Stoichiometry 1

Chem 120 Sec. 03 Fall 2003

me: C. Chieh Ph. D. (Dr. Jay, Prof. J, or Peter if U ps c120)

Cyberspace Chemistry (CaCt)http://www. science.uwaterloo.ca/~cchieh/cact/

Office: C2-263Phone: 888-4567 ext. 5816e-mail: cchieh@uwaterloo.caPlease provide your ID number in your e-mail.

Try to know me and want me to know you, then we probably will be friends.

Stoichiometry 2

Students ResponsibilityPlan to fit the class – survive in society.

Solve suggested problems - do them regularly.

Ask for help before it’s too late – we are here to help.

Read all instructions – distinguish assumptions and fact.

Learn basics and techniques of science, and apply them to solve problems and achieve your goals.

Know what to learn, how to organize, what to ask, where to find answers, know if you know your stuff, and how to learn in general.

Stoichiometry 3

University Students

You are a proud student of the university – very different fromyour past, very difficult at first, and very independent eventually. You are responsible for everything you do, and no one else but you are responsible for you.

You have to do all things on your own – no one tells you what to learn or what to do.

You have to know the right from wrong – think independently, do not completely trust anyone, have sound judgments, and make wise decisions.

Get not just the look, but what under the hood.

Stoichiometry 4

Announcements

Labs starts this week for odd-numbered lab sections (1, 3, 5, … 27)

Labs starts this week for odd-numbered lab sections (2, 4, 6, … 28)

Purchase safety goggles and lab manual before first lab from ChemStores, ESC 109, across from the first-year lab.

Must have a valid WHMIS sticker to work in the lab. If not, attend a WHMIS training session before your second lab period.

Tutorials will start next week, no tutorials this week.

Sign up for a Study Skill (4 weekly sessions: time management, note taking, reading, preparation for exams) workshop at http://www. adm.uwaterloo.ca/infocs/

Stoichiometry 5

Announcements cont.Science 101 – Student Handbook (useful information and survival tips) is available for pick up in the Science Undergrad Office, ESC 253.

Unlike years in the past, students who will fail Chem120 will not be allowed to take Chem123. They can repeat Chem120 in the Winter term by DE courses or the next fall. (serious)

Stoichiometry 6

Properties of MatterMatter, energy & man-made things

Physical properties: b.p., m.p., density, color, hardness, dielectric constant, heat of fusion, heat of vaporization, heat of sublimationmalleability, brittle …

Chemical properties:composition, bonding formula, structural, 3-D structureoxidation, reduction, combustion, chemical changes,decomposition, acidity, stability

Biological properties:toxicity, chemical changes in biological systems

Stoichiometry 7

Classification of MatterMixture (substances)

homogeneous: gasoline, perfume, wine, drinkshetrogeneous: milk, orange juice

Compounds (pure) water H2O, methane CH4, carbon dioxide CO2, …

Elementshydrogen, helium, lithium, nitrogen, oxygen, … uranium

Make a diagram to classify matter

Stoichiometry 8

SI Units and Measurements

7 Base Quantities

Length: mmass: kgtime: s

temperature: Kamount of substance: mol

electric current: Aluminous intensity: cd

Derived Units

1000 mm = 1 m

1 L = 1 dm3

Concentration: mol L – 1

Density: g cm – 3

speed: km/hr

Always use units to specify quantities!

Stoichiometry 9

Unit Conversions

1 m1 km

---------------1000000 mm

1000 mm ------------

1 m= 0.001 km

What’s the mass of 250 mL ethanol, whose density = 0.789 g / mL?

0.789 g ------------

1 mL= ____ g ethanol 250 mL ethanol

Calculate volume of 250 g ethanol

Stoichiometry 10

Unit Conversions – cont.

Convert 50 miles per hour to meters per second (1 mile = 1.609 km)

50 mile---------

1 hr

1.609 km ------------

1 mile

1000 m ------------

1 km

1 hr ------------60 min

1 min ------------

60 s

= 22 m s – 1

Pay some attention to significant figures, precision, and accuracy.

1 hr--------------

3600 s

1609 km--------------

1 mile

Stoichiometry 11

Atoms and Atomic Theory (1808)

1. Each element is made up of tiny individual particles called atoms.

2. Atoms are indivisible; they cannot be created or destroyed.

3. All atoms of each element are identical in every respect.

4. Atoms of one element are different from atoms of any other element.

5. Atoms of one element may combine with atoms of another element, usually in the ratio of small whole numbers, to form chemical compounds.

Picture atoms in your mind!

Stoichiometry 12

Rutherford AtomsFollowing the discovery of electrons and radioactivity Rutherford concluded that atoms consists of heavy dense nuclei whose diameters are 10 – 5 times that of the atom from his alpha scattering experiment.

Following the discovery of protons and neutrons, we know that atomic nuclei consists of neutrons and protons.

Explain the following: you need to understand them to solve problems

chemical elementsatomic structureisotopesatomic mass, atomic weight, atomic radiusmolar mass, molecular weight

Stoichiometry 13

Natural Potassium

Isotope 39K19 40K19 41K19

atomic mass38.9637069 39.9639987 40.9618260 abundance 93.2581% 0.0117% 6.7302%

stable Half-life: (radioactive) stable1.277e9 years

Beta Decay Energy 1311.093 keV

Evaluate the atomic weight from the given data

Stoichiometry 14

The Mole (mol)

How many 40K (0.012% of natural K) atoms are in 500 mg of KCl?

Avogadro’s proposed the concept of molecules

NA = 6.022e23 (x1023)

0.5 g KCl0.00012 mol 40K --------------------

1 mol K

1 mol K ---------------

(39.1+35.5) g

6.022e23 atoms--------------

1 mol

= 4.8e17 40K atoms

Stoichiometry 15

Chemical Formulas

Molecular formula: H2O H2O2 C6H12

Empirical formula: - HO CH2

Structural formula: H-O-H H-O-O-H

Structure

Explain these terms yourself

Stoichiometry 16

Determine Chemical FormulasThe experiment to determine chemical composition (formula)

CxH2y =(burned in O2)=> x CO2 + y H2O

After the combustion, CO2 and H2O are absorbed by some chemicals and their mass determined. From their masses, relative compositions of H and C are determined.

Other method is used to determine N and O.

When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of CO2 and 1.29 g of H2O is produced. What is the empirical formula?

Stoichiometry 17

When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of CO2 and 1.29 g of H2O is produced. What is the empirical formula?

Solution:

12 g C 1 mol C 3.14 g CO2 ------------- ------------- = 0.0714 mol C

44 g CO2 12 g C

2 g H 1 mol H 1.29 g H2O ------------- ------------- = 0.143 mol H

18 g H2O 1 g H

Thus, mole ratio of C : H is 0.0714 : 0.143 = 1 : 2. Therefore, the empirical formula is CH2

What quantities do you need in order to find the molecular formula?

What is the weight % of C and H in this compound?

Stoichiometry 18

Oxidation States1. The oxidation state of any element such as Fe, H2, O2, P4, S8 is zero (0).

2. The oxidation state of oxygen in its compounds is -2, except for peroxides like H2O2, and Na2O2, in which the oxidation state for O is -1.

3. The oxidation state of hydrogen is +1 in its compounds, except for metal hydrides, such as NaH, LiH, etc., in which the oxidation state for H is -1.

4. The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion.

5. The following elements usually have the same oxidation states in their compounds:

+1 for alkali metals - Li, Na, K, Rb, Cs

+2 for alkaline earth metals - Be, Mg, Ca, Sr, Ba

-1 for halogens except when they form compounds with oxygen or one another

Stoichiometry 19

Oxidation states – cont.

What are the oxidation states of N in these compounds?

NH3 N2H4 NH2OH N2 N2O NO NO2 – NO2 NO3

–

Determine oxidation states of all elements of any formula you see anywhere.

What are the oxidation states of Cl in these compounds?

Cl – Cl2 ClO – ClO2 – ClO2 ClO2 – ClO3

– ClO4

–

Stoichiometry 20

Chemical NamesChemicals are named in a systematic way according to certain rules.

Knowing the rules allows you to name a substance, and lets you understand the substance from the name.

Thus, it’s a good idea to review, or learn, the rules for naming organic and inorganic compounds.

Its not fun to teach these rules, but you may have fun reading them to find out how much you already know.

Stoichiometry 21

Chemical & Physical Reactions

Chemical reactions:

4 NH3 + 3 O2 2 N2 + 6 H2O

Balance: (atoms cannot be destroyed nor created) __ Al + __ O2 __ Al2O3

__ C6H14O4 + __ O2 __ CO2 + __ H2O__ H3PO4 + __ CaO __ Ca3(PO4)2 + __ H2O

Physical reactions: (s = solid; l = liquid; g = gas)

H2O(s) H2O(l) (energy is involved in

H2O(s) H2O(g) phase transitions)

Stoichiometric coefficient

Study states of matter & phase transition

Stoichiometry 22

Reaction StoichiometryStoichiometry: quantitative relationship among reactants and products in a balanced reaction equation. Quantities may be in mole, mass, weight, or volume.

How much KCl and O2 are produced by decomposing 123.0 g of KClO3?

2 KClO3(s) 2 KCl (s) + 3 O2 (g)

245.2 g 149.2 g 96 g123.0 x y

use proportionality to get x = 74.8 g KCl; y = 48.2 g O2

Stoichiometry 23

Stoichiometry Calculations

123.0 g KClO3

How much KCl and O2 are produced by decomposing 123.0 g of KClO3?

2 KClO3(s) 2 KCl (s) + 3 O2 (g)

245.2 g 149.2 g 96 g123.0 x y

use proportionality to get x = 74.8 g KCl; y = 48.2 g O2

96.0 g O2

245.2 g KClO3

1 mol O2

32.0 g O2

22.4 L O2

1 mol O2

Versatile , getting the amount in various units.

Find amount of KCl in g, mol, and volume (specific gravity = 1.984)

123.0 g KClO3

Stoichiometry 24

Solutions and Concentrations

Concentrations are expressed in many ways, g / 100 mL, g / L, mol / L (= M), mole fraction, weight fraction, percentage etc.

Molarity is the expression of concentration in mole per liter.

Explain solvent, solute, solution, and mixture

For quantities in reactions:

Amount = concentration * volume

C1 V1 = C2 V2

Stoichiometry 25

Excess and Limiting Reagent

The reactant that is completely reacted is the limiting reagent or reactant. Reactants left behind due to a limiting reagent are excess reagents or reactants.

How much AgNO3 should be added in order to completely precipitate all the chloride ions in 25.00 mL of 0.1000 M NaCl solution?

25.00 mL 0.1000 mol NaCl1000 mL

1 mol AgCl1 mol NaCl

169.9 g AgNO3

1 mol AgCl

= 0.4248 g AgNO3

Practice this method

Stoichiometry 26

How much AgNO3 should be added in order to completely precipitate all the chloride ions in 25.00 mL of 0.1000 M NaCl solution?

Excess and Limiting Reagent - cont

If less than 0.4248 g AgNO3 is used, AgNO3, what is the limiting reagent.?

What if more than 0.4248 g AgNO3 is used?

If 0.3000 g AgNO3 is used, how much AgCl is formed?

= 0.4248 g AgNO3

0.3000 g AgNO3143.4 g AgCl

169.9 g AgNO3= 0.2532 g AgCl

Explain excess and limiting reagent

Stoichiometry 27

Theoretical, actual, and percent yields

= 0.2532 g AgCl0.3000 g AgNO3

From the previous example,

+ 25.00 mL 0.1000 M NaCl This amount is the

theoretical yield, what you get, usually less, is the actual yield.

Percent yield =Actual yield * 100 %

Theoretical yield

Explain

Stoichiometry 28

Percent in a Mixture

When 2.0 g of NaCl and NaNO3 mixture is dissolved in water, sufficient amount of AgNO3 is added. The amount of dry AgCl is1.0 g. What is the percentage of NaCl in the mixture?

Chemistry: Na+, Cl–, and NO3

– ions are present in solution; so do Ag+ and NO3–

Ag+ + Cl– AgCl (s, a white ppt)

1.0 g AgCl 58.5 g NaCl143.5 g AgCl

100 % NaCl2.0 g sample = 20.4 % NaCl

1.0 g AgCl

Is it possible to get more than 5.0 g AgCl in this experiment? Why or why not?