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)
Nomenclature
 The –ates:
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Sulfate (SO4-)
Chlorate (ClO3-)
Phosphate (PO43-)
Carbonate (CO32-)
Nitrate (NO3-)
 Acetate (C2H3O2- =
CH3COO-)
 Chromate (CrO42-)
 Dichromate (Cr2O72-)
 Permanganate (MnO4-)
 Hydride = HHydrogen ion = H+
ammonium ion = NH4+
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
A
Z
X
q
 A = mass # = protons + neutrons
 Z = atomic # = protons
 q = protons – electrons
Reactions and Other Stuff
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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
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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!