Heat Capacity (C) in J/C
C=q
∆ T
Specific Heat (c) in J/gC
c=q
m ∆ T
Percent Abundance:
1. [(xmass element1) + ((1-x)mass element2)]
solve for x
2. Element1 percent abundance = (x 100) (e.g. 0.4 x 100 =
4%)
3. Element2 percent abundance = [(1 – x) 100]
Number of moles:
¿mol=mass of element
molar mass
# mol from concentration
and volume:
(
mol
L
)
∙ L given ∙
(
mol desired ¿eq¿ ¿mol given ¿eq ¿
)
Enthalpy (H) example:
100 mL of 1.00 M Ba(NO3)2, mixed with 100 mL of 1.00 M Na2SO4, both with initial
temperature of 25C and final temperature of 28.1C. Heat capacity of 4.18 J/gC
and density of 1.0 g/mL. Change in enthalpy for formation of 1 mole of sodium
nitrate?
Ba(N O3)2+Na2SO4→ BaS O 4+2NaN O3
1. (100 mL + 100 mL) = 200 mL
2.
m=200 mL∙
(
1.0 g
mL
)
=200 g ∙ 10−3=¿0.2 g
3.
qsolution=
(
0.2 g
)
∙
(
4.18 J
g℃
)
∙
(
28.1 ℃−25 ℃
)
=¿2.59
4.
qsoution=−qreaction , e . g . q reaction=−2.59 kJ
5.
∆ H =qreaction
ntotal
=¿−2.59 kJ
0.2 mol ≈−13 kJ /mol
Change in Heat example:
25 g glucose, 40 g oxygen, molar mass glucose: 180.16 g/mol
C6H12 O6+6O2→6CO2+6H2O ∆ H =−2480 kJ
1.
25 g C6H12 O6∙
(
1mol C6H12 O6
180.16 g
)
∙
(
6mol C O2
1mol C6H12 O6
)
=0.83 mol C O2←limit
2.
40 g O2∙
(
1mol O2
32 g
)
∙
(
6mol C O2
6mol O2
)
=1.25 mol C O2
3.
q=∆ H ∙ n C6H12O6=¿
(
−2480 kJ
)
∙
(
0.139 mol C6H12 O6
)
=−344 kJ
Number of atoms/molecules:
¿molecules=
(
mass of element
molar mass
)
(6.022 x1023 )
Volume of product with pressure example
300.15 K, 0.951 atm, 8.88 g Ga, V of H?
2Ga
(
s
)
+6HCl
(
aq
)
→2Ga Cl3
(
aq
)
+3H2(g)
8.88 g Ga ∙
(
1mol Ga
69.72 g Ga
)
=0.127 mol Ga
0.127 mol Ga ∙
(
3mol H
2mol Ga
)
=0.19 mol H
V=nRT
P
V=
(0.19 mol H )∙
(
0.08206 atmL
molK
)
∙(300.15 K)
0.951 atm ≈4.9 L
Steric Number (central atom only):
# double bonds + # single bonds + # lone pairs
Example: SF4 = S has 4 single bonds + 1 lone pair = 5 steric number
~ trigonal bipyramidal and seesaw ~
ENAmount Bond
0.4
<
0 Pure
covalent
0.4-
1.8
intermediat
e
Polar
covalent
1.8
>
large ionic
Sigma and Pi Bonds
Single bond: 1 sigma bond, 0 pi
bonds
Double bond: 1 sigma bond, 1 pi
bond
Triple bond: 1 sigma bond, 2 pi
bonds
Percent composition:
%element=
(
mass of element ∈grams
mass of sample/compound ∈grams
)
100
Atomic number (Z): # of protons
Mass number (A): # of neutrons +
protons
Number of Neutrons: A – Z
Atomic charge:
(# of protons) – (# of electrons)
COMMON POLYATOMIC COMPOUNDS
Compound Name Compound Formula
Acetate C2H3O2-
Carbonate CO32-
Hydrogen carbonate HCO3-
Hydroxide OH-
Nitrite NO2-
Nitrate NO3-
Chromate CrO42-
Dichromate Cr2O72-
Phosphate PO43-
Hydrogen phosphate HPO42-
Dihydrogen phosphate H2PO4-
Ammonium NH4+
Hypochlorite ClO-
Chlorite ClO2-
Chlorate ClO3-
Perchlorate ClO4-
Permanganate MnO4-
Sulfite SO32-
Hydrogen sulfite HSO3
Sulfate SO42-
Hydrogen sulfate HSO4-
Cyanide CN-
Peroxide O22-
Theoretical Yield
mol desired=mol given ∙
(
mol desired ¿eq¿ ¿ mol given ¿eq ¿
)
** the smallest moles between the reactants to produce specified product will be the limiting
reactant and the highest moles is the excess reactant ** ~ multiply by the desired compounds
molar mass to convert to grams prn
Percent Yield
Acid+Base → Salt +H2O
ex: HCl + NaOH
Water + NaCl
1. mass start
(
g
)
=theoretical yield ∙
(
1mol
molar mass
(
g
)
)
∙
(
mol start with
mol produced ∈reaction
)
∙
(
molar mass(g)
)
2. % yield =
(
mass start with
(
g
)
theoretical yield
(
g
)
)
∙100
Excess Reactant
1. theoretical yield ∙
(
mol excess reactant ¿eq¿¿ mol product ¿eq¿
)
=mol reactant used
2. mol reactant start −mol reactant used=mol excess
g/L prior to precipitate
g
(
precip
)
∙
(
g precip
molar mass precip
)
∙
(
mol o . g .
mol precip
)
∙¿
Ionic Compound:
Metal + Non-metal
(2 or more elements: -ide)
(3 or more elements: -ate)
Molecular Compound:
Non-metals
DIATOMIC MOLECULES
H2 O2 N2 F2 Cl2 Br2 I2
POLYATOMIC MOLECULES:
P4 Se8 S8
Steric
#
# bonding
pairs
# lone
pairs
Electron
geometry
Molecular
geometry
Hybridization
2 2 0 Linear linear sp
3 3 0 Trigonal
planar
Trigonal
planar
sp2
3 2 1 Trigonal
planar
Bent sp2
4 4 0 Tetrahedral Tetrahedral sp3
4 3 1 Tetrahedral Trigonal
pyramidal
sp3
4 2 2 Tetrahedral Bent sp3
5 5 0 Trigonal
bipyramidal
Trigonal
bipyramidal
sp3d
5 4 1 Trigonal
bipyramidal
seesaw sp3d
5 3 2 Trigonal
bipyramidal
t-shaped sp3d
5 2 3 Trigonal
bipyramidal
linear sp3d
6 6 0 Octahedral Octahedral sp3d2
6 5 1 Octahedral Square
pyramidal
sp3d2
6 4 2 Octahedral Square
planar
sp3d2
Compound/ion
Type
Compound/Ion Exceptions
Soluble Ionic
Compounds
N H 4
+¿¿
None
Soluble Cations
Li+¿, Na+¿, K+¿, Rb+¿, Cs+¿¿¿¿¿¿
None
Soluble Anions
Cl−¿, Br−¿,I −¿¿ ¿¿
Ag+¿, Hg2
2+¿,Pb2+¿ ¿ ¿¿
Soluble Anions
F−¿¿
Group 2 metals and
Pb2+¿, Fe3+¿ ¿ ¿
Soluble Ionic
Compounds
C2H3O2
−¿, HC O3
−¿,N O 3
−¿,ClO3
−¿¿¿¿¿
None
Insoluble Exceptions
C O3
2−¿,Cr O 4
2−¿,P O4
3−¿,S2−¿¿ ¿¿¿
Group 1 cations and
NH 4
+¿¿
OH−¿ ¿
Group 1 cations and
Ba2+¿ ¿
Heat (q) in J or kJ Change in Enthalpy (
∆
H)
q=mc ∆ T
qsolution=−qreaction
+q = endothermic reaction
(heat absorbed)
-q = exothermic reaction
(heat produced)
C12 H22 O11+8KCl O3→12 C O 2+11H2O+8KCl
1. Convert to moles to find limiting reactant
2. Using
∆ H =q
n
find the enthalpy change
3. Multiply enthalpy by the coefficient in front of the limiting reactant
(e.g.
8∙ ∆ H =∆ H
for one mole of sucrose to react with
8 moles of potassium chlorate)
Internal Energy (u) (e.g. function)
∆ u=q+w
−∆ ureaction=∆ usurroundings
“Two-fold increase” in enthalpy:
Original:
H2
(
g
)
+
(
1
2
)
O2
(
g
)
→ H 2O
(
l
)
∆ H =−286 kJ
Two-fold increase: (multiply eq and
∆ H
by 2)
2H2
(
g
)
+O2
(
g
)
→2H2O
(
l
)
∆ H =
(
−286 kJ ∙2
)
=−572 kJ
Enthalpy (H): (e.g. function)
H=u+PV
∆ H =q
n
+
∆ H
= endothermic reaction
-
∆ H
= exothermic
reaction
“Two-fold decrease” in enthalpy:
Original:
H2
(
g
)
+
(
1
2
)
O2
(
g
)
→ H 2O
(
l
)
∆ H =−286 kJ
Two-fold decrease: (multiply eq and
∆ H
by 1/2)
1
2H2
(
g
)
+1
4O
2
(
g
)
→1
2H2O
(
l
)
∆ H =
(
−286
2kJ
)
=−143 kJ
Internal Energy (u) with
constant pressure/volume:
∆ H =∆ u
w=−P ∆ V
e.g.
bomb calorimeter
Standard Enthalpy of Combustion
Volume
(
L
)
∙
(
103mL
1L
)
∙
(
density
(
g
mL
)
)
∙
(
1mole
molar mass
(
g
)
)
=n
∆ HC∙ n=−q
∆ HC:enthalpy of combustion
Enthalpy Principles: Standard Enthalpy of Formation example
Breaking chemical bond = endothermic (e.g. adding O gas to solid
carbon)
Forming chemical bond = exothermic (e.g. formation of CO2)
Enthalpy from Bond Energy
∆ H =
[
bonds broken−formed
]
∆ H =[
(
C ≡ O+2H−H
)
−(3C−H+C−O+O−H)]
Reaction Types Basic Formulas
Combustion
A+B → CO2+H2O
Decomposition
AB → A +B
Synthesis/combination
A+B → AB
Single replacement
A+BC → AC +B
Double replacement
AB+CD → AD +BC
