SI units for time, distance, and mass are as follows:

SI Units for Time, Distance, and MassQuantity SI UnitTime SecondLength MeterMass Kilogram

In common usage, the SI units may be combined with a metric prefix, which shows a multiple of the base unit.

Prefix Order of MagnitudeKilo 103 1000Hecto 102 100Deca 101 10(Base Unit – no prefix) 100 1Deci 10-1 1/10Centi 10-2 1/100Milli 10-3 1/1000

It is often necessary in physics to convert from one unit to another in order to perform a calculation (converting millimeters to meters, for example).

A conversion can be done by multiplying a number by a conversion factor:

1 kilogram * 1000 grams/1 kilogram = 1000 grams

1 milligram * 1 gram/1000 milligrams = 1/1000 grams

The following table has useful conversion factors:

Starting from…

Going to…Kilo Hecto Deka Base Deci Centi Milli

Kilo 1 10 100 1000 10,000 100,000 1,000,000Hecto 1/10 1 10 100 1000 10,000 100,000Deca 1/100 1/10 1 10 100 1000 10,000Base 1/1000 1/100 1/10 1 10 100 1000Deci 1/10,000 1/1000 1/100 1/10 1 10 100Centi 1/100,000 1/10,000 1/1000 1/100 1/10 1 10Milli 1/1,000,000 1/100,000 1/10,000 1/1000 1/100 1/10 1

To use this table, you must know the metric prefix that you are starting from and the prefix that you are going to. Here’s an example:

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Convert 25 milligrams to hectograms.

We are starting with milligrams, and we’re going to hectograms. First, we find the row of the table for milli – the bottom row – and then we go across the table to the column for hector. We find the conversion factor to be 1/100,000, so we multiply to get our answer:

25 milligrams * 1 hectogram/100,000 milligrams = 0.000025 hectograms

Starting from…

Going to…Kilo Hecto Deka Base Deci Centi Milli

Kilo 1 10 100 1000 10,000 100,000 1,000,000Hecto 1/10 1 10 100 1000 10,000 100,000Deca 1/100 1/10 1 10 100 1000 10,000Base 1/1000 1/100 1/10 1 10 100 1000Deci 1/10,000 1/1000 1/100 1/10 1 10 100Centi 1/100,000 1/10,000 1/1000 1/100 1/10 1 10Milli 1/1,000,000 1/100,000 1/10,000 1/1000 1/100 1/10 1

Practice Problems:

1. Convert the following to centimeters:a. 35 millimeters b. 230 meters c. 41 decameters

2. Convert the following to hectometersa. 21 kilometers b. 505 centimeters c. 32 decimeters

3. Convert the following to grams:a. 37 centigrams b. 44 hectograms c. 465 decagrams

Section 3: Atomic Structure

At a very, very small level – much smaller than can be seen with the eye – everything is made of atoms. An atom is the smallest unit of an element that has all of the characteristics of that element. Put another way, an atom is the smallest bit of matter that can’t be cut into smaller pieces that have the same properties. All atoms of an element are the same; atoms of different elements are different from one another.

Atoms are made of smaller units called sub-atomic particles – pieces smaller than an atom. These particles – called protons, neutrons, and electrons – do not have the same characteristics as an atom or an element. Protons are positively charged. Neutrons have no charge – think ‘neutral’. Electrons are negatively charged.

Protons and neutrons are about the same size, and stick together in the center of the atom. The ‘ball’ of protons and neutrons is called the nucleus. Electrons are almost 2000 times smaller than protons and neutrons, and they move around the nucleus. The electrons are said to be in an electron cloud.

For every atom, the number of protons and the number of neutrons is equal, and the charge of the atom is zero.

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3.1 Protons

The number of protons is different from element to element. Hydrogen as one proton; Iron has twenty-six. It’s the number of protons that make a particular element; change the number of protons, and the element changes. The number of atoms is called the atomic number. Remember this number – it will be important in later sections.

3.2 Electrons

Electrons are tiny particles that move around the nucleus of an atom. They have a negative charge – a charge that is opposite to the charge of the proton. They are very small, having almost no mass. It was mentioned above that electrons are in an electron cloud around the atom.

Electrons occupy space in the cloud according to their energy level; the energy levels correspond to the rows of the Periodic Table. Each energy level is further divided into shells, called the s-, p-, d-, and f- orbitals.

3.3 Electron Configuration

Electrons are found in shells. For each of the seven energy levels, there are shells, as in the following chart:

Energy Level S p d f1 1s2 2s 2p3 3s 3p4 4s 4p 3d5 5s 5p 4d6 6s 6p 5d 5f7 7s 7p 6d 6f

Each shell can hold some number of electrons, based on the number of orbitals in the shell; each orbital can hold two electrons. There is only one s orbital, so s shells can hold two electrons. There are three p orbitals, so p shells can hold six electrons. There are five d orbitals, so d shells can hold ten electrons. Seven f orbitals give the f shells a capacity of 14 electrons.

There are two rules that guide how electrons fill in orbitals: the Aufbau Principle and Hund’s Rule.

Hund’s Rule states that electrons will try to get their own orbital – two electrons will not fill the same orbital if there is an empty orbital available. It’s like chairs – each electron wants its own chair, but it will sit with another electron if there aren’t any empty chairs.

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The Aufbau Principle tells us that electrons will fill an energy level before they move to the next one; that is, 2s will fill before 2p, and 2p will fill before 3s. There are exceptions – if each orbital can have one electron (meaning the orbital is half-full), that is preferred.

The position of electrons can be plotted on a diagram:

Carbon – 6 electrons

↑↓ ↑_ ↑_ ↑_ ↑_

1s 2s 2p 2p 2p

The above configuration is preferred to the configuration that would be expected based on the rules (below). Each of the 2p and 2s orbitals has a single electron – is half-full – in the above example.

↑↓ ↑↓ ↑_ ↑_ ___

1s 2s 2p 2p 2p

Let’s look at the electron configuration of Potassium, with 19 electrons.

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑_

1s 2s 2p 2p 2p 3s 3p 3p 3p 4s

3.4 Writing Electron Configuration in Shorthand

Electron configuration can be written in shorthand by listing the energy level, the shell, and the number of electrons within the orbital. For example, Hydrogen is written as 1s 1. The first number indicates the energy level; the letter indicates the shell, and the second number – in superscript – tells how many electrons are in the shell.

Let’s look at the notation for Oxygen. Oxygen has 8 electrons – two are in the s orbital of the first energy level, two are in the s orbital of the second energy level, and four are in the p orbital of the second energy level. Therefore, Oxygen can be written as 1s22s22p4.

As you might expect, the electron configuration of an element lower on the periodic table can be quite long. The Strontium, Sr, has the electron configuration 1s22s22p63s23p64s23d104p65s2. It can be shortened to [Kr]5s2. This technique uses the configuration of the noble gases to represent large parts of the configuration of another element.

To use this technique, look at the Periodic Table. Locate the element you want to write the electron configuration for; go to the row above that element and identify the noble gas (the last element in the row). Place the symbol of the noble gas in brackets and then write the rest of the element’s configuration.

Example: Write the short configuration of Magnesium.

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1. Look at the Periodic Table and find magnesium – it is the second element of the third row of the periodic table.

2. Go up one row and identify the noble gas – in this case, the noble gas is neon.3. Put the symbol of the noble gas in brackets – [Ne]4. Write the rest of the electron configuration – [Ne]3s2

Practice:

1. Name the following elements and write the electron configuration of the following:a. Nb. Fc. Cad. Fee. Ar

2. Write shorthand electron configurations for the following:a. Cab. Kc. Cld. Cs

Section 4: The Periodic Table

There are currently 118 known elements, and they are organized for reference on the Periodic Table of the Elements. The Periodic Table arranges elements in order of their atomic number, and groups elements with similar characteristics together.

The rows of the Periodic Table are called periods; there are seven periods. The columns of the Periodic Table are called groups, and most elements fit into 18 groups (the exceptions are the Lanthanides and the Actinides, which will be discussed later).

The Periodic Table can also be divided into blocks of elements; the first two groups form the s-block; the final six are the p-block; the middle ten are the d-block. The last two rows (again, the lanthanides and actinides) are called the f-block.

A standard Periodic Table will have three pieces of information for each element: the atomic number, the atomic mass, and the element’s symbol. Recall that the atomic number is the number of protons in a single atom of a particular element.

The atomic mass is the number of protons plus the number of neutrons. On the Periodic Table, the atomic mass is a decimal (Oxygen has an atomic mass of 15.99). The reason for the decimal is that the number of neutrons in an atom may be different. One atom of oxygen may have 8 neutrons,

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while another has 7; the atomic mass is a weighted average of the different possible masses of atoms of an element.

The element’s symbol is one or two letters that are a short way of identifying an element, rather than having to write the element’s full name. Instead of writing Hydrogen, the symbol “H” may be used. Following is a table of selected elements and their symbols.

Element Symbol Element SymbolHydrogen H Helium HeCarbon C Nitrogen NOxygen O Fluorine FNeon Ne Sodium NaPotassium K Phosphorous PSulfur S Chlorine ClIron Fe Copper CuBromine Br Silver AgGold Au Lead Pb

Practice:

1. Give the atomic mass and atomic number of the following elements (use a Periodic Table):a. Nab. Agc. Cld. Mg

Section 5: Bonding

Atoms from elements can bond to form molecules. A molecule is a set of atoms that are bonded together; they may be atoms of the same element or of different elements. The main types of bonding are ionic bonds and covalent bonds. Metallic bonds and hydrogen bonds will also be discussed.

5.1 Ionic Bonds

Ionic bonds are formed by ions – atoms that have a charge. Positively charged ions and negatively charged ions will come together to have a total charge of zero (called a “net charge”). Ionic bonds occur between a metal and a nonmetal.

Ions are created when an atom gains or loses electrons. Recall that an atom has an equal number of protons and electrons, and a zero charge. If an atom gains an electron, it will have one more negative charge (from the electron) than positive charges (from the protons) - and will be a negative

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ion. Conversely, if an atom loses electrons, it will have more positive charges than negative charges and will be a positive ion.

Table salt – Sodium Chloride - is an ionic compound. The sodium atom loses one electron, becoming the Na+ ion. Chlorine gains one electron, becoming the Cl- ion. The two ions will be attracted to one another, forming an ionic bond.

Na+ + Cl- = NaCl

5.2 Covalent Bonds

Covalent bonds are the primary bonds formed between nonmetals. Atoms want to have a full outer shell of electrons – and they can get to a full shell by sharing electrons with another atom. A covalent bond is a bond made by the sharing of electrons between two nonmetal atoms.

For example, oxygen gas is made of two oxygen atoms. Each oxygen atom has six electrons in its outer shell, but wants to have eight. If each oxygen atom shares two electrons with the other, then both atoms will be able to count eight electrons in their outer shell.

An atom of a nonmetal can usually form as many bonds as it needs to get to eight electrons in its outer shell. The group 18 elements, the noble gases, already have a full shell and so do not bond. The group 17 elements (such as Chlorine) need one more electron to have a full shell, so they form a single covalent bond. Group 16 elements (like Oxygen) form two bonds. Group 15 elements (like Nitrogen) can form three bonds. Group 14 elements (Carbon) can form four bonds. Group 13 elements (such as Boron) can form three bonds.

Examples:

BF3

In this example, Boron has three electrons in its outer shell. It shares all three of them – one to each of the three Fluorine atoms.

CO2

In this example, the Carbon atom has four electrons to share. It shares two with each of the Oxygen atoms, forming two covalent bonds to each Oxygen atom.

HCN

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In this example, hydrogen has one electron to share, carbon has four, and nitrogen has three electrons to share (plus a pair of electrons). One bond forms between the hydrogen and carbon atoms, and three form between the carbon and nitrogen atoms.

5.3 Hydrogen Bonds

Hydrogen bonds are made by the interaction between hydrogen atoms on one molecule with negatively charged atoms on a second molecule. For example, hydrogen bonding occurs in water; recall that water is H2O. The hydrogen atoms of one molecule of water will have a slight positive charge, while the oxygen atom carries a slight negative charge. Opposites attract, so a hydrogen atom of one molecule pulls on the oxygen atom of a second, creating a weak force that is called a hydrogen bond.

[Illustration]

Hydrogen bonds are much weaker than ionic bonds or covalent bonds.

Practice:

1. Which type of bond occurs between metals and nonmetals?2. Describe a covalent bond.3. Predict what type of bonds are found in the following molecules (a Periodic Table may be

helpful):a. CH4

b. NaBrc. NH3

d. Na2SO4

Section 6: Calculations

6.1 Formula Weight

The formula weight of a compound is simply the weight of all of the atoms in a single molecule of a compound added together. For example, the formula mass of H2O is 18 atomic mass units (amu). One oxygen atom has a weight of 16 amu, and each of the two hydrogen atoms has a mass of 1 amu. 16 amu + 1 amu + 1 amu = 18 amu

Example: Find the formula weight of sulfuric acid, H2SO4.

Make a table:

Element Atomic Mass Number in Molecule Mass * Number

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Hydrogen 1 amu 2 2 amuSulfur 32 amu 1 32 amuOxygen 16 amu 4 64 amu

Add the weights in the last column: 2 amu + 32 amu + 64 amu = 98 amu. Therefore, a single molecule of H2SO4 has a weight of 98 amu.

6.2 Percent Composition

Percent composition tells how much of a specific molecule’s weight is made of a specific element. For example, H2SO4 is about 2% Hydrogen, 33% Sulfur, and 65% Oxygen (note that numbers have been rounded). Here’s how the percent composition is found:

First, the formula weight needs to be calculated, as in the last section. The formula weight of sulfuric acid was found to be about 98 amu.

Add another column to the table from the previous section:

Element Atomic Mass Number in Molecule

Mass * Number Element Weight / Formula Weight * 100

Hydrogen 1 amu 2 2 amu (2 / 98) * 100 = 2%Sulfur 32 amu 1 32 amu (32 / 98) * 100 = 32.6%Oxygen 16 amu 4 64 amu (64 / 98) * 100 = 65.3%

By dividing the total weight of each element by the formula weight of the whole molecule and multiplying by 100, the percent composition of the molecule can be found.

6.3 Moles and Molar Mass

Moles are a convenient measure of the amount of an element or compound present. It’s usually more useful for chemists to work in terms of moles than in individual elements. One mole of a substance is equivalent to 6.022 * 1023 atoms (This number is called Avogadro’s Number). One mole of a substance has a weight that is equal to the formula mass of the substance, except the unit is grams instead of atomic mass units.

Example: How many moles of sulfuric acid are present if there are 1.200 * 10 24 molecules of H2SO4 in solution?

The answer is found by dividing the number of molecules present by Avogadro’s Number.

1.200 * 1024 molecules H2SO4 / 6.022 * 1023 = 1.99 moles H2SO4

Molar mass can be found by adding together the masses of the atoms present in a compound. For example, the molar mass of Carbon Dioxide (CO2) is 44 grams/mol.

Element Molar Mass Number in Molecule Mass PresentCarbon 12 grams/mol 1 1 * 12 = 12 grams/molOxygen 16 grams/mol 2 2 * 16 = 32 grams/molTotal 12 + 32 = 44 grams/mol

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Practice:

Calculate the following:

1. The formula mass of HNO3, AgCl3, NaOH, and I2

2. What percent of the mass of HNO3 is from nitrogen?3. The formula mass of CH3COOH4. What percent of the mass of CH3COOH is from carbon?5. How many moles of HNO3 are present if there are 4.6 * 1024 molecules of HNO3?

Section 6: Basic Chemical Reactions

There are 118 elements on the Periodic Table, but how do we get all of the different types of material around us? The answer to that is in chemical reactions, where the elements of the Periodic Table combine in different ways to form different compounds. As we learn about reactions, keep in mind the general form of a reaction:

Reactants Products

Reactants, the elements or compounds that are going into the reaction, are on the left. The arrow is like an = sign in math; on the right side of the arrow are the products, or what comes out of the reaction.

6.1 Synthesis Reactions

Synthesis reactions are reactions in which two elements combine to form a single compound. The basic form of the synthesis reaction is

A + B AB

That is, element A reacts with element B to form compound AB.

6.2 Decomposition Reactions

A decomposition reaction is the opposite of a synthesis reaction. One compound splits into two in a decomposition reaction.

AB A + B

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6.3 Single Replacement Reactions

In some reactions, a reactant will replace an element that is already part of a compound. Such a reaction is called a single replacement reaction.

AB + C AC + B

Elements B and C have switched places. C is now bonded to A, and B is free.

6.4 Double Replacement Reactions

Double replacement reactions are reactions where two compounds trade elements.

AC + BD AD + BC

Elements C and D have switched places. Element A is bonded to element D in the products, and B is bonded to C.

Section 7: Gas Laws

7.1 Kinetic Molecular Theory

Molecules of gases are free to move, and chemists have developed a theory that describes the behavior of gases. The Kinetic Molecular Theory of Gases has five postulates that describe how gas molecules act, and these postulates give a useful look at how real gases interact.

Postulate 1: Gas molecules move in constant straight-line motion.

This means that gas molecules don’t curve, and don’t stop. They keep going in a line until they hit something, and change direction when they bounce off of an atom.

Postulate 2: Gas molecules are points – they have no mass.

Gas molecules are treated as small points that don’t have any mass – basically, they don’t weigh anything.

Postulate 3: Collisions between gas molecules are perfectly elastic – energy is not lost when gas molecules hit each other.

Gas molecules will bounce off of one another and keep going with the same energy as before the collision.

Postulate 4: Gas molecules do not attract or repel one another.

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Gas molecules don’t push or pull on one another.

Postulate 5: The kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas.

The higher the temperature of the gas, the faster the gas molecules move and their higher their kinetic energy is. Higher temperature = higher energy. Lower temperature = lower energy.

The postulates of kinetic theory describe ideal gases – real gases behave slightly differently, but the postulates are a useful approximation.

7.2 Useful Units for Gases

Volume: The volume of a gas is usually given in units of Liters.

Pressure: Pressure, a measure of the force gas exerts on a container, is given in several units. Remember the following conversions:

1 atmosphere (atm) = 760 millimeters of Mercury (mmHg) = 760 torr = 101.3 Kilopascals (KPa)

Temperature: Temperature is given as an absolute temperature, in units of Kelvin. The conversion from Celsius to Kelvin is given as

K = oC + 273

7.3 Boyle’s Law

Boyle’s Law is a gas law that relates pressure and volume; it assumes that the gas remains at a constant temperature.

The relationship is written in math terms as PV = k, where k is a constant; or as P1V1 = P2V2

In PV = k, if pressure increases, volume must decrease in order for k to stay the same. The same can be seen in the second case:

Given P1V1 = P2V2, where P1 = 4 atm; V1 = 4 L; P2 = 8 atm; V2 = 2 L

16 = 4 atm * 4 L = 8 atm * 2 L = 16

The pressure increased from 4 atmospheres to 8 atmospheres, and the volume decreased. As a result, both sides are equal to 16.

7.4 Charles’ Law

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Charles’ Law states that volume is directly proportional to temperature under constant pressure. Therefore, if volume increases, temperature does as well. If temperature increases, so does volume.

Symbolically, Charles’ Law is V / T = k, or V1 / T1 = V2 / T2.

Example: V / T = 2. For V1 / T1 = V2 / T2 =2, any values that divide to equal 2 will work.

V1 = 10 Liters; T1 = 5 K; V2 = 600 Liters; T2 = 300 K

2 = 10 Liters / 5 K = 600 Liters / 300 K = 2

7.5 Combined Gas Law

The Combined Gas Law puts Boyle’s Law and Charles’ Law together. Symbolically it is written as P 1V1

/ T1 = P2V2 / T2

7.6 Avogadro’s Law

Avogadro’s Law holds that moles of a gas will occupy a certain amount of space. If more moles of gas are present, a larger volume will be occupied.

It can be written as n1 / V1 = n2 / V2, where nx is some number of moles of gas and Vx is the volume of gas in liters.

7.6 Ideal Gas Law

The Ideal Gas Law brings together pressure, volume, temperature, and the number of moles of gas present. It can be mathematically manipulated using algebra to find any one of those quantities.

The Ideal Gas Law is PV = nRT

P = Pressure V = VolumeN = number of moles of gasR = A Gas ConstantT = absolute temperature

The value of R changes depending on the units used. Most often, its value is 0.0821 L atm -1 K-1

Example: What is the pressure of 4 moles of gas that occupies a volume of 3 Liters at 273 K?

The equation PV = nRT can be algebraically rewritten as P = nRT / V. Replacing the variables with numbers and solving will yield the answer.

P = (4 moles) * (0.0821 L atm-1 K-1) * (273 K) / 3 L

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= 29.88 atm

7.7 Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures says that the total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases.

Suppose that you have a mix of three gases – hydrogen, oxygen, and nitrogen. The pressure of hydrogen is one atmosphere, while the pressure of oxygen is two atmospheres, and the pressure of nitrogen is three atmospheres. The total pressure will be 1 atm + 2 atm + 3 atm = 6 atm.

Symbolically, PTotal = P1 + P2 + P3 + P4 + P5 + …Px

For any mixture of gases, simply add the pressures of each individual gas together, no matter how many gases are in the mixture.

Practice:

1. Identify which gas laws would be useful for solving a problem with the provided information:a. P1 = X, P2 = Y, V2 = Zb. V1 = X, T1 = Y, T2 = Mc. P = Y, T = X, n = Z, R = Wd. P1 = X, P2 = Y, P3 = Z

2. Convert the following:a. 122 oC to Kb. 45 oC to Kc. 515 K to oCd. 298 K to oCe. 10 atmospheres to torrf. 202.6 kPa to torrg. 1760 torr to atmospheres

3. If a gas is initially at a pressure of 225 torr and a volume of 3 Liters, what will the pressure of the gas be if its volume decreases to 1 Liter? Assume that temperature is constant.

4. If a gas initially occupies a volume of gas that is equal to 2 Liters at a temperature of 298 K, at what temperature would the gas have a volume of 6 Liters? Assume that pressure is constant.

5. What is the volume of 5 moles of gas that is at a pressure of 10 atm and a temperature of 320 K? Use the gas constant R = 0.0821 L atm K-1 mol-1.

6. What is the total pressure of a gas mixture that contains oxygen at 1.5 atm, nitrogen at 380 torr, and argon at 202.6 kPa?

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Section 8: Solutions

8.1 Solution Terminology

A solution is a homogenous (same throughout) mixture of solvents and solutes. A solvent is a substance that allows another substance to dissolve in it; a solute is anything that dissolves in a solvent.

Solutions do not have a fixed composition – the amount of solute in a solution can be different from solution. For example, a solution of water and sulfuric acid might be 80% water and 20% sulfuric acid. It could also be 20% water and 80% sulfuric acid.

A solution that has a lot of solute is said to be concentrated. A solution with less solute and lots of solvent is said to be dilute. A concentrated solution may be diluted by adding more solvent to the solution.

The terms unsaturated, saturated, and supersaturated are used to describe the amount of solute in a solution. A solution to which more solute can be added is unsaturated. If no more solute can be added, the solution is saturated. And if more solute than normal is in the solution, it is supersaturated.

8.2 Molarity

Molarity is a measure of the concentration of a solution. It is found by dividing the number of moles of solute by the volume of the solution.

Molarity (M) = [moles solute] / Liters of solution

Example:

What is the molarity of a solution that has 10 moles of H2SO4 in 5 liters of an aqueous solution?

Molarity (M) = [moles solute] / Liters of solution

= 10 moles H2SO4 / 5 Liters

= 2 M

8.3 Molality

Molality is another measure of the concentration of a solution; instead of using the volume of solution, it uses kilograms of solvent.

Molality (m) = [moles solute] / kg solvent

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Example:

What is the molality of a solution of 12 moles of H2SO4 in 3 kg of H20?

Molality (m) = [moles solute] / kg solvent

= 12 moles H2SO4 / 3 kg H2O

= 4 m

8.4 Mole fraction

Mole fraction is a way of determining how much of a solution is a given substance. It compares the moles of one part of the solution to the total number of moles of solution. For example, one mole of NaCl is mixed into one mole of H2O. The NaCl splits into one mole of Na+ and one mole of Cl-. Therefore there are one mol H2O + 1 mol Na+ + 1 mol Cl- = 3 mol solution.

The mole fraction of Na+ is thus 1 mol Na+ / 3 mol total = 1/3

8.5 Colligative Properties

There are certain properties of solutions that can be altered just by the presence of solute – any solute. These properties are called colligative properties. Only the amount of solute matters, not the identity of the solute.

The two main colligative properties are freezing point depression and boiling point elevation.

In freezing point depression, the temperature at which a liquid changes phase to solid is decreased. For example, water freezes at 0 oC. If salt is added to the water, the temperature at which the salt water freezes will be less than 0 oC.

The answer to how much the freezing point changes is calculated using the formula ∆T = K f * i * moles of solute. In the equation, Kf is a freezing point constant, and it is different for each solvent. i is called the van’t Hoff factor, which is the number of pieces a solute molecule breaks into in solution.

For example, the van’t Hoff factor of NaCl is 2 – in solution, NaCl splits into Na + and Cl-, which is two pieces. The van’t Hoff factor of CaBr2 is 3 – in solution, it is Ca2+ and 2 Br-, for a total of three pieces.

Example:

What is the new freezing point of a solution if the original freezing point of the solvent is 10 oC? Three moles of solute A are added, and solute A breaks into three pieces in solution. Assume that Kf for the solvent is -2 OC.

The equation is ∆T = Kf * i * moles of solute.

There are three moles of solute, and the van’t Hoff factor i is three. Kf = -2 OC. Replace the variables.

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∆T = -2 OC * 3 *3 moles = -18 OC.

The original freezing point is 10 oC, and the calculated temperature change is -18 OC. Add the two together to find the new freezing point of -8 OC.

Boiling point elevation works the same way, but the constant Kf is replaced by Kb, the boiling point elevation constant.

Practice:

1. Calculate the molarity of the following solutions:a. A solution that has 4 moles of hydrofluoric acid in 6 liters of waterb. A 8 liter solution containing 2 moles of nitric acidc. A solution of 16 moles of carbonic acid in 2 liters of water

2. Calculate the molality of the following:a. A solution of 16 moles of acid in 32 kilograms of solvent.b. A solution of 9 moles of sulfuric acid in 27 kilograms of solvent.c. A solution of 6 moles of acid in 0.5 kilograms of solvent.

3. Calculate the mole fraction of each component of the following:a. A solution of 6 moles of NaCl, an ionic solute, in 7 moles of waterb. A solution of 3 moles of sugar in 6 moles of waterc. A solution of 3 moles of Na2CO3, an ionic solute, in 5 moles of water

4. Calculate the boiling point elevation and freezing point depression of the following:a. A solution containing 3 moles of NaCl; Kb = ; Kf =b. A solution containing 2 moles of K2CO3; Kb = ; Kf = c. A solution containing 2 moles of sucrose, a sugar; Kb = ; Kf =

Section 9: Acid-Base Chemistry

9.1 General Properties of Acids and Bases

Acids have a few general properties:

1. Taste sour2. React with certain metals like Zinc and Iron to give off hydrogen gas3. Cause certain dyes to change color4. React with bases to form salts and water

By comparison, bases have these general properties:

1. Taste bitter2. Feel slippery/soapy3. Dissolve oils and grease

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4. Cause certain dyes to change color5. React with acids to form salts and water

9.2 Arrhenius Acids and Bases

According to the Arrhenius Theory of Acids and Bases, acids are substances that dissociate in water to produce H+ ions. Bases are substances that dissociate to produce the OH- ion.

Hydrochloric acid – HCl – is an Arrhenius acid. In solution, the molecule will dissociate into an H + and a Cl- ion.

Sodium hydroxide – NaOH – is an Arrhenius base. In solution, it dissociates to Na+ and OH-.

9.3 Bronsted-Lowry Theory of Acids and Bases

The Bronsted-Lowry Theory of Acids and Bases states that acids are proton donors; in solution, they give up an H+ ion. Bases are proton acceptors, meaning that they can accept a proton.

9.4 Lewis Theory of Acids and Bases

According to the Lewis Theory of Acids and Bases, acids are electron-pair acceptors. Bases are electron-pair donors. The Lewis Theory of Acids and Bases is the most general definition of acids and bases, and allows for molecules that do not have hydrogen atoms to still be classed as acids and bases.

9.5 Conjugate pairs

An acid has a molecule called a conjugate base. The conjugate base is the same molecule as the acid, minus a proton. For example, the molecule H2SO4, sulfuric acid, has the conjugate base HSO4

-. Similarly, a base has a conjugate acid.

9.6 Strong vs. Weak

Acids and bases can be classified as strong or weak based on how completely they dissociate in solution. A strong acid – like HCl – will completely dissociate in solution. Similarly, a strong base – NaOH, for example – will completely dissociate in solution. Weak acids and weak bases will not completely dissociate in solution.

9.7 Formation of Salts

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When an acid and base react together, a double replacement reaction may occur. This reaction is called a neutralization reaction and the products are water and a salt.

For example, sulfuric acid and sodium hydroxide react to form sodium sulfate and water:

H2SO4 + 2NaOH Na2SO4 + 2H2O

9.8 pH Scale

The pH scale is a measure of the acidity or basicity of a solution. It is calculated by taking the negative logarithm of the concentration of the Hydronium ion, H3O+ (sometimes shown simply as H+). Acids will have lower pH because they increase the concentration of H3O+

in solution. Bases have higher pH because there is relatively little H3O+ in basic solutions.

The pH scale runs from 0 to 14. A stronger acid will have a pH nearer to 0, while a stronger base will have a pH nearer to 14. The pH of water is 7; water is of neutral pH.

Practice:

1. Explain the difference between Arrhenius, Bronsted-Lowry, and Lewis Acids and Bases.2. Identify the following as an acid or a base:

a. HClb. HIc. Ca(OH)2

d. H2CO3

e. NaOHf. Mg(OH)2

3. What does it mean to be a ‘proton donor’?4. How many protons can be removed from H3PO4?5. Predict the products of the neutralization reaction of H3PO4 and KOH.6. Coffee is acidic. Would you expect the pH of coffee to higher or lower than 7? Explain.7. Stomach acid is more acidic than citric acid (acid found in oranges). Which would have the

lower pH? Explain.

Section 10: Introduction to Thermodynamics

Thermodynamics is study of changes in the energy contained in reactions.

10.1 Enthalpy

Enthalpy is a measure of the heat content of a system. A system is what is involved in the reaction – the chemicals involved – while everything outside of the reaction is called the surroundings.

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Enthalpy can’t be measured directly, but the change in enthalpy can be measured, and is represented by the quantity ∆H.

∆H is usually given in units of Kilojoules/mole (KJ/mol).

10.2 Entropy

Entropy is a measure of the disorder in a system – basically, how messy is the system? An easier way to think of entropy, however, is to think of it as the tendency of energy to spread out. Instead of all of the energy in a reaction staying in one place, it spreads out, or decentralizes.

The change in entropy is represented by ∆S, and the usual units of ∆S are Joules/Kelvin (J/K).

10.3 Gibbs Free Energy

Gibbs Free Energy combines enthalpy and entropy, along with the absolute temperature of a system, to show whether a reaction will occur without the input of energy. A value for the change in Gibbs Free Energy is calculated according to the formula ∆G = ∆H - T∆S. If the value of ∆G is negative, the reaction will proceed spontaneously (happens by itself) and is said to be an exergonic reaction. If a reaction has a ∆G value that is greater than zero, it will not occur spontaneously and is called an endergonic reaction.

There are four possible conditions of ∆H and ∆S, and we can predict if a reaction will be spontaneous or non-spontaneous for each set of conditions:

∆H ∆S ∆G Notes+ + Depends on T Spontaneous if T is high- - Depends on T Spontaneous if T is low- + <0 Always spontaneous+ - >0 Never spontaneous

Let’s look at an example:

Calculate the change in Gibbs Free Energy of a reaction if ∆H is 100 kJ and ∆S is 100 J/K at a temperature of 500 K. Is the reaction spontaneous or nonspontaneous at 500 K? Is the reaction always spontaneous or nonspontaneous?

This question asks for three different things – the change in Gibbs Free Energy, ∆G, whether or not the reaction is spontaneous, and whether the reaction’s spontaneity is dependent on temperature.

We can answer the third question just by looking at the numbers. ∆H and ∆S are both positive, so we can say that the reaction is spontaneous if the temperature is high.

Let’s calculate ∆G:

∆G = ∆H - T∆S -This is the formula for ∆G

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∆G = 100 kJ – 500 K (100 J/K) -Plug in the values from the question

∆G = 100 kJ – 50,000 J -Multiply T∆S

∆G = 100 kJ – 50 kJ -Convert Joules to kilojoules and subtract

∆G = 50 kJ -- ∆G is equal to 50 kJ

We know that if ∆G is positive, the reaction is not spontaneous. Thus, at a temperature of 500 K, this reaction is not spontaneous.

Practice:

1. Explain what is meant by ‘spontaneous’.2. What is the difference between an endergonic and an exergonic reaction?3. Predict whether the following reactions will be spontaneous or nonspontaneous:

a. ∆H = -100 kJ, ∆S = 50 J/K, T = 100 Kb. ∆H = 100 kJ, ∆S = 50 J/K, T = 500 Kc. ∆H = -100 kJ, ∆S = -500 J/K, T = 200 Kd. ∆H = 100 kJ, ∆S = -500 J/K, T = 300 K

4. Calculate ∆G for the following reactions:a. ∆H = -100 kJ, ∆S = 50 J/K, T = 100 Kb. ∆H = 100 kJ, ∆S = 50 J/K, T = 500 Kc. ∆H = -100 kJ, ∆S = -500 J/K, T = 200 Kd. ∆H = 100 kJ, ∆S = -500 J/K, T = 300 K

Section 11: Reaction Mechanics

Why do reactions occur? Why do reactions happen as quickly as they do?

11.1 Collision Theory

A major theory of reactions is that reactions occur when reactants run into one another. The collision of molecules with enough energy allows for chemical bonds to be broken – which allows new bonds to form.

11.2 Graphs of Reaction

A reaction can be graphed in terms of energy vs. reaction progress. Energy is on the vertical axis, while the reaction proceeds to the right along the horizontal axis.

11.3 Activation Energy

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The amount of energy necessary for a reaction to occur is called the activation energy. Most reaction diagrams (see above) have a curve; the top of the curve represents the energy present in the system at what is called the activated complex. The activated complex is essentially the molecules of the reactants colliding.

The difference between the energy level of the activated complex and the reactants is the amount of energy that must be added to the system for the reaction to proceed. A reaction with a high activation energy is not likely to start, while a reaction with a low activation energy starts quickly.

11.4 Exothermic and Endothermic Reactions

When looking at reaction graphs, it is easy to see whether the reactants or the products have a higher energy level – one side of the graph is higher than the other!

If the reactants have higher energy than the products, the reaction is said to be exothermic. An exothermic reaction results in energy leaving the system as heat.

If the products have more energy than the reactants, then energy has to have been taken from the surroundings (and in fact, endothermic reactions feel cold to the touch). Such reactions are called endothermic reactions.

11.5 Equilibrium and Reversible Reactions

Some reactions are said to be reversible reactions. In reversible reactions, the reaction can proceed both forward and back – from reactants to products, and from products to reactants. Such reactions use a double arrow, indicating the reaction can go back and forth.

For example, the reaction A + B C is a reversible reaction. A and B are reactants for the forward reaction, with the product of C. In the reverse reaction, C acts as a reactant and A and B become products.

In reversible reactions, the forward and reverse reactions will both occur. Depending on concentrations, the reactions will occur at different rates; however, at certain concentrations, the reaction rates will be equal, and so products from the forward reaction and reactants from the reverse reaction will form at equal rates. The reaction is said to be at equilibrium.

11.6 Le Chatelier’s Principle

What happens to the equilibrium if something in the system changes? If more product is added, what happens to the reaction rates?

Le Chatelier’s Principle tells us that the equilibrium will shift toward the products or toward the reactants to restore equilibrium in the system.

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For example, if a system is at equilibrium and more product is added, the equilibrium position will shift toward the reactants. More product means that the reverse reaction is more likely to occur.

Practice:

1. Draw a graph of an exothermic reaction. Label the reactants, products, activated complex, and activation energy of the reaction.

2. Describe what would be expected to happen to the equilibrium position of a reaction if more product is added to a system.

3. Explain the difference between and exothermic reaction and an endothermic reaction.

Section 12: Nuclear Chemistry

12.1 Radioactivity

Elements can give off particles or bits of energy (called radiation). These elements are said to be radioactive. Radiation is caused by the nuclear decay of atoms of element.

Atoms decay to become more stable elements, with stability determined by the number of protons and neutrons in the nucleus of an atom. Light elements tend to have about twice as many protons as neutrons, while heavier elements usually have more neutrons than protons. If the balance of protons and neutrons is off, decay will occur to create a more stable nucleus.

12.2 Types of Decay

Alpha decay is emission of an alpha particle by an atom. An alpha particle is a helium nucleus – two protons and two neutrons, with no electrons. Alpha particles are the heaviest and most damaging type of radiation.

An alpha particle can be written using the symbol α, or as 42He2+.

If an element emits an alpha particle, it will become a different element because it has lost two protons.

Example: The alpha decay of uranium to thorium

23892U 234

90Th + 4

2He2+

A second type of decay is called Beta decay, in which a neutron turns into a proton and an electron is emitted. The electron is called a beta (β) particle. Since a neutron has changed into a proton, the element changes.

Example: The beta decay of iodine to xenon

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13153I 131

54Xe + 0-1e

Gamma decay is the emission of energy by an atom in a high-energy state. The atom gets rid of some energy without changing to a different element. Gamma radiation is represented by the symbol γ, the Greek letter for gamma.

Example: The Gamma decay of cobalt

6027Co 60

27Co + γ

12.3 Half-life

The half-life of a compound is the amount of time it takes for half of a sample of the compound to undergo nuclear decay. For some elements, this time is much less than a second. For others, the half-life is billions of years.

Some basic calculations can tell us how much of a given sample of a radioactive element is left after some number of half-lives.

Example: How much of a sample of element X is left after 4 half-lives if there were 100 grams of element X?

After each half-life, half of the initial sample is gone, so the answer can be found by dividing the initial mass of the sample by 2 repeatedly (dividing by 2 gives half of the sample).

After one half-life 100/2 = 50 grams

After two half-lives 50/2 = 25 grams

After three half-lives 25/2 = 12.5 grams

After four half-lives 12.5/2 = 6.25 grams

So, after 4 half-lives, there are 6.25 grams of element X remaining.

Practice:

1. Give the alpha decay product of 24797Bk.

2. Give the beta decay product of 8035Br.

3. Write the alpha decay product of 22688Ra.

4. How much of 200 gram sample of an element will remain after 2 half-lives? After 5 half-lives?

5. If a compound has a 200 year half-life, how many half-lives will pass in 300 years? In 600 years? If there were initially 100 grams of the compound, how many grams will remain after 600 years?

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Section 13: Introduction to Organic Chemistry

Organic chemistry is the chemistry of complex carbon compounds. More than that, it is the chemistry of life. All life depends on organic compounds in order to function. Organic chemistry focuses on just a handful of elements – Carbon, Hydrogen, Oxygen, and Nitrogen are the most common – but there are more than six million known organic compounds.

13.1 Prefixes

In organic chemistry a set of prefixes is used that tells how many carbon atoms are in the longest chain of an organic compound.

The basic prefixes:

Prefix # of Carbon Atoms Prefix # of Carbon AtomsMeth- 1 Hex- 6Eth- 2 Hept- 7Prop- 3 Oct- 8But- 4 Non- 9Pent- 5 Dec- 10

If you see a compound with a name like heptanol, you should immediately recognize that there are seven carbon atoms in the compound.

13.2 Alkanes

Alkanes are the simplest type of organic molecule. They are made entirely of carbon and hydrogen, and they have simple, single bonds between atoms.

The simplest alkane is methane, CH4:

Much of basic organic chemistry comes down to remembering basic formulas and rules. For alkanes, we can predict how many hydrogen atoms are present if we know the number of carbon atoms. There’s a simple formula for alkanes:

CNH2N+2

That is, the number of hydrogen atoms is two times the number of carbon atoms, N, plus 2. Think about methane. There is one carbon atom in methane, so by the formula, C (1)H(2*1+2) = CH4

Similarly, a 5 carbon alkane will have 12 hydrogen atoms: C(5)H(2*5+2) = C5H12

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Alkanes are called by simple names, made of the prefix that gives the number of carbon atoms, and the ending ‘ane’, which indicates single bonds.

C5H12 is thus called pent + ane = pentane

13.3 Alkenes

Alkenes are a class of simple organic molecules that contain a double bond. Recall that double bonds are formed when two covalent bonds form between two atoms. Carbon can form double bonds with carbon atoms, and so an alkene has at least two carbon atoms that are bonded with a double bond.

The simplest alkene is called ethane, C2H4:

Alkenes have a formula, much like alkanes do; it is CNH2N; each additional bond between carbon atoms removes two hydrogen atoms. A 5 carbon alkene has 10 hydrogen atoms: C (5)H(2*5) = C5H10

Alkenes are named using a prefix and the ending ‘ene’. The 5 carbon alkene is thus pent + ene = pentene.

13.4 Alkynes

Alkynes are organic compounds that have a triple bond between two carbon atoms.

The simplest alkyne is ethyne, C2H2.

Alkynes follow the formula CNH2N-2; each additional bond removes two carbon atoms from the formula for alkanes. A 5 carbon alkyne is C(5)H(2*5-2) = C5H8

Alkynes are named with a prefix for the number of carbons, and the ending ‘yne’. Therefore, a 5 carbon chain with a single triple bond would be called pent + yne = pentyne.

13.5 Homologous Series

A key concept in organic chemistry is the idea of homologous series, or organic molecules that differ only in the number of CH2 groups in the carbon chain. For example, the molecules propane and pentane are part of a homologous series:

Propane: H3C-CH2-CH3

Pentane: H3C-CH2-CH2-CH2-CH3

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The groups on the ends are the same – here, both molecules have CH3 groups at both ends. The difference is that propane has 1 CH2 group in the middle, and pentane has 3 CH2 groups in the middle.

For basic organic chemistry, it is usually sufficient to check and see if the ends of two molecules are the same and then count the number of CH2 carbons in the middle.

13.5 Isomers

Organic molecules don’t have to be in a straight chain, leading to many ways to put together the same number of atoms. These different arrangements of the same atoms are called isomers. Let’s look at the example of C5H12 – which we have seen before as pentane.

Pentane: H3C – CH2 – CH2 – CH2 – CH3

2-methylbutane: H3C – CH – CH2 – CH3

CH3

CH3

2,2-dimethylpropane: H3C – C – CH3

CH3

Counting atoms, each one of the above molecules has five carbon atoms and twelve hydrogen atoms, so they are all isomers. Additionally, we can call 2-methylbutane and 2,2 – dimethylpropane branched chain hydrocarbons.

To name a branched chain molecule, follow these steps:

1. Find the longest carbon chain and count how many carbons are in the chain. 2. Identify the branches – these are called substituents. Count the number of carbons in the

substituents and name them according to the number of carbons. A one carbon substituent is called ‘methyl’; three carbons is ‘propyl’; a prefix +’yl’ names the substituent.

3. Number the carbons in the longest chain, starting from the end that is closest to a substituent.

4. Assign numbers to the substituents.5. Put everything together. Place substituents in alphabetical order (Butyl before ethyl, ethyl

before methyl, methyl before propyl, and so on).

Example: Name the following molecule: H3C – CH2 – CH – CH2 – CH2 – CH3

CH3

1. Find the longest carbon chain: H3C – CH2 – CH – CH2 – CH2 – CH3

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CH3

The longest chain has 6 carbon atoms, so the chain is hexane.

2. Identify the branches: H3C – CH2 – CH – CH2 – CH2 – CH3

CH3

This is a one-carbon substituent, or a methyl group.

1 2 3 4 5 6

3. Number the carbons in the longest chain: H3C – CH2 – CH – CH2 – CH2 – CH3

CH3

4. Assign numbers to the substituents: 1 2 3 4 5 6

H3C – CH2 – CH – CH2 – CH2 – CH3

CH3

The methyl group is connected to the third carbon in the chain, so it is called ‘3-methyl’

5. Put the pieces together:

3-methyl + hexane = 3-methylhexane.

Remember the following rules:

1. Give substituents the lowest numbers possible – if from one end you have a substituent on the second carbon, and at the other end the first substituent is on the third carbon, start at the end with the closest substituent.

2. Arrange substituents in alphabetical order – it would be 4-ethyl, 2-methylheptane, not 2-methyl, 4-ethylheptane.

3. If you have more than one of a substituent, list all of the numbers and then use a prefix before the substituent. For example, if there are two methyl groups on the second carbon and a third on the fourth carbon of an eight-carbon chain, it would be called “2,2,4-trimethyloctane”

Practice:

1. Name the following:a. A six-carbon molecule with a double bondb. A six-carbon molecule with a triple bondc. A five-carbon molecule with single bondsd. A four –carbon molecule with a double bond

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2. Give the names of the homologous series of single-bonded alkanes from one carbon to nine carbons.

3. Give the names of the homologous series of alkynes from two carbons to seven carbons.4. Draw and name two isomers of a five-carbon alkane.5. Name the following compounds:

a. H3C – CH2 – CH2 – CH3

b. H3C – CH2 – CH(CH3) – CH3

c. H2C = CH – CH2 – CH2 – CH3

Section 14: Hydrocarbon Derivatives

Basic hydrocarbons are composed of just hydrogen and carbon. Organic molecules can incorporate other elements, like nitrogen, oxygen, and halogens. These elements can be classified as various types of hydrocarbon derivatives. These molecules include alcohols, haloalkanes, aldehydes, ketones, carboxylic acids, esters, ethers, amines, and others.

14.1 Alcohols

Alcohols are molecules that contain an –OH group. Naming an alcohol is done similarly to naming alkanes, but with the ending ‘ol’. If the molecule has only single bonds, it will be called an ‘anol’. If there are double bonds, it is an ‘enol’. And if there is a triple bond, the molecule is called an ‘ynol’.

The alcohol in Club Beer is called ‘ethanol’, or ‘ethyl alcohol’.

14.1.1 Primary, Secondary, and Tertiary Alcohols

Alcohols are classified as primary, secondary, or tertiary alcohols by the number of carbons attached to the carbon atom with which the –OH group bonds. If there is only one carbon atom attached, it is a primary alcohol.

Example: The primary alcohol propanol

H3C – CH2 – CH2OH

If there is a second carbon atom attached, the molecule is a secondary alcohol.

Example: The secondary alcohol butan-2-ol

H3C – CH2 – CHCH3OH

If there are three carbon atoms attached, the molecule is a tertiary alcohol.

Example: The tertiary alcohol 2-methylbutan-2-ol

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H3C – CH2 – C(CH3)2OH

Example: Is the following alcohol a primary, secondary, or tertiary alcohol? Name the molecule.

H3C – CH2 – CH – CH2 – CH3

OH

First, look at the carbon that the –OH group is attached to (the third carbon). What else is attached to it? You should see that it is connected to two other carbon atoms, and a single hydrogen atom. Therefore, it is a secondary alcohol.

Now, for naming:

1. How long is the chain? It has five carbon atoms, making it a pentane chain.

2. Are there are any substituents? There are no substituents, just the alcohol group, which means the molecule with end in ‘ol’.

3. At which carbons are substituents attached?The alcohol group is attached to the third carbon.

4. Put it all togetherPentan-3-ol is the name of the molecule

14.2 Aldehydes

Aldehydes are a class of organic molecule that have a specific arrangement of atoms attached to the chain. That arrangement is –COH, where the carbon atom has a double bond to the oxygen atom and a single bond to the hydrogen atom. A carbon that has a double bond to a single oxygen atom is called a carbonyl carbon.

Aldehydes are named using the ending ‘al’. The following molecule is called ‘propanal’.

H3C – CH2 – COH

14.3 Ketones

Ketones are a class of organic molecules that have a carbonyl carbon that is bonded to two other carbon atoms. Ketones are named with the ending ‘one’.

This molecule is called ‘butanone’.

H3C – CH2 – CO – CH3

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14.4 Carboxylic Acids

Carboxylic acids are yet another class of organic molecules containing a carbonyl carbon. The carbonyl carbon is bonded to a second carbon, as well as an –OH group. The carboxylic acid functional group is written as –COOH.

Carboxylic acids are named with the ending ‘oic acid’. The following molecule is called “propanoic acid”.

H3C – CH2 – COOH

14.5 Esters

Esters are similar to carboxylic acids, but instead of having an –OH group attached to the carbonyl, they have an –OR group, where R represents a carbon chain. Esters are written as –COOR.

Esters are tricky to name – the name of the R group comes first, and is the same as naming a substituent. Then the main chain is named, with the ending ‘oate’. The following molecule is ‘ethyl butanoate’.

H3C – CH2 – CH2 – COOCH2CH3

14.6 Acid Chlorides

In acid chlorides, the carbonyl carbon is bonded to a chlorine atom; this functional group is written –COCl.

An acid chloride:

14.7 Amides

Amides are molecules in which a carbonyl carbon is bonded to a nitrogen atom. Such molecules usually are named with the ending ‘amide’.

The following molecule is butanamide.

H3C – CH2 – CH2 – CONH3

14.8 Ethers

Ethers are a class of hydrocarbon derivatives that have an oxygen atom in the middle of the carbon chain.

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Naming ethers is simple – count the carbons on one side of the oxygen, and name that chain according to the number of carbons. Do the same on the opposite side of the oxygen. Put the names together, and follow it with ‘ether’.

The following molecule is called ethyl propyl ether.

H3C – CH2 – O – CH2 – CH2 – CH3

14.9 Alkyl Halides

Alkyl halides are organic molecules that have a halogen atom as a substituent (recall that fluorine, chlorine, bromine, iodine, and astatine are halogens). An alkyl halide is named similarly to a substituent, using the number of the carbon atom that it is attached to, and then a version of the element name – fluoro, chloro, bromo, iodo – followed by the name of the chain.

The following is 2-bromo, 3-chloropentane.

H3C – CHBr – CHCl – CH2 – CH2 – CH3

14.10 Review of Functional Groups

Functional groups are a set of atoms that show up repeatedly in organic molecules and give their molecules a certain set of characteristics. Sections 14.1 to 14.10 go through some of the functional groups.

The following chart summarizes the functional groups. R represents a carbon chain. X represents a halogen.

Functional Group Formula StructureAlcohol R – OH R – OHAlkyl halide R – X R – XAcid chloride R – COClAldehyde R - COHAmide R – CONH3

Carboxylic Acid R – COOHKetone R – CO – REster R – COOREther R – O – R

Practice:

1. Draw and name the following unbranched hydrocarbons:a. A seven-carbon chain containing a double bond and an aldehyde functional group.b. A four-carbon ketonec. A nine-carbon carboxylic acidd. A five-carbon amide

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e. A three-carbon alcoholf. A four-carbon secondary alcoholg. A six-carbon alkyl halideh. A five-carbon acid chloride

2. Explain how to name an ester, and give an example.3. Explain how to name ether, and give an example.

Answers to Practice Problems

Section 1:

1. Tell whether the following is a physical change or a chemical change:a. A piece of paper is burned. This is a combustion reaction – the paper is burned to

produce water and carbon dioxide, which are different from the starting material. This is a chemical change.

b. Cassava is pounded to make GB – The cassava hasn’t chemically changed, just been beaten. This is a physical change.

c. A piece of paper is folded. – Still a piece of paper, so this is physical change.d. Dry ice (solid carbon dioxide) changes to carbon dioxide gas – The carbon dioxide has

simply changed state from solid to gas, making this a physical change.

Section 2

1. Convert the following to centimeters:a. 35 millimeters b. 230 meters c. 41 decametersb. 35 millimeters * 1 cm/10 mm = 3.5 cmc. 230 meters * 100 cm/1 m = 23,000 cmd. 41 decameters * 1000 cm/1 decameter = 41,000 cm

2. Convert the following to hectometersa. 21 kilometers b. 505 centimeters c. 32 decimetersb. 21 km * 10 hm/1 km = 210 hectometersc. 505 cm * 1 hm/10,000 centimeters = 0.0505 hectometersd. 32 decimeters * 1 hm/1000 decimeters = 0.032 hectometers

3. Convert the following to grams:a. 37 centigrams b. 44 hectograms c. 465 decagramsb. 37 cg * 1 g/10 cg = 3.7 gramsc. 44 hectograms * 100 g/1 hg = 4400 gramsd. 465 decagrams * 10 grams/1 decagram = 4,650 grams

Section 3:

1. Name the following elements and write the electron configuration of the following:

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a. N – Nitrogen ↑↓ ↑↓ ↑_ ↑_ ↑_ b. F – Fluorine ↑↓ ↑↓ ↑↓ ↑↓ ↑_ c. Ca – Calcium ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ d. Fe – Iron ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ e. Ar – Argon ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

2. Write shorthand electron configurations for the following:a. Ca --- [Ar]4s2

b. K --- [Ar]4s1

c. Cl --- [Ne]3s23p5

d. Cs --- [Xe]6s1

Section 4:

1. Give the atomic mass and atomic number of the following elements (use a Periodic Table):a. Na – Sodium has an atomic mass of 22.99 g/mol and an atomic number of 11.b. Ag – Silver has an atomic mass of 107.87 g/mol and an atomic number of 47.c. Cl – Chlorine has an atomic mass of 35.45 g/mol and an atomic number 17.d. Mg – Magnesium has an atomic mass of 24.31 g/mol and an atomic number of 12.

Section 5:

Calculate the following:

1. The formula mass of HNO3, AgCl3, NaOH, and I2

a. The formula mass of HNO3 is 1 amu + 14 amu + 3(16 amu) = 63 amu.b. The formula mass of AgCl3 is 108 amu + 3(35 amu) = 213 amuc. The formula mass of NaOH is 23 amu + 16 amu + 1 amu = 40 amud. The formula mass of I2 is 2(127 amu) = 254 amu

2. What percent of the mass of HNO3 is from nitrogen?a. The formula mass of HNO3 is 63 amu; 14 amu is from nitrogen. 14/63 = 0.22 or

22%3. The formula mass of CH3COOH

a. The formula mass is 2(12 amu) + 2(16 amu) + 4(1 amu) = 60 amu 4. What percent of the mass of CH3COOH is from carbon?

a. 24/60 = 0.4 or 40%5. How many moles of HNO3 are present if there are 4.6 * 1024 molecules of HNO3?

a. One mole is 6.02 * 1023 molecules, so 4.6 * 1024/6.02*1023 = 7.6 moles

Section 7:

1. Identify which gas laws would be useful for solving a problem with the provided information:

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a. P1 = X, P2 = Y, V2 = Z There are two pressures and a volume, which suggests Boyle’s Law

b. V1 = X, T1 = Y, T2 = M There are two temperatures and a volume, which suggests Charles’ Law

c. P = Y, T = X, n = Z, R = W The given values suggest that the ideal gas law should be used.

d. P1 = X, P2 = Y, P3 = Z Multiple pressures with no other information should suggest Dalton’s Law of Partial Pressures.

2. Convert the following:a. 122 oC to K 122 + 273 = 395 Kb. 45 oC to K 45 + 273 = 318 Kc. 515 K to oC 515 – 273 = 242 oCd. 298 K to oC 298 – 273 = 25 oCe. 10 atmospheres to torr = 10 atm * 760 torr/1 atm = 7600 torrf. 202.6 kPa to torr = 202.6 kPa * 760 torr/101.3 kPa = 1520 torrg. 1760 torr to atmospheres = 1760 torr * 1 atm/760 torr = 2.32 atm

3. If a gas is initially at a pressure of 225 torr and a volume of 3 Liters, what will the pressure of the gas be if its volume decreases to 1 Liter? Assume that temperature is constant.

a. To solve this problem, use Boyle’s Law: P1V1 = P2V2

b. P1 = 225 torr, V1 = 3 Liters, V2 = 1 Liter, P2 = ? Since we don’t know P2, we will solve for it.

c. Dividing both sides of the equation by V2 will get P2 by itself. i. (P1V1)/V2 = (P2V2)/V2

d. Substitute the numbers into the equation and solve:i. (225 torr * 3 Liters) / 1 Liter = P2 = 675 torr

4. If a gas initially occupies a volume of gas that is equal to 2 Liters at a temperature of 298 K, at what temperature would the gas have a volume of 6 Liters? Assume that pressure is constant.

a. To solve this problem, use Charles’ Law: V1/T1 = V2/T2

b. V1 = 2 Liters, T1 = 298 K, V2 = 6 Liters, T2 = ? Since we don’t know T2, we will solve for it.

c. To solve this problem, start by cross-multiplying. That yields V1T2 = V2T1

d. Next, divide both sites by V1. (V1T2)/V1 = (V2T1)/V1 e. Substitute the numbers into the equation and solve:

i. T2 = (6 Liters * 298 K)/ 2 Liters = 894 K5. What is the volume of 5 moles of gas that is at a pressure of 10 atm and a temperature of

320 K? Use the gas constant R = 0.0821 L atm K-1 mol-1.a. The Ideal Gas Law is PV = nRT. Using algebra (divide both sides by P), it can be

written as V = nRT / Pb. Substitute in the numbers: V = (5 moles * 0.0821 L atm K-1 mol-1 * 320 K) / 10 atmc. The answer is 13.136 liters

6. What is the total pressure of a gas mixture that contains oxygen at 1.5 atm, nitrogen at 380 torr, and argon at 202.6 kPa?

a. This is a Dalton’s Law problem. PTotal = P1 + P2 + P3…b. In this case, PTotal = POxygen + PNitrogen + PArgon

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c. In order to add the different pressures, they all be in the same unit. For simplicity, we will convert to atmospheres.

i. Oxygen is already in atmospheresii. Nitrogen is in torr. 1 atm = 760 torr, so 380 torr * 1 atm/760 torr = 0.5 atm

iii. Argon is in kPa. 1 atm = 101.3 kPa, so 202.6 kPa * 1 atm/101.3 kPa = 2 atmd. Add all the pressures together: PTotal = 1.5 atm + 0.5 atm + 2 atm = 4 atm.

Section 8:

1. Calculate the molarity of the following solutions:a. A solution that has 4 moles of hydrofluoric acid in 6 liters of water

i. Molarity is a measure of the moles of solute divided by the liters of solution – M = moles solute / liters solution

ii. There are 4 moles of the solute, hydrofluoric acid (HF), and 6 liters of solution. M = 4 mol HF / 6 liters H20 = 0.67 M

b. A 8 liter solution containing 2 moles of nitric acidi. M = moles solute / liters solution

ii. There are 2 moles of nitric acid, HNO3, and 8 liters of solution. M = 2 mol HNO3 / 8 liters = 0.25 M

c. A solution of 16 moles of carbonic acid in 2 liters of wateri. There are 16 moles of carbonic acid, the solute, and 2 liters of solution.

M = 16 mol H2CO3 / 2 liters = 8 M 2. Calculate the molality of the following:

a. A solution of 16 moles of acid in 32 kilograms of solvent.i. Molality is calculated by dividing the moles of solute by the kilograms of

solvent. m = moles solute / kg solventii. m = 16 moles acid / 32 kg solvent = 0.5 m

b. A solution of 9 moles of sulfuric acid in 27 kilograms of solvent.i. m = moles solute / kg solvent

ii. There are 9 moles of sulfuric acid, H2SO4, and 27 kilograms of solvent.m = 9 moles H2SO4 / 27 kg solvent = 0.33 m

c. A solution of 6 moles of acid in 0.5 kilograms of solvent.i. m = moles solute / kg solvent

ii. There are 6 moles of acid in 0.5 kg of solvent m = 6 moles acid / 0.5 kg of solvent = 12 m

3. Calculate the mole fraction of each component of the following:a. A solution of 6 moles of NaCl, an ionic solute, in 7 moles of water

i. Ionic solvents dissolve in solution to form ions. NaCl will split in water into the Na+ and Cl- ions. Since there were 6 moles of NaCl, there will be 6 moles of Na+ and 6 moles of Cl-. Add in that there are 7 moles of water, and there are 6 + 6 + 7 = 19 moles present in the solution.

Mole fraction is the moles of a particular substance divided by the total moles in the solution. The mole fractions are thus:

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6 mol Na+ / 19 mols = 6/19 or 0.326 mol Cl- / 19 mol = 6/19 or 0.327 mol water / 19 mol = 7/19 or 0.37

b. A solution of 3 moles of sugar in 6 moles of wateri. 3 moles of sugar + 6 moles of water = 9 moles

ii. The mole fractions are 3 moles sugar / 9 moles = 0.33 and 6 moles water / 9 moles = 0.67

c. A solution of 3 moles of Na2CO3, an ionic solute, in 5 moles of wateri. Na2CO3 will break up in Na+, Na+, and CO3

in solution. If there were three moles of Na2CO3, there will be 6 moles of Na+ and 3 moles of CO3. Combined with 5 moles of water there are 6 + 3 + 5 = 14 moles in total.

ii. The mole fractions are 6 mol Na+ / 14 mol = 0.43 3 mol CO3

2- /14 mol = 0.215 mol H2O / 14 mol = 0.36

4. Calculate the boiling point elevation and freezing point depression of the following:a. A solution containing 3 moles of NaCl in water; Kb = 0.51 ; Kf = -1.86

i. First, identify the van’t Hoff factor, i. In the NaCl, i = 2ii. Use the formulas:

1. ∆T = Kf * i * moles of solute2. ∆T = Kb * i * moles of solute

iii. Put in numbers: ∆T = -1.86 * 2 *3 = 11.16 and ∆T = 0.51 *2 * 3 = 3.06b. A solution containing 2 moles of K2CO3 in acetic acid; Kb = 3.07 ; Kf = -3.90

i. Identify the van’t Hoff factor, i. K2CO3 will break into three pieces, so i=3.ii. Use the formulas:

1. ∆T = Kf * i * moles of solute2. ∆T = Kb * i * moles of solute

iii. Plug in numbers: : ∆T = -3.90 * 3 *2 = -23.4 and ∆T = 3.07 * 3 *2 = 25.86c. A solution containing 2 moles of sucrose, a sugar, in benzene; Kb = 2.65 ; Kf = -5.12

i. Identify the van’t Hoff factor, i. Sugar will not break into pieces, so i=1.ii. Use the formulas:

1. ∆T = Kf * i * moles of solute2. ∆T = Kb * i * moles of solute

iii. Plug in numbers: : ∆T = -5.12 * 1 *2 = -10.24 and ∆T = 2.65 * 1 *2 = 5.3

Section 9:

1. Explain the difference between Arrhenius, Bronsted-Lowry, and Lewis Acids and Bases.a. Arrhenius Acids are compounds that generate the H+ ion in solution; Arrhenius bases

are compounds that generate the hydroxide (OH-) ion in solution. Bronsted-Lowry acids are compounds that can donate a proton – that is, give up a proton. Bronsted-

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Lowry bases accept protons. Lewis acids are electron-pair acceptors, and Lewis bases are electron-pair donors.

2. Identify the following as an acid or a base:a. HCl – this is hydrochloric acidb. HI – hydroiodic acidc. Ca(OH)2 – Calcium hydroxide is a based. H2CO3 – Carbonic acide. NaOH – Sodium hydroxide is a basef. Mg(OH)2 – Magnesium hydroxide is a base

3. What does it mean to be a ‘proton donor’? a. A proton donor is a compound that can give away a hydrogen ion, which is also

called a proton.4. How many protons can be removed from H3PO4?

a. There are 3 hydrogen atoms in H3PO4, so three protons can be removed from the compound.

5. Predict the products of the neutralization reaction of H3PO4 and KOH.a. Neutralization reactions always result in water and a salt. Therefore, the products

will be water and K3PO4, a salt.6. Coffee is acidic. Would you expect the pH of coffee to higher or lower than 7? Explain.

a. Acids have pHs under 7, so if coffee is acidic it will have a pH that is less than 7.7. Stomach acid is more acidic than citric acid (acid found in oranges). Which would have the

lower pH? Explain.a. The more acidic compound will have a pH that is closer to 0. In this case, that’s

stomach acid.

Section 10:

1. Explain what is meant by ‘spontaneous’.a. Spontaneous means ‘will happen by itself’

2. What is the difference between an endergonic and an exergonic reaction?a. Exergonic reactions happen spontaneously; endergonic reactions are

nonspontaneous.3. Predict whether the following reactions will be spontaneous or nonspontaneous:

a. ∆H = -100 kJ, ∆S = 50 J/K, T = 100 Ki. ∆H is negative and ∆S is positive, so we can predict that ∆G is negative and

the reaction is spontaneous.b. ∆H = 100 kJ, ∆S = 50 J/K, T = 500 K

i. ∆H is positive and ∆S is positive, so we can predict that ∆G is negative at some temperatures, so the reaction may be spontaneous or nonspontaneous depending on the temperature.

c. ∆H = -100 kJ, ∆S = -500 J/K, T = 200 Ki. ∆H is negative and ∆S is negative, so we can predict that ∆G is negative at

some temperatures, so the reaction may be spontaneous or nonspontaneous depending on the temperature.

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d. ∆H = 100 kJ, ∆S = -500 J/K, T = 300 Ki. ∆H is positive and ∆S is negative, so we can predict that ∆G is always positive

and the reaction is nonspontaneous.4. Calculate ∆G for the following reactions:

a. ∆H = -100 kJ, ∆S = 50 J/K, T = 100 Ki. ∆G = ∆H - T∆S

ii. ∆G = -100 kJ – 100K(0.05 kJ/K)iii. ∆G = -105 kJ

b. ∆H = 100 kJ, ∆S = 50 J/K, T = 500 Ki. ∆G = ∆H - T∆S

ii. ∆G = 100 kJ – 500K(0.05 kJ/K)iii. ∆G = 75 kJ

c. ∆H = -100 kJ, ∆S = -500 J/K, T = 200 Ki. ∆G = ∆H - T∆S

ii. ∆G = -100 kJ – 200K(0.5 kJ/K)iii. ∆G = -200 kJ

d. ∆H = 100 kJ, ∆S = -500 J/K, T = 300 Ki. ∆G = ∆H - T∆S

ii. ∆G = 100 kJ – 300K(-0.5 kJ/K)iii. ∆G = 50 kJ

Section 11:

1. Draw a graph of an exothermic reaction. Label the reactants, products, activated complex, and activation energy of the reaction.

2. Describe what would be expected to happen to the equilibrium position of a reaction if more product is added to a system.

a. Le Chatelier’s Principle tells us that the equilibrium position moves depending on changes to a system. If more product is added to the system, the equilibrium position shifts toward the reactants. If more reactant were added, the equilibrium position would shift back toward the products.

3. Explain the difference between and exothermic reaction and an endothermic reaction.a. In an exothermic reaction there is more energy in the reactants than in the products,

and energy is given off – often as heat. In endothermic reactions, there is more energy in the products than in the reactions, so energy is taken from the system.

Section 12:

1. Give the alpha decay product of 24797Bk.

a. In alpha decay, a helium nucleus is emitted, meaning the mass number will decrease by 4 and the atomic number by 2.

i. 24797Bk 243

95Am + 42He2+

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2. Give the beta decay product of 8035Br.

a. In beta decay, a neutron is converted to a proton. The mass number stays the same and the atomic number goes up by 1.

i. 8035Br 80

35Kr3. Write the alpha decay product of 226

88Ra.a. In alpha decay, a helium nucleus is emitted, meaning the mass number will decrease

by 4 and the atomic number by 2. i. 226

86Ra 22284Rn + 4

2He2+

4. How much of 200 gram sample of an element will remain after 2 half-lives? After 5 half-lives?

a. With each half-life, the amount of the sample remaining decreases by half.i. After 1 half-life, 200/2 =100 grams

ii. After 2 half-lives, 100/2 = 50 gramsiii. After 3 half-lives, 50/2 = 25 gramsiv. After 4 half-lives, 25/2 = 12.5 gramsv. After 5 half-lives, 12.5/2 = 6.25 grams

5. If a compound has a 200 year half-life, how many half-lives will pass in 300 years? In 600 years? If there were initially 100 grams of the compound, how many grams will remain after 600 years?

a. The number of half-lives can be figured about by dividing the time period by the known half-life. In this case, the known-half life is 200 years. Therefore, in 300 years, 300/200 = 1.5 half-lives will pass. In 600 years, 600/200 = 3 half-lives will pass.

b. If there were initially 100 grams of the compound, then after three half-lives there will be 12.5 grams of the compound remaining:

i. After 1 half-life, 100/2 = 50 gramsii. After 2 half-lives, 50/2 = 25 grams

iii. After 3 half-lives, 25/2 = 12.5 grams

Section 13:

1. Name the following:a. A six-carbon molecule with a double bond

i. Six carbons in a chain is ‘hex’, and a double bond is given by ‘ene’. Hex+ene = hexene.

b. A six-carbon molecule with a triple bondi. Six carbons in a chain is ‘hex’, and a triple bond is given by ‘yne’. Hex+yne =

hexyne.c. A five-carbon molecule with single bonds

i. Five-carbon chains are ‘pent’, and single bonds are given by ‘ane’. Pent+ane = pentane

d. A four –carbon molecule with a double bondi. A four-carbon chain is ‘but’ and a double bond is given by ‘ene’. But+ene =

butene.

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2. Give the names of the homologous series of single-bonded alkanes from one carbon to nine carbons.

a. The names of the alkanes from one carbon to nine carbons are methane, ethane, propane, butane, pentane, hexane, heptane, octane, and nonane.

3. Give the names of the homologous series of alkynes from two carbons to seven carbons.a. The alkyne series is ethyne, propyne, butyne, pentyne, hexyne, heptyne

4. Draw and name two isomers of a five-carbon alkane.

5. Name the following compounds:a. H3C – CH2 – CH2 – CH3

i. This is a four carbon alkane - butaneb. H3C – CH2 – CH(CH3) – CH3

i. This is 2-methylbutane.c. H2C = CH – CH2 – CH2 – CH3

i. This is pentene.

Section 14:

1. Draw and name the following unbranched hydrocarbons:a. A seven-carbon chain containing a double bond and an aldehyde functional group.b. A four-carbon ketonec. A nine-carbon carboxylic acidd. A five-carbon amidee. A three-carbon alcoholf. A four-carbon secondary alcoholg. A six-carbon alkyl halideh. A five-carbon acid chloride

2. Explain how to name an ester, and give an example.a. Esters are named by naming the chain that is bonded to the oxygen atom, followed

by the name of the chain containing the carbonyl, and given the ending ‘oate’. An example would be ethyl pentanoate.

3. Explain how to name ether, and give an example.a. Ethers are named by naming the chain on either side of the oxygen atom, followed

by the word ether. An example is ethyl propyl ether.