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Unit 1: ReviewContent Outline: Lab Safety 1.1

I. Safety Rules (10 Commandments of Safety) 1. Know what equipment is being used and what it is used for. 2. Never do anything not in the lab without instructions or without your teacher’s permission.3. Never eat or drink in the lab or eat, drink, or sniff the lab chemicals.4. Know your safety symbols and identify all possible dangers.5. No open shoes in the lab. Pull long hair up when needed.6. Make sure your lab area is clean and uncluttered.7. Dispose of materials properly. 8. In case of any accident, inform your teacher. 9. Know where the nurse and emergency equipment are located.10. Only 1 person per group at a time may go get supplies or use equipment.

II. Safety Symbols

Unit 1: ReviewContent Outline: Scientific Measurement and the Metric System

I. QuantityA. This term is used to describe something that has magnitude, size, or amount.B. This is not the same thing as measurement.

1. Measurement is a process that scientists perform to represent a specific unit of some object. For example, you measured the length of a piece of paper to be 11 inches, or you measured out 3 cups of salt.

2. A measurement nearly always has a number plus a unit.

II. The SI System of measurement used in science

A. SI stands for the French Le Système International d’Unités (International System of Measurement) that was globally accepted in 1960 at the General Conference on Weights and Measures in Sèvres, France.1. It is used and recognized by all scientists around the world, despite the

reluctance of Americans to adopt the system over the old English system of measurement.

B. The SI system is based upon 7 Fundamental Units of Measurement. They are:1. Length (l)

a. Length is measured in meters (m).2. Mass (m)

a. Mass is measured in grams (g).i. Mass is measured using a scale or balance.

b. Mass is different from weight.i. Weight a measure of the gravitational pull on matter (an object).ii. Weight is measured on a spring scale and measure in Newtons after

the great scientist Isaac Newton, who worked with gravity.c. Weight can change from location to location (earth vs. moon); but mass does

not change.3. Time (t)

a. Time is measured in seconds (s).4. Temperature (T)

a. It is measured in Kelvin (K).i. To convert degrees Celsius (°C) to Kelvin:

273 K + °C ; for example 273 + 27 °C = 300K

ii. To convert degrees Fahrenheit (°F) to degrees Celsius (°C):

(°F – 32) X 5/9; for example (78° F - 32) X 5/9 = 46 X 5/9 = 25.6°C

5. Amount of a given substance (n)a. It is measured in moles (mol)b. A mole is a quantity equal to the Formula Weight of a molecule but

measured out in grams. 6. Electric Current (I)

a. Electric current is measured in Amps (A).7. Luminosity (IV)

a. Luminosity is measured in candelas (cd) Sounds like candles.b. You can see this one on light bulb packages in stores. The more…the

brighter.C. Prefixes (Additions at the front of a word) for Magnitude (greater than 1): getting

larger1. Deka (da) = 10; therefore 1 dekameter = 10 meters2. Hector (h) = 100; therefore 1 hectometer = 100 meters3. Kilo (k) = 1,000; 1therefore 1 kilometer = 1,000 meters4. Mega (M) = 1,000,000 ( 1 million); therefore 1 Megameter = 1,000,000 meters5. Giga (G) = 1,000,000,000 (1 billion); therefore 1 Gigameter = 1,000,000,000

meters

6. Tera (T) = 1,000,000,000,000 (1 Trillion); therefore 1 Terameter = 1 Trillion meters

D. Prefixes for portions (pieces) of a whole: getting smallerUnit 1: ReviewContent Outline: Basic Math Concepts

I. Significant Figures (Important numbers in measurements)A. Consists of all digits known with certainty and one final digit that is

estimated/uncertain.B. All digits 1 – 9 are considered significant. For example:

I.95 centimeters = 3 significant figures; 1 textbook = 1 significant figure

C. The significance of zeros depends on the location within the number. The four rules of zeros apply:1. Zeros appearing between non-zero digits are significant. For example:

305 grams = 3 significant figures; 40.06 meters = 4 significant figures

2. Zeros appearing at the end of a number and to the right of the decimal point are significant. For example:

25.00 liters = 4 significant figures; 3,000,000.00 milligrams = 9 significant figures

3. Zeros at the end of a number but to the left of the decimal point may or may not be significant . It depends on if a decimal point appears. For example:

4000 grams = 1 significant figure; 4000.0 grams = 5 significant figures

The zeros in the first number act as “space holders” (they may or may not be known). The presence of the decimal point indicates that they are known, as in the second example.

4. Zeros appearing at the front of all non-zero digits are not significant, as they too are considered “place holders”. For example:

0.0086 meters = 2 significant figures; 0.0000002346 candelas = 4 significant figures

D. Addition or Subtraction involving significant figures1. When adding or subtracting decimals, the answer must have the same number of

digits to the right of the decimal point as there are in the measurement having the fewest digits to the right of the decimal point. For example:

14.3 g + 2.64 g = 16.9 g; 14.3 g – 2.645 g = 11.7 g

E. Multiplication or Division involving significant figures

1. When multiplying or dividing, the answer must have no more significant figures than are in the measurement with the fewest number of significant figures. For example:

25.6 g = 6.1g/mL (6.09 rounds up to 6.1) ; 25.6 g X 14.765 = 378 (377.9 rounds up to 378)

4.2 mL

F. The rules for D and E do not apply when using conversion factors as those conversions are considered as standards that do not change as they express an established relationship. For example:

2.065 g X 1000mg = 2065 mg 1 g

II. Scientific Notation A. This is essentially a way of writing numbers with large amounts of digits in a

condensed form.B. Only significant figures are written when using Scientific Notation.C. It is also based on the powers of 10; but as exponents.

1. Exponents are whole numbers written in superscript to represent a specific number of places the decimal point has moved.a. If the exponent is a positive whole number, the decimal point has been moved

to the left. This would be a larger than 1 number.b. If the exponent is a negative whole number, the decimal point has been

moved to the right. This would be a smaller than 1 number.D. Numbers written in scientific notation have a basic format:

M.N X 10Z ; M = First Significant digit in the number (always followed by the decimal point)

N = Second Significant digit in the number Z = a whole number representing the number of places the decimal

point has moved.

For example: 1,000,000.0 g = 1.0 X 106 g250.0 L = 2.5 X 102 L0.000465 m = 4.65 X 10-4 m

E. Addition and subtraction using Scientific Notation:1. These mathematical operations can only be performed if they possess the same

exponent value.For example:

2.4 X 106 + 5.3 X 106 = 7.7 X 106 OR 5.3 X 106 – 2.4 X 106 = 2.9 X 106

a. If they do not have the same exponent, then one of the numbers will need to be converted so that they do match.

2.4 X 105 + 3.1 X 103 = 2.4 X 105 + 0.031 X 105 = 2.431 X 105

OR

2.4 X 105 + 3.1 X 103 = 240.0 X 103 + 3.1 X 103 = 243.1 X 103

F. Multiplication using Scientific Notation:1. The significant digits, of each number, are multiplied first.2. Then the exponents are added together.

For example:

(2.4 X 105) X (3.6 X 103) = 8.64 X 108

G. Division using Scientific Notation:1. The significant digits are divided first.2. Then the exponents are subtracted.

For example:2.45 X 10 23 = 4.3 X 1010

5.65 X 1012

Step one: 2.45 /5.65 = 0.433 (round to 0.43)Step two: 23 – 12 = 11Step three: Move the decimal to the right to turn 0.43 into 4.3Step four: Since you had to move the decimal to the right, you need to correct your

exponent number to reflect that 11 becomes 10

*If you move the decimal to the right; then subtract that number of moves to the exponent.

* If you move the decimal to the left; the add that number of moves to the exponent

Unit 1: ReviewContent Outline: Dimensional Analysis & Percent Error

II. Dimensional Analysis (Changing on unit/dimension into another unit/dimension)A. This is the mathematical expression of relationship between two different sets of

units, that are related; but in the form of a ratio/fraction. For example, you know than $1 is equal to 4 quarters (0.25) and that a quarter (0.25) is equal to 5 nickels (0.05). See the relationships below:

$1.00 = 4 Quarters; 1 quarter = 5 nickels; therefore $1 = 20 nickels (4 x 5)

4 quarters(0.25) 5 nickels (0.05) 20 nickels (0.05)1 dollar (1.00) 1 quarter (0.25) 1 dollar (1.00)

Notice that each part of the fraction numerator and denominator add up to the to the same quantity. So each ratio equals 1.

1. Numerator – this term refers to the top number in the fraction.2. Denominator – this term refers to the bottom number in the fraction.

3. Always start by listing your units (they must have a direct relationship or it will not work) and then put in your numbers.

III. Conversion Factor A. This is a mathematical technique that allows you to use units to solve problems

involving measurements (remember measurements need units). The basic premise is this:

Unit given = Unit wanted = Unit wanted Unit given

This relationship allows us to cancel out the given unit of measurement and convert/replace it with the wanted unit of measurement. This is an extremely important relationship you must understand as you will use it in the entire course!

For exampleYou know that your desk is 3 feet wide. How many inches is that?

3 feet (unit given) = 12 inches (unit wanted) = 36 inches (unit wanted) (3X 12) = 36

1 foot (unit given) 1

Notice we used the conversion factor of 1 foot being equal to 12 inches; BUT notice the unit placement so that we could convert/replace the given unit with the wanted unit. The units must work in converting or you cannot solve the problem.

IV. Percentage error – the mathematical difference (value) between what you observed and what was expected. Can be thought of as “How far from perfect were we?”A. Measured using the below equation:

% Error = (VObserved - VExpected)/ VExpected X 100

Here, “V” represents any value that is being measured. The closer your percent error gets to zero, the more it becomes a perfect outcome; not a perfect experiment. The farther away from zero, the worse your results are.

Unit 1: ReviewContent Outline: Chemistry Basics

I. AtomA. The smallest unit of matter that has chemical properties because of it having all the subatomic

parts.B. Atoms still maintain their original properties of that element, because the subatomic parts are all

present.

II. Subatomic Particles (Small parts that make up atoms.) “sub” means “below” or “lower”A. Proton - These particles carry a positive charge. (They are located in the nucleus of an atom.)

1. The number of protons never changes in an element. This allowed the Periodic Table to be created.

B. Neutron -These particles carry NO charge, which is called neutral. (They are also located in the nucleus of an atom.)

1. The number of neutrons can change. Atoms with different numbers of neutrons than the normal amount for that element are called Isotopes.

C. Electrons -These particles carry a negative charge. (They are located in the “Electron cloud”. The cloud is created because electrons move at the speed of light, which creates a blur around the atom.)

1. The electrons moving, which is called kinetic energy, is why they are associated with energy and batteries. It is potential energy when they are “bonded”. The symbol is e-.

2. The number of electrons can change. (Atoms with different numbers of electrons than the normal amount for that element are called Ions.)

III. MoleculeA. Two or more atoms bonded together. (They maybe the same type of atom or they maybe different

atoms.)

IV. Energy (represented by the symbol “E”)A. Energy comes primarily from the rapid movement of electrons (e-).B. Potential Energy (PE) – Energy of position. (Usually refers to electrons “locked” in a chemical

bond.)C. Kinetic Energy (KE) – Energy of movement. (Usually refers to electrons that can move freely.)D. E levels or e- shells – Where the electrons or E is located within an atom or molecule.E. Adding energy to electrons makes them move farther out; losing energy causes them to move

inward.F. Valence Shell- Where the outer most electrons are located on an atom.G. Valence e- - Refers to the outer most electrons. (These are the most important for chemical

bonds and the chemical properties of an element or molecule.)H. Valence – Refers to the bonding capacity of an atom. (Depends on the number of valence

electrons.)

V. Chemical Bonds (These occur between elements or molecules.)A. These are attempts to fill the outer most shell (valence shell) so as to become stable molecules

with lower energy states. (Lower Energy = more stable… Second Law of Thermodynamics supports this claim.)

B. Covalent Bonds 1. This type is the strongest type of chemical “bond”.

a. Results from sharing electrons between elements or molecules to fill both outer shells.

2. They always create a molecule. (The size of the molecule may differ though.)a. Two or more atoms together of any kind.

3. Polar molecules carry an electrical charge at opposite poles(poles refers to the “ends” of the molecule) and non-polar molecules do not have an electrical charge

4. Electronegativitya. Refers to the element’s or molecule’s desire to acquire or release electrons.b. Hydrogen atoms (The LEAST electronegative biological element. It wants to

RELEASE e-)c. Fluorine atoms (The MOST electronegative element. It wants to ACQUIRE

e-)C. Structural Formula is used to show the shape of the molecule.D. Molecular Formula are used to tell the elements, and number of atoms of each, that make up a

molecule.1. A.K.A. Chemical Formula

E. Ionic Bonds 1. These are fairly strong bonds while dry – but are weak in water so they dissolve into

ions.

2. These bonds are created by swapping electrons between elements so that each element can fill it’s outer most shell.

3. When dissolved in water Ions (charged particles) are created. (Gatorade is a ion loaded drink.)

a. Cations – possess a positive charge because it has more protons than electrons.b. Anions – possess a negative charge because it has more electrons than protons.c. THESE LOVE WATER. (Because water is a polar molecule too.)

4. Ionic Compounds a. A cation bonded to an anion to make a salt when dry.

F. Hydrogen Bonds 1. Fairly weak bonds. (It is “like” a magnet) (A positive Hydrogen attracted to a negative

“Substance”…USUALLY oxygen.)2. These are not really bonds but instead intermolecular attractions.

(“Inter” means “between”…between molecules)G. Van der Waals Interactions or London Dispersion Forces

1. These are temporary bonds. (Usually a fraction of a second.)(Involves enzymes mostly.)

2. These interactions are “created” when electrons clump on one side of an atom making that side temporarily “negative” and the other side “positive” so that charged particles can attach momentarily and then they unclump, because electrons are moving, and the “interaction” disappears because of loss of the charge.

Unit 1: ReviewContent Outline: Nomenclature

Please pay close attention to the suffixes (endings) as these can help you ID the molecule.

I. Naming Monatomic, Binary, and Polyatomic Ions This is the easiest group!A. Naming (nomenclature) Monatomic Ions:

1. These are single charged (+ or -) atoms.2. Positive ions:

a. They are named using the elements normal name.b. They can possess charges of +1, +2, +3, or even +4.

Looking at the element’s column on the Periodic Table can be helpful for this.

3. Negative ions:a. Step 1: Drop the elements normal ending on the name.

Step 2: Replace the ending with the ending of “ide”.For example, Oxygen becomes Oxide; Fluorine becomes Fluoride

4. Ion names with Roman Numerals: (Mainly used with “d Block elements”)a. These Roman Numerals are part of the Stock System of naming ions and

elements.b. These elements can form 2 or more cations only, each possessing a different

charge.i. There are no elements can form more than 1 anion.

c. The Roman Numeral indicates the charge of the ion.For example, Fe(II) 2+, Fe (III) 3+ OR Pb(II) 2+, Pb (IV) 4+

B. Writing Binary Ionic Compounds1. The total compound charges must equal out to be neutral.2. Step 1: Cation symbol is written first.

Step 2: Write the element’s ionic charge (1+, 2+) above the name.Step 3: Anion symbol is written next to the positive ion.

Single or Polyatomic Step 4: Write the element’s ionic charge (-1,-2) above the name. Step 5: Balance the charges, using the subscripts, so as to equal out to neutral.

The “crossover” method can be used for this step.For example, Al2O3

Aluminum has a charge of +3 so Al+3

Oxygen has a charge of -2 so O-2

Now “crossover” by cross multiplication; Al (+3 x 2 = +6); O (-2 x 3 = -6)

So the formula is written as Al2O3

C. Naming Binary Ionic Compounds1. Step 1: Use your chemical formula.

Step 2: The cation name is written first. (Include Roman numerals if shown.)Step 3: The anion name goes second.

For example, Al2O3 Aluminum Oxide; NaF Sodium Fluoride.a. The ratio of the ions is not indicated in the name because it is understood

based upon the relative charges of the compound’s ions to balance out to be neutral.

D. Naming Polyatomic Ionic Compounds1. Most polyatomic ions with a negative charge contain Oxygen and are referred to

as Oxyanions.(“oxy” refers to “Oxygen” and Anion for negation ions.)

a. The number of Oxygen atoms affects the name of the polyatomic ion.i. The oxygen is not written in the name though.

b. Oxyanions with two forms:i. Ion with the greater number of Oxygens ends with “ate”. Like Nitrate

NO3

ii. Ion with the lesser number of Oxygens ends with “ite”. Like Nitrite NO2

c. Oxyanions with more than 2 forms:i. Same as above applies, but the following 2 also.ii. Ion with the least number of Oxygen begins with the prefix “Hypo”

(means “least”).Like Hypochlorite. ClO-

Then Chlorite. (See above) ClO2-

Then Chlorate. (See above) ClO3-

iii. Ion with the most number of Oxygens begins with the prefix “Per” (means “most”).Like Perchlorate ClO4

-

iv. The most common for this are Chlorine, Phosphorus, and Manganesed. The name is associated with the first element if there is no Hydrogen also

present.Like Bromate BrO3

- Exceptions: Cyanide CN-, Acetate CH3COO-

i. If Hydrogen is present, it is written first, then the next element is given the primary name.

Like Hydrogen Carbonate HCO3- Exception: Hydroxide OH-

II. Complex Ions (Hydrated Cations)

A. They are transitional (d-block) metal ions that have coordinate-covalent bonds with water molecules.1. Coordinate-covalent bonds

a. These are interatomic attractions resulting from the sharing of a lone pair of electrons from one atom with another atom.For example, Fe2+ + 6H2O [Fe(H2O)6]2+

b. These ions are associated with acids.c. These solutions also tend to be very colorful.

III. Naming of Molecular Compound These ones can be a little trickier.A. This is a compound composed of individually covalently bonded atoms, units, or

molecules.B. These can be named (nomenclature) in one of two different ways:

1. Stock Methoda. This system uses the elements or polyatomic molecules oxidation

numbers/state.2. Prefix system

a. This system uses prefixes to indicate the number of atoms, of each kind, within a molecule.

b. Prefixes: 1 –Mono 4 – Tetra 7 – Hepta 10 - Deca 2 – Di 5 – Penta 8 – Octa

3 – Tri 6 – Hexa 9 – Nona

IV. Stock System (Oxidation number)A. Oxidation number/state

1. This term refers to the number of electrons that must be added to or removed from an atom in a combined state to convert the atom into the elemental form.a. Indicated the distribution of electrons among bonded atoms.

B. Assigning Oxidation numbers to atoms or polyatomic groups.1. As a rule, shared electrons go to the more electronegative atom.2. Additional Rules:

a. Atoms in a pure element have oxidation numbers = 0.b. The more electronegative element in a binary (2) compound is assigned it’s

anion charge; the least electronegative element gets assigned it’s cation charge.

c. Fluorine, the most electronegative element, is always assigned an oxidation number = -1.Remember, your Periodic Trends for electronegativity on the Periodic Table? As you go across a period they increase; as you go down a column/group they decrease.

d. Oxygen is always assigned -2 unless with Fluorine. If Fluorine is present, the Oxygen is assigned +2.You must always check for Fluorine, if you see Oxygen.

e. Hydrogen is always assigned +1 unless it is paired with a metal. If it is with a metal, then it is assigned -1.Know your metals by being able to look at a Periodic Table quickly. Remember, there are metals, metalloids, and non-metals.

f. Neutral compounds equal zero.g. Polyatomic ions are assigned oxidation numbers equal to the charge of the

group.

Know those common polyatomic ions to make your life in Chemistry class easier.

h. This system can also be used with ions.C. Naming compounds (molecular or ionic) using oxidation numbers:

1. This method of nomenclature will use Roman numerals for cations2. Step 1: Use your chemical Formula.

Step 2: Write the cation name. First letter is capitalizedStep 3: Look at the subscript of the anion.Step 4: Look at the subscript of the cation.Step 5: Make the charges of each equal out to zero net charge (neutral). It may involve a little math… adding or dividing.Step 6: Write the numeral used to balance the cation charge with the anion charge as a Roman numeral after the cation name. Such as Iron (II) or Lead (IV)Step 7: Write the anion name after the Roman numeral. First letter is lower case.

Remember, to drop the ending and add the ending “ide”.For example, PCl3 Cl = -1 so P must be +3 to equal zero; therefore, it is Phosphorus (III) chloride.

N2O O = -2 (no Fluorine present) so N2 (each must be 1); therefore, it is

Nitrogen (I) oxide. Mo2O3 O = -2 (no Fluorine present) so -2 x 3 = -6; 2 Mo so 6/2 =

3 therefore Molybdenum (III) oxide

V. Prefix System (numbers)A. This system of naming uses the prefix to identify the number of atoms present.B. Step 1: The Element with the smaller group (column) goes first.

If both elements come from the same group, then the highest energy level (row) goes first.

Step 2: If only 1 atom is present, just use the element’s name.If more than 1 atom is present, you must add the prefix to the elements

name.For example, 1 Hydrogen present Hydrogen Chloride (HCl)

3 Hydrogens present Trihydrogen Antimonide (H3Sb) 2 Sodium & 2 Oxygen Disodium Dioxide (Na2O2)

Step 3: Drop the ending on the anion and replace with “ide”. DO NOT confuse the steps with your polyatomic ion rules.

C. If a prefix ends with an “o” or an “a” and the word following begins with a vowel, drop the “o” or “a” from the prefix.

D. If both elements in the compound are non-metals the first named element is decide based upon this sequence: (First)C P N H S I Br Cl O F (last possible)This is an easy way to remember the sequence:C.P. Newton & H.S. Igor brought Cauliflower to the party.

VI. Naming AcidsA. Acid

1. This is a substance that generates Hydrogen Ions (H+) when placed in a solution.a. The Chemical Formula usually begins with H. Exception Acetic Acid

CH3OOH

b. There are basically two types of acids:i. Binary Acids (2 elements present) . One element is α Hydrogen. b. The second element is a Halogen(Group 17).

ii. Oxyacids (3 elements present) . One element is Hydrogen.α b. The second element is Oxygen. c. The third element is a non-metal or polyatomic ion.

B. Naming:1. Step 1: Do you have a monatomic anion or a polyatomic anion. Step 2: If monatomic, - write “hydro” in front of the anion name.

For example, HCl Hydrochloric If polyatomic, - use the polyatomic ion naming rules for the anion.

For example, H3PO4 Phosphoric * The Hydrogen is not mentioned.

Step 3: If monatomic, replace “ide” with “ic” and add the word “acid”. If polyatomic, replace “ite” with “ous” OR “ate” with “ic”

* Don’t forget your prefix, if needed… Hypo or PerFor example, HClO Hypochlorous Acid

HClO2 Chlorous Acid HClO3 Chloric Acid HClO4 Perchloric Acid

VII. Naming Organic Molecule BasicsA. Carbon to Carbon bonds

1. This type of bonding is referred to as Catenation.2. Carbon molecules can form chains or rings.3. Types of Carbon Bonds

a. Single bonds (-)i. These molecules with all single bonds are called Alkanes. (end with

“ane”)ii. If circular, they are called cycloalkanes.

b. Double bonds (=)i. These molecules are called Alkenes. (end with “ene”)

c. Triple bonds ( )Ξi. These molecules are called Alkynes. (end with “yne”)

4. Carbon loves Hydrogen. The two together are referred to as Hydrocarbons.a. These are energy sources, such as oil and gasoline and also oils (plants) and

fats (animals).b. Types of Hydrocarbons

i. Saturated Hydrocarbons. These molecules have α all single bonds in the hydrocarbon area.

* These chains tend to be straight.b. These are solids at room temperature.For example, animal fat

ii. Unsaturated Hydrocarbons. These molecules can have a α double or triple bond in the

hydrocarbon area.* These cause the chains to “kink” or bend” at the double or

triple bond.

b. These are liquids at room temperature.For example, plant oils such as vegetable oil or corn oil.

iii. Polyunsaturated Hydrocarbons. These molecules can have α multiple double or triple bonds in the

hydrocarbon area.* These have many “kinks” or “bends”.

b. These are also liquids at room temperature.

VII. Naming Basic Hydrocarbons A. Uses a Prefix naming system

1 – Meth 4 – But 7 – Hept 10 - Dec2 – Eth 5 – Pent 8 - Oct3 –Prop 6 – Hex 9 – Non

B. Step 1: Count the carbons present in the chain or circle.Step 2: Use the number of carbons to get the appropriate prefix.Step 3: Look at the number of Hydrogens in the compound.

“ane” – 2n + 2 (n = number of Carbons) “ene” – 2n “yne” – 2n-2

Step 4: Write the appropriate ending after the prefix.Step 5: If circular, add “cyclo” in front of the prefix.

Unit 1: ReviewContent Outline: Atomic Mass and the Mole Concept calculations

I. Mole (mol OR n)A. This is a SI (Le Système International d’ Unités) unit (remember from Unit 1) that is

used to represent the amount of a substance.B. It can be written to show the number of atoms or molecules in a working sample of

some element, compound, or molecule, such as sucrose (table sugar) – C6H12O6

1. This concept is very important because scientists, teachers and students cannot work with individual atoms or molecules because they are very, very small and can’t be handled one at a time.

2. So the mole was conceived to represent a working amount of a substance.C. How to calculate a mole:

1. Determine the total atomic or molecular mass of the substance you are working with, using the chemical formula and Periodic Table. (Remember, how to find the Atomic Mass? – Hint…subscript.)

2. Then weigh out, using an electronic balance and weigh boat, that calculated amount.

3. Congratulations, you have just weighed out 1 mole of that substance!

For example: Atomic MassAluminum has an atomic mass of 26.98 AMUs. So you would weigh out 26.98

grams of Aluminum to get a workable amount called a mole.

Molecular MassSalt (NaCl) has an atomic mass of: Sodium – 23.00 AMUs

Chlorine – 35.5 AMUs

Total AMUs = 23.00 + 35.5 = 58.5 AMUsWeigh out 58.5 grams of salt

The unit for of measurement is g/mol.D. Molarity

1. Take your 1 mole of a substance and dissolve it in a small amount of distilled water, if it will dissolve, inside a volumetric flask. Then add distilled water to bring the volume to 1 L of distilled water. You now have a 1 Molar (1 M) solution.

2. Molarity is used for liquid solutions.E. Amedeo Avogrado (1811)

1. He was an Italian physicist.2. He proposed that the volume of a gas (at a given temperature & pressure) is

proportional to the number of atoms, regardless of the type of gaseous substance used.a. This eventually was modified to state: That in 1 mole of a substance there

will always be 6.022 x 1023 atoms or molecules present. (That is a massive amount!)

b. This number became known as Avogadro’s constant when the French physicist Jean Perrin confirmed and proposed this in 1909 in honor of Avogadro’s work.i. Jean Perrin would win the Nobel Prize for his work in 1926.ii. The Nobel Prize is Sciences’ highest Award, The Super Bowl trophy

in Pro Football.c. Perhaps your class will celebrate Mole Day on October (10th month) 23 at

6:02 am.

II. Basic Measurements or Unit Conversions involving the Mole concept:A. More than a mole: the basic concept is: amount you have/ amount of a mole = # of

molesYou have 21.6 grams of Boron (B). How many moles do you have?1 mole of Boron = 10.8 grams so… 21.6/10.8 = 2.0 moles.

You have 77.25 grams of Phosphorus (P). How many moles do you have?1 mole of Phosphorus = 30.9 grams so… 77.25/30.9 = 2.5 moles.

B. Less than a mole: the basic concept is: amount you have/ amount of a mole = # of molesYou have 16.03 grams of Sulfur (S). How many moles do you have?1 mole of Sulfur = 32.06 grams so… 16.03/32.06 = 0.5 moles

You have 2.43 grams of Magnesium (Mg). How many moles do you have?1 mole of Magnesium = 24.30 grams so… 2.43/24.30 = 0.1 moles

C. Conversions from one unit to another unit involving the mole concept:

The basic concept is: Unit given x unit wanted = Unit wanted Unit given

The given unit cancels out and leaves you with the unit wanted.

1. Moles Atoms/Moleculesa. You have 2.0 moles of Copper. How many atoms of Copper do you have?

1 mole = 6.022 x 1023 atoms so: 2 moles x 6.022 x 10 23 atoms = 12.044 x 1023 atoms

1 moleBut using your rules for scientific notation, it becomes 1.2044 x 1024 atoms

b. You have 0.25 moles of Oxygen. How many atoms of Oxygen do you have?

1 mole = 6.022 x 1023 atoms so: 0.25 moles x 6.022 x 10 23 atoms = 1.505 x 1023 atoms

1 mole

2. Atoms/Molecules molesa. You have 1.806 x 1024 atoms of Zinc (Zn). How many moles of Zinc do you

have?Step 1: Convert 1.806 x 1024 to 18.06 x 1023 (It must have the Avogadro exponent of 23.)

Step 2: 18.06 x 1023 Atoms x 1 mole = 3 moles 6.022 x 1023 atoms

b. You have 5.9 x 1022 atoms of Titanium (Ti) How many moles of Titanium do you have?Step 1: Convert 5.9 x 1022 to 0.59 x 1023 (It must have the Avogadro exponent of 23.)

Step 2: 0.59 x 1023 Atoms x 1 mole = 0.098 moles 6.022 x 1023 atoms

3. Grams Molesa. You have 54.0 grams of Carbon (C). How many moles of Carbon do you have?

54.0 grams x 1 mole = 4.5 moles 12.0 grams

b. You have 10.0 grams of Nickel (Ni). How many moles of Nickel do you have?


4. Moles Gramsa. You have 8.5 moles of Fluorine (F) gas. How grams of Fluorine do you have?

8.5 moles x 19.00 grams = 161.5 grams1 mole

b. You have 0.45 moles of Scandium (Sc) gas. How grams of Scandium do you have?


Unit 1: ReviewContent Outline: Advanced Mole Concept Problems

I. In most chemistry equations, students need to be able to perform numerous conversions, often referred to as Dimensional Analysis.

A. Complex Mole Concept problems.1. The underlying fundamental concept map (often shown as a mole map)

Atoms/particles moles grams

We can go from one end of the map to the other end remembering:

Unit given X Unit Wanted 1 x Unit Wanted 2 = Unit Wanted 2 Unit Given Unit Wanted 1

2. Atoms/Particles gramsa. You have 9.45 x 1023 atoms of Cesium (Cs). How many grams of Cesium do you have?

9.45 x 1023 atoms x 1 mole x 132.91 grams = 208.57 grams 6.022 x 1023 atoms 1 mole

Step 1: Divide 9.45 by 6.022 (MAKE SURE YOUR Exponents are in 23) This equals 1.569 moles

Step 2: Multiple 1.569 x 132.91 = 208.54 grams

b. You have 1.25 x 1023 atoms of Tin (Sn). How many grams of Tin do you have?

1.25 x 1023 atoms x 1 mole x 118.71 grams = 24.69 grams 6.022 x 1023 atoms 1 mole

Step 1: Divide 1.25 by 6.022 (MAKE SURE YOUR Exponents are in 23) This equals 0.208 moles

Step 2: Multiple 0.208 x 118.71 = 24.69 grams

3. Grams Atoms/Particlesa. You have 156.90 grams of Gallium (Ga). How many atoms do you have?

156.90 grams x 1 mole x 6.022 x 10 23 atoms = 13.549 x 1023 atoms 69.72 grams 1 mole

Remember your rules for scientific notation: 1.355 x 1024 atoms

Step 1: Divide 156.90 by 69.72 This equals 2.250 moles

Step 2: Multiple 2.250 x 6.022 x 1023 = 13.549 x 1023 OR 1.355 x1024 (move the decimal point to the left… number got bigger)

b. You have 380.00 grams of Glucose (C6H12O6). How many molecules do you have?

380.00 grams x 1 mole x 6.022 x 10 23 atoms = 12.706 x 1023 molecules

180.06 grams 1 mole

Remember your rules for scientific notation: 1.271 x 1024 molecules

Step 1: Determine the Molecular mass of glucose: C6 = 12.01 x 6 = 72.06 AMUsH12 = 1.00 x 12 = 12.00 AMUsO6 = 16.00 x 6 = 96 AMUs

Molecular Mass = 72.06 + 12.00 + 96.00 = 180.06 AMUs

Step 2: Divide 380.00 by 180.06 This equals 2.110 moles

Step 3: Multiple 2.110 x 6.022 x 1023 = 12.706 x 1023 OR 1.271 x1024 (move the decimal point to the left… number got bigger)

Unit 1: ReviewContent Outline: Composition Stoichiometry

I. Stoichiometry (“stoich” means “element”; “metry” means “measure”… measuring elements)A. This is the careful quantitative (numbers) analysis of substances involved in chemical

reactions.B. There are 2 basic types of stoichiometry problems Think about what each is

referring to in name.1. Composition Stoichiometry

a. This type deals with the mass relationships of elements in compounds.2. Reaction Stoichiometry

a. This type deals with the mass relationships between reactants and products in chemical reactions.

b. To solve these problems, you must start with a correctly balanced chemical reaction equation.i. This tells you the relative number of moles (coefficients) for each

reactant & products.

II. Review of Calculating Atomic Mass & Molar Mass for a compoundF. How to calculate a mole:

4. Determine the total atomic or molecular mass of the substance you are working with, using the chemical formula and Periodic Table. (Remember, how to find the Atomic Mass? – Hint…subscript.)

5. Then weigh out, using an electronic balance and weigh boat, that calculated amount.

6. Congratulations, you have just weighed out 1 mole of that substance!

For example: Atomic MassAluminum has an atomic mass of 26.98 AMUs. So you would

weigh out 26.98 grams of Aluminum to get a workable amount called a mole.

Molecular MassSalt (NaCl) has an atomic mass of: Sodium – 23.00 AMUs

Chlorine – 35.5 AMUsTotal AMUs = 23.00 + 35.5 = 17.00 AMUsWeigh out 58.5 grams of salt

7. The unit for of measurement is g/mol.

III. Basic Measurements or Unit Conversions involving the Mole concept:D. More than a mole: the basic concept is amount you have/ amount of a mole = # of

molesYou have 21.6 grams of Boron (B). How many moles do you have?1 mole of Boron = 10.8 grams so… 21.6/10.8 = 2.0 moles.

E. Less than a mole: the basic concept is amount you have/ amount of a mole = # of molesYou have 16.03 grams of Sulfur (S). How many moles do you have?1 mole of Sulfur = 32.06 grams so… 16.03/32.06 = 0.5 moles

F. Conversions from one unit to another unit involving the mole concept:

the basic concept is: Unit given x unit wanted = Unit wanted unit given

The given unit cancels out and leaves you with the unit wanted.

5. Moles Atoms/Moleculesa. You have 2.0 moles of Copper. How many atoms of Copper do you have?

1 mole = 6.022 x 1023 atoms so: 2 moles x 6.022 x 10 23 atoms = 12.044 x 1023 atoms

1 moleBut using your rules for significant figures, it becomes 1.2044 x 1024 atoms

6. Atoms/Molecules molesc. You have 1.806 x 1024 atoms of Zinc (Zn). How many moles of Zinc do you

have?Step 1: Convert 1.806 x 1024 to 18.06 x 1023 (It must the Avogadro exponent of 23.)

Step 2: 18.06 x 1023 Atoms x 1 mole = 3 moles 6.022 x 1023 atoms

7. Grams Molesc. You have 54.0 grams of Carbon (C). How many moles of Carbon do you have?


8. Moles Gramsc. You have 8.5 moles of Fluorine (F) gas. How grams of Fluorine do you have?


IV. Mole RatioA. A conversion factor that relates the amounts, in moles, of any two substances

involved in a chemical reaction.B. All Stoichiometry problems require a mol to mol (mol:mol)ratio be used to convert

substance A information into substance B information.1. This information is found in the substance coefficients; not the subscripts.2. The Law of Conservation of Matter must be satisfied before you can begin to

work the problem.For example: 2Al2O3(l) 4Al(s) + 3O2(g)

You can compare: 2 mol/4 mol OR 4 mol/2 molORYou can compare: 2mol/3 mol OR 3 mol/2 molORYou can compare: 4 mol/3 mol OR 3 mol/4mol

Unit 1: ReviewContent Outline: Review of Reaction Stoichiometry

Students need to pay close attention to the colors and follow the colors to “see” the math concepts.

I. There are basically 4 types of reaction stoichiometry problems: (Think in terms of “given” & “wanted”)A. Given: Moles; Wanted: MolesB. Given: Moles; Wanted: GramsC. Given: Grams; Wanted: MolesD. Given: Grams; Wanted: GramsE. All revolve around this central concept: (You must remember this!)

Grams Substance A Moles Substance A Moles Substance B Grams Substance B

1. You will only Multiple or Divide in these calculations.a. Multiple = Increase quantityb. Divide = Decrease quantityc. So think logically about your answers using the above central concept.

II. Solving Given: moles; Wanted: moles problemsA. This type uses only the middle portion of the central concept:

Moles Substance A Moles Substance B B. Think of it as in terms of scientific units:

Given X Wanted = Wanted Given

Given units cancel out and leave you with wanted units

For example A: How many moles of S2Cl2 are required to react with 7.00 moles of SO2, according to the given chemical reaction:

2 SO2 + S2Cl2 + 3Cl2 4SOCl2

Step 1: Identify the type of chemical reaction that is occurring. (It is a synthesis reaction.)Step 2: Make sure the equation is balanced.

S atoms 4 S atoms 4Cl atoms 8 Cl atoms 8O atoms 4 O atoms 4 (It is balanced)

Step 3: Identify given: 7.00 moles of SO2

Step 4: Identify wanted: moles of S2Cl2

Step 5: Write the needed conversion for units to cancel from given to wanted using coefficients;1 mol of S 2Cl2

2 mol of SO2

Step 6: Solve by putting it all together:

7.00 moles of SO2 X 1 mol of S 2Cl2 = 3.5 moles of S2Cl2

2 mol of SO2

Math worked out: 7 X 1 = 7; Then 7 divided by 2 = 3.5

For example B: How many moles of SOCl2 can be produced from 1.95 moles of Cl2, according to the given chemical reaction:

2 SO2 + S2Cl2 + 3Cl2 4SOCl2

Step 1: Identify the type of chemical reaction that is occurring. (It is a synthesis reaction.)Step 2: Make sure the equation is balanced.

S atoms 4 S atoms 4Cl atoms 8 Cl atoms 8O atoms 4 O atoms 4 (It is balanced)

Step 3: Identify given: 1.95 moles of Cl2

Step 4: Identify wanted: moles of SOCl2

Step 5: Write the needed conversion for units to cancel from given to wanted using coefficients;4 mol of SOCl 2

3 mol of Cl2


1.95 moles of Cl2 X 4 mol of SOCl 2 = 2.60 moles of SOCl2

3 mol of Cl2

Math worked out: 1.95 X 4 = 7.80; Then 7.80 divided by 3 = 2.60

III. Solving Given: moles; Wanted: Grams problemsA. This type uses only this portion of the central concept:

Moles Substance A Moles Substance B Grams Substance BB. Think of it as in terms of scientific units:

Given X Wanted 1 X Wanted 2 = Wanted 2 Given Wanted 1

Given units and Wanted 1 units cancel out and leave you with Wanted 2 units

For example A: How many grams of Ca3P2 are required to produce 6.67 moles of H3P, according to the given chemical reaction:

Ca3P2 + 6 H20 2 H3P + 3 Ca(OH)2

Step 1: Identify the type of chemical reaction that is occurring. (It is a Double Replacement reaction.)Step 2: Make sure the equation is balanced.

Ca atoms 3 Ca atoms 3P atoms 2 P atoms 2O atoms 6 O atoms 6

H atoms 12 H atoms 12 (It is balanced)Step 3: Identify given: 6.67 moles of H3PStep 4: Identify wanted: grams of Ca3P2

Step 5: Write the needed conversion 1 for units to cancel from given to wanted 1 using coefficients;1 mol of Ca 3P2

2 mol of H3P

Step 6: Calculated the Molecular Mass of Ca3P2 as we will need to convert from moles to grams.Ca has a mass of 40.0 g x 3 = 120.0 gP has a mass of 30.9 g x 2 = 61.8 gTotal molecular mass for Ca3P2 = 181.8 g/mol

Step 7: Write the needed conversion 2 for units to cancel from wanted 1 to wanted 2;181.8 grams of Ca 3P2

1 mol of Ca3P2


6.67 moles of H3P X 1 mol of Ca 3P2 X 181.8 grams of Ca 3P2= 606.3 grams of Ca3P2

2 mol of H3P 1 mol of Ca3P2

Math worked out: 6.67 X 1 = 6.77; Then 6.77 divided by 2 = 3.335; 3.335 x 181.8 = 606.3

For example B: Use the same equation. Assuming that Ca3P2 is in excess, how many grams of Ca(OH)2 can be produced from 4.23 moles of H2O, according to the given chemical reaction:

Ca3P2 + 6 H20 2 H3P + 3 Ca(OH)2

Step 1: Identify the type of chemical reaction that is occurring. (It is a Double Replacement

reaction.)Step 2: Make sure the equation is balanced.

Ca atoms 3 Ca atoms 3P atoms 2 P atoms 2O atoms 6 O atoms 6

H atoms 12 H atoms 12 (It is balanced)Step 3: Identify given: 4.23 moles of H2OStep 4: Identify wanted: grams of Ca(OH)2

Step 5: Write the needed conversion 1 for units to cancel from given to wanted 1 using coefficients;3 mol of Ca(OH) 2

6 mol of H2O

Step 6: Calculated the Molecular Mass of Ca(OH)2 as we will need to convert from moles to grams.Ca has a mass of 40.0 g x 1 = 40.0 g0 has a mass of 16.0 g x 2 = 32.0 gH has a mass of 1.0 g x 2 = 2.0 gTotal molecular mass for Ca3P2 = 74.0 g/mol

Step 7: Write the needed conversion 2 for units to cancel from wanted 1 to wanted 2;74.0 grams of Ca(OH) 2

1 mol of Ca(OH)2


4.23 moles of H2O X 3 mol of Ca(OH) 2 X 74.0 grams of Ca(OH) 2= 156.5 grams of Ca(OH)2

6 mol of H2O 1 mol of Ca(OH)2

Math worked out: 4.23 X 3 = 12.69; Then 12.69 divided by 6 = 2.115; 2.115 x 74.0 = 156.5

IV. IV. Solving Given: Grams; Wanted: Moles problemsA. This type uses only this portion of the central concept:

Grams of Substance A Moles Substance A Moles Substance B

B. Think of it as in terms of scientific units:

Given X Wanted 1 X Wanted 2 = Wanted 2 Given Wanted 1

Given units and Wanted 1 units cancel out and leave you with Wanted 2 units

For example A: How many moles of SO3 are required to produce 367.3 grams of BF3, according to the given chemical reaction:

Na2B4O7 + 6 CaF2 + 7 SO3 4 BF3 + 6CaSO4 + Na2SO4


Ca atoms 6 Ca atoms 6F atoms 12 F atoms 12Na atoms 2 Na atoms 2

B atoms 4 B atoms 4S atoms 7 S atoms 7O atoms 28 O atoms 28 (It is balanced)

Step 3: Identify given: 367.3 grams of BF3

Step 4: Identify wanted: moles of SO3

Step 5: Calculate the Molecular Mass of BF3 as we will need to convert from grams to moles.B has a mass of 10.8 g x 1 = 10.8 gF has a mass of 19.0 g x 3 = 57.0 gTotal molecular mass for BF3 = 67.8 g/mol

Step 6: Write the needed conversion 1 for units to cancel from given to wanted 1; 1 mol of BF 3

67.8 grams of BF3

Step 7: Write the needed conversion 2 for units to cancel from wanted 1 to wanted 2 using coefficients;

7 mol of SO 3

4 mol of BF3


367.3 grams of BF3 X 1 mol of BF 3 X 7 mol of SO 3 = 9.485 mol of SO3

67.8 grams of BF3 4 mol of BF3

Math worked out: 367.3 X 1 = 367.3; Then 367.3 divided by 67.8 = 5.42; Then 5.42 x 7 = 37.94; then 37.94 divided by 4 = 9.485

For example B: Using the same equation and assuming CaF2 and SO3 are in excess, How many moles of CaSO4 can be produced from 523.7 grams of Na2B4O7, according to the given chemical reaction:

Na2B4O7 + 6 CaF2 + 7 SO3 4 BF3 + 6CaSO4 + Na2SO4


Ca atoms 6 Ca atoms 6F atoms 12 F atoms 12Na atoms 2 Na atoms 2B atoms 4 B atoms 4S atoms 7 S atoms 7O atoms 28 O atoms 28 (It is balanced)

Step 3: Identify given: 523.7 grams of Na2B4O7

Step 4: Identify wanted: moles of CaSO4

Step 5: Calculate the Molecular Mass of Na2B4O7 as we will need to convert from grams to moles.B has a mass of 10.8 g x 4 = 43.2 gNa has a mass of 22.9 g x 2 = 45.8 gO has a mass of 16.0 g x 7 = 112.0 gTotal molecular mass for Na2B4O7 = 201.0 g/mol

Step 6: Write the needed conversion 1 for units to cancel from given to wanted 1; 1 mol of Na 2B4O7

201.0 grams of Na2B4O7


6 mol of CaSO 4

1 mol of Na2B4O7


523.7 grams of Na2B4O7 X 1 mol of Na 2B4O7 X 6 mol of CaSO 4 = 15.63 mol of CaSO4

201.0 grams of Na2B4O7 1 mol of Na2B4O7

Math worked out: 523.7 X 1 = 523.7; Then 523.7 divided by 201.0 = 2.605; Then 2.605 x 6 = 15.63; then 15.63 divided by 1 = 15.63

C. Solving Given: Grams; Wanted: Grams problemsD. This type uses only this portion of the central concept:

Grams of Substance A Moles Substance A Moles Substance B Grams of Substance B

E. Think of it as in terms of scientific units:

Given X Wanted 1 X Wanted 2 X Wanted 3 = Wanted 3 Given Wanted 1 Wanted 2

Given units and Wanted 1 units and Wanted 2 cancel out and leave you with Wanted 3 units

For example A: When NaCl reacts with AgNO3, AgCl precipitates out of the solution. What mass of AgCl is produced from 75.0 grams of AgNO3 based upon the following chemical equation?

NaCl + AgNO3 AgCl + NaNO3


Cl atoms 1 Cl atoms 1Ag atoms 1 Ag atoms 1Na atoms 1 Na atoms 1N atoms 1 N atoms 1O atoms 3 O atoms 3 (It is balanced)

Step 3: Identify given: 75.0 grams of AgNO3

Step 4: Identify wanted: grams of AgClStep 5: Calculate the Molecular Mass of AgNO3 as we will need to convert from grams to moles.

Ag has a mass of 107.8 g x 1 = 107.8 gN has a mass of 14.0 g x 1 = 14.0 gO has a mass of 16.0 g x 3 = 48.0 gTotal molecular mass for AgNO3 = 169.8 g/mol

Step 6: Write the needed conversion 1 for units to cancel from given to wanted 1; 1 mol of AgNO 3

169.8 grams of AgNO3


1 mol of AgCl 1 mol of AgNO3

Step 8: Calculate the Molecular Mass of AgCl as we will need to convert from moles to grams.Ag has a mass of 107.8 g x 1 = 107.8 gCl has a mass of 35.5 g x 1 = 35.5 gTotal molecular mass for AgCl = 143.3 g/mol

Step 9: Write the needed conversion 3 for units to cancel from wanted 2 to wanted 3143.3 grams of AgCl 1 mol of AgCl


75.0 grams of AgNO3 X 1 mol of AgNO 3 X 1 mol of AgCl X 143.3 grams of AgCl = 63.3 g of AgCl

169.8 grams of AgNO3 1 mol of AgNO3 1 mol of AgCl

Math worked out: 75.0 X 1 = 75.0; Then 75.0 divided by 169.8 = 0.442; Then 0.442 x 1 = 0.442; then 0.442 divided by 1 = 0.442: Then 0.442 x 143.3 = 63.3; then divide by 1 = 63.3

Unit 1: ReviewContent Outline: Review of Limiting Reactants & Yield

I. Limiting ReactantA. This is a reactant that limits the amount of the other reactant(s) from reacting to

make product.1. Once it is completely consumed; no more product can be formed…unless you add

more of it.2. It is also called Limiting Reagent.

II. Excess ReactantA. These is/are a reactant(s) that are not completely consumed in the reaction.

1. They will still be present after the reaction has stopped.

III. Solving Limiting Reactant problems

For example A: Using the given chemical reaction below:

IF5 (s) + 3 H2O (l) HIO3 (aq) + 5 HF (g)

If you mixed 7.00 grams of IF5 with 2.00 grams of H2O, which substance would be the limiting reactant?

Step 1: Identify the type of chemical reaction. (It is a Double replacement…I and H exchange)Step 2: Make sure the equation is balanced.

I atoms 1 I atoms 1F atoms 5 F atoms 5H atoms 6 H atoms 6O atoms 3 O atoms 3 (It is balanced.)

Step 3: Compare the given amounts to the reaction values. You are given grams & equation is molesStep 4: Calculate the Molecular Mass of IF5

I has a mass of 126.9 gF has a mass of 19.0 g x 5 = 95 gTotal Molecular Mass = 221.9 g/mol

Step 5: Calculate the Molecular Mass of H2O

H has a mass of 1.0 x 2 = 2.0O has a mass of 16.0 x 1 = 16.0Total Molecular Mass = 18.0 g/mol

Step 6: Convert to each to # moles by dividing the given mass by the molecular mass, so we can use our equation; which is in moles.

IF5 = 7.00 grams = 0.0316 mol of IF5

221.9 gram/mol

H2O = 2.00 grams = 0.111 mol of H2O 18.0 gram/mol

Step 7: Convert each to a percentage by dividing the result of step 6 by the # of moles from the equation so that we can compare percentages to see which is least – limiting

IF5 = 0.0316 mol = 0.0316%1 mol IF5

H2O = 0.111 mol = 0.0370%3 mol H2O

Step 8: Lowest percentage (closest to zero) is the limiting reactant. (In this case, it is IF5 as 0.0316 is less than 0.0370)

For example B: Using the given chemical reaction below:

SiO2 (s) + 4HF(g) SiF4(g) + 2 H2O(l)

If you mixed 6.00 moles of HF with 4.50 moles of SiO2, which substance would be the limiting reactant?

Step 1: Identify the type of chemical reaction. (It is a Double replacement…Si and H exchange)Step 2: Make sure the equation is balanced.

Si atoms 1 Si atoms 1F atoms 4 F atoms 4H atoms 4 H atoms 4O atoms 2 O atoms 2 (It is balanced.)

Step 3: Compare the given amounts to the reaction values. You are given moles & equation is moles (so skip to step 7 from above procedure.)

Step 7: Convert each to a percentage by dividing the result of step 6 by the # of moles from the equation so that we can compare percentages to see which is least – limiting

HF = 6.00 mol HF = 1.5 % 4 mol HF

SiO2 = 4.5 mol SiO2 = 4.5 % 1 mol SiO2

Step 8: Lowest percentage (closest to zero) is the limiting reactant. (In this case, it is HF as 1.5 is less than 4.5)

IV. YieldA. This is the amount of product that is generated by a given chemical reaction.B. There are 2 types of yield for every chemical reaction:

1. Theoretical Yielda. This is the maximum amount of product that could be produced from a

100% consumption of given reactant (It is calculated using stoichiometry)b. This is almost always more than what will actually be generated… hence

theoretical.2. Actual Yield

a. This is the actual amount of product produced and measured from a chemical reaction (Measured amount of product in the laboratory experiment.)

V. Percent YieldA. This is the efficiency of the chemical reaction.B. It is a comparison between the actual yield and the theoretical yield.

1. 100% Percent Yield would be a perfect chemical reaction, but highly unlikely.C. Formula for calculating is:

Percent Yield = Actual Yield (AY) x 100% Theoretical Yield (TY)

VI. Calculating Theoretical Yield:

For example: Calculate the Theoretical Yield for the given chemical reaction:

C6H6(l) + Cl2(g) C6H5Cl(l) + HCl(g)

If you are given 36.8 grams of C6H6 and Cl2 in excess, calculate the theoretical yield of C6H5Cl.

Step 1: As this is a chemical reaction, you must determine which type of reaction stoichiometry problem this is. (It is a Given: grams; Wanted: grams type of problem)

C. This type uses only this portion of the central concept:

Grams of Substance A Moles Substance A Moles Substance B Grams of Substance B

D. Think of it as in terms of scientific units:

Given X Wanted 1 X Wanted 2 X Wanted 3 = Wanted 3 Given Wanted 1 Wanted 2

Given units and Wanted 1 units and Wanted 2 cancel out and leave you with Wanted 3 units

Step 2: Make sure the equation is balanced.C atoms 6 C atoms 6Cl atoms 2 Cl atoms 2H atoms 6 H atoms 6 (It is balanced)

Step 3: Identify given: 36.8 grams of C6H6

Step 4: Identify wanted: grams of C6H5ClStep 5: Calculate the Molecular Mass of C6H6 as we will need to convert from grams to moles.

C has a mass of 12.0 g x 6 = 72.0 gH has a mass of 1.0 g x 6 = 6.0 gTotal molecular mass for AgNO3 = 78.0 g/mol

Step 6: Write the needed conversion 1 for units to cancel from given to wanted 1; 1 mol of C 6H6

78.0 grams of C6H6


1 mol of C 6H5Cl1 mol of C6H6

Step 8: Calculate the Molecular Mass of C6H5Cl as we will need to convert from moles to grams.

C has a mass of 12.0 g x 6 = 72.0 gCl has a mass of 35.5 g x 1 = 35.5 gH has a mass of 1.0 x 5 = 5.0 gTotal molecular mass for C6H5Cl = 112.5 g/mol

Step 9: Write the needed conversion 3 for units to cancel from wanted 2 to wanted 3112.5 grams of C 6H5Cl1 mol of C6H5cl


36.8 grams of C6H6 X 1 mol of C 6H6 X 1 mol of C 6H5Cl X 112.5 grams of C 6H5Cl = 63.3 g of C6H5Cl

78.0 grams of C6H6 1 mol of C6H6 1 mol of C6H5Cl

Math worked out: 36.8 X 1 = 36.8; Then 36.8 divided by 78.0 = 0.472; Then 0.472 x 1 = 0.472; then 0.472 divided by 1 = 0.472: Then 0.472 x 112.5 = 53.1; then divide by 1 = 53.1 is the theoretical yield

VII. Calculating percentage yieldA. % yield = AY/TY x 100%

For example: Using the data above, you perform the same chemical reaction, measure your product but only get 38.8 grams. What is the percentage yield for this experiment?

Step 1: Calculate Theoretical yield using above process (We know it to be 53.1 grams.)Step 2: Identify the actual yield (It must be given to you, if you are not actually performing the experiment.)Step 3: Calculate using the formula:

38.8 grams x 100% = 61.29%63.3 grams

Your top number should always be less than the lower, as there is no such thing as a perfect experiment. If the actual yield is higher than the theoretical yield this indicates an impure product.

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