Unit 1: Stoichiometry and Reactions

Unit 1: Stoichiometry and Reactions Sarah & Sara

Nomenclature • Cations—electron-deficient, (+) charge, charge is group # • transition metals-charge shown with Roman Numeral in name • exceptions: Zn2+ Ni2+ 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

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-

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]

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 Chlorite= chlorous acid

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)

The –ates: Sulfate (SO4-) Chlorate (ClO3-) Phosphate (PO43-) Carbonate (CO32-) Nitrate (NO3-) Acetate (C2H3O2- = CH3COO-) Chromate (CrO42-) Dichromate (Cr2O72-) Permanganate (MnO4-) Hydride = H- Hydrogen ion = H+ ammonium ion = NH4+ Nomenclature

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

Isotope Notation X A q • A = mass # = protons + neutrons • Z = atomic # = protons • q = protons – electrons Z

Reactions and Other Stuff • Decomposition reaction: Y  C + D • Synthesis/composition: A + B  X • Combustion: CxHyOz CO2 + H2O • Avogadro’s Number = NA = 6.02 x 1023 • Spectrometer  average mass – weighted average • Diatomic elements = H, O, N, Cl, Br, I, F

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

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

Example Problem 1 • C6H6 + Br2 C6H5Br +H2 • You have 30.0g of C6H6 and 65.0g of Br2 present for a reaction. • A) Find the limiting reactant, and state how much C6H5Br will be produced? • B) Calculate the percent yield if 56.7g of C6H5Br are produced by the reaction.

Solution to Example 1 • A) First calculate the moles of each reactant you have present at the beginning of the reaction. • 30.0g C6H6 x (1 mol C6H6 / 78.11g C6H6) = .384 mol C6H6 • 65.0g Br2 x (1 mol Br2 / 159.8g Br2) = .407 mol Br2 • Because C6H6 and Br2 are in a 1:1 ratio, C6H6 is the limiting reactant and determines the theoretical yield. • .384 mol C6H6 x (1 mol C6H5Br / 1 mol C6H6) x (157.0g C6H5Br / 1 mol C6H5Br) = 60.3g C6H5Br

Solution fo Example 1 • B) % yield = actual or experimental/ theoretical • % yield = (56.7g C6H5Br actual / 60.3g C6H5Br theoretical) x 100 = 94.0%

Example Problem 2 • Calculate the percent composition of Carbon in each of the following molecules. A) C7H6O B )C8H8O3 C) C7H14O2

Solution to Example 2 • A) C7H6O 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%

Solution to Example 2 • B )C8H8O3 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%

Solution to Example 2 • C) C7H14O2 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%

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.

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 C4H5ON2

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 C8H10O2N4.

ZE END!

Unit 1: Stoichiometry and Reactions