1. Unit 1: Stoichiometry and Reactions Sarah & Sara

2. Nomenclature  Cations—electron-deficient, (+) charge, charge is group #  transition metals-charge shown with Roman Numeral in name  exceptions: Zn 2+ Ni 2+ Ag +  the less charged cation of each atom, such as Copper uses the –ous ending, while the one with greater charge uses the –ic ending  Copper (II) ion can also be called cuprous

3. Nomenclature  Anions—excess electrons, (-) charge, charge is # of columns from noble gases (**exceptions),  monatomic –ide suffix  polyatomic oxyanions –ate suffix most common (-ite with fewer O)  oxyanions modified by H +  **cyanide = CN - hydroxide = OH -

4. Nomenclature  Neutral ionic compound = a salt = metal cation + nonmetal anion  [empirical formula with net charge of zero]  to name: modify cation and/or anion with subscripts [name the cation then the anion, the subscript is not stated, write transition metals with Roman Numerals for charge]

5. Nomenclature  Acids—have H + cation, name is based on anion -ide = hydro_______ic acid -ate = ____________ic acid -ite = ____________ous acid EX: Chloride= hydrocholric acid Chlorate= chloric acid Chlorate= chloric acid Chlorite= chlorous acid Chlorite= chlorous acid

6. Nomenclature  Covalent (or binary) compounds = 2 nonmetals (name farthest left first with subscripts stated in name, exceptions—Fluorine is always last, Oxygen is last unless with F)  Organic compounds = hydrocarbons, only C and H; others can include O, N, and/or S  alkanes – C “backbone,” all single bonds, as many H as necessary to fill bonds, end with –ane  Name with # of C: 1=meth 2=eth 3=prop 4=but 5+ = binary prefixes  alcohols – O-H “functional group” in place of 1 or more H [-ane becomes -anol]  Number position in functional group in front (or to which C is it bonded)

7. Nomenclature  The –ates:  Sulfate (SO 4 - )  Chlorate (ClO 3 - )  Phosphate (PO 4 3- )  Carbonate (CO 3 2- ) Nitrate (NO 3 - )  Acetate (C 2 H 3 O 2 - = CH 3 COO - )  Chromate (CrO 4 2- )  Dichromate (Cr 2 O 7 2- )  Permanganate (MnO 4 - )  Hydride = H - Hydrogen ion = H + ammonium ion = NH 4 +

8. Atomic Structure  angstrom = 10 -10 m = Å = size of atom in meters  Weight of 1 electron = (1/1800) proton  size of the electron cloud is the full atomic size and is about (1/4000) of an atom’s mass  1 x 10 -10 to 5 x 10 -10 m = 1 to 5 Å =.1 to.5 nm = 100 to 500 pm  nucleus contains 99.9% mass of an atom;  1 x 10-15 to 1 x 10-14 m =.00001 to.0001 Å

9. Isotope Notation  A = mass # = protons + neutrons  Z = atomic # = protons  q = protons – electrons X A Z q

10. Reactions and Other Stuff  Decomposition reaction: Y  C + D  Synthesis/composition: A + B  X  Combustion: C x H y O z  CO 2 + H 2 O  Avogadro’s Number = NA = 6.02 x 1023  Spectrometer  average mass – weighted average  Diatomic elements = H, O, N, Cl, Br, I, F

11. Stoichiometry  Stoichiometric Relationships:  grams of A  moles of A  moles of B  grams of B  Limiting reactants: practical solutions (amounts of reactants are mismatched)  all calculations are based on limiting reactant  limiting reactant is completely consumed  Method A: use one reactant to find how much of other reactant is needed  fewer calculations, most efficient if only finding limiting reactant  Method B: use both to find out how much product is made  more calculations, fool proof strategic planning

12. Yield!  Theoretical Yield = grams of limiting reactant  moles of limiting reactant  moles of product  grams of product  % Yield=(actual or experimental) / theoretical

13. Example Problem 1  C 6 H 6 + Br 2  C 6 H 5 Br +H 2  You have 30.0g of C 6 H 6 and 65.0g of Br 2 present for a reaction.  A) Find the limiting reactant, and state how much C 6 H 5 Br will be produced?  B) Calculate the percent yield if 56.7g of C 6 H 5 Br are produced by the reaction.

14. Solution to Example 1  A) First calculate the moles of each reactant you have present at the beginning of the reaction.  30.0g C 6 H 6 x (1 mol C 6 H 6 / 78.11g C 6 H 6 ) =.384 mol C 6 H 6  65.0g Br 2 x (1 mol Br 2 / 159.8g Br 2 ) =.407 mol Br 2  Because C 6 H 6 and Br 2 are in a 1:1 ratio, C 6 H 6 is the limiting reactant and determines the theoretical yield. .384 mol C 6 H 6 x (1 mol C 6 H 5 Br / 1 mol C 6 H 6 ) x (157.0g C 6 H 5 Br / 1 mol C 6 H 5 Br) = 60.3g C 6 H 5 Br

15. Solution fo Example 1  B) % yield = actual or experimental/ theoretical  % yield = (56.7g C 6 H 5 Br actual / 60.3g C 6 H 5 Br theoretical) x 100 = 94.0%

16. Example Problem 2  Calculate the percent composition of Carbon in each of the following molecules. A) C 7 H 6 O B )C 8 H 8 O 3 C) C 7 H 14 O 2

17. Solution to Example 2  A) C 7 H 6 O FW: 7(12.0 amu) + 6(1.0 amu) + 1(16.0 amu) = 106.0 amu %C = ( 7(12.0 amu) / 106.0 amu ) x 100 = 79.2%

18. Solution to Example 2  B )C 8 H 8 O 3 FW: 8(12.0 amu) + 8(1.0 amu) + 3(16.0 amu) = 152.0 amu %C = ( 8(12.0 amu) / 152.0 amu ) x 100 = 63.2%

19. Solution to Example 2  C) C 7 H 14 O 2 FW: 7(12.0 amu) + 14(1.0 amu) + 2(16.0 amu) = 130.0 amu %C = ( 7(12.0 amu) / 130.0 amu ) x 100 = 64.6%

20. Example Problem 3  Caffeine is readily available in common foods and drinks. By mass, caffeine contains 49.48% C, 5.15% H, 16.49% O, and 28.87% N. Its molar mass is 194.2g/mol.  A) Determine the empirical formula for caffeine.  B) Determine the molecular formula for caffeine.

21. Solution to Example 3  A) Assume the sample is 100g  49.48g C x ( 1 mol C/ 12.011 g C) = 4.1195 mol C  5.15g H x (1 mol H/ 1.00794g H) = 5.1096 mol H  16.49g O x (1 mol O/ 15.9994g O) = 1.03066 mol O  28.87g N x (1 mol N/14.00674g N) = 2.06115 mol N  DIVIDE ALL BY THE LOWEST MOLE AMOUNT (1.03066 mole)  Gives us: ~4 mol C, 5 mol H, 1 mol O, and 2 mol N  Therefore, our empirical formula is C 4 H 5 ON 2

22. Solution fo Example 3  B) molecular formula  4(12.011g) + 5(1.00794g) + 15.9994g + 2(14.00674g)= 97.09658 g/ mol empirical  194.2 g/mol / 97.09658 g/mol = 2  Therefore, the empirical equation needs to be double to get the proper molecular formula of caffeine, which is C 8 H 10 O 2 N 4.

23. ZE END!