IVC Chemistry 100

Laboratory Manual

Name______

- 1 - Table of Contents

Lab 1: DIMENSIONAL ANALYSIS ...... 4 Data for Lab 1: Dimensional Analysis...... 9 ADDITION AND SUBTRACTION ...... 9 MULTIPLICATION & DIVISION OF SIGNIFICANT DIGITS ...... 9 CALCULATORS AND SIGNIFICANT DIGIETS ...... 10 MATH CONCEPTS IN CHEMISTRY ...... 14 SIGNIFICANT DIGITS ...... 14 ROUNDING OFF NONSIGNIECANT DIGITS ...... 14 SCIENTIFIC NOTATION ...... 14 TEMPERATURE ...... 15 METRIC - METRIC CONVERSIONS ...... 16 METRIC - ENGLISH CONVERSIONS ...... 16 GENERAL EXERCISES...... 16 Lab 2: Atomic Structure ...... 17 Discussion...... 17 Experimental Procedures ...... 18 Data for Lab 2: Atomic Structure ...... 19 Lab 3: Periodic Properties ...... 25 Data for Lab 3: Electronic Configurations and Periodic Trends ...... 29 Lab 4 Nomenclature ...... 33 Group 1 Ionic Compounds ...... 33 Class 1. Fixed Oxidation State (O.S.) ...... 33 Class 2. Fixed Oxidation States ...... 34 Group 2: Covalent ...... 36 Class 3. Covalent Non-acid ...... 36 Class 4. Binary Acids ...... 36 Oxyoacids ...... 36 Data for Lab 4: Nomenclature ...... 39 Lab 5 Lewis Dot ...... 41 VSEPR: Valence shell electron-pair repulsion theory ...... 45 Data for Lab 5: Lewis Structure ...... 49 Lab 6 Heat Capacity of Metals ...... 51 Heat and Enthalpy ...... 52 Experimental Procedure ...... 53 Calculations ...... 53 Data for Lab 6: Heat Capacity ...... 55 Lab 7: Chemical EQUATION: WRITING BALANCED EQUATIONS ...... 57 INSTRUCTIONS: ...... 57 Lab Procedure ...... 58 Data for Lab 7: Writing Chemical Equations ...... 59 Part 2. Rewrite the following word equations into balanced chemical equations...... 61 PART 4: ...... 61 Lab 8 Formula of a Hydrate...... 63 Data for Lab 8: Formula of a Hydrate ...... 65

- 2 - Lab 9 Equilibrium ...... 69 Data Sheet for Lab 9 Equilbrium ...... 73 Lab 10 Standard Molar Volume of a Gas ...... 75 Procedure ...... 76 Data for Lab 10: Electronic Configurations and Periodic Trends ...... 77 Lab 11 Net Ionic Equations (NIE) ...... 79 Data Sheet Lab 11 NIE ...... 83 Lab 12 Titration ...... 87 Titration General Information ...... 88 Data Sheet Lab 12 Titration ...... 91 30 points Extra Credit: Nomenclature (due when your instructor tells you :-) ...... 93

- 3 - Lab 1: DIMENSIONAL ANALYSIS

Rules for Addition: An answer obtained by adding or subtracting has the same number of decimal places as the measurement with the fewest decimal places (look to the right of the decimal).

25.2 one decimal place + 1.34 two decimal places 26.54 calculated answer

Final Answer 26.5 answer with one decimal place

Another problem: 14400000 235.05 Your final answer goes to the furthes right a + 72000000 +19.6 86000000 as your least significant number + 2. 256.65 rounds to 257 86000000

Rules for Multiplication: An answer obtained by multiplying or dividing has the same number of significant figures as the measurement with the fewest significant figures (look for the fewest significant figures).

Use rounding to limit the number of digits in the answer. 110.5 x 0.048 = 5.304 (calculator) 4 SF 2 SF The final answer is rounded off to give 2 significant figures = 5.3 (2 SF).

Another problem: 40.311 ÷ 0.007 = 60

Mixed Mutliplication and Addition: Do each operation seperately.

Dimensional analysis

This is a very important lab. You will use what you learn in this lab EVERYDAY during lecture and lab. Dimensional analysis is a problem-solving technique.

What will you learn: A Mathematical Technique, to solve EVERY problem we encounter.

numerator numerator Every Problem You Solve Will Look Like:  ...  denomeator denomeator

4 The Math Principle is a Beginning Algebra concept of cross cancellation:

A B C    A B C 1

You are expected to solve ALL of your problems using this method, hence the reason for spending 2 days on this lab.

What Problems Can You Solve Using This Method?

You'll learn to solve many types of problems. For example how many mL’s are in 2.83qts, or how many inches are in 259 cm? You'll learn to use the steps of DIMENSIONAL ANALYSIS as a model to solve these types of problems.

To solve a dimensional analysis problem we must learn what a conversion factor is first.

Here are 2 examples of conversion factors.

1 ft 12 in 1 ft = 12in  or 12 in 1 ft There are millions of conversion factors!

2.54 cm 1 in 2.54 cm = 1 in  or 1 in 2.54 cm

Note: these are exact relationships between two totally different units i.e. ft  in or cm  in

A conversion factor is important in math because they are equal to 1 and are therefore useful in chemistry to help us do calculations.

For example, we know that 1 ft = 12 in, how is this equal to 1.

We know 1 ft = 12 in To obtain the conversion factors or we had to divide by 12 in. and 1 ft respective 1 ft 12 in 1 ft = 12 in, now divide both sides by 12 in or 1 12 in 12 in

12 in 1 ft 1 ft = 12 in now divide both sides by 1ft or 1 1 ft 1 ft

We use these conversion factors when solving Dimensional Analysis Problems. Dimensional Analysis 5

Let’s examine the problem already asked: How many inches are in 259 cm?

Use the conversion factor found on the first page that shows the relationship between inches and centimeters. The answer would look like the following:

259 cm 1 inch ? inches =  = 102 inches 1 2.54 cm

Use unit analysis to solve the following problems.

1. 4 dollars = dimes Conversion factor

Hint: there are 10 dimes = 1 dollar

4 dollars 10 dimes Solution: × = 40 dimes 1 1 dollar

Notice how the dollar is canceled in the numerator and dominator!

Try the following: SHOW ALL OF YOUR WORK!

2. 5 ft = inches

Conversion Factor Hint: there are 12 inches = 1 ft. You fill in the numerator and denominator with the listed conversion factor, and then do the multiplication.

inches? = 5 ft  = ______inches 3. 24 hrs = minutes

What is the Conversion Factor? How many minutes in an hour?

What is Dimensional Analysis? This is what it looks like.

new unit old unit  = answer in new units old unit Notice how the “old unit” cancels in the numerator and denominator. This is why Dimensional Analysis is so easy, you do NOT have to think in advance of all the steps. Dimensional Analysis is a VISUAL 6 MATEMATICAL METHOD; you have a visual way of looking at your solution by looking at what units cancel.

If you are still trying to solve these problems in your head I hope the next problem convinces you to give up your old ways and try Dimensional Analysis!

In the fictitious country of Nolenville, 1.00 nolenzels equals 0.50 US dollars. How many nolenzels are equal to 700 US dollars?

Solution using Dimensional Analysis:

1.00 nolenzels ?nolenzles $700   1400nolenzels $0.50

Notice that the “$” cancel. Only “nolenzels” remains in the numerator and since it does not cancel so that, is your final unit.

7

8 Data for Lab 1: Dimensional Analysis Name ______

Date ______

Labmates ______

ADDITION AND SUBTRACTION 1. Perform the indicated mathematical operation. Express the answer with the proper units and significant digits.

5000. cm 1.0254 mL 45.3 g a. .0256 cm b. 172.0 mL c. - 41.119 g + 0.00050 cm + 15.99 mL

MULTIPLICATION & DIVISION OF SIGNIFICANT DIGITS

2. Perform the indicated mathematical operation. Express the answer with the proper units and significant digits. a. 21.1 cm × 20 cm

b. 5.15 cm × 2.55 cm × 1.00 cm

131.78 cm3 c. 19.26 cm

d.

9

CALCULATORS AND SIGNIFICANT DIGIETS

Using a Scientific Calculator

Students often purchase calculators that cost upwards of $200.00 and yet they do not know how to use them. In this section, you should learn to MASTER your scientific calculator.

All Calculators have different key setups ….so you will have to figure out where you SCIENTIFIC NOTATION button is on your calculator. Locate ONE of the following scientific notation buttons on your calculator (it may not even be listed here): 1. EE (Some Sharp) 2. EXP (Casio, TI, Shape, HP) 3. 2nd Function EE (many TI’s)

At this point you will want to practice with you calculator.

First, enter the following

1  102 Times 5  103 =

You should get either 500,000 or 5  105 and either is correct. IF YOU DID NOT GET THAT ANSWER THEN…….read on……….

DO NOT’s of Scientific calculators. 1. DO NOT USE the 10X key. DO NOT enter the above numbers as 1 times 102 times 5 times 103 even though you get the same answer….IT IS A WASTE OF KEY STROKES…you do not have time to waste typing extra keystrokes. USE THE EE OR EXP BUTTON---THAT’S WHAT YOUR VERY EXPENSIVE CALCULATOR IS MADE FOR. 2. DO NOT use the X2 or X3 buttons, again a waste of time. 3. You MUST see ME if you do not get this answer because your Chemistry 2A life will be a living hell and it is very SIMPLE for me to show you how to work your calculator.

Try this one 1  10-12 Times 5  10-5 = 10

You can only get ONE ANSWER 5  10-17…… IF YOU DID NOT GET THAT ANSWER THEN…….read on………. You MUST ALSO learn to switch your calculator into SCIENTIFIC MODE. 1. Get your manual out and figure out how to get your calculator into SCI Mode. 2. You MUST see ME if you do not get this answer because your Chemistry 100 life will be a living hell and it is very SIMPLE for me to show you how to work your calculator.

I hope this Helps!? If Not…SEE Me or ask someone for help!

Perform the indicated mathematical operation. Express the answer in decimal form (not scientific notation) with the correct number of significant digits. a. (4.65  105) (9.5  102)

b. (1. 11  108) / (2.05  10-11)

c. (54.9  10-6) / (7.87  104)

Perform the indicated mathematical operation. Express the answer in scientific notation with the correct number of significant digits.

4.97 1084 3.24 10  96.77 1078 0.00151 10  a. b. 56.71 102  3.49 1010 

Solve the following: SHOW ALL OF YOUR WORK FOR FULL CREDIT!

How many feet in 365.0 yards?

11

You threw the Frisbee 190. ft. How many yards did you throw the Frisbee?

What is the cost of 1.5 pound of sugar if sugar costs $1.37 per 5.0 pounds?

Dimensional Analysis: Problems with more then one conversion factor. Often more then one conversion factor is needed to solve a problem. SHOW ALL WORK and do not forget to Cancel Units.

How many aspirin tablets can be made from 100. g of aspirin if each tablet contains 5.00 grains of aspirin? (7.00 × 103 grains is equal to one pound) (454g = 1 lb)

12 What is the cost of 300.0 grams of aspirin if each tablet contains 5.00 grains of aspirin and 100 tablets cost $2.35? (7.00 × 103 grains is equal to one pound, 454 g is equal one pound.)

The following relationships describe an antiquated set of British liquid units. 1 hogshead = 7 firkin 18 pottle = 1 firkin 140 pottle = 1 puncheon 504 pottle = 1 tun Use dimensional analysis to determine the number of tuns in 144 hogshead.

Find the number of cm3 in 2.60 cubic foot----this is a very hard problem and requires much thought.

13 MATH CONCEPTS IN CHEMISTRY State the basic unit and symbol for the following quantizes. a. length = b: volume = c, mass = d. temperature = e. time = f. density =

SIGNIFICANT DIGITS State the number of significant digits in the following measurements. a. 0.707 g = b. 0.10000 cm =

c. 2500 mL = d. 0.0000110 g =

e. 100,000. cm = f. 100,000 mL =

ROUNDING OFF NONSIGNIECANT DIGITS Round off the following measurements to three significant digits- a. 12.59 mL = b. 0.03662 g =

c. 31559745 cm = d. 202577.94 mL =

SCIENTIFIC NOTATION Convert the following exponential numbers into scientific notation. a. 352 × 104

b. 416 × 103

c. 0.170 × 102

14 TEMPERATURE Learn the following 3 formulas for temperature formulas for your next exam

9 F = C + 32 5

5 °C = (°F - 32) 9

K = C + 273

Express the following temperatures in degrees Celsius.

(1) 1000. F

(2) -40. F

Convert the following temperatures in degrees Fahrenheit.

(1) 420.0 C

(2) -200. C

Express the following in Kelvin units.

(1) 190.0 F

(2) 980 F

15

METRIC - METRIC CONVERSIONS Convert the following metric system measurements using two conversion factors. a. 120 mm to cm

a. 1.55cm to km

METRIC - ENGLISH CONVERSIONS Convert the given measurements into the units indicated. a. 1.01 lb to g

b. 360 cm to ft

GENERAL EXERCISES Consider an atom of 64Zn, calculate the density of the nucleus in grams per cubic centimeter, knowing

-6 64 -22 4 3 that the nuclear radius is 4.8  10 nm and the mass of the Zn atom is 1.06  10 g. Vsphere = r , 3 2.54 cm = 1 in, d = 2r.

If there are 2.62  1022 atoms in 1.00 g of sodium and they are lined up side by side, what is the length of the line of sodium atoms in miles? Assume that the atoms are spheres of radius 0.186 nm.

16 Lab 2: Atomic Structure Goals •Write the correct symbols or names of some elements. •Describe some physical properties of the elements you observe. •Categorize an element as a metal , nonmetal or metalloid from its physical properties. •Given the complete symbol of an atom, determine its mass number, atomic number, and the number of protons, neutrons, and electrons.

Discussion Primary substances, called elements, build all the materials about you. Some look similar, but others look unlike anything else. In this experiment, you will describe the physical properties of elements in a laboratory display and determine the location of elements on a blank periodic table.

A. Physical Properties of Elements Metals are elements that are usually shiny or have a metallic luster. They are usually good conductors of heat and electricity, ductile (can be drawn into a wire), and malleable (can be molded into a shape). Some metals such as sodium or calcium may have a white coating of oxide formed by reacting with oxygen in the air. If these are cut, you can see the fresh shiny metal underneath. In contrast, nonmet- als are not good conductors of heat and electricity, are brittle (not ductile), and appear dull, not shiny.

B. Periodic Table The periodic table, shown on the inside front cover of this lab manual and your textbook, contains information about each of the elements. On the table, the horizontal rows are periods, and the vertical columns are groups. Each group contains elements that have similar physical and chemical properties. The groups are numbered across the top of the chart. Elements in Group I are the alkali metals, elements in Group 2 are the alkaline earths, and Group 17 contains the halogens. Group 18 contains the noble gases, which are elements that are not very reactive compared to other elements. A dark zigzag line that looks like a staircase separates the metals on the left side from the nonmetals on the right side.

C. Subatomic Particles There are different kinds of atoms for each of the elements. Atoms are made up of smaller bits of matter called subatomic particles. Protons are positively charged particles, electrons are negatively charged, and neutrons are neutral (no charge). In an atom, the protons and neutrons are tightly packed in the tiny center called the nucleus. Most of the atom is empty space, which contains fast-moving electrons. Electrons are so small that their mass is considered to be negligible compared to the mass of the proton or neutron. The atomic number is equal to the number of protons. The mass number of an atom is the number of protons and neutrons. atomic number = number of protons (p+) mass number = number of protons plus number of neutrons (p+ + n°)

17 D. Isotopes Isotopes are atoms of the same element that differ in the number of neutrons. This means that isotopes of an element have the same number of protons, but different mass numbers. The following example represents the symbol of a sulfur isotope that has 16 protons and 18 neutrons.

Complete Symbol of an Isotope Meaning mass number (p+ and n°)  34 34 This atom has 16 protons and 18 neutrons. symbol of element  S 16S The element is sulfur. atomic number (p+= e-)  16p+  16e- The atom has 16 protons and 16 electrons

Experimental Procedures A. Physical Properties of Elements Identify each element as a metal (M) , a nonmetal (NM), or Metalloid (MT)

B. Periodic Table Materials: Periodic table, colored pencils, display of elements

B.1 On the incomplete periodic table provided in the report sheet, write the atomic numbers and sym- bols of the elements you observed in part A. Write the group number at the top of each column of the elements (Groups 1-18). Write the period numbers for each of the horizontal rows shown (1-5). Shade (with a pencil) & lable the alkali metals, alkaline earths, halogens, and noble gases, and the transition elements. Draw a heavy zig-zag line to separate the metals and nonmetals.

B.2 Use the periodic table to decide whether the elements listed on the report sheet would be metals or nonmetals.

C. Subatomic Particles For each of the neutral atoms described in the table, write the atomic number, mass number, and number of protons, neutrons, and electrons.

D. Isotopes Complete the information for each of the isotopes of calcium: the complete nuclear symbol and the number of protons, neutrons, and electrons.

E. Isotopic mass calculations.

18 Data for Lab 2: Atomic Structure Name ______

Date ______

Labmates ______

1. On the following list of elements, circle the symbols of the transition elements and underline the symbols of the halogens:

Mg Cu Br Ag Ni Cl Fe

2.Complete the list of names of elements and symbols:

Name of Element Symbol Name of Element Symbol Potassium Na

Sulfur P

Nitrogen Fe

Magnesium Cl

Copper Ag

19

A. Properties of Elements Element Symbol Number Metal/Nonmetal/Metalloid

Aluminum ______

Carbon ______

Copper ______

Iron ______

Magnesium ______

Nickel ______

Nitrogen ______

Oxygen ______

Phosphorus ______

Silicon ______

Silver ______

Sulfur ______

Tin ______

Zinc ______

20 B. Periodic Table B.1 (Note Read B.1 in Experimental Procedure on what to do here)

Questions and Problems

Q.1 From their positions on the periodic table, categorize the following elements as metals (M) or nonmetals (NM).

Na S Cu F Fe C Ca

______

Q.2 Give the name of each of the following elements: a. Noble gas in Period 2 ______

b. Halogen in Period 2 ______

c. Alkali metal in Period 3 ______

d. Halogen in Period 3 ______

e. Alkali metal in Period 4 ______

f. The first metal in Group 14 ______

g. The second nonmetal in Group 16 ______

21 B.2

Element Identify as either a Metal/Nonmetal

Chromium Gold Carbon Cadmium Sulfur

C. Subatomic Particles Atomic Mass Element Protons Neutrons Electrons Number Number Iron 30

27 13

19 20

Bromine 80

Gold 197

53 74

D. Isotopes Nuclear Symbol Protons Neutrons Electrons 40 Ca 20 20 22

43 Ca 20 24 20

46 Ca 20

22 E. Isotopic Mass Calculations

Isotopes are the same element…with different numbers of neutrons (see above).

If you look at the periodic table you notice that the atomic mass of lithium is 6.94 amu.

Mass  % Abundance = Mass  (% Abundance/100) 6Li 6.015 7.42 0.4463 7Li 7.016 92.58 6.4954 Total 100.00 % 6.94 Li

To calculate the Atomic Mass or Atomic Weight of Li, as seen on a Periodic Table we must find the total mass of all the Isotopes of Li as we find them in nature. The general formula is as follows:

mass of isotope #1  % abundence #1 mass of isotope #2 % abundence #2  + ... 100 100

= Ave AMU

Calculating Atomic Weight for Lithium totalLi = Mass 6Li  (% Abundance 6Li /100) + Mass 7Li  (% Abundance 7Li /100)

or

totalLi = 6.015  0.0742 + 7.016  0.9258 = 6.94

1. A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.11 amu) makes up 75.8% of the sample, and the isotope Cl-37 (mass = 37.69 amu) makes up 24.3% of the sample. What is the average atomic mass for chlorine?

63 2. Copper has two naturally occurring isotopes. A typical sample consists of 69.17% 29 Cu (62.939598 65 amu) and 30.83% 29 Cu (64.927793 amu). Calculate the atomic mass of copper.

23

24 Lab 3: Periodic Properties

Goals Draw a graph of atomic diameter against atomic number. Interpret the trends in atomic radii within a family and a period.

Electron Configuration In an electron configuration, electrons are arranged by subshells starting with the lowest energy. The number of electrons in each subshell is written as a superscript. The electron arrangement of an element is related to its position in the periodic table. The electron configuration can be written by following the subshell blocks across the periodic table starting with period 1. The s- block is formed by Groups 1 and 2. The p-block includes the elements in Groups 13 to 18. The d- block includes the elements in Groups 3 to 12.

Examples of Groups 1, 16 and 18 with Periods 2, 3 and 4 are as follows: Li 1s22s1 O 1s22s22p4 Ne ls22s22p6 Na ls22s22p63s1 S ls22s22p63s23p4 Ar ls22s22p63s23p6 K ls22s22p63s23p64s1 Se ls22s22p63s23p64s23d104p4 Kr ls22s22p63s23p64s23d104p6

A. Graphing a Periodic Property: Atomic Radius Since the 1800s scientists have recognized that chemical and physical properties of certain groups of elements tend to be similar. A Russian scientist, Dmitri Mendeleev, found that the chemical properties of elements tended to recur when the elements were arranged in order of increasing atomic mass. This repetition of similar characteristics is called periodic behavior. He used this periodic pattern to predict the characteristics of elements that were not yet discovered. Later, H. G. Moseley established that the similarities in properties were associated with the atomic number.

In the electron arrangement of an element, the electrons in the highest or outermost energy level are called the valence electrons; note in the above electron configurations the bolded shells, subshells and electrons. The valence electrons determine the chemical properties of the elements. If the elements are grouped according to the number of valence electrons, their chemical and physical properties are similar. The similarities of behavior occur periodically as the number of valence electrons is repeated.

In this exercise, you will graph the relationship between the atomic radius of an atom and its atomic number. Such a graph will show a repeating or periodic trend. Observe the graph in Figure 1, which was obtained by plotting the average temperature of the seasons. The graph shows that a cycle of high and low temperatures repeats each year. Such a tendency is known as a periodic property. There are three cycles on this particular graph, one full cycle occurring every year. When such cycles are known, the average temperatures for the next year could be predicted.

25

Figure 1 A graph of average seasonal temperatures.

Lab Information Time: 1 1/2 hr

Comments: Obtain a periodic table or use the inside cover of your textbook. Tear out the Lab 3 report sheets and place them beside the procedures. In neutral atoms, the number of electrons is equal to the number of protons. Related Topics: Electrons and protons, energy levels, and electron arrangement

Experimental Procedures:

A. Electron Configuration Write the electron configuration of each atom listed on the laboratory report. Indicate the number of valence electrons and the group number for the element.

B. Graphing a Periodic Property: Atomic Radius The atomic radii for elements with atomic numbers 1-25 are listed in Table 1. On the graph, plot the atomic radius of each element against the atomic number of the element on the graph. Be sure to connect the points. Use the completed graph to answer questions in the report sheet about valence electrons and group number.

26 Table 1 Atomic Radii for the Elements with Atomic Numbers 1-25

Element Symbol Atomic Number Atomic Radius (pm*) first period hydrogen H 1 37 helium He 2 50 second period lithium Li 3 152 beryllium Be 4 111 boron B 5 88 carbon C 6 77 nitrogen N 7 70 oxygen O 8 66 fluorine F 9 64 neon Ne 10 70 third period sodium Na 11 186 magnesium Mg 12 160 aluminum Al 13 143 silicon Si 14 117 phosphorus P 15 110 sulfur S 16 104 chlorine Cl 17 99 argon Ar 18 94 fourth period potassium K 19 231 calcium Ca 20 197 scandium Sc 21 160 titanium Ti 22 150 vanadium V 23 135 chromium Cr 24 125 manganese Mn 25 125 *(picometer = 10-12 m)

Electron Configuration and Periodic Properties

27

28 Data for Lab 3: Electronic Configurations and Periodic Trends

Name ______

Date ______

Labmates ______

B. Electron Configuration

Number of Valence Atom Electron Configuration Group Electrons

O

Na

Ca

Ti 4 4

V 5

Br

Sr

Cs 1

Xe 8

Bi 5 15

Pb 4

Ra

29 Questions and Problems

# orbitals 1 3 5 7 Subshells s p d f Max # electrons 2 6 10 14

Q.2 Complete the following electron shells, subshells, and configurations:

Number of subshells in 2nd energy level

Number of orbitals in the 2p subshell

Number of total electrons in only a 3d subshell

Number of total electrons in only a single 3p orbital

Period and Group number of carbon

First element that that begins to fill after the 4s is filled

Number of valence electrons in As

Q.3 Give the symbol of the element that meets the following information:

Element that fills the 3s sublevel

Period 4 element in the same group as F (fluorine)

Element with 3d6 electron configuration

Element with a filled 5p level

Element with five 3p electrons

The element with the electron configuration: 1s22s22p63s23p64s23d104p3

Element that completes 2nd energy level

Period 6 element in the same group as Mg

30 Graphing a Periodic Property: Atomic Radius

Atomic Radius vs. Atomic Number

Describe the change in the atomic radii for the elements in Period 2 from lithium to neon.

Why are the atomic radii of of elements in Period 3 larger then in Period 2?

31

32 Lab 4 Nomenclature

There are 5 (main) types/classes of Molecules in Inorganic Nomenclature…See Flow-Chart Handout

There are 2 Groups of Inorganic Compounds 1. Ionic 2. Covalent

Group 1: Ionic

There are 2 types of compounds in this group: 1. Fixed Oxidation States 2. Variable Oxidation States

Group 1 Ionic Compounds

Class 1. Fixed Oxidation State (O.S.)

Certain Metals have Fixed O.S., like Na, is always +1, and Sulfur is Always –2, so the names are written using their metal and non-metal-ide name.

Full metal name + Nonmetal with an –ide ending.

NaCl = sodium chloride

Bromine bromide Br- Carbon carbide C4- Chlorine chloride Cl- Fluorine Fluoride F- Hydrogen Hydride H-

Same rules applies to other main group ions Iodine, Nitrogen, Oxygen, Phosphorus, Sulfur

Examples:

Na2S sodium sulfide not disodium sulfide CaBr2 calcium bromide not Calcium dibromide Ca4C calcium carbide not tetracalcium carbide

33 Class 2. Fixed Oxidation States

Metals with Fixed Oxidation States do not need any special Prefix because the O.S. IMPLIES how many non-metals are around it.

HOWEVER------Transition Metals….need Special endings because they have Variable O.S.

Variable Oxidation states for Transition Metals

Chromium Cr (II) Cr2+, Cr (III) Cr3+ Iron Fe (II) Fe2+, Fe (III) Fe3+ Copper Cu+, Cu2+ Zinc Zn2+ Tin Sn2+, Sn4+ Lead Pb (II) Pb2+, Pb (IV) Pb4+

We will not use the –ous (lowest oxidation state) and –ic (highest oxidation state) [Fe (II) Fe2+ ferrous, Fe (III) Fe3+ ferric or Cu (I) Cu+ cuprous, Cu2+ cupric]

Examples PbS = PbO2 =

Titanium (II) Chloride = = TiCl3 Titanium (IV) Chloride = Metals with Polyatomic Ions

(Hint---1st determine if the metal has a fixed or variable OS before naming the compound)

The Anion is going to have a negative charge (except ammonium & Hydronium).

Nitride N3- - Nitrite NO2 - Nitrate NO3 -ide = no oxygen atoms -ite = least # of oxygen atoms (helpful hint: little) -ate = maximum # of oxygen atoms (helpful hint: lots )

Sulfide S2- 2- Sulfite SO3 2- Sulfate SO4

- Hydrogen Sulfate HSO4 - Hydrogen Sulfite HSO3 34 Phosphide P3- 3- Phosphite PO3 3- Phosphate PO4

2- Hydrogen Phosphate HPO4 - Dihydrogen Phosphate H2PO4

2- Carbonate CO3 Cyanide CN- + Ammonium NH4 + Hydronium H3O - Permanganate MnO4 2- Chromate CrO4 2- Dichromate Cr2O7 Writing Polyatomic Ions

{cation} + {anion} {transition metal (oxidation state)} + {suffix of anion}

+ - NH4 Cl NH4Cl ammonium chloride

2+ - Ca OH written CaOH2 or Ca(OH)2 calcium hydroxide

2+ 2- Ca SO4 not Ca2(SO4)2 but CaSO4 calcium sulfate

Examples:

Fe (II) with sulfite, Fe (III) with sulfite Cr(IV) with sulfate, Cr (III) with sulfite

Hydrate

“ “ = hydrate Greek Prefix 1-mono” “ 6-hexa 2-di 7-hepta 3-tri 8-octa 4-tetra 9-nona 5-penta 10-deca

Ba3(PO3)26 H2O = calcium sulfate trihydrate =

35 Group 2: Covalent

There are 3 types of inorganic compounds in this group 1. Covalent Non-acid 2. Binary non-oxyo-acid and Oxyo-acids

Class 3. Covalent Non-acid

Use Greek Prefixes except….Never use mono if there is only one 1st element

Name the following:

CH4

BF3

P2O5

SO3 (note…no charge…this is NOT an ion)

Class 4. Binary Acids

When a molecular formula starts with an H then this is an acid. The H is called hydro- followed by the anion-ic acid

Note: Never use the Greek Prefixes to describe the number of Hydrogen’s…the number of hydrogen’s is assumed based on the OS of the anion

Name the following:

HF (aq) HCl (aq) H2S (aq)

Oxyoacids

Oxyoacids are formed from F, Cl, Br, and I. Do not confuse these with the Phosphates and Sulfates already mentioned.

Polyatomic names ending with -ate  -ic

And names ending with –ite  ous

36 An example of Chlorine

Polyatomic Polyatomic Potassium Name Acid Acid Name Name salt - ClO4 perchlorate KClO4 potassium HClO4 perchlorate perchlorate acid - ClO3 chlorate KClO3 potassium HClO3 chlorate acid chlorate - ClO2 chlorite KClO2 potassium HClO2 chlorite acid chlorite ClO- hypochlorite KClO potassium HClO hypochlorite hypochlorite acid

So…it is possible to have—only the oxygenated halogen do this!

Polyatomic Polyatomic Polyatomic Polyatomic Polyatomic Polyatomic Name Name Name - - - FO4 Perfluoate BrO4 Perbromate IO4 periodate

- - - FO3 Fluorate BrO3 Bromate IO3 iodate

- - - FO2 Fluorite BrO2 Bromite IO2 iodite

FO- Hypofluorite BrO- hypobromite IO- hypoiodite

End Discussion

How do I look at nomenclature? This is how my mind works out nomenclature. 1st. Always Look at the First Element 1. Is it a metal 2. Is it a non-metal 2nd. If it is a metal 1. Does it have fixed OS 2. Does it have a variable OS 3. If it is a variable OS I figure out the OS 4. Figure out the anion 3rd. At this point it must be a covalent compound 1. Look for oxygen 2. If no oxygen is it a. If a Hydrogen comes first it must be an acid b. Otherwise it is Covalent c. Figure out the anion 3. If it has oxygen it is an oxyo-acid

The following Chart covers the rules just covered. Put the appropriate “Class Number” next to the molecule. This is always the first step in determining the name of your molecule. 37 38 Data for Lab 4: Nomenclature Name ______

Date ______

Labmates ______

Practice Ionic Nomenclature Classes 1 & 2.

barium sulfate

Co(NO3)3

calcium permanganate

Al2(HPO4)3

Iron (III) oxalate

Na2S

Manganese (II) nitride

Practice Covalent Nomenclature Classes 3, 4 and 5.

HBr(aq)

Lithium bromide

CS2

Disulfur pentafluoride

HFO4

Fluoric acid

HFO2

Hypofluorous acid

H2SO4

Sulfurous acid

39 Give an appropriate name for the following

1. SF2 9. P2O5

2. HgC2H3O2 10. Ni(OH)2

3. H2 11. ZnCO3

4. KClO3 12. HF(aq)

5. H3PO4 (aq) 13. HFO4

6. KCN 14. Co2(C2O4)3

7. H2S (aq) 15. Cu(CN)2

8. LiBrO4 16. CS2

Give an approprate formula:

17. Lithium nitrite 23. hypoiodous acid

18. Dinitrogen heptaphosphide 24. Sodium chloride

19. Ammonium chloride 25. Boron trifluoride

20. Chlorine 26. Aluminum chloride

21. Iron (IV) carbonate 27. Titanium (IV) iodate

22. dicarbon tetrasulfide 28. Titanium (III) iodate

40 Lab 5 Lewis Dot

G. N. Lewis, 1916 published a paper in JACS introducing the theory that a bond between two atoms resulted from the sharing of 2 electrons. (Lewis was from Berkeley and is also known for his theory of acid-base chemistry.) Lewis introduced this topic to his students to help them visualize how bonding produced molecular shapes.

Lewis Structures are based on the idea that: 1. A bond consist of sharing 2 electrons 2. Atoms with partially filled valance strive to fill their valance shell

A simple example of the first idea “A bond consist of sharing 2 electrons” is illustrated below between two p-orbitals atoms.

p-orbital p-orbital

To describe the second idea “Atoms with partially filled valance strive to fill their valance shell” a new term Octet-Rule is introduced. This rule applies to nonmetals usually in the first two periods, and roughly applies to the remaining 4 periods as well. 1. Atoms continue to form bonds until all their vacant orbitals are filled. Octet refers to the number 8, which is the maximum number of electrons it takes to fill ns2 np6 period. 2. Once the ns2 np6 orbitals are filled the atoms has the same number of electrons as the Noble or Inert Gasses. The word Inert doesn’t necessarily imply unreactive, more then it implies containing the maximum number of electrons.

41 An example of fluorine: 19F 2s2 2p5

Fluorine is one electron from having a full octet. Adding a second electron fills the octet for Fluoride. No more electrons can be added to fluorine, fluorine will NOT continue to form bonds.

+e-

2p5 2p6

2s2 2s2

- F F

RULES FOR WRITING SIMPLE LEWIS STRUCTURES l. Write the chemical formula for the molecule (or ion) and determine the total number of valence electrons in the molecule.

2. Draw the skeletal arrangement of the molecule showing single bonds connecting the atoms.

3. Assume that each bond in the skeleton requires two valence electrons (an electron-pair bond). After subtracting two electrons for each bond from the total number of valence electrons, assign the remaining electrons to give each atom an octet, or share of eight electrons.

4. If after each atom has been given a share of eight electrons, additional electrons remain, assign the extra electrons to the central atom of the molecule.

5. If there are not enough electrons to give each atom a share of eight electrons, then form multiple bonds between atoms by moving electron pairs to form double (or triple) bonds.

-3 - Examples of these rules with PF3, PO4 and NO3 l. Write the chemical formula for the molecule (or ion) and determine the total number of valence electrons in the molecule. Valance Electrons for P Valance Electrons for F Total Valance Electrons

PF3: 1 5 3  7 26

-3 PO4 : (1 5) + (4 6) + (3  1) = 32 - NO3 : (1 5) + (3 6) + (1  1) = 24

42

2. Draw the skeletal arrangement of the molecule showing single bonds connecting the atoms.

F F O O O O P P N

O O F O

3. Assume that each bond in the skeleton requires two valence electrons (an electron-pair bond). After subtracting two electrons for each bond from the total number of valence electrons, assign the remaining electrons to give each atom an octet, or share of eight electrons.

4. If after each atom has been given a share of eight electrons, additional electrons remain, assign the extra electrons to the central atom of the molecule.

F F O O O O P P N O O F O

- - - 24 e 32 e 24 e

5. If there are not enough electrons to give each atom a share of eight electrons, then form multiple bonds between atoms by moving electron pairs to form double (or triple) bonds.

43 F F O O O O P P N

O O F O

- - - 26 e 32 e 24 e

Resonance Structures.

1 2 1 2 1 2 O O O O O O N N N

O O O 3 3 3

- Possible Resonance Structures for NO3

1 2 1 2 1 2 O O O O O O N N N

O O O 3 3 3

44 VSEPR: Valence shell electron-pair repulsion theory

To derive the VSEPR Electron Pair Name and VSEPR Molecular Shape Name you must first find the Lewis Structure.

Lewis Structure  VSEPR Electron Pair Shape Name  VSEPR Molecular Shape Name

Electron pair is 1. Bonding Pair or 2. Lone Pair Number Electron Pair AXmEn Molecular Shape Shape electron pair Name Name Groups 2 Linear AX2 Linear X A X

3 Trigonal Planer AX3 Trigonal Planer X X A

X

Trigonal Planer AX2E Bent (V-Bent, X X Angular) A

4 Tetrahedral AX4 Tetrahedral X

A X X X

Tetrahedral AX3E Trigonal pyramidal A X X X

Tetrahedral AX2E2 Bent (V-Bent, Angular) A X X

45

Bond Polarity: based on Electronegativity Pauling Electronegativity scale: Linus Pauling devised a scale of Electronegativity based on the relative ability of a bonded atom to attract the shared electrons.

Pauling gave Fluorine a EN value of 4.0 and Lithium 1.0. All other atoms fall between 0.8 and 4.0. The element with the smallest EN value is Cesium and Francium = 0.8.

H 2.1 Li 1.0 Be 1.5 …. B 2.0 C 2.5 N 3.0 O 3.5 F 4.0 Na 0.9 Mg 1.2 …. Al 1.5 Si 1.8 P 2.1 S 2.5 Cl. 3.0 K 0.8 Ca 1.0 …. Br 2.8

For reference: Period 2 elements Fluorine, 4.0, has the GREATEST ability to Attract Electrons. Li has the LEAST ability to Attract Electrons. i.e. C-F, C-Cl, C-Br, C-I….rank from most to least EN.

Two types of bonds a. Ionic---metal & nonmetal b. Covalent---nonmetal & nonmetal

Bonds are further classified based on electronegativity. 1. Ionic (transfer of electrons) 2. Polar covalent (unequal sharing of electrons) 3. Nonpolar covalent (equal sharing of electrons)

Examples of Polar Covalent using Partial Positive + and Partial Negative -

    C F C F

Which is a Polar/Nonpolar: SO3 -vs- SO2 O O O O S S

O

46

Polarity is determined by: 1. Electronegativity charge calculation 2. Presence of Lone Pairs (except in Triganol bipyramidal –Linear and Octahedral-square planer) 3. Symmetry of molecule

***Polarity is draw Polarity based on the Molecular Shape

VSEPR provides 3D arrangement of Groups around the central atom…directly related to bond angle 1st draw the Lewis dot structure 2nd determine the molecular 3D shape of the electron-pair name (Linear, Trigonal Planar, Tetrahedral) 3rd determine the VSEPR molecular shape name (Linear, Trigonal Planar, Angular (V-Bent), Trigonal pyramidal, Tetrahedral)

Repulsion Ordering (deviations from expected bond angle) Lone pair-Lone pair >> Lone pair-Bonding pair >> Bonding pair-Bonding pair VESPR electron-pair name/molecular shape name Trigonal Angular Linear Trigonal Planar pyramidal Tetrahedral (V-Bent) molecular molecular molecular molecular molecular shape, shape shape shape, Always shape Always Polar Polar Linear X CS , HCN electron-pair 2 180

Trigonal Planar X X electron-pair SO3, BF3 SO2, O3  120 <120

X Tetrahedral X X electron-pair CH4, SiCl4 H2O, OF2 NH3, PF3 109.5 <<<109.5 <109.5

47 NOTE: The VSEPR GEOMETRY/Molecular shape NAME only takes into account the Shape between the Atoms (bonding pair) attached to the central atom and disregards the lone pair of electrons: REMEMBER THIS CHART!

48 Data for Lab 5: Lewis Structure Name ______

Date ______

Labmates ______

Molecule # V.E. Lewis Dot Diagram Electron pair Molecular shape Bond Indicate with shape name name Angles polarity arrows if polar or nonpolar SF2

COCl2

H2O

NH3

+ NH4

49 N2O (the central atom is the first N)

O3

PCl3

CH2Cl2

- BrO3

SCN-

CS2

CH2O

50 Lab 6 Heat Capacity of Metals

Equipment: 1. 2 Styrofoam cups 2. thermometer 3. 100 mL graduated cylinder 4. Ring stand, iron ring, Bunsen burner 5. 600 mL beaker 6. 2 large test-tubes with fitted rubber stoppers 7. unknown metal

Today you will determine the identify of an unknown metal by calculating the heat capacity of your metal; all substances have a different heat capacity.

Background: Today’s lab may seem little like a physics lab. However, you will be calculating a physical characteristic of a metal hence, the connection to chemistry. The physical characteristic you will be measuring is called heat capacity. Again, all substances have a different heat capacity. This lab causes problems for all beginning chemistry students because we measure heat capacity indirectly. That is we measure everything but heat capacity to make this determination. Do not be discouraged, all formulas are carefully laid out for your convenience. If you can walk away from this lab understanding how heat works you will be well on your way to understanding the principle of how heat works.

Often students don’t see all of the interconnectivity of what the lab is composed of. So lets take a look at what heat is. Heat simply is how fast a molecule is moving. The faster a molecule moves the “hotter” or more heat is posses. Think of a pool table setup before the queue ball breaks the triangle of balls. The triangle of billiard balls are stationary. You aim the queue ball at the billiard balls, use the queue stick to send the queue ball down the table to “break” apart the billiard balls. The force of the queue ball sets all of the stationary billiard balls into motion. That is exactly how heat works. Fast moving molecules come into contact with slower molecules and the faster moving molecules sets in motion the slower moving molecules. This is why heat “flows” into cold. If you grab a glass of water with ice, the heat flows out of your hand into the cool glass of water. In today’s lab we use the exact same principle. We put a hot object into a cold object and measure the temperature change of the cold object; pretty simple.

What is Heat Capacity?

energy C p  temperature change

Since energy is measured in Joules (J) the typical units for Cp are J/K or J/C. Another unit of energy is the Calorie (cal).

51 Two types of Cp: J 1. Molar Heat Capacity: C  pm mol K

J C  2. Specific Heat: p g  K , we’ll be measuring this today in lab.

In a typical experiment to measure Cp, energy is measured using a water bath and temperature change is measured with a thermometer. The reason for the water bath is joules and calories, units of energy, are defined as the quantity of energy to increase the temperature of 1 g of water 1C. Therefore, if the appropriate experimental conditions are chosen carefully, measuring Cp experiments can be preformed using simple equipment….a Styrofoam cup for example.

Heat and Enthalpy Heat is defined as:

Heat = q = Enthalpy + work

In chemistry, we often ignore the work term. Imagine a car motor where the pistons are allowed to move up and down. The pistons turn a crankshaft hence the car moves. We know pistons move up and down because gas gets in the cylinder, is ignited, and expands to move the pistons….the pistons must move against the pressure of the atmosphere. Under these conditions, the volume of the system is constantly changing and pressure is constant. However, in chemistry, we like to keep the volume constant and let pressure change because we don’t “like” our experiments to do we work, we just want to observe the reaction take place…we want all the heat to go into the molecules themselves. So chemist often define Heat as: Heat = q = Enthalpy (H)

We can now write our Heat or Enthalpy equation as we see it in the chemistry laboratory:

H = q = mCp T

For any lab experiment it is easy to identify the Cp of any substance if we know it’s mass and it’s change in temperature (T). T is change in temperature and always written as T = Tf – Ti.

Our final statement about Heat is the principle called “Conservation of Energy.” This says, in the simplest terms, energy in is equal to energy out, or you have to account for all the energy in the system and it must equal 0. Today’s lab we will measure the

Heat in (cold object) + Heat out (hot object) = 0

Heat in (cold object) = -Heat out (hot object)

52 Experimental Procedure

1. Obtain a 600 mL beaker (hot water bath) and fill it ¾ full with deionized water (DI) and begin heating the water to boiling 2. Obtain an unknown metal, record its number in your lab report. Using a scale weigh, recording the EXACT weight, approximately 80 g of metal by weighing it into a beaker. Transfer the metal to a large test tubes and repeat for the second test tube. Place the two test tubes with metal into the 400 mL beaker. LOOSLEY fit a rubber stopper on top of each test tube. 3. Allow the test tubes of metal to heat in the hot water bath (boiling water) for at least 40 minutes. 4. Dry the inside and outside of each Styrofoam cup, placing one cup inside the other, forming a double-walled insulated cup; both cups are nestled inside each other. 5. Measure Exactly 50.0 mL of water, USING a 50.0 mL Pipet with Pipet Bulb into the Styrofoam cup setup (2 Styrofoam Cups/dry/nestled inside each other). 6. Carefully pour one test tube of hot metal into the cold water, and while constantly stirring record the Highest temperature in the Styrofoam cups…it will take between 30 seconds to 2 minutes for this to happen. After recording your highest temperature achieved carefully pour out the water into the sink, then dump the metal back on a piece of paper. 7. Repeat step 4-6 for the second hot metal USING fresh water…you do not need to record the temperature of the cold water again…use the same temperature you recorded for the first experiment. 8. Dry metal before putting back into it’s jar

Calculations You must first measure the Heat capacity of the Styrofoam cups. Remember we are adding hot water to cold water. While most of the heat from the hot water is transferred to the cold water, some of the heat goes into the Styrofoam cups; it only makes sense. To account for all the heat we write the following.

Hcal + HCW + HHW = 0 Or qcal + qCW + qHW = 0

Where cal = Styrofoam cups, cw = cold water, and hw = hot water

Since q = mCp T, we can be rewrite the above as:

mCp T + mCp T + mCp T = 0

Styrofoam cool hot Cup water water

So referring back to the previous equations we see that heat is being absorbed by the Styrofoam cup and according to the Law of Conservation of Energy, we need to keep track of heat lost to the Styrofoam cup. Of course any heat added to a Styrofoam cup with cold water, heat is going to be lost to the glass thermometer and air of the surroundings, however these are negligible for our purposes.

53 Also note what T is. This is the change in temperature of the cold and hot water. In other words the cold water starts cold but warms up once the hot water is put into it. The hot water starts hot but cools down when put in the cold water. Since T = Tf – Ti and since we mix BOTH the cold and hot water the FINAL TEMPURATURE or Tf are BOTH THE SAME! Keep this in mind.

Careful inspection shows that the only constant we don’t know is Cp for the Styrofoam cups; Cp for cool and hot water (4.184 J/gK).

We change mCp T for the Styrofoam cup to BTCW to account for mass of Styrofoam and glass thermometer, this is just an effective value.

From before: Hcal + HCW + HHW = 0

Final form for the calorimeter is: Hcal = -(HCW + HHW) or BTCW + mCWCp TCW + mHWCp THW = 0

Styrofoam cool hot Cup water water Or

H cal B = Tcw Here, you do not calculate “B.” For two 16 oz Styrofoam cups you can use the value of “25 J/g.” NOTE: Measure using your Graduated Cylinders the 40.0 mL remember that the density of water is as follows: 1 gH2O = 1mLH2O, so 40.0 mL of H2O = 40.0 g H2O.

Your job is to calculate the Cp of your metal. In the following formula you know every variable expect Cpmetal

BTCW + mCWCp TCW + mmetalCpmetal Tmetal = 0

Styrofoam cool hot Cup water water

Compare your heat capacity (Cp) with the following to identify your unknown metal; all of units of J/gK

H2O = 4.18 Fe = 0.473 Mg = 1.04 Ni = 0.444 Co = 0.456 Cu = 0.387 Cd = 0.232 Zn = 0.386 Sn = 0.217 Pb = 0.128 Al = 0.904 Au = 0.129

54 Data for Lab 6: Heat Capacity Name ______

Date ______

Labmates ______

Data: Mass RECORD ALL DIGITS FROM THE DIGITAL DEVICE

1. Unknown Metal number ______

2. Mass of metal in test tube #1 ______

3. Mass of metal in test tube #2 ______

Data: Temperatures

4. Temperature of Cold water (in cup) ______

5. Temperature of hot water (boiling water in beaker) ______

6. Final Temperature (cold water and metal) of Metal experiment #1 ______

7. Final Temperature (cold water and metal) of Metal experiment #2 ______

Calculations Part 1.

1. Referring to your data tables above

8. Experiment #1 TCW = Tf – Ti = #6 - #4 = ______

9. Experiment #1 THM = Tf – Ti = #6 - #5 = ______(this is a negative #)

10. Experiment #2 TCW = Tf – Ti = #7 - #4 = ______11. Experiment #2 THM = Tf – Ti = #7 - #5 = ______(this is a negative #) *Since you used exactly 50.0 mL of cold water = 50.0 g

55 WATCH YOUR SIG FIGS!!!!! **Note: an average “B” is = 25 J/g

BTCW + mCWCp TCW + mmetalCpmetal Tmetal = 0

Part 2.

12: Exp #1: BTCW = 25 J/g  ______(8 above) = ______

13: Exp #1: mCWCp CWTCW = 50.0 g  4.186 J/(gC)  ______(8 above) = ______

14: Exp #2: BTCW = 25 J/g  ______(10 above) = ______

15: Exp #2: mCWCp CWTCW = 50.0 g  4.186 J/(gC)  ______(10 above) = ______

BTH   -   CW  Exp #3 Cpmetal = mTmetal HM

#3 & #7 found on the first data page.

BTH   -   CW  Exp #4 Cpmetal = mTmetal HM

Average your two values of Cpmetal. Show work. Cpmetal =

UNKNOWN Metal = ______

Now, take your Cpmetal and divide it into 24.9, this is known as the Delong Petit formula. This will find the molar mass of your metal.

24.9 Also try:  ______(molar mass of metal) ; where Cp is your metals calculated heat capcity. Cp

56 Lab 7: Chemical EQUATION: WRITING BALANCED EQUATIONS

OBJECTIVES: Given the names of the reactants and products, you should be able to write the correct formulas for these in equation form.

You should also be able to write the balanced equations given the above information, and to identify the following types of reactions.

1. Combination or Synthesis 2. Decomposition 3. Complete oxidation or combustion (burning of organic compounds) 4. Single Replacement (a-type of Redox or oxidation / reduction reaction). 5. Double Rep1acement (ion combination) of the following types a, Precipitation Rxn. b. Neutralization Rxn.

INSTRUCTIONS:

Part 1. Perform the experiments for Combination, Decomposition, Combustion, Single and Double replacements.

Parts 2-4. Write the balanced chemical equation for each reaction described below. Remember that the process of writing equations involves two steps:

1. Write the correct formula for each reactant and product. The subscripts must be correct. If necessary, consult your ion chart and periodic table.

2. Ba1ance the number of atoms of each element on each side of the equation only by using coefficients (DO NOT CHANGE SUBSCRIPTS) If the coefficient is one, no coefficient is necessary.

3. For these reactions assume the oxidation number of the metal does not change upon reactions to form products

Examples 1. Combination or Synthesis Rxn. Ammonia (g) and su1furic acid combine to form ammonium sulfate.

2NH3 (g) + H2SO4  (NH4)2SO4

2. Decomposition Rxns. When heated, potassium chlorate(s) decomposes into oxygen gas and potassium chloride solid.

KClO4  2O2 + KCl

57 3. Combustion or oxidation of organic compounds. Propane(g), C3H8 burns in air (with O2). C3H8 + 5O2  3CO2 + 4H2O

4. Single Replacement (Oxidation-Reduction) Hydrogen(g) is released when aluminum metal reacts with hydrochloric acid. Al + HCl  AlCl3 + H2

5. Double Replacement (neutralization or precipitate formation) Barium carbonate precipitates from the reaction of barium chloride and sodium carbonate solutions. BaCl2 + Na2CO3  BaCO3 + 2NaCl

Lab Procedure

DO EXPERIMENT 4 1st! It takes 30 Minutes.

Part 1. 1. Combination Reaction 2+ a. Obtain 4 mL of Ni(H2O)6 in a 50mL beaker b. To this solution add about 20 mL of water c. Next add about 2 g NH4NO3 d. Then add 5 mL of con. NH3

2. Decomposition a. In a small test tube add about 1 gram of copper(II) sulfate pentahydrate (CuSO4•5H2O) b. Holding the test tube with test tube tongs gently warm the test tube until you begin to see water droplets form at the top of the test tube

3. Complete oxidation or combustion (burning of organic compounds) Use extreme caution, open flames are hazardous. a. Obtain a watch glass and a dropper of ethanol b. Place about 3 drops of ethanol on the watch glass c. Ignite a wood splint using Bunsen burner d. Using the burning wood split ignite the ethanol on the watch glass.

4. Single Replacement (a type of Redox or oxidation / reduction reaction). a. Obtain 20mL of 1.0M CuSO4 solution in a 50mL beaker, And 2 mL of 1.0M CuSO4 in a test tube; keep the test tube for the end of the 30 min time period to compare the color of the original color of the solution (in your test tube) the color of the beaker at the end of the experiment. b. To the beaker with solution add a single piece of zinc metal c. At the end of the experiment, recover the zinc metal, wash and dry it.

5. Double Rep1acement (ion combination) of the following a. Obtain about 5mL of 1.0M iron(III) chloride (FeCl3) in a small beaker b. Obtain about 5mL of 1.0M potassium thiocyanate (KSCN) in a small beaker c. Pour the two reagents (reactants) together 58 Data for Lab 7: Writing Chemical Equations Name ______

Date ______

Labmates ______

Observation:

Part 1. Combination Reaction

2+ 2+ Ni (aq) + NH3(ag)  Ni(NH3)6 (aq)

What is the color of the original Ni2+ solution ______

2+ What is the color of the final Ni(NH3)6 solution ______

Part 2. Decomposition

CuSO4•5H2O(s)  CuSO4(s) + 5H2O(g)

What is the color of the original CuSO4•5H2O solid ______

What is the color of the final CuSO4 solid ______

Part 3. Complete oxidation or combustion (burning of organic compounds)

Use extreme caution when handling ethanol (C2H5OH), do allow it to become wet (contact with wet fingers or wet test tube), and do NOT dispose of remaining CaC2 in the sink.

The combustion reaction of acetylene is as is as follows: Balance the equation.

_____ C2H5OH(g) + _____O2(g)  _____CO2(g) + ____H2O(l)

59 What observation did you notice (smell, sound, color, light etc…) when you ignited the acetylene gas (be brief).

Part 4. Single Replacement (a type of Redox or oxidation / reduction reaction).

Make observations as follows:

1. Initial color of solution and metal:

2. Color of solution and metal after 15 minutes:

3. Color of solution compared to your small test tube and appearance of new metal after 30 minutes:

Write a Single Replacement reaction between zinc metal and CuSO4 below

Part 5. Double Replacement (ion combination) of the following

Observations:

1. Color of each original solution FeCl3 ______& KSCN ______

2. Final color of mixed solution ______

Write the double replacement reaction below

60 Part 2. Rewrite the following word equations into balanced chemical equations.

1. Sulfur trioxide combines with water to form sulfuric acid.

2. Copper metal and a solution of silver nitrate will react to form silver metal and a blue solution.

3. Sodium metal reacts violently with water to form hydrogen and sodium hydroxide

PART 3: Beneath each word equation write the formula equation and balance it

4. Zinc metal + sulfuric acid  Zinc sulfate + hydrogen (g)

5. Aluminum metal + hydrochloric acid  aluminum chloride + hydrogen

PART 4: Complete and balance the following double replacement reaction equation. Assume all reactions go to completion

7. Na2CO3 + HCl 

8. NH4Cl + NaOH 

9. H2SO3 + NH4OH 

61

62 Lab 8 Formula of a Hydrate

A. Introduction Many compounds are formed in reactions that take place in water solutions. The water is then evaporated to obtain the crystalline compound. In some cases water molecules are weakly attracted to the ions or molecules that make up the compound and are retained within the crystal structure. Crystalline compounds that retain water during evaporation are referred to as being hydrated or are said to contain water of hydration. The ratio of moles of water to moles of compound is a small whole number. The formula for the hydrated compound copper (II) sulfate is

CuSO45H2O

The dot shows that for every mole of CuSO4 in the crystal, there are 5 moles of H2O. The ratio of moles of H2O to moles of compound can be determined experimentally in most cases by heating to remove the water. The compound with the water removed is anhydrous. In this experiment you will determine the formula for a hydrated unknown compound. The formula is determined by comparing the mass of the hydrated and anhydrous forms of the compound.

B. Procedure and Data

1. Clean the crucible first, remove excess residue 1st before heating. After this point NEVER touch your crucible and lid with your fingers, the oils on your fingers will actually affect your mass.

2. Obtain a clean crucible and lid, support there on a clay triangle, and heat with an intense flame for 5 minutes. Allow cooling. Weigh the crucible and lid together. Handle the crucible and lid with the crucible tongs (or a test tube holder) for the rest of the experiment; do not use your fingers.

Note: Use a ring stand, and Bunsen burner. The Blue Inner-Cone is the hottest part of the Flame. Position your crucible a few cm above the Inner-Cone.

3. Add a maximum of 1.5 (1-1.5g) g of the BaCl2 hydrate (BaCl22H2O) to the crucible and weigh the crucible, lid, and sample together. Record.

Note: Use the Analytical Scales for the entire lab.

4. Place the crucible with the cover on a clay triangle over a laboratory burner.

5. At first, heat the sample slowly and then gradually intensify the heat. Do not allow the crucible to become red-hot. This can cause the anhydrous salt to decompose. Heat the sample for 15 minutes. Cool to room temperature, and weigh the crucible, lid, and sample together.

6. Repeat with a second sample---2 Trial

7. Collect ALL Ba solid in waste container

63

64 Data for Lab 8: Formula of a Hydrate Name ______

Date ______

Labmates ______RECORD ALL DIGITS FROM THE DIGITAL DEVICE

Trial 1 1. Empty Crucible & Cover mass after heating ______

2. Crucible & Cover with sample (before heating = hydrate) ______

3. Crucible & Cover with sample after 20 min heating ______

4. Mass of hydrated compound used (2-1) ______

5. Mass of anhydrous compound produced (3- 1 ) ______

6. Mass of water lost (4 - 5 ) ______

Trial 2 7. Empty Crucible & Cover mass after heating ______

8. Crucible & Cover with sample (before heating = hydrate) ______

9. Crucible & Cover with sample after 20 min heating ______

10. Mass of hydrated compound used (8-7) ______

11. Mass of anhydrous compound produced (9 - 7 ) ______

12. Mass of water lost (10 - 11 ) ______

65 C. Calculations and Results

1. Determine the percentage of water in the hydrated compound for each trial and the average percentage of water.

mass H O %H O  2  100 2 mass of hydrate

mass H O %H O  2 100 Trial #1: 2 mass of hydrate

mass H O %H O  2 100 Trial #2: 2 mass of hydrate

Average %H2O: Show Work. %H2O =

2. Calculate the number of moles of anhydrous compound produced and the number of moles of water lost. ___g BaCl 1 mole BaCl moles BaCl  2  2 # moles BaCl 2 2 1 208.3 g BaCl 2

___g H O 1 mole H O moles H O  2  2 # moles H O 2 2 1 18.0 g H2O

Show your work for Trial #1 & Trial #2

66 3. Determine the formula for the hydrate

For example if you find you get 0.0010 moles of BaCl2 and 0.0022 moles of H2O do the following by taking the smaller number into the larger number:

0.0022/0.0010 = 2.2 moles of H2O 0.0010/0.0010 = 1.0 mole BaCl2

Your molecular formula would be BaCl22.2H2O. You do NOT need to get whole numbers---for This Experiment ONLY---so do not round.

Show your formulas from each trial below

4. Calculate the theoretical % error.

What is the % error? As in 4 above the example showed the formula to be BaCl22.2H2O. The experimental value is supposed to be: BaCl22H2O:

Base on the value obtained in 4 (for illustrational purposes only) my error would be

Exp-Theory % Error  100 Theory

2.20-2.00 % Error  100  10.% error 2.00

Do this for each trial, then find the average error.

67

68 Lab 9 Equilibrium

PURPOSE To observe a number of interesting and colorful chemical reactions that are examples of chemical systems at equilibrium. To see how these systems respond to changes in the concentrations of reactants or products or to changes in temperature. To see that the direction of the shift in the equilibrium tends to at least partially offset the change in conditions, a principle first dearly stated by Le Châtelier.

PRE-LAB PREPARATION

Any chemical equation that describes an actual chemical reaction sums up the result of an experiment (more often, several experiments). Someone had to measure out the reactants, carry out the reaction, identify the products, and measure quantitatively the number of moles of products produced per mole of reactant consumed. The last step is the writing of a balanced chemical equation that concisely summarizes the experimental observations.

Many chemical reactions go essentially to completion, but some do not, stopping at some point of equilibrium that lies between no reaction and an essentially complete reaction. A state of equilibrium is the point of balance where the rate of the forward reaction equals the rate of the backward reaction. For these reactions, the same point of equilibrium can be reached from either end of the reaction, by mixing together either the reactants or the products, dearly indicating that chemical reactions can go either forward or backward.

In principle, every chemical reaction is a two-way reaction, but if the point of equilibrium greatly favors the reactants, we say that there is no reaction. If the point of equilibrium greatly favors the products, we say the reaction is complete. For either of these two extremes it is difficult to measure experimentally the concentrations of all of the reactants and products at equilibrium. In the first case (reactants greatly favored), the concentration of products is practically zero. In the second case (products greatly favored), no significant quantity of the reactants will be produced when the products are mixed together. Thus, when the equilibrium greatly favors the products, the reaction appears to go in only one direction: reactants  products.

Nearly every chemical reaction consumes or releases energy. (We call these, respectively, endothermic or exothermic reactions.) For these reactions, we can regard energy as if it were a reactant or product, and by adding energy to a system (by heating it) or removing energy from a system (by cooling it), we can produce a shift in the point of chemical equilibrium. The concentrations of reactants and products change to reflect the new equilibrium.

There is an added complication that must be considered. Not all chemical reactions are rapid In fact, some are so slow that you might be fooled into thinking that no reaction occurs, whereas the equilibrium point, when it is finally reached, might greatly favor the products.'rhe rate at which a chemical reaction reaches equilibrium will be taken up when you study the kinetics of chemical reactions.

For reactions that rapidly reach equilibrium, the point of equilibrium may be approached from either the side of the reactants or the side of the products, which emphasizes the dynamic nature of chemical 69 reactions. The reactions you will study in this experiment are all rapid reactions, so you will be able to observe almost immediately the effects of changing the concentration of either reactants or products. In particular you will observe that when you change the concentration of a reactant or product, the point of equilibrium shifts in a direction that tends to offset the change. This behavior can he summarized in a general principle that was first htuy stated in 1884 by the French chemist Henri Louis Le Chateher: A chemical reaction thin is displaced from equilibrium by a change ht conditions (concewratiorts, temperature, pressure, volume) proceeds toward a new equilibrium state in the direction that cat least partially o/]sets the change in conditions.

This introductory experiment is designed to show qualitatively several important features of chemical equilibria-ln subsequent experiments you will learn about the quantitative aspects of chemical equilibria, measuring the equilibrium constants for a variety of reactions, and making calculations of equilibrium concentrations from previously measured equilibrium constants.

COMPLEX ION EOUILIBRIA. It is common for cations (especially those with +2 or +3 charge) to attract negatively charged ions (or neutral molecules with lone pairs of electrons) to form aggregates called complexes. If the resulting aggregate has a net charge, it is called a complex ion. The composition of these complexes may vary with the proportion and concentration of reactants.

Part 1: The Thiocyanatoiron (III) Complex Ion. This ion, sometimes called the ferric thiocyanate complex ion, is formed as a blood-red substance described by the following equilibrium equation.

Fe3+ + SCN- ⇄ Fe(SCN)2+ (1)

In a 100-ml, beaker add 2 ml of 0.1 M Fe(NO3)3, to 2 mL of 0.1 M KSCN. Dilute by adding 60-70 mL of water until the deep-red color is reduced in intensity, making further changes (either increases or decreases in color) easy to see. Split this solution into 4 test tubes, so each test tube is about 2/3 full. To test tube #1 add 1 mL (about 25 drops) of 0.1 M Fe(NO3)3 (Equation #1 in Data Section). To the second test tube, add I mL of 0.1 M KSCN (Equation #2 in Data Section). To a third add 5 to 6 drops of 6 M NaOH (Equation #3 in Data Section). NOTE: Fe(OH)3, is very insoluble. Use the fourth portion as a control (use this test tube to compare color with 1-3). Compare the relative intensity of the red color of the thiocyanato complex in each of the first three test tubes to that of the original solution in the fourth tube. Interpret your observations, using Le Châtelier’s principle and considering the equilibrium shown in Equation (1) above.

Part 2: The temperature-Dependent Equilibrium of Co(II) Complex Ions. The chloro complex that consists 2- 2- of cobalt(II), CoCl4 is tetrahedral and has a blue color. The aquo complex of cobalt(II) Co(H2O)6 , is octahedral and has a pink color. (Figure 1 shows the geometry of the tetrahedral and octahedral complexes.) There is an equilibrium between the two forms in aqueous solution, and because the conversion of one form to another involves a considerable energy change, the equilibrium is temperature dependent:

2- - 2- CoCl4 + 6H2O ⇄ 4Cl + Co(H2O)6 + energy (2)

70 Le Chatelier's principle applied to Equation (2) predicts that if energy is removed (by cooling the system), the equilibrium tends to shift toward the aquo complex, because a shift in this direction produces some energy, thus partly offsetting the change.

Put 2 mL of 0.10 M CoCl2 (in methanol) into a 13 x 100 mm test tube (medium sized test tube); methanol is used as a solvent so that you can observe the effects of adding water. Using a dropper, add just enough water to the purple (it may be light pink to begin with) methanol solution to change the color to that of the pink aquo complex. Next, add concentrated (12 M) HCl (Equation #4 in Data Section) one drop at a time until you observe a color change to a blue color. Record your observations. Cool this test tube (Equation #5 in Data Section) in an ice bath and record your observations. Heat the test tube (Equation #6 in Data Section) in a beaker of hot water (65 to 70°C). (Caution: Methanol vapors are toxic and flammable.) Use EXTREME CAUTION to not let the methanol solution come into contact with your open flame. You should note a color change. (If you do not, you probably did not add enough water to the original methanol solution. Try again.) The color change is reversible. Cooling the solution in an ice bath will restore the original pink color. Repeat the cycle of heating and cooling to verify this. Record your observations, and interpret them by applying Le Châtelier’s principle described below.

H O 2 Cl- OH2 H2O Co2+ - Cl Co2+ OH H O 2 2 Cl- - H2O Cl

Pink Octahedral Blue Tetrahedral Co(II) Co(II) (cold) (hot)

FIGURE 1

The cobalt(II) complexes with chloride ion and water have different molecular geometries and different colors.

Procedures: 1. EXTREME CAUTION to not let the methanol solution come into contact with your open flame 2. Addition of reagents is approximate, no need to measure reagents i.e. use about a mL of reagent 3. Add enough reagent until you see a chemical reaction take place 4. All solutions are in the hood 5. Dispose of the FeSCN and any Co+2 solutions in appropriate, separate waste containers in the fume hood

71 Applying Le Châtelier’s Principle

1. Write the equilibrium equation for the system.

2. Write the equation for the added substance

3. Use arrow(s) to show the stress. Up arrows,  , for increased concentrations and down arrows,  , for decreased concentrations.

4. Use an arrow to show the shift in position of equilibrium.  or 

5. Use arrows to show the resulting change in concentrations due to the shift described in step 3

6. Similar to 4, though use an arrow  or  to show the change in color or other noted change Example of how to write Reaction #1. Fe(SCN)+2 plus Fe+3

Test Tube #1 Fe(SCN)+2 plus Fe+3 +3 - +2 1. Fe + SCN Fe(SCN)

+3 - 2. Fe(NO3)3 Fe + 3NO3

3 – 6 +3 - +2 Fe + SCN Fe(SCN)

3 5 5

4

6 ~colorless dark red

72 Data Sheet for Lab 9 Equilbrium Name ______

Date ______

Labmates ______

Test Tube #2 Fe(SCN)+2 plus SCN-

+3 - +2 1. Fe + SCN Fe(SCN)

2. KSCN K+ + SCN- common Ion

3 – 6

#3 Fe(SCN)+2 plus OH-

1. Fe+3 + SCN- Fe(SCN)+2

+3 - 3+ 2. Fe + 3OH Fe(OH)3 (s) notice, Fe is removed through a precipitation reaction.

3 – 6

73 #4 Co+2 plus HCl

-2 - +2 1. CoCl4 + 6H2O 4Cl + Co(H2O)6 + energy

2. HCl  H+ + Cl-

3 – 6

#5 Co+2 minus heat (cool)

1.

2. N/A

3 – 6

#6 Co+2 plus heat

1.

2. N/A

3 – 6

74 Lab 10 Standard Molar Volume of a Gas

Purpose and Objectives Determine the volume and moles of H2 (g) collected at room temperature and pressure. To apply Dalton's Partial pressure Law, the combined gas law, and the Ideal Gas Law. To determine the molar mass of the unknown metal.

Discussion The Ideal Gas Law relates the variables associated with gases through the equation:

PV = nRT where P is measured in mmHg, but is converted to atmospheres by converting 760 mmHg/1 atm before using the equation. Volume is measured in mL in the laboratory, but is converted to liters before using in the equation. Moles, n, is calculated as n = mass/ MM (MM is the molar mass and defined as grams/mole). T is measured in °C in the lab, but must be converted to Kelvin:

C + 273.15 = K before using in the equation. R is the gas law constant,

0.0821 L atm R = mol  K

The Combined Gas Law relates the P, V, and T for a gas in which one or more of these three parameters (measurable quantities) have changed, but in which the moles, n, have remained the same. The law is really a simplified version of the ideal gas law:

PVPV 1 1 = 2 2 TT12

Note that “1’s” and “2’s” represent initial (1) and final (2) conditions.

Dalton's Law of Partial Pressures, PT = P1 + P2 + P3 + ...Pn, relates the moles of a mixture of gases to the pressure of the total amount of gases. If a gas mixture was 50% O2(g) and 50% H2(g), the total pressure would be the sum of the two pressures. H2O(g), usually called vapor pressure of water is the gas we usually make corrections for in laboratory work. In dealing with gases, it is convenient to establish a standard reference point, called Standard Temperature and Pressure, (STP). At STP, any gas occupies exactly 22.4 L for exactly 1 mole. Since this is exactly one mole, it contains one Avogadro number of molecules or atoms of gas. Therefore, the Molecular Volume given contains exactly one gram-molecular mass (or one gram- atomic mass) of the gas.

In this experiment we will collect H2(g) from the reaction of a metal which reacts according to the following equation: M(s) + 2 HCl(aq)  H2(g) + MCl2(aq)

75

Note: this is one of the 4 special “Gas Forming” reactions in chemistry; metals react with strong acids to produce hydrogen gas (H2 (g)) and a metal salt.

The metal we will use will be an unknown metal that will ionize to the +2 state. Thus one mole of the metal will form from one mole of HZ(e). The molar mass of the metal and the identity of the metal will be determined.

Procedure 1. Fill a large beaker to about 3/4 full with water.

2. Weigh no more then 0.03 g of magnesium ribbon and record to three decimal places, then roll it and tie a copper wire through it and leave a two inch tail of copper. Wrap the tail side of the copper through a one hole rubber stopper. The metal ribbon should be suspended about 1 cm from the small end of the stopper.

3. Fill to about the 20mL mark on the gas buret with 3M hydrochloric acid (HCI). Carefully fill the buret with water using your wash bottle to the brim. Any air left will be mistaken for the H2 gas.

4. Carefully fit the rubber stopper into the top, place your finger over the hole in the stopper, invert the stopper end of the buret into the large beaker containing water, release your finger, and hold the buret upright. The metal will then react with the acid. (The acid is denser than the pure water and slowly moves to the bottom of the tube where it can react with the metal). The reaction is complete when there are no more bubbles coming from the bottom, and the metal is completely gone. If some metal breaks off and floats to the top, wait until it completely reacts.

5. Accurately read and record to one decimal place the volume of gas collected, holding the buret vertically, and reading the bottom of the meniscus.

6. Read the barometer, and record the pressure in mmHg. Read the thermometer on the barometer and record the room temperature to the tenths place. This temperature is taken as the gas temperature T1. Record the vapor pressure of water at room temperature.

Vapor Pressure of Water Temp mm g Temp mmHg 19 16.5 24 22.4 20 7.5 25 23.8 21 18.6 26 25.2 22 19.8 27 26.7 23 21.2 28 28.3

7. Repeat steps 2-5 for second piece of metal for a second trial.

76 Data for Lab 10: Electronic Configurations and Periodic Trends

Name ______

Date ______

Labmates ______

RECORD ALL DIGITS FROM THE DIGITAL DEVICE

Trial 1 Trial 2 Metal Mass (g)

V1 = Volume Hydrogen collected in L

T1 = Temperature of air (close enough) in K

Atmospheric Pressure (PT) i.e. barometric pressure in mmHg

Vapor Pressure H2O (PH2O) found in table above in mmHg

Partial pressure of H2 is: PT = PH2O + PH2, therefore PH2 = PT –PH2O

P1 = Partial Pressure H2 (PH2) =

77 Calculations: WATCH YOUR SIG FIGS 1. Calculate the volume (V2) of hydrogen gas produced, at STP conditions for each run. Use the Combined Gas Law where P2 =760 mmHg and T2 = 273 K. Calculate the average volume.

PVPV 1 1 = 2 2 TT12

Trial 1 Trial 2

V2 = Volume H2 (STP) in L

Average Volume (V2) in L =

2. Calculate the moles of H2 produced used:

Mg + 2HCl  MgCl2 + H2

____g Mg 1 mole Mg 1 mol H2    ____mol H2 1 24.3g Mg 1 mol Mg

Trial 1 Trial 2

Moles H2

Average mol =

V 3. Calculate the molar volume at STP from your averaged numbers. Use 2 . mol

78 Lab 11 Net Ionic Equations (NIE) There are two Instructor demonstrations.

1. The Instructor will demonstrate, with a conductivity apparatus, the following: 1. Di-ionized water 2. Tap water 3. Di-ionized water with NaCl(s) added 4. Di-ionized water with C6H12O6 (sugar) added

Your instructor will explain the results of these experiments.

I. OBJECTIVES:

1. To become familiar with the use, of a Table of Solubility Rules and the application of a Table of Electrolytes. 2. To practice identifying and drawing the form that a chemical species is present in aqueous solution in a beaker. 3. To become familiar with writing chemical equations as a molecular, ionic, and net ionic equations.

II. THEORY: Read in the text sections discussing electrolytes, net ionic & ionic equations.

REMEMBER! 1. Molecular Equation: Shows all substances present before and after the reaction- Write these in "molecular form"- 2. Ionic Equation: Shows what actual species (ions or molecules) are present before and after the reaction, including all participating and "spectator" ions- 3. Net Ionic Equation Shows which species; took part in the solution reaction (eliminating all "spectator ions").

Here are some HINTS, for writing correct "ionic" and "net ionic equations".

A. Molecular equations 1. First, write a "normal" molecular equation (as we have been doing) and BALANCE it!! 2. Check the solubility chart and identify each species as (s), (aq), (g), or (l).

B- Ionic equations 1. For each soluble substance (aq), check the electrolyte chart to see which should be written as ions: a. If it is a strong electrolyte write the cation and anion as separate ions + 2- example: Na2SO4 (aq) : 2Na (aq) + SO4 (aq)

79 b. If it is a weak electrolyte write it as a "molecule" example: NH4OH (aq) : NH4OH (aq) HC2H3O2 (aq) : HC2H3O2 (aq)

c. If it is a nonelectrolyte write it as a molecule. example: methanol (aq) : CH3OH (aq) water (l) : H2O (l)

2. For insoluble substances ; precipitates, and gases and pure liquids write these as molecules example: BaCrO4(s), AgCl (s), HCl(g), H2O(1), CO2(g)

3. Write the full ionic equation, being sure to show ALL CHARGES for ions and correct coefficients

C. Net ionic Equations: 1. Cross out all "spectator ions" (those which do not participate and do not change the reaction).

2. Show only the species which actually participated in the reaction. These are the species which underwent a chemical change and are not identical on both sides of the equation.

*****All equations must be balanced, both atoms and charges!!

Example:

Molecular: NaBr(aq) + AgNO3(aq)  AgBr(s) + NaNO3(aq)

+ - + - + - Ionic: Na (aq) + Br (aq) + Ag (aq) + NO3 (aq)  AgBr(s) + Na (aq) + NO3 (aq)

Net ionic: Ag+(aq) + Br-(aq)  AgBr (s)

2. The Instructor will demonstrate the NaBr and AgNO3 reaction for the students after writing the equation on the board to demonstrate the reliability of the solubility rules.

So how do I write net ionic equations???

I ask - What am I starting with? To answer this, I consider the reactants: Is each one a solid (s), liquid (1), or a gas (g) ? Or is it in solution (aq) ?

A. If it is a (s), (1), or a (g), leave it in molecular form. Be sure the formula is correct.

80 B. If it is an (aq), then consider what type of electrolyte it is: STRONG - Write it as separated ions consult the ion chart if necessary weak - most are molecules, so I write them as molecules in my equation. non - leave it as molecules because all are in molecular (aq) form.

I ask - What am I forming? To answer this, I must consider each product:

A. If it is a precipitate, it must NOT be soluble in water. 1. Consult the solubility rules to check if this is so. 2. Write the formula correctly, including (s). 3. Leave the formula in MOLECULAR form - all ions stay together.

B. If it is soluble, I ask - is it a strong, weak or non-electrolyte? HC1, HBr, HI, H2SO4, HNO3, and HCIO4 are all strong acids. 1. All others are WEAK 2. Almost all soluble salts are strong electrolytes, separated into ions with an (aq). 3. If the species is a weak electrolyte or a non-electrolyte, leave as molecules-

+ - REMEMBER: H (aq) + OH (aq)  H2O(l)

C. If it is a gas, what is it?

4 common gas forming reactions are:

If you see the following 3 as products, replace those molecules with the appropriate gas and H2O (l).

1. H2CO3  H2O(l) + CO2(g) 2. H2SO3  H2O(l) + SO2(g) 3. NH4OH  NH3(g) + H2O(l)

This is a common method for making hydrogen gas + 4. H + some metals  metal salt (aq) + H2(g)

Spectator ions undergo ABSOLLUTLY NO CHANGE - include in ( ) after the balanced net ionic equation is written.

An equation is balanced only if both atoms and charges are conserved (the same number on both sides - total).

81

82 Data Sheet Lab 11 NIE Name ______

Date ______

Labmates ______

This is a Paper and Pencil Lab. Students predict the results of these reactions using solulbility rules.

1. sodium carbonate reactions with phosphoric acid.

Molecular

Ionic

Net Ionic

2. Potassium iodide reacts with lead (II) nitrate to form a precipitate.

Molecular

Ionic

Net Ionic

83 3. Copper (II) sulfate reacts with silver nitrate.

Molecular

Ionic

Net Ionic

4. Lithium sulfite reacts with a solution of hydrochloric acid.

Molecular

Ionic

Net Ionic

5. Solutions of aluminum sulfate and calcium iodide reacts.

Molecular

Ionic

Net Ionic

84 6. Solutions of potassium acetate and lithium bromide are mixed

Molecular

Ionic

Net Ionic

7. Sulfuric acid solution neutralizes a solution of potassium hydroxide.

Molecular

Ionic

Net Ionic

8. When baking a cake we sometimes combine baking soda (NaHCO3) and vinegar (HC2H3O2) to create bubbles that cause the cake to rise. Write an equation for this reaction.

Molecular

Ionic

Net Ionic

85 9. calcium hydrogen sulfite + sulfuric acid

Molecular

Ionic

Net Ionic

10. Silver hydrogen carbonate + sulfuric acid

Molecular

Ionic

Net Ionic

11. Aluminum metal reactions with phosphouric acid to make aluminum sulfate and hydrogen gas

Molecular

Ionic

Net Ionic

86 Lab 12 Titration

Demo by Instructor 1. Buret cleaning, usage 2. adding solids to flask, when the flask is on a scale

This is a Quantitative Analysis Lab, meaning the aim of the lab is to measure precision and accuracy.

The reaction for this lab is an acid base reaction. All acid base reactions produce a salt and water

Acid + Base Salt + H2O

For example: H3PO4 + NaOH Na3PO4 + H2O

Looking at the above example, we can write the NIE H3PO4(aq) + NaOH(aq) Na3PO4(aq) + H2O(l) which leads to: + - H + OH H2O(l)

This is referred to as a neutralization reaction.

The point at which mol H+ = mol OH- is referred to as the Equivalency Point. In today’s lab we us an indicator to let us know when the EQ point has been reached. Phenolphthalein, our indicator, is clear under acidic conditions and turns a light pink under basic conditions which is at the point after the H+ is consumed.

+ + - H H OH - H+ H+ -OH + H2O OH- OH H OH-

87 I. Purpose: The purpose of this experiment is to: 1. Prepare a 0.1 M sodium hydroxide solution 2. Standardize it with a primary standard acid

II. Theory: In the neutralization of an acid with a base, the end point of the titration is reached when the number of moles of hydrogen ion exactly equals the number of moles of hydroxide ion. Calculations can be make using balanced chemical equations and standard stoichiometric methods.

III. Procedure: Titration of H2C2O4 with NaOH Solution. (*DEMO) Set up and clean a 50 mL buret as demonstrated by your instructor. Be sure that the buret is clean, then rinse three times with 5 ml, portions of the NaOH solution you will be doing the titrations with; approximate 0.1 M NaOH solution. Fill the buret, and be sure all air bubbles are eliminated from the tip. Start by transfering approximately 0.01 gram (record accurately) of oxalic acid dehydrate (your sample), H2C2O42H2O, to a clean, dry 125 ml, or 250 ml, Erlenmeyer flasks. Add about 30 mL of distilled water and two-four drops of phenolphthalein indictor to each sample. Titrate the samples with the approximately 0.1M NaOH solution to the first appearance of faint pink end point that persists for at least 30 seconds. NOTE: You should use approximately 18 mL of NaOH for each titration…in other words, adjust your sample size so you USE between 18-25 mL of NaOH. The 0.01 g is JUST a starting point, you WILL need to add more sample.

Initial and final buret readings are to made to the nearest 0.01 mL. Do a minimum of 4 titrations with "good" end points. Calculate the molarity of the NaOH solution for each titration.

Note: H2C2O42H2O is a diprotic acid, it takes 2 moles of NaOH for every one mole of oxalic acid.

Dispose of all solutions into the sink, run water as you do this.

Titration General Information 1. This experiment is to be done in pairs. 2. The grade for this experiment is based entirely on the results. 3. All measurements should be made to the appropriate number of figures to be acceptable. Mass measurements should be made to the nearest 0.001 g and volume measurements should be made to the nearest 0.01 mL with the exception of the volumetric flask, which should be measured to the nearest 0.1 mL. 4. The data section must contain only data! Note the special way of setting up the data and the calculation sections for this experiment.

Buret readings 3-4 sig figs with 2 decimal places Mass readings 4 sig figs with 3 decimal places

88 ***Critical

In order to be more accurate and to avoid losing precious time during the lab periods, beware of these common mistakes: Buret: 1. not rinsing the buret one time with deionized water three times with your base solution 2. not checking to see if the buret is clean 3. not watching for leaks around the stopcock 4. not removing all the air from the dropper tip

Making Solutions: 1. not weighting solids accurately (i.e. dropping crystals, not weighing to the nearest 0.0001 g) 2. trying to get all the solid samples the same size 3. Using dirty equipment

Titrating: 1. trying to make the initial volume read exactly 0.00 mL 2. not reading the initial and final volumes carefully and consistently to the nearest 0.01 mL 3. forgetting the indicator 4. not washing down tip of the buret and the neck of the flask with deionized water 5. not swirling flasks over white paper while titrating 6. not running enough "good" trials 7. misreading labels on chemicals used 8. misrecording data 9. making math errors

89

90 Data Sheet Lab 12 Titration Name ______

Date ______

Labmates ______#1 mass acid volume reading RECORD THE MASS READ THE BURRET CORRECTLY CORRECLTY ______final______

start______

Total Volume _____

#2 mass acid volume reading

______final______

start______

Total Volume ______

#3 mass acid volume reading

______final______

start______

Total Volume ______

#4 mass acid volume reading

______final______

start______

Total Volume ______

91 Molarity Calculations:

Molarity is defined as Mol/L. Here you will be determining the molarity of your NaOH solution.

1 mole oxylic acid ____ mol NaOH 1 ____gM oxylic acid     ______NaOH ______g oxylic acid _____ mol oxylic acid ____L NaOH

Show your work for ALL 4 titrations.

What is your Average Molarity of NaOH =

92 30 points Extra Credit: Nomenclature (due when your instructor tells you :-)

Give an appropriate name: Name______

1. NaIO4 15. HIO3

2. NaIO3 16. HIO2

3. NaIO2 17. H2O

4. NaIO 18. FeCl3

5. KBrO 19. Fe2(SO4)3

6. KBrO2 20. FeS

7. KBrO3 21. HCl (aq)

8. KBrO4 22. HClO3

9. HBrO4 23. H2CO3 (aq)

10. HBrO3 24. CaCO3

11. HBrO2 25. Be2C

12. HBrO 26. SnSO3

13. HIO 27. (NH4)2S

14. HIO4 28. N2O4

93 29. K2HPO4 43. Pb(Cr2O7)2

30. Fe(NO3)3 44. Ca(OH)2

31. P4O6 45. AlP

32. NH4Cl 46. Ni2Cr2O7

33. K2CrO4 47. Be(NO2)2

34. Na2SO4 48. CuSO3

35. KMnO4 49. Fe(MnO4)3

36. WO3 50. CuCN

37. Ca(ClO2)2 51. N2O

38. FeSO4 52. Al2S3

39. P4O10 53. NaH2PO4

40. Sn(ClO3)4 54. Ti2O

41. K2Cr2O7 55. TiO

42. PbCr2O7 56. TiO2

94 57. Sodium iodite 73. Aluminum carbide

58. Chromium (VI) oxide 74. Magnesium phosphate

59. Potassium bromate 75. Nitrogen dioxide

60. Potassium perbromate 76. Ammonium sulfate

61. Sodium hypofluorite 77. Barium carbonate

62. Potassium fluorate 78. Tungsten (IV) carbonate

63. Dinitrogen tetrasulfide 79. Sodium hydrogen carbonate

64. Fluorous acid 80. Calcium dihydrogen phosphate

65. Hypobromous acid 81. Disulfur dichloride

66. Fluoric acid 82. Sodium iodite

67. Perfluoric acid 83. Sodium hypoiodite

68. Iodic acid 84. Sodium iodate

69. Peroidic acid 85. Sodium periodate

70. Barium chloride 86. Potassium bromate,

71. Tin (II) nitrate 87. Potassium perbromate

72. Tin (IV) nitrate 88. Sodium hypofluorite

95 89. Potassium fluorate 105. Iron (III) hydrogen phosphate

90. Potassium fluoride 106. Iron (III) dihydrogen phosphate

91. Hydrofluoric acid 107. Magnesium iodide

92. Lithium fluoride 108. Hypoiodous acid

93. Hypobromous acid 109. Periodic acid

94. Perfluoric acid 110. Hypobromous acid

95. Iodic acid 111. Magnesium periodate

96. hydroiodic acid 112. Zinc iodate

97. Sulfuric acid 113. Copper (I) iodite

98. Hydrosulfuric acid 114. Copper (II) iodite

99. Lithium sulfide 115. Hypofluorous acid

100. Sulfourous acid 116. Periodic Acid

101. Iron (II) phosphate 117. Tin (IV) hypoiodite

102. Iron (II) hydrogen phosphate 118. Tin (II) hypoiodite

103. Iron (II) dihydrogen phosphate 119. Calcium hypoiodite

104. Iron (III) phosphate 120. hydrocyanic acid

96