Transcript Chemistry 20 Final Review

Chemistry 20 Final Review
Bonding Unit
Gases Unit
Solutions, Acids and Bases Unit
Stoichiometry Unit
Bonding Unit Topics:
Lewis Structures:
These are your electron diagrams for
individual elements (showing valence e-)
Ex.
Mg
S
Intramolecular Forces:
Remember these are the forces WITHIN
a molecule
What forces hold a molecule together?
ANS: BONDS
What type of bonds are there?
ANS: COVALENT AND IONIC
Ionic Bond Structures
Remember how to draw electron dot
diagrams for ionic compounds
They DO NOT SHARE electrons – the
metal looses its outer shell electrons
and the non-metal gains to a full 8 eEx:
Covalent Bond Structures
These are the bonds holding
MOLECULAR compounds together
They DO SHARE the electrons
These were also called Lewis Structures
The element that goes in the middle is
the one with the most BONDING eExamples:
eg) PH3
H
H
H
P
H
H P
H
Remember double and triple
bonds:
Each element except hydrogen needs 8
electrons around it and there should be
NO LONE PAIRS
This is when double and triple bonds
form
Ex.
O O
N
N
Structural Diagrams and Shape
Diagrams
When there are 2 shared electrons
between two elements in a molecule
draw a line to show this bond
Ex.
H
H
C
C
H
H
Remember the shapes and shape codes
Ex. tetrahedral, trigonal planar, pyramidal, linear,
bent
the code has two numbers:
1. the number of atoms attached to the central atom
2. the number of lone pairs on the central atom
CH4
eg) NH3(g)
 
H
N
H
H
H
3-1
pyramidal
C
H
H
H
4-0
tetrahedral
Code
4–0
Shape
tetrahedral
3 – 0 trigonal planar
Example
CH4
CH2O
3–1
pyramidal
NH3
2–1
bent
HNO
2–2
bent
H2O
***all other codes are linear
Molecules take on these shapes due to the VSEPR
theory - valence shell electron pair repulsion
molecules adjust their shapes so that valence eare as far away from each other as possible
Polar vs. Nonpolar
Remember electronegativities:
The number in each element box above
the element
It shows how badly an elements wants eThe higher the number, the stronger it
pulls
When two elements are bonded together
and there is a difference in electroneg.
then you have a polar bond
• Ex. H – F (see next slide)
Bond Dipole Arrows
“+” at the end that is +
+
H–F
“arrow” points
towards element with
higher
electronegativity (-)
-
you can use the difference in electronegativity
between two atoms to determine the bond
Difference in Electronegativity
3.3
1.7
mostly ionic
0.5 0
polar
covalent
slightly
polar
covalent
non-polar
covalent
Polar vs. Nonpolar Molecules
 tetrahedral: nonpolar if all atoms attached
have the same pull (in or out), polar if
different atoms attached
 trigonal planar: nonpolar if all atoms
attached have the same pull (in or out), polar
if different atoms attached
 pyramidal: polar as long as there is a
difference in electronegativity between the
atoms
 bent: polar
 linear: polar or nonpolar …look at
electronegativity difference
Examples
1. H2O
2. HCl
O
H
H
H
polar
polar
3. C2H2
4. C2HI
np
np
H
C
Cl
C
nonpolar
H
H
C
C
polar
I
Intermolecular Forces
These are the forces that cause attraction
BETWEEN molecules
They are weaker then bonding within a
molecule
They are responsible for the bp and mp of
compounds since when you boil/ melt a
molecule you are ONLY breaking these forces
BETWEEN molecules
The three intermolecular forces we talked
about the occur between MOLECULAR
compounds
HB, DD, LD
DD - Dipole - Dipole
These attractions occur in POLAR molecular
compouds ONLY
The slightly positive end of one molecule is
attracted to the slightly negative end of
another molecule
+

-


+
+

+


-


-
LD: London Dispersion Forces
These attractive force occurs between ALL
molecular compounds
It is caused by electrons in atoms and
molecules constantly being in motion
So sometimes one side of a molecule can have
more electron then the other side
This creates a temporary polar molecule
An attraction then forms between the ends of
these polar molecules
Remember you have stronger LD forces as the
molecule becomes larger or has more
electrons
HB: Hydrogen Bonding
These attractive forces occur in molecular
compounds that H bonded to either N, O or F
Draw the structural diagram of the molecular
compound to make sure the H is actually
bonded to the N, O or F


the hydrogen has such a low electroneg. in
comparison to N, O and F so it has its electrons
pulled so far away from it. This makes it able to be
attracted not only to the  pole but also to the lone pairs


O
H
H
Other melting/ boiling point’s
Remember intermolecular forces only occur
between molecular compounds and are
weaker forces then intramolecular forces
(bonds)
So when melting molecular compounds only
the LD, DD, and HB need to be overcome
Metals and ionic compounds are attracted to
one another by the bonds holding them
together
Metallic structures and ionic compounds
therefore have high bp/ mp due to having to
overcome their intramolecular forces
(bonding)
MP of Metals?
Metals are solid at room temp. because metal
atoms have very strong forces between them
I.e. metallic bonding
So in order to melt them you need to add
LOTS of energy (high temp) to overcome
these strong forces
Metallic Bond Model
metal
cations
“sea” of
delocalized
electrons
Ionic Compounds
Ionic compounds are also attracted to one
another by strong forces (not quite a strong
as metallic though)
I.e. ionic crystals
So in order to melt them you need to add
quite a bit of energy (high temp) to overcome
these forces
Ionic Crystals
ionic compounds have crystal structure
they form so that oppositely charged ions are
as close together as possible
this 3-D array of alternating positive and
negative ions is called a crystal lattice
Scale of Forces
very
high
very
low
LD
DD
HB
covalent
Intermolecular Forces
(between)
London Dispersion
Dipole – Dipole
Hydrogen Bonding
ionic
network
covalent
Intramolecular Forces
(within)
metallic ** wide range
ionic
covalent
network covalent
eg) diamond, SiC, SiO2
Order of bp’s
Using the scale of forces you can order
compounds based on their relative bp’s
Ex. From Highest to Lowest
Network covalent compound (ex. SiO2)
Ionic compound
Molecular compound with HB, DD, LD
Molecular compound with DD, LD
Molecular compound with LD (if 2 molecular
compounds have LD only then bigger molecule or
molecule with more electrons has higher bp)
Gases Unit
Remember your formulas!!!!!
When to use what?
Use Boyle’s Law when temp. is constant
P1V1 = P2V2
Use Charles’ Law when pressure is constant
V1/T1 = V2/T2
Use Combined Gas Law when all three
variables change
P1V1/T1 = P2V2/T2
Use PV=nRT if given a mass or number of
moles
Some others
If you are only given info about pressure and
tempurature and its in a sealed container
then V1 = V2 so using Combined Gas Law
cancel out the volumes to get left with:
P1V1/T1 = P2V2/T2
P1/T1 = P2/V2
Law of Combining Volumes
You can use a balanced equation and multiply by
coefficients wanted/given to get the volume of
one gas if you know the volume of another
What volume of oxygen is used up if 100 mL of steam is
formed in a composition reaction?
What
What
O2(g) + 2H2(g)  2H2O(g)  you
are 
are
solving x mL
100 mL
given
for
100 mL x
1
2
x mL = 50.0 mL
Convert 650 mmHg to kPa.
Ratio
of
known
values
101.325 kPa
760 mmHg
=
x
650 mmHg
x = 86.6… kPa
Ratio
of
what
you
are
trying
to find
Solutions, Acids and Bases
Remember your formula’s here
too:
c=n/v
n=m/M
Remember this is the dilution formula
V1C1=VfCf
pH=-log[H30+]
pOH=-log[OH-]
[H30+]=10^-pH
[OH-]=10^-pOH
pH+pOH=14
Experiments
Remember in experiments there are
always three variables:
Manipulated
Responding
Controlled
what you are changing
the response to the change
what you keep the same
Ex: What effect does eating
carrots have on eyesight?
Manipulated: amount of carrots eaten
Responding: how well you can see
Controlled: Same types of carrots, not
eating any other food that could effect
eyesight
Electrolytes?
Compounds that conduct electricity in
water because they break apart into
ions (ex. ionic compounds, acids)
Ex. NaCl  Na+(aq) + Cl-(aq)
Molecular compounds DO NOT break
down into ions so they are nonelectrolytes
Solubility
The ability to dissolve
If the solution is holding as many solutes as
possible the solution is SATURATED and
adding anymore solute will NOT be able to
dissolve
A saturated solution usually has a small
amount of UNDISSOLVED solute at the
bottom. This is in constant EQUILIBRIUM
with the solute that is dissolved in the
solution (they switch places with each other
all the time)
Dissolved
Undissolved
Equilibrium
Standard Solution
Remember how to prepare a standard
solution
Use formula’s n=m/M and c=n/v to get the
mass you need for the certain volume and
concentration
Steps:
Weigh out solute
Dissolve in ~half amount of water in a beaker
Pour into final volume volumetric flask
Fill flask, and invert to mix
Dilution
When you have a solution that has too high
of a concentration you can add water to
dilute it (water it down so its not as strong)
Ex. You have 100mL of a 5.0 mol/L solution. You
add 500mL of water. What is the new
concentration?
Ci = 5.0 mol/L
Vi = 0.1L
Vf = 0.6L
Cf = ?
ViCi=VfCf
Solve for Cf
Dissociation and dissociation equations
This is the same as ‘dissolving’
When you have a compound and put it in water 4
situation to know:
It doesn’t dissolve
• Ex. C25H52(s)  C25H52(s)
It does dissolve and its ionic
• Ex. Ca(OH)2(s)  Ca2+(aq) + 2OH-(aq)
It does dissolve and its molecular
• Ex. C12H22O11(s)  C12H22O11(aq)
It does dissolve and its an acid (this is a special case
because it is a molecular compound but it acts as an ionic
compound)
• Ex. H2SO4(aq)  2H+(aq) + SO42-(aq)
Concentration of Ions
If asked to calculate the concentration of an ion in a solution
1st write the dissociation equation then treat it like a solution
stoich. Question (no volumes are needed since they all have the
same volume so you don’t need to calculate n first)
Ex. Calculate the ion concentrations when you have 0.500 mol/L
H2SO4(aq) ?
g
w
w
H2SO4(aq)  2H+(aq) + SO42-(aq)
c=0.500 mol/L c= ?
c=?
c of H+is = 0.500 mol/L x 2/1 = 1.00 mol/L
c of SO42- is = 0.500 mol/L x 1/1 = 0.500 mol/L
Remember your properties of
acids/ bases
Acids
Neutral
Substances
Bases
 sour taste
 electrolytes
 bitter taste
 electrolytes
 electrolytes,
neutralize bases
nonelectrolytes
 neutralize acids
react with indicators  react with indicators  do not
litmus - red
bromothymol blue - yellow
litmus - blue
bromothymol blue - blue
affect indicators
the same way
phenolphthalein phenolphthalein - pink
colourless
 react with metals
to produce
H2(g)
 pH less than 7
eg) HCl(aq),
H2SO4(aq)
pH greater than 7  pH of 7
eg) Ba(OH)2(aq)
NH3(aq)
eg) NaCl(aq),
Pb(NO3)2(aq)
What is an acid?
The Arrhenius definition of an acid is that it has H+
at the beginning of the compound and is (aq)
Ex. HF(aq)
The modified Arrhenius definition of an acid is it
reacts with water to form H3O+ ions
Ex. HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
What is a base?
The Arrhenius definition of a base is that it has
OH- ions at the end of an IONIC compound
Ex. NaOH(aq)
The modified Arrhenius definition of a base is
that it reacts with water to form OH- ions.
Ex. Na2CO3(aq) + HOH(l)  NaOH(aq) + H2CO3(aq)
OHions
Strong acids/ bases
Weak acids and bases don’t 100% break down to
form H3O+ ions and OH- ions (strong one’s DO)
Strong acids are listed on the back of your periodic
table
If they are NOT on that list they are a weak acid
Strong bases have OH- ions in them OR a metal with
oxygen
Ex. NaOH and MgO
Every other base is a weak base
Ex. NH3 and Na2CO3
Monoprotic vs. Polyprotic
Monoprotic acids only have 1 H+ ion to give away (to
water)
Ex. HCl, HF
Similarly monoprotic bases can only accept one H+ ion
(from water) or has 1 OH- ion
Ex. NaOH, ions with only 1- charge (ex. F-)
Polyprotic acids have more than 1 H+ to give away
Ex. H2SO4, H3PO4
Similarly polyprotic bases can accept more than 1 H+
ion
Ex. compound with ions that have more then 1charge (ex. CO32-, PO43-)
Remember the pH scale and the
pOH scale
pH scale is 0-14
0 = strong acid, 14 strong base
pOH is opposite
0 = strong base, 14 strong acid
…and the calculations
Ex. what is pH if [H30+] = 0.05 mol/L
pH=-log[H30+]
= - log[0.05]
= 1.3 (remember SD – only 1 SD = 1 after decimal
place on pH and pOH)
What is pOH if [H30+] is 0.90 mol/L
pOH=-log[OH-]
But we don’t know [OH-]
pOH also = 14 – pH so lets find pH
…
pH=-log[H30+]
pH=-log[0.9]
= 0.04575749…
pOH = 14 - 0.045749…
pOH = 13.95 (2 SD = 2 after decimal)
What is [OH-] if pOH = 6.7
[OH-]=10^-pOH
[OH-]=10^-6.7
= 2 x 10-7 mol/L (1 after decimal = 1 SD)
Stoich!!!
Now put it all together
If dealing with MASS its gravimetric stoich…use
n=m/M
If dealing with GASES its gas stoich…use n=m/M
or PV=nRT (or if just volumes use volumes
directly and x wanted/given)
If dealing with SOLUTIONS and
CONCENTRATIONS its solution stoich…use
n=m/M and c=n/v
LR vs. ER
Remember that during a stoichiometric reaction
there are always 2 reactants – one is being used all
up during the reaction (LR) and one will have some
leftovers (ER)
You need to figure these out so you know how
much product will be produced
You do this by calculating n of each reactant and
then dividing them by their coefficient = the
SMALLER # is the LR
The LR’s n (#of moles before you divided by the coefficient) is
then what you use to calculate the product
Example 1
When 80.0 g copper and 25.0 g of sulphur react,
which reactant is limiting and what is the maximum
amount of copper(I) sulphide that can be
produced?
16 Cu(s)
+
1 S8(s)

8Cu2S(s)
m = 80.0 g
M = 63.55 g/mol
n = 80.0 g
63.55g/mol
= 1.25… mol
n/16 = 0.0786…mol
\ limiting
m = 25.0 g
M = 256.56 g/mol
n = 25.0 g
256.56g/mol
= 0.0974… mol
xg
M = 159.17 g/mol
n = 1.25…mol 8/16
= 0.629… mol
n/1 = 0.0974… mol
\ excess
m = (0.629…mol )  (159.17 g/mol)
= 100.17… g
= 100 g
% Yield and % Error
Remember these formulas:
Predicted = calculated amount from a stoich. calculation
Actual amount = amount you weigh after experiment
% error = actual – predicted x 100
predicted
% yield = actual x 100
predicted
Calculate the % error and % yield for the
following:
predicted mass of ppt = 100 g
actual mass of ppt = 93.5 g
% error = 93.5 g - 100 g x 100
100 g
= -6.50 %
% yield = 93.5 g x 100
100 g
= 93.5 %
Titration's
These where just a type of experiment in which you
use solution stoichiometry and you find the volume
of one of your reactants to calculate the
concentration.
In order to do a solution stoich. question, you need
to know 3 variables in order to solve for the 4th
Ex. The 4: the c and v of one reactant, and the c and v of
the other reactant (in order to solve the c of one you
must know the other 3)
In the question however you are only given 2 variables –
the c and v of only one reactant – after the titration
experiment you will have a volume to use giving you all 3
Ex. Calculate the concentration of HCl when it is
titrated into 10.0 mL of 0.50 mol/L Ca(OH)2?
w
g
2HCl(aq) + Ca(OH)2(aq)  CaCl2(aq) + 2HOH(l)
c= x
v=
c=0.50 mol/L
find during
v=10.0 mL
titration
n= cv = (0.50)(0.01) = 0.005 mol x 2/1 = 0.01 mol
c=n/v
c = 0.01 mol/v Lets do the titration and see what v is.
Titration:
When doing this experiment you want
to make sure you have the correct
volume – so we do the experiment
several times to make sure the volumes
we are getting are all the same (or very
close – within 0.2 mL of each other)
This is why you have several trials.
Results:
Exclude these results
Trial 1
0.0 mL
Initial
Reading
Final
5.9 mL
Reading
Volume 5.9 mL
Used
5.3 + 5.2 + 5.3 = 5.3 mL
3
Trial 2
5.9 mL
Trial 3
Trial 3
11.2 mL 16.4 mL
11.2 mL 16.4 mL 21.7 mL
5.3 mL
5.2 mL
5.3 mL
These 3 are closest together so use them
(they are within 0.2 mL of each other)
Ex. Calculate the concentration of HCl when it is
titrated into 10.0 mL of 0.50 mol/L NaOH?
w
g
2HCl(aq) + Ca(OH)2(aq)  CaCl2(aq) + 2HOH(l)
c= x
v=
c=0.50 mol/L
find during
v=10.0 mL
titration
n= cv = (0.50)(0.01) = 0.005 mol x 2/1 = 0.01 mol
c=n/v
c = 0.01 mol/v Lets do the titration and see what v is…
c = 0.01/0.0053
= 1.88679…
= 1.9 mol/L
Lastly….
Remember titration curves and where
the equivalence point is and what it
means (when the acid and base fully
react to create a neutral solution)
Strong Acid Titrated with Strong Base
14
pH
7
Endpoint

0
volume of titrant added (mL)
Strong Base Titrated with Strong Acid
14
pH
7
0

Endpoint
volume of titrant added (mL)