Welcome to AP Chemistry!

This course is for the student desiring a first year college-level chemistry course. It is a second year high school course completed after Chemistry. The topics covered are solution and gas chemistry, oxidation-reduction, electrochemistry, equilibrium relations, acid-base/buffer systems, thermodynamics, kinetics, and descriptive chemistry. This course also includes numerous “hands-on” laboratories which come from a variety of sources utilizing standard lab procedures, micro-scale, calculator-based labs, etc. You will be required to collect, process, and manipulate data taken from physical observations, develop and formally report your conclusion.

AP Chemistry is a math intensive course and students must have a strong grasp of the concepts learned in Algebra 2 their sophomore year. It is recommended that students are taking Pre-Calculus concurrently, possibly Statistics, and have taken or taking AP Physics. This is a junior or senior level class in high school. Chemistry or Pre-AP Chemistry is a pre-requisite for AP Chemistry.

Students should expect to spend about 30 minutes every day on reading and homework(or approximately 2 hours a week), with an additional 2 to 3 hours writing formal laboratory reports once a 9 weeks. (Approximately 6 minor labs will be performed every 9 weeks of class).

Everyone taking this class is expected to take the AP Chemistry test. We will spend all year preparing for this test and you will be ready to take it in May. If finances will hinder you from taking the AP test, let me know so we can get that covered by the time of the test.

All tests will be composed of test questions from the released AP Chemistry tests. A scale will be used to grade the tests based on the AP grading scale from the previous year. This will cause the test grades to be higher than the raw score. They will be given scores on a 100 point scale as well as the AP scores 1-5. This will be good preparation for the real AP exam.

Thank you for taking a risk and signing up for AP chemistry next year.  It’s a hard class, but it will teach you to look at your world differently.  Our focus is not only to pass the AP test, but to give you a good foundation in Chemistry which is essential to any health or engineering related career you might be considering.  And, to be a good artist you must be a good scientist, too.  This is a pretty short list because I also want you to enjoy your summer. 

School starts on Wednesday, August 22nd. Your summer assignment is to come to school prepared to take a test(major grade) on Monday August 27th. This will be a test based on prior knowledge from 1st year Chemistry. I have attached documents to study as well as links to online resources such as Quizlets and videos.Here are the topics you need to review: polyatomic ions, nomenclature, molar conversions, types of reactions, balancing reactions, etc. You have already covered these topics in Pre-AP Chemistry just for this purpose! There is NO assignment you have to do or turn in!

1.Polyatomic Ions - I have marked the polyatomic ions (*) you must know. You will NOT be allowed to use a polyatomic ion chart throughout AP Chemistry. You will also need to be able to determine a polyatomic ion that is a variation of an “-ate” ion. The highlighted ions are the ones

that you need to be able to name the various forms for: ex. perchlorate ion – ClO4-1, chlorate ion – ClO3-1, chlorite ion – ClO2-1, hypochlorite ion – ClO1-1. We used them often in Pre-AP Chemistry. Now you must have them memorized! Making flash cards is handy to help memorize the polyatomic ions. Here’s a Quizlet to study.

https://quizlet.com/_3gqu16

2. Nomenclature - Naming compounds and writing formulas. You should be able to name and write formulas for metal and transition metal compounds, covalently bonded compounds, acids, and hydrates.

3. Stoichiometry & Reactions- You should be able to write a balanced chemical reaction from the names as well as predict the products. You will need to convert the mass of compound into moles of the compound and vice versa. You should also be able to determine the molar mass of a substance using a periodic table. Also, mole to mole, mole to mass, mass to mole, and mass to mass stoichiometric calculations using limiting reactant and percent yield are necessary conversions to know.

4. Scientific Mathematics – You should be able to make calculations using scientific notation, dimensional analysis, and correct use of significant figures.

For video help on Topics #2-4, here are a list of last year topics & explanations. http://www.bozemanscience.com/chemistry/

Do you need help with any of this? I will be at the school in my room on these days to help! I can meet at other times as well if these days don’t work for you…just let me know. Also, don’t hesitate to e-mail me at any time over the summer! I am ready to help you be successful!

Also, videos to help you with these topics are loaded on my school website: https://www.marblefallsisd.org/Domain/2905

Tuesday August 7th – 8:30am – 10:30am Wednesday August 8th – 8:30am-10:30am

Tuesday August 14th – 4-6pm Thursday August 16th – 4-6pm

I hope you have a great summer! See you in the fall.Mrs. Natalie Allen – Room 402nallen@mfisd.txed.netPart 1 – Polyatomic Ions

Charges of Common Polyatomic Ions1+

*Ammonium NH4+

2+

*Mercury (I) Hg22+

1-

*Acetate Bicarbonate*Bromate*Chlorate*Cyanide *Hydroxide*Iodate*Nitrate*Permanganate*Thiocyanate

C2H3O2-

HCO3-

BrO3-

ClO3-

CN-

OH-

IO3-

NO3-

MnO4-

SCN-

2-

*Carbonate*Chromate*Dichromate Oxalate*Peroxide Selenate Silicate*Sulfate Thiosulfate

CO32-

CrO42-

Cr2O72-

C2O42-

O22-

SeO42-

SiO32-

SO42-

S2O32-

3-

Arsenate Borate Citrate Hexacyanoferrate (III)*Phosphate

AsO43-

BO33-

C6H5O73-

Fe(CN)63-

PO43-

4-

Hexacyanoferrate (II)

Fe(CN)64-

Determining the Ion or Acid – A Polyatomic ApproachPolyatomic

AnionAcid Examples

per______ate ion

Gains 1 Oxygen per______ic acid

perchlorate ionClO4-1

perchloric acidThis is your

starting point. It is all based off

of the “-ate” ion

______ate ion ______ic acid chlorate ionClO3-1

chloric acid

______ite ion Loses 1 Oxygen ______ous acid chlorite ionClO2-1

chlorous acidAll the polyatomic ions in the group will have the exact same charge.

hypo______ite ion

Loses 1 Oxygen hypo______ous acid

hypochlorite ion

ClO1-1

hypochlorous acid

They only have a different number

of oxygen (and name)

______ide ion * Has no oxygen (Just an element from the periodic

table)

hydro______ic acid

chloride ionCl-1

hydrochloric acid

Part 2 – Nomenclature

IONIC COMPOUNDS WHICH CONTAIN A METAL WITH ONLY ONE OXIDATION STATEIonic compounds are made up of a cation (positive ion) and an anion (negative ion). The cation is written first in the formula.Metals that have only one oxidation state are: Li, Na, K, Rb, Cs, Mg, Ba, Ca, Sr, Al, Zn, Ag, and Cd.

Rules for Naming These Compounds:1.) Name the metal with its complete name.2.) If the anion is composed of single element, use the root of the non-metal and add the ending “ide”.Roots for Some Non-Metals:hydrogen = hydr oxygen = ox fluorine = fluornitrogen = nitr iodine = iod chlorine = chlorbromine = brom sulfur = sulf phosphorus = phosphcarbon = carb selenium = selen arsenic = arsen

IONIC COMPOUNDS WHICH CONTAIN METALS THAT HAVE MULTIPLE OXIDATION STATESMetals with multiple oxidation states are commonly transition and post-transition metals.

Rules for Naming Compounds Containing Metals With Multiple Oxidation States1.) Give the full (total) name of the metal.2.) Follow the name with the Roman numerals indicating the oxidation state. Be sure the

Roman numeral is in parenthesis and that the parenthesis are close to the metal name.3.) If the anion is a polyatomic ion, use the purple rules to complete the naming of the compound.4.) If the anion is composed of a single element, use the root of the non-metal and add the ending “ide”.

COMPOUNDS WHICH CONTAIN POLYATOMIC ATOMSThese compounds contain more than two elements. In order to successfully name these compounds, it will be necessary to be proficient at using the list of polyatomic ions.

Rules for Naming Ionic Compounds With Polyatomic Ions1.) If the polyatomic ion is an anion (negative ion):

a. Name the metal (orange or green rules) or polyatomic cation with the complete name.

b. Use the name of the polyatomic anion. (Pro Tip: Omit the word “ion” in the names.)2.) If the polyatomic ion is ammonium, NH4+1:

a. Write the name ammonium.b. If the anion is composed of a single element, use the root of the non-metal and

add the ending “ide”. If the anion is a polyatomic ion, use the name of the polyatomic anion. (Pro Tip: Omit the word “ion” in the names.)

Positive Polyatomic IonsIon Formula Name of Ion Ion Formula Name of Ion

NH4+1 Ammonium Ion Hg2+2 Mercury (I) Ion

Negative Polyatomic Ions

Ion Formula Name of Ion Ion Formula Name of Ion

C2H3O2-1 Acetate Ion OH-1 Hydroxide Ion

BrO3-1 Bromate Ion IO3-1 Iodate Ion

ClO-1 Hypochlorite Ion NO2-1 Nitrite Ion

ClO2-1 Chlorite Ion NO3-1 Nitrate Ion

ClO3-1 Chlorate Ion MnO4-1 Permanganate Ion

ClO4-1 Perchlorate Ion PO3-3 Phosphite Ion

CN-1 Cyanide Ion PO4-3 Phosphate Ion

SCN-1 Thiocyanate Ion HPO4-2Monohydrogen Phosphate Ion

CO3-2 Carbonate Ion H2PO4-1Dihydrogen Phophate Ion

HCO3-1Hydrogen Carbonate or

Bicarbonate Ion SO3-2 Sulfite Ion

CrO4-2 Chromate Ion SO4-2 Sulfate Ion

Cr2O7-2 Dichromate Ion HSO3-1Hydrogen Sulfite or

Bisulfite Ion

C2O4-2 Oxalate Ion HSO4-1Hydrogen Sulfate or

Bisulfate Ion

BINARY COVALENT COMPOUNDSThese compounds are usually combinations of two non-metals. Both elements in binary covalent compounds are found to the right of the zig-zag line which divides the metals and non-metals.

Rules for Naming Binary Covalent Compounds1.) If there is only one atom of the first element in the compound, name the element using its complete name. If there is more than one atom of the first element in the compound, use a Greek prefix to indicate the number of atoms of the element in the compound.

2.) Always use a Greek prefix with the root of the second element and the ending “ide”.

Greek Prefixes and Meanings

Number Greek Prefix Number Greek Prefix

One Mono Six Hexa

Two Di Seven Hepta

Three Tri Eight Octa

Four Tetra Nine Nona

Five Penta Ten Deca

ACIDS WITHOUT OXYGEN (“BINARY” ACIDS)Many covalent compounds are made of hydrogen and some other non-metal break up in water and become acids. Examples of binary acids: HCl, HF, and H2S. Exception: HCN is hydrocyanic acid.

Rules for Naming Binary Acids1.) They all begin with “hydro”.2.) Use the root of the second element.3.) Add the ending “ic” and the word “acid”.

TERNARY ACID (OXYACIDS)Ternary acids are composed mainly of hydrogen and a polyatomic ion which contains the element oxygen.

Rules for Naming Ternary Acids1.) Use the root of the polyatomic for the basis of the name.2.) If the polyatomic ion ends in “ate” or “ide”, use the ending “ic” and add the word “acid”.3.) If the polyatomic ion ends in “ite”, use the ending “ous” and add the word “acid”.

IONIC COMPOUNDS WITH WATER (HYDRATES)These compounds contain water molecules that have adhered to the ions as the solid forms. The water molecules become part of the crystals, and are called hydrates – ionic compounds with a specific number of water molecules bound to the atoms.

Rules for Naming Hydrates

In the formula of a hydrate, the number of water molecules associated with each formula unit of the compound is written following a dot.1.) Name the ionic compound first.

a. Name the metal or polyatomic cation with the complete name.b. If the anion is composed of a single element, use the root of the non-metal and

add the ending “ide”. If the anion is a polyatomic ion, use the name of the polyatomic anion. (Pro Tip: Omit the word “ion” in the names.)

2.) Always use a Greek prefix for the number of water molecules attached to the compound and the ending “hydrate”. Ex. 10H2O = decahydrate

Your turn - NomenclatureChemical Name Covalent, Ionic, or

Acid?Chemical Formula

1 Epsom Salt (Magnesium Sulfate)

2 Hydrobromic Acid

3 Ammonia

4 Baking Soda (Sodium Bicarbonate)

5 Gallium Bromide

6 Bleach (Sodium Hypochlorite)

7 Sulfuric Acid

8 Manganese(II) Sulfide

9 Carbon Monoxide

10

Lead(II) Nitrate

11

Barium Nitride

12

Hydrosulfuric Acid

13

Potassium Oxide

14

Methane (Carbon Tetrahydride)

15

Phosphoric Acid

16

Tin(IV) Arsenide

17

Gold(III) Oxalate

18

Laughing Gas (Dinitrogen Monoxide)

19

Nitric Acid

20

Carbon Dioxide

21

HClO3

22

H2O2

23

HCl

24

Al2(CO3)3

25

AgC2H3O2

26

Ca2C

27

Cu3(PO4)2

28

Cr(OH)3

29

H2O

30

SF6

31

HClO4

32

NO2

33

Fe2O3

34

SrSO3

35

HCN

36

CCl4

37

Sc(BrO3)2

38

HI

39

Ni(NO2)3

40

NaCl

Part 3 – Stoichiometry & Reactions

Chemical Equations:Defined as the symbolic representation of the chemicals involved in a chemical reaction. The reactants are written first, each reactant is separated by an addition sign. An arrow symbolizes the chemical conversion to new a substance or substances. On the other side of the arrow the new substances, the products, are written and again separated by an addition sign. The parts of a chemical equation are as follows:1. Reactant chemical(s) and Product chemical(s)2. States of each of the chemicals3. Must obey Law of Conservation of Matter – Balanced4. Arrow may show a catalyst or environmental condition such as pressure5. Heat may be shown as a reactant or product

The states of matter will be represented by a subscript in parenthesis using the following major four abbreviations:

gas (g), solid (s), liquid (l) - lower case cursive ‘L’, aqueous (aq) only found in a water medium

Here is an example of the reaction of methane gas and oxygen gas to yield carbon dioxide gas and water gas.

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)The 2 in front of the O2 and H2O actually means two of those molecules are needed or are generated in the reaction. The reaction could be written as follows:

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

But it makes more sense to combine like terms, which is what has been done in the first equation.

Another example is:2 Na(s) + 2 H2O(l) 2 NaOH(s) + H2(g)

Balancing Equations (Stoichiometry):The law of conservation of mass states that mass can neither be created nor destroyed, it may only change form. When a chemical reaction occurs, sometimes it seems like mass has been destroyed, but in fact it has not. When a tree burns, it seems as if the tree disappears, where does all that mass or stuff go? It is converted into another form. These new forms are gases. There is a lot of carbon, oxygen and hydrogen in living organisms. When a living organism burns, these atoms will now be found in the following molecules H2O and CO2. The remaining soot is mostly carbon, plain carbon atoms.

C6H12O6(s) + O2(g) CO2(g) + H2O(g) C6H12O6 is the general formula for carbohydrates, which is the main component of a tree. The problem with the above chemical equation is that it is not balanced. We need to have the same number of each atom on each side of the equation. We must balance the equation.

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)

The above equation is now balanced. There are 6 carbons on each side of the arrow. There are 12 hydrogens on each side and there are 18 oxygens on each side of the arrow.

Mass of Atoms & Molecules:Atomic Mass is the mass in amu’s of a particular atom, the average value of which will be found on the periodic table. Formula or Molecular Mass is the sum of all the atoms in the molecule. Some may refer to this as molecular weight, but this is not correct. For example, what if you are an astronaut doing an experiment on the international space station? Weight does not apply but mass does.

Calculating the molecular mass of a molecule is a simple process. Add the masses of each atom in the molecule. Masses are found on the periodic table.

Examples:H2O (1.008 x 2) + (16.00 x 1) = 18.02 amuCO2 (12.01 x 1) + (16.00 x 2) = 44.01 amu

The Mole:By definition an amu is 1/12th the mass of a C-12 atom. The unit of mass needs a reference point and a specific amount of matter to which all other matter can be referenced. This is a standard, as C-12 is very abundant, but this mass is very small, too small to work with. Generally, you will work with quantities of atoms greater than one. A new unit was developed from work by the Italian Chemist Lorenzo Romano Amedeo Carlo Avogadro. The unit is named the “mole” or Avogadro’s number.

This unit is nothing more than a number, a very big number. The mole works no differently than the following units, a dozen equals 12, a baker’s dozen equals 13 or a gross equals 144. The mole is simply a number equal to 6.02 x 1023. Written in decimal format it is equal to 602,000,000,000,000,000,000,000. Can you comprehend a number this big?

Why would you need to use a number this big? The mass of atoms is so small that you need many of them to have a mass significant enough for people to weigh. The mass of an atom is measured in atomic mass units, AMU. So, scientists picked a number equal to a certain mass of a certain element. The mass is 12.000 grams and the element is carbon-12. Carbon-12 has 6

protons and 6 neutrons. The masses of all other elements are standardized against the carbon-12 atom.

The mole’s ease of use is that the mass of one atom of an element in amu’s is also equivalent to the mass of one mole of that element in grams. So, looking at the periodic table the number below the symbol has two meanings. Look at helium. The number below the He symbol is 4.00260. This means that the mass of one helium atom is 4.00260 amu but it also means that if we have 1 mole of helium atoms those helium atoms would weigh 4.00260 grams.

By definition a mole of carbon 12 atoms has a mass of 12 grams.If you had 6.02214199 x 1023 carbon 12 atoms, they would have a mass of 12 grams.

Examples:H2O 1.008 x 2 + 16.00 x 1 = 18.02 g/molCO2 12.01 x 1 + 16.00 x 2 = 44.01 g/mol

Below are few simple conversion examples:If you are given two moles of iron, its mass would be what?

2 mol Fe 55.85 g Fe = 2 x 55.85 = 111.7 g Fe1 mol Fe 1

If you are given 50 grams of sulfur, how many moles would you have?

50 g S 1 mol S = 50 x 1 = 1.78 mol S28.09 g S 28.09

How many atoms are found in a 7 gram gold ring?

7 g Au 1 mol Au 6.022x1023 atoms

Au = 7 x 1 x 6.022 x 1023 = 2.14 x 1022 atoms Au

197.97 g Au

1 mol Au 197.97 x 1

Given 2.8 grams of NaCl, calculate the number of moles of NaCl present.

2.8 g NaCl 1 mol NaCl = 2.8 x 1 = 0.048 mol NaCl58.44 g NaCl 58.44

In the lab we cannot count up all the atoms or molecules we will be using in an experiment, we mass the substances out with a balance or scale. We must then calculate the amount of objects, atoms or molecules, present. Because, in a chemical reaction the atoms combine in simple whole number ratios to from compounds. That was Dalton’s theory, and it is correct. When making water, you need 2 hydrogens and 1 oxygen, that is all that matters. It does not matter how much each atom weighs.

Reactions Types:Decomposition – 1 chemical becomes 2 chemicals

A B + C NH4NO2(s) N2(g) + H2O(g)

Synthesis/Combination – 2 chemicals become 1 chemicalA + B C P4(s) + O2(g) P4O10(s)

Single Replacement – 1 chemical replaces another chemical in a compoundA + BC B + AC Zn(s) + CuCl2(aq) Cu(s) + ZnCl2(aq)

Double Replacement – cations switch compoundsAB + CD AD + CB NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

Combustion – a chemical is combined with oxygen in a rapid reaction.A + O2 AxOy (+ Z…) C3H8(g) + O2 CO2(g) + H2O(g)

Redox - at least 2 chemicals in a reaction change their oxidation number/charge For every oxidation there must be a reduction All reactions except double replacement reactions are redox reactions.

Ca(s) + S(s) CaS(s) o The reactants Ca and S are both elements in their elemental state.o This means their oxidation numbers are zero.o If an element is in a compound it must have an oxidation number.o So, if the products started with no oxidation number and now have an oxidation number, then

this must be a redox reaction.o This works the same going the other direction. If two elements are combined in a compound,

like H2O, then are decomposed to H2 and O2. The hydrogen and oxygen used to have an oxidation number and now do not, therefore a redox reaction occurred.

Stoichiometry Math:The real usefulness of the mole is that you can calculate the number of grams of a substance needed in a chemical reaction. The mole is needed because the mole represents a specific number of atoms or molecules.

If we wanted to react sodium metal and chlorine gas together to make sodium chloride (a combination reaction) we would want to use equal numbers of sodium atoms and chlorine atoms; that does not mean equal masses of sodium metal and chlorine gas. We would not react 10 grams of each as the chlorine is much more massive than the chlorine atom. 10 grams of sodium would contain many billons more atoms than 10 grams of chlorine.

2 Na(s) + Cl2(g) 2 NaCl(s)

If we use the stoichiometry of the chemical equation and the molar masses of each of the components in the reaction we can determine the amount of each reactant to add to the reaction container.

If we start with 10 grams of sodium metal we would need to follow the next few steps to determine how much chlorine to add to the sodium metal.

10 g Na |1 mol Na22 .9898 g Na

|1 mol Cl2

2 mol Na|70 .906 g Cl2

1 mol Cl2=15 . 42 g Cl2

The use of mole in stoichiometry calculations follows a predictable pattern. You begin with a mass in grams of a substance. You covert that substance’s mass into moles. You next use the moles of the first substance and convert them into moles of the desired substance. Finally convert the moles of the desired substance into grams of the desired substance.

The following can be used as a template for practically all stoichiometry calculations.

given mass in grams of A |1 mol Amolar mass in grams of A

|Y mol BX mol A

|molar mass in grams of B1 mol B

= mass in grams of BIf we use the above calculation with sodium and chloride, we begin with a given mass of A, 10 grams of sodium. We convert the mass of A into moles of A, 22.9898 g of Na per mole of Na. Next we convert X moles of A into Y moles of B, 2 moles of Na into 1 mole of Cl2. We finish by converting moles of B into grams of B, 1 mole of Cl2 equals 70.906 grams of Cl2.

Limiting Reagent, and the like, Problems:When two or more substances react, one of these substances will run out first. Just as when you make a bunch of apple pies you are bound to run out of apples or pie crust, leaving leftover apples or crust. You may be able to fudge this in the baking world but in the atomic world you cannot.

There is no way around running out of one substance first. No matter how accurate you measure, we do not have scales that have 23 decimal places, and even if we did, if you add one grain to such a scale you would change the mass to the 6 place simply by adding a single grain of sand.

So, whichever substance runs out first, that is your limiting reagent. These types of stoichometric problems are very common to everyday life. If you are making steel you need to know how much carbon to add to every ton of steel or your steel will not be as strong as you need it to be or may be more brittle than you desire.

For instance, if you mix sodium metal and chlorine gas, you make table salt. You would not want to have any sodium or chlorine left over, so you would want your measurements to be very accurate.

There are three ways these problems work. Either you are given the amounts of each reactant and asked how much product you could make. Or you are given one reactant and asked how much of the other reactant(s) you need to complete the reaction. Or you are told you need to make this much product, how much of each reactant will you need.

See the following examples:

Given 87.5 grams of silver and 75.0 grams of bromine liquid, how much silver bromide could be made? Remember bromine is one of the 7 diatomics.

First you must write a balance chemical equation:

2 Ag(s) + Br2(l) → 2 AgBr(s)

87.5 g Ag 1 mol Ag 2 mol AgBr 187.78 g AgBr = 152.31 g AgBr 107.87 g Ag 2 mol Ag 1 mol AgBrThis calculation must be completed twice, once for each reactant.

75.0 g Br2 1 mol Br2 2 mol AgBr 187.78 g AgBr = 176.24 g AgBr 159.8 g Br2 1 mol Br2 1 mol AgBr

Given these two calculations, we can say the limiting reagent is silver and the most AgBr that could be made with 87.5 g of Ag and 75.0 g of Br2 would be 152.31 grams. We could not make 176.24 grams because we would run out of silver once 152.31 grams were made.

Another example, if you wanted to make 58 grams of copper (II) oxide, how much copper and oxygen would you need?

2 Cu(s) + O2(g) → 2 CuO(s)

58.0 g CuO 1 mol CuO 2 mol Cu 63.54 g Cu = 46.33 g Cu 79.54 g CuO 2 mol CuO 1 mol Cu

58.0 g CuO 1 mol CuO 1 mol O2 32 g O2 = 11.67 g O2 79.54 g CuO 2 mol CuO 1 mol O2

So, to make 58 g of CuO you need 46.33 grams of Cu and 11.67 g of O2. Predicting Products:The ability to predict what products will result by mixing substances together is a very important skill a chemist learns. We will discuss only the most basic of these predictions. For now, we will only deal with single and double replacement reactions.

When dealing with a double replacement reaction, you simply swap our the cations, the positively charge atoms, basically, the metals. But you must remember, double replacement reactions are not redox reactions, so the oxidation number for each element must be the same on each side of the equation.

Example:

MgCl2(aq) + AgNO3(aq)

If you swap out the metals, Mg and Ag, you get the following

AgCl(s) + Mg(NO3)2(aq)

Notice the charges for all are the same, meaning not redox.

In the single replacement reactions, you will take the metal, sitting as an element, by itself and replace it with the metal in the compound.

Example:

ZnCl2(aq) + Ca(s) →

If you swap out the metals, Zn and Ca, you get the following:

CaCl2(aq) + Zn(s) Notice the charges for all are different, meaning this is a redox reaction. Zn had a +2 charge and now it has a 0 charge. Ca had a 0 charge and now has a +2 charge.

Stoichiometry Sample Problems:In each of the equations below, write the correct products and then balance the equation. Identify the type of chemical reaction before writing the products.

1. CaCO3

2. Al + O2

3. Fe + CuSO4

4. C6H12 + O2

5. NaOH + H3PO4

6. (NH4)2SO4 + Ca(OH)2

7. Al2(SO4)3 + Ca3(PO4)2

8. SO2 + H2O

9. (NH4)3PO4 + Ba(OH)2

10.C3H8 + O2 Solve the following stoichiometric calculations.11. Mercury has an atomic mass of 200.59 amu. Calculate the a. Mass of 3.0 x 1010 atoms

b. Moles of Hg

12. Calculate the molar masses (g/mol) ofa. Ammonia (NH3)

b. Baking soda (NaHCO3)

c. Osmium Metal (Os)

13. Convert the following to molesa. 3.86 grams of Carbon dioxide.

b. 6.0 x 10 g of Hydrazine (N2H4), a rocket propellant.

14. The molecular formula of morphine, a pain-killing narcotic, is C17H19NO3.a. What is the molar mass?

b. What fraction of atoms in morphine is accounted for by carbon?

c. Which element contributes least to the molar mass?

15. Calculate the percentage by mass of the following compounds:a. SO3

b. CH3COOCH3

c. Ammonium Nitrate

16. Determine the empirical formula of the compounds with the following compositions by mass:a. 10.4 % C, 27. 8% S, 61.7 % Cl

b. 21.7 % C, 9.6 % O, and 68.7 % F

17. What mass of copper is required to replace silver from 4.00g of silver nitrate dissolved in water?

Cu(s) + AgNO3 → Cu(NO3)2 + Ag

18. Limestone, coral, and seashells are composed primarily of solid calcium carbonate. The test for the identification of a carbonate is to use a few drops of aqueous hydrochloric acid. The products are calcium chloride solution, carbon dioxide gas, and water.

(a) Write a balanced chemical equation for this reaction, including states for each reactant and product.

(b) How many moles of carbon dioxide is released by the reaction of 0.150 moles of calcium carbonate?

(c) What mass of carbon dioxide gas is produced by the reaction of 0.150 moles of calcium carbonate?

19. How many grams of ammonia, NH3, can be prepared from 85.78 g of nitrogen gas and 18.17 g of hydrogen gas?

20. Silver nitrate and barium chloride solutions are mixed.

(a) Using solubility rules, write a balanced chemical equation for the reaction of silver nitrate and barium chloride. Include states for each reactant and product.

(b) If a solution that contains 41.6 g silver nitrate is mixed with a solution containing 35.4 g barium chloride, which is the limiting reactant?

(c) How many grams of each product are formed?

21. A 45.0-g sample of iron ore, containing Fe3O4, is reacted with carbon to form purified iron. The other product of the reaction is carbon dioxide gas. 1.56 g of iron is recovered from one such trial.

(a) Write a balanced chemical equation for this reaction, including states for each reactant and product.

(b) What is the theoretical yield of iron from this reaction?

(c) What is the percent yield?

Part 4 – Scientific MathematicsChemistry: Significant Digits

1.

Significant numbers are always measurements and thus should always be accompanied by the measurement's unit. For simplicity, units are not included in the following examples.

In an attempt to get away from the mathematical burden of uncertainties, scientists have gone to the use of established rules for significant digits that have greatly simplified calculations. These rules are:

2. Any numbers (that are measurements) other than zero are significant. (Many times the zeros are also significant as you will see below.) Thus 123.45 contains five significant digits.3. Any zeros between numbers are significant, thus 1002.05 contains six significant digits.4. Unless told differently, all zeros to the left of an understood decimal point (a decimal that is not

printed) but to the right of the last number are not significant. The number 921000 contains

three significant digits.5. Any zeros to the left of a number but to the right of a decimal point are not significant. 921000. has

six significant digits.

6. These zeros are present merely to indicate the presence of a decimal point (they are used as place holders), (these zeros are not part of the measurement). The number 0.00123 has three significant digits. The reason that these zeros are not significant is that the measurement 0.00123 grams is equal in magnitude to the measurement 1.23 milligrams. 1.23 has three significant digits, thus 0.0123 must also have three significant digits.

7. Any zeros to the right of a number and the right of a decimal point are significant. The value 0.012300 and 25.000 both contain five significant digits. The reason for this is that significant figures indicate to what place a measurement is made. Thus the measurement 25.0 grams tells us that the measurement was made to the tenths place. (The accuracy of the scale is to the tenths place.)

Significant figures in derived quantities (Calculations)In all calculations, the answer must be governed by the least significant figure employed.

ADDITION AND SUBTRACTION: The answer should be rounded off so as to contain the same number of decimal places as the number with the least number of decimal places. In other words, an answer can be only as accurate as the number with the least accuracy.Thus: 11.31 + 33.264 + 4.1 = 48.674 Rounded off to 48.7

MULTIPLICATION AND DIVISION: The answer should be rounded off to contain the same number of digits as found in the LEAST accurate of the values.Thus: 5.282 x 3.42 = 18.06444 Rounded off to 18.1

Prefixes to memorizePrefix Symbol Numerical Exponenti

al giga G 1,000,000,000 109 mega M 1,000,000 106 kilo k 1,000 103 deca da 10 101

no prefix means: 1 100

deci d 0.1 10¯1 centi c 0.01 10¯2 milli m 0.001 10¯3 micro m 0.000001 10¯6

nano n 0.000000001 10¯9 pico p 0.00000000000

1 10¯12

Sample Problems:1. How many significant figures are in each of the following?

a. 1.92 mmb. 0.030100 kJc. 6.022 x1023 atomsd. 460.00 Le. 0.00036 cm3f. 100 g. 1001h. 0.001

2. Calculate the following to the correct number of significant figures.a. 1.27 g / 5.296 cm3

b. 12.235 g / 1.01 Lc. 17.3 g + 2.785 gd. 2.1 x 3.21e. 200.1 x 120f. 17.6 + 2.838 + 2.3 + 110.77

3. The actual density of a certain material is 7.44 g/cm3. A student measures the density of the same material as 7.30 g/cm3. What is the percent error of the measurement? (Show all work round answer to correct number of sig figs!)

4. Convert the following:a. 20 gallons into mlb. 3 meters into centimeters c. 10 kilometers into metersd. 15,050 milligrams into gramse. 3,264 milliliters into litersf. 9,674,444 grams into kilograms